Chapter - 1 Power Electronics Control - WordPress.com

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Chapter - 1 Power Electronics Control

Power control analog/digital

elect roni c device circuit

power component static retrtary

electronic

Application of Power Electronics Power electronics combine power electronics and control. Control deals with the steady state and dynamic characteristics of the clone loop system. Power deals with the static and rotating power equipment. for the generation, transmission and distribution of electronic power. Electronic deals with the solid state device and circuits for signal processing to meet the desire control objective. Power electronics may be defined as the application electronics for the control and conversion of electric power, the inter-relationship of Potential Energy with power, electronics and control is shown in figure above. Some applications of power electronics are as follows: 1. Aerospace : Example – aircraft power system 2. Commercial : Example – advertising, heating 3. Industrial : Example – cement mill, welding 4. Residential : Example – cooking, lighting 5. Telecommunication : Example – Battery chargers 6. Transportation : Example – street car, trolley buses 7. Utility System : Example – high voltage direct current (HVDC), transmission system History of Power Electronics 1948 → Bell Lab (Silicon Transistor) 1956 → PNPN traggiration

Downloaded from www.jayaram.com.np /- 1

Downloaded from www.jayaram.com.np The history of power electronics begin with the mercury arc rectifier in 1900. Then metal tank rectifier, grid control, vacuum tube rectifier, ignitron, pkahotron & Hydraten were introduced gradually. These devices were applied for power control until 1950. The first electronics revolution begin in 1948 with the invention of silicon transistor at a bell telephone library by Bardeen, Brattain, Schookely. The next break through in 1956 was also f aboratory, the invention of PNPN triggering transistor which was defined a thyrastor or rom Bell Silicon Control Rectifier (SCR). The second electronics revolution begin in 1956 with the development of commercial thyrastor by the general electric company. That was the beginning of a new era of power electronics. Since then many different types of power semi-conductor devices and conversion techniques have been introudced. Power Semi-conductor Devices: Power Semi-conductor Devices can be broadly divided into 5 types. They are: 1. Power diode 2. Thyristor 3. Power BJT 4. Power MOSKET 5. Insulated Gate Bi-polar Transistor (IGBT) or Static Induction Transistor (SIT) Thyristor can be divided into 8 groups. They are: 1. Forced Commutated Thyristor 2. Line Commutated Thyristor 3. Gate Turn off Thyristor (GTO) 4. Reverse Conducting Thyristor (RCT) 5. Static Induction Thyristor (SITH) 6. Gate Assisted Turn off Thyristor (GATT) 7. Light Activated Silicon Control Rectifier (LASCR) 8. MOS Control Thyristor Power diode are of two types: 1. High Speed Power Diode 2. Schotty Power Diode

Control Characteristics of Power Devices

-By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 2

Downloaded from www.jayaram.com.np Vg

Vg

+ VS

Thyristor or SCR

V R

V

Vs

-

t

SITH Vg

.

.

A

+

+

K GTO

Vo

R

VS

-

K MCT

ON Fig : GTO/SITH/MCT Switch

t1

OFF T 1 V B

+

+ V B

V0

VS

-

Fig : Transistor Switch

t1 T

t

t1 T

t

Vo

d) MOSFET/IGBT Switch Downloaded from www.jayaram.com.np /- 3


C

D

+

vgs 1 t1 T

vs

IGBT G

+

G

S

+

V0

Vgs

-

1) 2) 3) 4)

t

vs -

t1 T

E

t

Uncontrolled turn on & off : e.g. diode Controlled turn on and uncontrolled turn off : e.g. SCR Controlled turn on and off : e.g. GTO Continuous Gate Signal Requirement : e.g. Transistor

Types of Power Electronics Circuit: 1. Diode Rectifiers 2. AC-DC Converter (Controlled Rectifier) 3. AC-AC Convert (AC Voltage Controllers) 4. DC-AC Converter (Inverter) 5. DC-DC Converter (DC Choppers) 6. Static Switches 1) Diode Rectifier This rectifier converts AC voltage into fixed DC voltage.

AC Supply

D1 + v s=v mSinwt load resistance + R vs a) Circuit Diagram

D2



Vm

Vs=Vm Sinwt

wt

2π

π

Vm

Vm

V0

π

wt

2π

b) Voltage waveform

2) AC-DC Converter (Controlled Rectifier) By this rectifier, average output voltage can be controlled by varying the conduction time of thyristor or firing angle α . Vm

Vs=Vm Sinwt

d

vs=Sinwt

AC Supply

T1 Lead

result

π

2π

wt

Vm

vs

Fig : Circuit Diagram

T2 Vm

V0

d

π

2π

wt

Fig : Voltage Waveform

3. AC-AC Converter • Such types of converter are used to obtain a variable ac voltage from fixed ac voltage. • Output Voltage is controlled by varying the conduction time or firing angle α .


Downloaded from www.jayaram.com.np 4. DC-AC Converter (Inverter) 5. DC-DC Converter 6. Static Switch Since the power devices can be operated as static switches, the supply to there switches could be either DC or AC and the switches are called as AC Static Switches or DC Switches. Design of Power Electronic Equipment The design of power electronics can be divided into four parts: 1. Design of power circuit 2. Protection of power devices 3. Determination of control strategy 4. Design of logic and gating circuit Peripheral Effects The operations of power converter are based mainly on the switching of the power save conductor devices and as a result, the converter introduces current and voltage harmonics into the supply system and on the output of converter. This can cause problem of distortion of output voltage, harmonic generation into the supply system and interference with the communication signaling circuit. It is normally necessary to introduce filters on the input and output of the converter system to reduce a harmonic level to an acceptable magnitude. Figure shows the block diagram of generalized power converter. Power converter can cause radio frequency interference due to electromagnetic radiation and the getting circuit may generate erroneous signal and this interference can be avoided by grounded shielding.



Chapter - 2 Semi-conductor Diodes •

A power diode is two terminal P-N Junction diode.

•

When anode potential is higher than that of cathode, the diode is said to be forward biased and the diode conducts.

•

When cathode junction is higher than that of anode, the diode is said to be reversed. Under reverse condition, a small reverse current (also known as leakage amount) is the range of micro or milli ampere flows and this leakage amount of current increases slowly in magnitude with reverse voltage until the avalanche or zenor voltage is reached. I

I

ID A V

K

VD

V 0

D V

Reverse l eakage current

b) Ideal

a) Practical Fig : V-I characteristics of diode

v-i characteristics can be expressed by Shockley diode equation as given below:

(

ID = Is e

V p nVT

)

−1

where, I D = Current through diode A V D = Diode voltage with anode positive w.r.t. cathode I s = Leakage current n = Emperical constant known as emission co - efficient or ideality factor whose value varies from 1 to 2

For Ge diode : For Si diode :

n=1 n=2 KT VT = Thermal Voltage = q

where, K = Boltzmann constant = 1.3806×10-23 J q = Electron charge = 1.6022×10-19 C T = Absolute temperature in Kelvin At a junction temperature of 25°c, we have VT =

KT 1.3806 × 10 −23 × 298 = ≈ 25.8mV q 1.6022 × 10 −19

The diode characteristics can be divided into three regions. They are: 1. Forward Biased Region, V D > 0 2. Reverse Biased Region, V D < 0 Downloaded from www.jayaram.com.np /- 7

Downloaded from www.jayaram.com.np 3. Breakdown Region, V D < −V zk Consider a small diode voltage, V D = 0.1V , n = 1 & VT = 25.8mV

(

I D = I s e VD

)

(

)

− 1 = I s e 0.1 (1×0.0288 ) − 1

nVR

= I s (48.23 − 1) [neglecting I s ] = 48.23 I s with 2.1% error

For VD>0.1 V, ID>>Is

(

I D = I s e VD

nVT

)

− 1 ≈ I s e VD

nVT

with 21% error

Reverse Biased Region, V D < 0 For VD < −0.1V , the exponential term becomes negligible as compared to unity.

(

ID = Is e

− VD nVT

)

− 1 ≈ −I s

In Breakdown Region, the reverse voltage is high usually greater than 1000 V, the magnitude of the reverse voltage exceeds the specified voltage known as breakdown voltage V Br . The reverse voltage increases rapidly with small change in reverse voltage beyond V Br . Reverse Recovery Characteristics The current in a forward leased junction diode is due to the net effect of majority and minority carriers. Once a diode is in forward conductor mode and then it’s forward current is reduced to zero, the diode continuity to conduct due to minority carrier which remain store in P-N Junction and bulk semi-conductor material. The minority carrier requires a certain time to recombine with opposite charges and to be neutralized. This time is called reverse recovery time of the diode. Figure shows two reverse recovery characteristics of a junction diode. IF

t rr

I

F

trr

t

ta

t

0.25 I

RR

I

RR

t1

IRR

a) Soft Recovery

Fig : Reverse Recovery Characteristics

− (1)

t rr = t a + t b

t6

b) Abrupt Recovery

t a = due to charge storage in depletion region t b = due to charge storage in bulk semiconductor material The ratio t a / t b is known as Softness Factor, denoted by SF. I RR = t a

di dt

− (2)


Downloaded from www.jayaram.com.np Reverse recovery time (TRR) may be defined as the time interval between the instant the current passes through zero during the change over from forward conduction to reverse blocking conduction and the movement, the reverse current has decayed to 25% of it’s peak reverse value I RR . Reverse Recovery Charge (θ RR ) It is the amount of charge carrier that flow across the diode in the reverse direction due to charge over from forward conduction to reverse blocking condition. 1 1 2 2 1 = I RR .t rr − (3) 2

θ RR ≅ t a .I RR + t b .I RR

I RR =

2θ RR t RR

− (4 )

From (2) and (4), we have: ta

di 2θ RR = dt t rr

− (5)

If t b is negligible as compared to t a , t rr ≅ t a t rr ≈

2θ RR di dt

t rr ≈

2θ RR di dt

2

I RR ≈ 2θ RR .

− (6) di − (7 ) dt

Types of Power Diode Depending upon the recovery characteristics and manufacturing technique, the power diode can be classified into three categories. They are: 1. General or Standard Purpose Diode 2. Fast Recovery Diode 3. Schottky Diode General Purpose Diode These diodes have relatively high reverse recovery time of the order of about 25 µ s. Their current ratings vary from 1A to several thousand amperes and range of voltage rating is from 50V to 5KV. Application of power diode of this type include battery charging, UPS, etc. Fast Recovery Diode The diodes with low reverse recovery time of about 5 µs or less are classified as fast recovery diode. These diode convert current rating from less than 1A to 100A with voltage rating from 50V to 3KV. They are used in DC to DC, DC to AC converter circuit.


Downloaded from www.jayaram.com.np Schottky Diode It uses metal to semi-conductor junction for rectification purpose instead of P-N junction. They have very fast recovery time and low forward voltage drop. The maximum available voltage is generally limited to 100V. The current rating of schottky diode varied from 1A to 300A and they are used in high frequency instrumentation system. Effects of forward & reverse recovery time I1 I0

Ls + I1 Sw

I2

D1 R

I0 Vs

R

t

IP=Is/R

I0

I2

Vs Dm

L

I2

I0

I3

t2

t1

-

a) Circuit Diagram

t1

t2

t

Due to reverse recovery of Dm

b) Wave Form

If the switch is turned on at t=0 and remains on long enough a steady state current of

I o = Vs / R would flow through the load and the free wheeling diode Dm will be reverse bias. If the switch is turned off at t=t1 diode Dm would circulate through Dm. Again, the switch is turned on at t=t2. The rate of rise of forward current of diode D1 and the rate of fall of forward current of diode Dm would be very high tending to be infinity. According to the equation, I RR = 2θ RR

di , the peak reverse current of diode would be very dt

high and diodes D1 and Dm may be damaged. This problem is normally overcome by connecting di/dt limiting inductor Ls as shown in the figure.


Downloaded from www.jayaram.com.np Ls + I1 Sw

I2

D1

I1 I0

R I1 I0

Vs Dm

Ip

L

I2

I2

t1

t

t2

I0

-

tm

a) Circuit Diagram t1

t2

t

b) Waveform

The practical diodes require a certain turn on time before the entire area of junction becomes conductive and di / dt must be kept low to meet the turn on time limit. This time is some time known as forward recovery time (tfr). The rate of rise of current through the diode D1 which should be same as the rate of fall of current through diode Dm is

di Vs = . If trr is the recovery time of Dm, the peak dt L

reverse current of Dm is

di dt V = t rr s Ls

I RR = t rr

and, the peak current through the inductor Ls would be I p = I o + I rr = I o + t rr

Vs Ls

When the inductor current becomes I p diode Dm turn off suddenly. Due to a highly inductive lead, the current cannot change suddenly from I o to I P . The excess energy stored in Ls would induce high reverse voltage across Dm and this may damage diode Dm . The excess energy stored as a result of reverse recovery time is 1 wR = Ls I p2 − I o2 2 1 2 = Ls (I o + I RR ) − I o2 2 1 2 = Ls (I o + t rr ) − I o2 2

[

[ [

]

]

]

This excess energy can be transfer from inductor Ls to a capacitor C s which is connected across diode Dm . 1 C sVc2 = wR 2 2w C s = 2R Vc where, VC is the allowable reverse voltage of the diode. A resistor Rs is connected in series with Cs to damp out any transient oscillation.


Downloaded from www.jayaram.com.np Series Connected Diode Let us consider two diode connected in series. i

I=-Is VD1

D1

+-

+

VD2 +

VD1 VD2

D2

V Is

Is

a) Circuit Diagram b) v-i chracteristics

In practical v-i characteristics for the same type of diode differ due to tolerances in their production process. In forward bias condition, both diodes conduct the same amount of current and the forward voltage drop of each diode would be all most equal. However in reverse blocking condition each diode has to carry the same leakage current and as a result the blocking voltage will be differ significantly. A simple solution to this problem is to force equal voltage sharing by connecting a resistor across a diode. − (i )

I s = I R + I s2

i=-I s

− (ii )

I R1 + I s1 = I s

VD1

From (i) and (ii),

I R1 + I s1 = I R2 + I s2 VD1 R1

+ I s1 =

VD2 R2

VD2 +

+ I s2

R

+ I s1 =

VD2 R

+ I s2

IR2

Is1 Is2

+

i

Is

If R 1 = R2 = R, then

VD1

IR1

-VD1=-V D2

− (iii )

V -Is1 -Is2

Parallel Connected Diode In high power application diodes are connected in parallel to increase the current carrying capability to meet the desire current requirement. The current sharing of diodes would in accord with their respective forward voltage drop. Uniform current sharing can be achieved by connecting current sharing resistor. Q. Two diodes are connected in series as shown in the figure to share a total voltage of V D = 5 KV , the reverse leakage current of the two diodes are I s1 = 30mA and I s2 = 35mA .


Downloaded from www.jayaram.com.np a. Find the diode voltages if the voltage sharing resistances are equal. R1 = R 2 = R = 100kΩ . b. Find the voltage sharing resistances R1 & R2 if the diode voltages are equal.

Solution: I s1 = 30mA

I s2 = 35mA

V D1 = ?

VD2 = ?

R1 = R2 = R = 100kΩ I s1 + I R1 = I s2 + I R2 VD1

I R1 =

I s1 +

I R2 =

R1

VD1

= I s2 +

R1

VD2 R2

VD2 R2

V D = V D1 + V D2

− (ii )

∴VD1 − VD2 = 500

V D1 + V D2 = 5000 ∴V D1 = 2750V & V D2 = 2250V

Q. The reverse recovery time of a diode is t rr = 5µs and the rate of fall of diode current di = 80 A / µs. If the softness factor, SF = 0.5, then determine storage charge. dt (a) θRR Solution: I RR = t a

(b) peak reverse current di dt

− (i )

θ RR =

1 I RR .t rr − (ii ) 2

SF =

tb = 0 .5 ta

t b = 0.5t a

− (i )

t rr = t a + t b = 5µs

− (ii )

t a = 3.33µs



Chapter - 3 Diodes Circuit & Rectifier Diodes with RC Load S1

+

D1

Vs

+ R C

VR1 + Vc -

-

When switch s1 is closed at t=0 V s = V R + Vc 1 idt + Vc (t = 0 ) c∫ With initial condition, Vc (t = 0) = 0 = Ri +

Vs I (s ) ≠ RI ( s ) + s s V 1⎞ ⎛ or , ⎜ R + ⎟ I ( s ) = s cs ⎠ s ⎝ V ⎛ Rcs + 1 ⎞ or , ⎜ ⎟ I (s ) = s s ⎝ cs ⎠ Vsc or , I (s ) = Rc( s + 1 Rc) Vs or , I (s ) = R(s + 1 / Rc ) Taking Inverse Laplace, V i (t ) = s e −t / Rc R Capacitor Voltage 1 t Vc (t ) = ∫ idt c 0 1 tV = ∫ R e −t Rc dt c 0 R

Vs ⎡ e − t / Rc ⎤ = ⎢ ⎥ Rc ⎣ − 1 / Rc ⎦

t

[ ] ] = V [1 − e = Vs 1 − e −t / Rc −t / T

s

T = Rc = Time constant of Rc load. The rate of change of capacitor voltage is dVc (t ) ⎛ 1 ⎞ −1 / Rc t Vs −t / Rc = −Vs ⎜ − = e ⎟e dt Rc ⎝ Rc ⎠ -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 14

Downloaded from www.jayaram.com.np & the initial rate of change of capacitor voltage (at t=0) is dVc (t ) V = s dt t =0 Rc

VS/

i R

Vc Vs 0.632Vs

0.36 VS/R

τ

t

τ τ=Rc Diode Circuit with RL load Ls

D1

+ R

VR -

Vs

+ L

VL -

a) Circuit Diagram When switch S is closed at t=0 Vs = VR + VL di z = Ri + L dt Vs = RI (s ) + L[sI (s ) ) − i (0 )] s With initial condition i(0) = 0 Vs = RI ( s ) + LsI ( s ) s (R + Ls )I s = Vs s Vs I (s ) = s (R + Ls ) Vs = Ls (s + R / L ) V ⎡1 1 ⎤L I ( s) = s ⎢ − L ⎣ s s + R / L ⎥⎦ R


t

Downloaded from www.jayaram.com.np I (s) =

Vs ⎡ 1 1 ⎤ − ⎢ R ⎣ s s + R / L ⎥⎦

By inversion, −R t ⎞ ⎛ V −Rt ⎜1 − e L ⎟ = s e L ⎜ ⎟ L ⎝ ⎠ V di = s dt t =0 L Voltage across inductor, R t − t − di V L = L = Vs e L = Vs e T dt L where, T = = time constant RL load R

i (t ) =

Vs R

VL VS

i Vs/R 0.63Vs

τ

0.36 VS

O

t

τ

t

b) waveform

Q. A diode circuit is shown in figure with R = 44 Ω and C = 0.1 µF . The capacitor has an initial voltage Vo = 220V . If switch s1 is closed at t=0, determine: a) Peaked diode current b) Energy dissipated in a resistor R c) The capacitor voltage at t=2 µs Solution: at t=0, s is closed 1 t Ri + ∫ idt + VC (t = 0 ) = 0 c 0 Vc (t = 0 ) = −Vo

i

VR

R +

C

-

V0 VC

Taking laplace transform, I (s ) Vo RI (s ) + − =0 Cs s


Downloaded from www.jayaram.com.np V 1 ⎞ ⎛ ⎜R + ⎟ I (s ) = o Cs ⎠ s ⎝ V ⎛ RCs + 1 ⎞ ⎜ ⎟ I (s ) = o s ⎝ Cs ⎠ CVo I (s ) = 1 ⎞ ⎛ RC ⎜ s + ⎟ RC ⎠ ⎝

I (s ) =

Vo 1 ⎞ ⎛ R⎜ s + ⎟ RC ⎠ ⎝

By inversion, i (t ) =

Vo −t / RC e R

a) Peak diode current =

Vo 220 = = 5A R 44

i(t) V0 /R

O

t

1 1 2 b) Energy dissipated in resistor = CVo2 = × 0.1 × 10 −6 × (220) = 2.42mJ 2 2 1 t c) Voltage across capacitor, VC (t ) = ∫ idt + Vc (t = 0 ) c 0 1 tV Vc(t) = ∫ o e −t / RC dt − Vo c 0 R O ⎤ ⎡ V ⎢ e −t / RC ⎥ = o ⎢ ⎥ − Vo RC ⎢ 1 ⎥ − ⎣⎢ RC ⎦⎥ = Vo (1 − e − t / RC ) − Vo

-Vo

= −Vo e −t / RC

At t = 2µs , Vc (t = 2µs ) = −220e − (2×10

−6

/ 4×0.1×10 −9

) = −139.64V

Diode with LC Load At t=0, switch s is closed. di 1 t Vs = L + ∫ idt + Vc (t =0) dt c 0 Downloaded from www.jayaram.com.np /- 17

Downloaded from www.jayaram.com.np Vc (t = 0) = 0 Using Laplace Transform: Vs I ( s) = L[sI ( s ) − i (0)] + s Cs i (0) = 0

+ L

V 1⎞ ⎛ ⎜ Ls + ⎟ I ( s ) = s cs ⎠ s ⎝ ⎛ Lcs 2 + 1 ⎞ V ⎜⎜ ⎟⎟ I (s ) = s s ⎝ cs ⎠

+

Vs

VL

- Vc

CVs

I (s ) ) =

I (s ) =

D

S

1 ⎞ ⎛ LC ⎜ s 2 + ⎟ LC ⎠ ⎝ Vs , L(s + ω 2 ) 2

where ω =

1 LC

By inversion, Vs 1 . . sin ωt L ω C = Vs . sin ωt L

i (t ) =

i (t ) = I p sin ωt where I p = Vs

C L

The rate of current is di C 1 = I p .ω cos ωt = Vs . . cos ωt L LC dt di Vs = . cos ωt − (2 ) dt L Initial rate of rise of current, di dt

= t =0

Vs L

− (3)

Voltage across capacitor t

Vc (t ) =

1 t 1 t 1 C ⎡ cos ωt ⎤ idt = ∫ I p sin ωtdt = .Vs − ∫ c 0 c 0 c L ⎢⎣ ω ⎥⎦ 0

=

Vs

.

1

LC ω

(1 − cos ωt )

= Vs (1 − cos ωt )

− (4)

At t = t 1 = π LC , the diode current i falls to zero and the capacitor is changed to 2Vs.


Downloaded from www.jayaram.com.np i(t)

O

t1/2

t1=

t 2c

Vc(t)

2Vs Vs O

t

Q. A diode circuit with Lc load is shown in figure with the capacitor having initial voltage Vo=220V, capacitance C=20 µF and inductance L = 80 µH . If switch s1 is closed at t=0, determine: a) Peak current through the diode b) Conduction time of the diode c) Steady state capacitor voltage Solution: At t=0, s1 is closed. di 1 t L + ∫ idt + Vc (t = 0 ) = 0 dt C 0 Vc (t = 0 ) = −Vo

S1 + i

Taking laplace transform, I (s ) Vo − =0 Cs s V 1 ⎞ ⎛ ⎜ Ls + ⎟ I (s ) = o Cs ⎠ s ⎝ L[SI (s ) − i (0 )] +

C = 110 A L

i (t ) = Vo

Conduction time, t1 = π LC = 125.66µs Vc (t ) = =

1 t idt + Vc (t = 0) C ∫0 C 1 t Vo sin ωtdt − Vo ∫ C 0 L

V = o C

=

Vo

t

C ⎡ cos ωt ⎤ − − Vo L ⎢⎣ ω ⎥⎦ 0

.

1

LC ω

(1 − cos ωt ) − Vo

Vc (t ) = −Vo cos ωt Steady-state Capacitor Voltage Downloaded from www.jayaram.com.np /- 19

VL Vc

L + -

V0

Downloaded from www.jayaram.com.np Vc (t = t1 ) = −Vo cos ωt = −Vo cos

O V0

1 LC

.π LC = −Vo cos π = Vo = 220V

t t1

V0

t1/2

t1

t

-V0

Q. In the diode LC H /W `is shown in the figure. The capacitor is initially charged to voltage Vc with upper plate +ve. Switch s is closed at t=0, device expressions for the current through and voltage across C. Find the conduction time of the diode, peak current through the diode and final steady state voltage across C. In case Vs = 400V , Vo = 100V , L = 100µH , C = 30µ . Determine also the voltage across the diode after stage conduction. Solution: At t=0, switch s is closed, di 1 t Vs = L + ∫ idt + Vc (t = 0 ) dt C 0 Vc (t = 0 ) = Vo

Ls

i

+

+ VL

L Vs

Using Laplace Transform,

Vs I (s ) V o = L[sI (s ) − i ( s )] + + s Cs s

C V0

+ -

-

V − Vo 1⎞ ⎛ ⎜ Ls + ⎟ I (s ) = s cs ⎠ s ⎝ ⎛ Lcs 2 + 1 ⎞ V − Vo ⎜⎜ ⎟⎟ I (s ) = s s ⎝ cs ⎠

I (s ) =

c(Vs − Vo ) 1 ⎞ ⎛ LC ⎜ s 2 + ⎟ LC ⎠ ⎝

1 ⎛ V − Vo ⎞ or , I (s ) = ⎜ s where ω = ⎟ 2 2 ⎝ L ⎠ s +ω

1 LC


Vc

Downloaded from www.jayaram.com.np By inversion i (t ) =

V s − Vo 1 C . sin ωt = (Vs − Vo ). sin ωt L L ω

i(t)

Voltage across capacitor, 1 t Vc (t ) = ∫ idt + Vc (t = 0) C 0

V1-V C ° /2 O

C 1 t = ∫ (Vs − Vo ). sin ωt + Vo C 0 L

t1/2

Vc(t)

t1

t

t

V − Vo C ⎡ cos ωt ⎤ . = s − + Vo C L ⎢⎣ ω ⎥⎦ 0

2Vs-V ° Vs

= (Vs − Vo )(1 − cos ωt ) + Vo

V ° Vs

a) t1 = π LC = 54.77 µs b) I p = (Vs − Vo )

t t -Vs+V

°

C = 164.32 A L

c) Steady state voltage across capacitor = 2Vs − Vo = 700V d) Vo = Vs − VL − Vc = Vs − 0 − (2Vs − Vo ) = Vs − 2Vs + Vo = −Vs + Vo = −300V Diode Circuit with RLC Load At t=0, switch s1 is closed. di 1 t Vs = Ri + L + ∫ idt + Vc (t = 0) dt C 0

S1

-

Vs

+ V

°

2

d i di i L 2 +R + =0 dt c dt 2 d i R di i + + =0 2 L dt LC dt Changing equation in s domain, R i s2 + s + =0 L LC

-

2

R 1 ⎛ R ⎞ ± ⎜ ⎟ − 2L LC ⎝ 2L ⎠ damping factor α =

VL

L

By differentiation,

We define,

+

+ VR -

With initial condition i (t = 0) = Vo

s1, 2 = −

R

+

R 2L

resonance frequency, ω o =

1 LC

s1, 2 = −α ± α 2 − ω o2

1st case Downloaded from www.jayaram.com.np /- 21

C

+ Vc

-

Downloaded from www.jayaram.com.np If α = ωο , the roots are equal s1 = s 2 and the circuit is called critically damp and the solution will be of the form i (t ) = ( A1 + A2 t )e s1t If α > ω o , the roots are real and the circuit is called over damped. The solution takes the form: i (t ) = A1e s1t + A2 e s2t If α < ω o , the roots are complex and the circuit is called under damped. The roots are s1, 2 = −α ± jω r where, ω r = ω12 − α 2 i (t ) = e −αt ( A1 cos ωr t + A2 sin ωr t ) Q. The 2nd order RLC circuit shown in the figure has the source voltage Vs = 220V , Inductance L = 2m + 1, Capacitance C = 0.05µF and Resistance R = 160Ω . The initial value of the capacitor voltage is Vo = 0 . If switch s1 is closed at t=0, determine: a) Expression for the current i(t) b) Conduction time of diode Solution: R 160 = = 40,000 rad / sec 2 L 2 × 2 × 10 −3 1 ωo = = 10 5 rad / sec LC

α=

1. α < ω o it is an under-damped circuit . The solution is i (t ) = e −αt ( A1 cos ωt + A2 sin ωt )

− (i )

ω r = ω o2 − α 2 = 91,652 rad / sec At t=0

i(t=0)=0 From (i), i (t ) = e

− αt

i(0)=A1=0 . A2 sin ωt

(

di = A2 e −αt .ω r cos ω r t − αe −αt sin ω r t dt di dt

)

= A2ω r t =0

Initial rate of rise of current is limited only by inductor L. di dt

= t =0

A2ω r = A2 =

Vs 220 = L 2 × 10 −3 Vs L

Vs = 1 .2 ωr L


i (t ) = 1.2e

−40 , 000 t

Downloaded from www.jayaram.com.np sin (91,652)t

b) Conduction time of diode, t1 =

π π = = 34.27 µs ω r 91,652

Freewheeling Diodes In the circuit steady state current after switch s1 is Vs . If switch s1 is now open the energy R

equal to

S1

i

+

stored R

2

1 ⎛ Vs ⎞ L⎜ ⎟ in inductance L will appear in the form of 2 ⎝R⎠

close is

D1

Vs

Dm if

arc

at

L

such the opening contact of switch s1. In order to avoid occurrence a diode called freewheeling diode is connected across RL as shown in the figure. The circuit operation can be divided into two mode. The circuit operation has two modes. Mode 1 begin when the switch is closed at t=0. Mode 2 begin when the switch is then opened. Mode-1 Diode Current

i1 =

Vs R

R t ⎞ ⎛ ⎜1 − e L ⎟ ⎜ ⎟ ⎝ ⎠

When the switch is open at t=t1, current at that time becomes R t1 ⎞ Vs ⎛ ⎜ i1 (t = t1 ) = ⎜1 − e L ⎟⎟ R⎝ ⎠ i2 R Mode-2 di 0 = L 2 + Ri2 L dt L[sI 2 (s ) − i2 (0 )] + RI 2 (s ) = 0 V i i2 (0 ) = s = I 1 R I1 i2 (Ls + R )I 2 (s ) = L. Vs R t1 t LVs R if I 2 (s ) = i2 RL(s + R / L ) R t2 t1 i2 (t ) = I 1e = t Waveform L Single Phase Half-wave Rectifier: Rectifier is a circuit that converts an AC signal into unidirectional signal.


Downloaded from www.jayaram.com.np Is Vm

+ AC, Supply Vp

Vs=Vm sinwt R

V2 o

VL

-

/2

2

Vm

Is Vm/R

o

/2

2

VD

o

/2

2

wt

wt wt

2

o -Vm

During +ve half cycle of i/p voltage diode D1 conduct and the i/p voltage appears across the lead. During –ve half cycle of input voltage, the diode is in blocking condition and the o/p voltage is zero. Performance Parameter A rectifier is a power processor that should give a DC output voltage with minimum amount of harmonic contains. There are different types of rectifier circuit. The performances of rectifier are normally evaluated in terms of following parameters. Average value of o/p (lead) voltage Vdc Average value of o/p current Idc Output DC Power Pdc = Vdc × Idc RMS value of o/p voltage Vrms RMS value of o/p current Irms Output AC Power Pac = Vrms . Irms

Efficiency (rectification ratio)

η=

Pdc Pac

Output Voltage can be considered as being composed of two components (1) DC value (2) AC component or ripple. Effective AC component of output voltage is Vac = V rms2 − Vdc2 From factor, measure shape of output voltage is FF =

Vrms Vdc

2 Vrms − Vdc2 ⎛V Vac Ripple factor measures ripple content is R F = = = ⎜⎜ rms Vdc Vdc ⎝ Vdc

Transformer Utilization Factor is TUF =

2

⎞ ⎟⎟ − 1 = FF 2 − 1 ⎠

Pdc P = dc Vs I s V A

where, Vs & Is are rms voltage and rms current of transformer secondary respectively. Let us consider the waveform


Downloaded from www.jayaram.com.np Vs

Ip

Is

Input Current Is 1

o

wt

-Ip Fundamental Component Input Voltage

Here, Vs = sinusoidal input voltage is = instantaneous input current is1 = fundamental component of instantaneous input current

If φ is the angle between fundamental component of input current and voltage, then the angle

φ is known as displacement angle. Displacement Factor is defined as cos φ i.e. DF = cosφ . The harmonic factor of input current is ⎛I −I defined as HF = ⎜ ⎜ I s2 1 ⎝ 2 s

⎞ ⎟ ⎟ ⎠

2 s1

1 2

1

⎡⎡ I ⎤ 2 ⎤ 2 = ⎢ ⎢ s ⎥ − 1⎥ ⎢ ⎣⎢ I s1 ⎦⎥ ⎥ ⎣ ⎦

Both I s and I s1 are expressed in rms. Input Power Factor is defined as: PF =

Vs I s1 cos φ Vs I s

=

I s1 P cos φ ⇒ ac Is VA

Crest Factor, CF is to specify the peak current ratings of devices and components CF of input current is defined as: CF =

I s ( peak ) Is

Q. The half wave rectifier has a purely resistive lead of R. Determine: a. Efficiency b. Foam Factor c. Ripple Factor d. Transformer Utilization Factor e. Peak Inverse Voltage of Diode D1 f. Crest Factor of i/p current Solution: P η = dc Pac 1 T VL (t )dt T ∫0 1 T2 = ∫ Vm sin ωtdt t T 0

Vdc =

V = m T =

T

⎡ cos ωt ⎤ 2 ⎢⎣− ω ⎥⎦ 0

Vm [1 − cos ωT / 2] Tω


Downloaded from www.jayaram.com.np Vm (1 − cos π ) 2π

=

Vm

= I dc =

Vrms

= 0.318Vm

π

Vdc 0.318Vm = R R

⎡1 =⎢ ⎣T ⎡1 =⎢ ⎣T

1

⎤2 ( ) V t dt ∫0 ⎥ ⎦ T

2 L

1

∫

T 2

0

⎡V 2 =⎢ m ⎣T

⎡V =⎢ ⎢⎣ 2T

2 m

⎤2 V sin ωtdt ⎥ ⎦ 2 m

1

∫

T 2

0

⎛ 1 − cos 2ωt ⎞ ⎤ 2 ⎜ ⎟dt ⎥ 2 ⎝ ⎠ ⎦ T 2

⎡ sin 2ωt ⎤ ⎢⎣t − 2ω ⎥⎦ 0

⎤ ⎥ ⎥⎦

1 2

1

⎡Vm2 ⎛ T sin ωT ⎞⎤ 2 =⎢ ⎜ − ⎟⎥ 2ω ⎠⎦ ⎣ 2T ⎝ 2 ⎡V T ⎤ =⎢ . ⎥ ⎣ 2T 2 ⎦ 2 m

1 2

Vrms =

Vm = 0.5Vm 2

I rms =

Vrms 0.5Vm = R R

Pac = Vrms I rms =

(0.5Vm )2 R

Pdc = Vdc I dc =

(0.31Vm )2 R

(0.318Vm ) P η = dc = = 40.5% 2 Pac R(0.5Vm ) 2

b) FF =

Vrms 0.5Vm = = 1.57 Vdc 0.318Vm

c) RF = FF 2 − 1 = 1.21 ⎡1 d) Vs = ⎢ ⎣T

1

∫

T

0

2 V (Vm sin ωt ) dt ⎤⎥ = m 2 ⎦ 2


Downloaded from www.jayaram.com.np Is =

0.5Vm = load current R

(0.318Vm ) . 2.R Pdc = = 0.286 Vs I s R.Vm × 0.5Vm 2

TUF =

e) PIV of diode = Vm f) I s ( peak ) = CF =

0.5Vm Vm ⇒ Is = R R

I s ( peak ) =2 Is

Single Phase Half wave Rectifier with RL load + is

io

D1

R

Vs=Vm sinwt AC, Supply Vp

VR

Dm VL

VS

VS=VS- VR VR =i° R

+6

2

is=i °

v°

VD D1 6

Current I o continuous to flow even after source voltage Vs has become –ve, this is because of presence of inductance L in the lead circuit. Voltage V R = I o R has the same wave shape as that of I o . Conduction period of D1 will extend beyound 180° until the current becomes zero at ωt = π + σ . Average Output Voltage, Vdc =

1 2π

Average Output Current, I dc =

Vdc R

∫

π +σ

0

Vm sin ωtdωt =

Vm [1 − cos(π + σ )] 2π


Downloaded from www.jayaram.com.np The average output voltage and current can be increased by making σ = 0 , which is possible by adding freewheeling diode Dm . The effect of this diode is to prevent a negative voltage appearing across the load and as a result the magnetic stored energy is increased. At t = t1 =

π , the current ω

from diode D1 is transferred to Dm and this process is called commutation of diode. Vs=Vmsinωt

Half-wave Rectifier as battery charges R +

+

Vs -

V1 -

Vs-E

D1

2π

π

3π

wt

Vm-E

i2 α

β

E Vm+E i

a) Circuit Diagram

Vm-E R

If the output is connected to a battery, the rectifier can be used as a battery charger. For Vs > E , the diode D1 conducts

α

Vm sin α = E sin α =

wt

β π

2π

E Vm ⎛ E ⎝ Vm

α = sin −1 ⎜⎜

⎞ ⎟⎟ ⎠

⎡ forα < ωt < β ⎤ ⎢β = π − α ⎥ ⎣ ⎦ Q. The battery shown in the figure is E = 12V and its capacity is 100 watt hour. The average Charging Current, I L =

Vs − E Vm sin ωt − E = R R

charging current should be I dc = 5 A . The primary input voltage V p = 120V ,60 Hz and transfer has a turn ratio of N = 2 : 1 . Then calculate a) The conduction angle δ of the diode b) Current turning resistance R c) Power ratings PR of R d) Charging time ho in hours e) Rectifier efficiency η f) Peak inverse voltage Solution: E = 12V V p = 120V Vp

120 = 60V 2 n Vm = V2Vs = 84.85V Vs =

=

R

+ Vp -

D1

+ Vs -


iL E

Downloaded from www.jayaram.com.np ⎛ E ⎝ Vm

⎞ ⎟⎟ = 8.13° ⎠ β = 180° − α = 171.87

α = sin −1 ⎜⎜

a) Conduction Angle, δ = β − α = 163.74° b) Average Charging Current, I dc =

1 2π

β

⎛ Vm sin ωt − E ⎞ ⎟dωt R ⎠

∫α ⎜⎝

1 [− Vm cos ωt - Eωt ]αβ 2πR 1 = − Vm cos β − E.β + Vm cos α + E.α 2πR 1 (− Vm cos(π − α ) + E (π − α ) + Vm cos α + Eα ) = 2πR 1 (2Vm cos α + 2 Eα − Eπ ) I dc = 2πR R = 4.26Ω =

c) RMS Battery Current, I

2 rms

1 = 2π

2

⎛ Vm sin ωt − E ⎞ ⎟ dω t ∫α ⎜⎝ R ⎠ β

1 = 2πR 2

⎡⎛ Vm2 ⎤ Vm2 2⎞ ⎟ ⎜ sin 2α − 4Vm E cos α ⎥ + E ⎟(π − 2α ) + ⎢⎜ 2 ⎠ ⎣⎢⎝ 2 ⎦⎥

= 67.4

I rms = 8.2 A

Power Rating of R, 2 PR = I rms .R = 67.4 × 4.26 = 286.4W

d) Power deliver to the battery, Pdc = EI dc = 12 × 5 = 60W ho Pdc = 100

100 = 1.667hrs 60 Power deliver to the battery e) Rectifier efficiency, η = total input power ho =

=

Pdc 60 = = 17.32% Pdc + PR 60 + 286.4

f) PIV diode = Vm + E = 84.85 + 12 = 96.85V Single Phase Full Wave Rectifier



VD1 + Vp -

+

+

+ Vs -

D1

-Vmo

i2

D2 +

Vm

i2

-

-

2

E V2 Vm

VD2 o

2

wt

VD

o

2 VD =0 1 VD 2

VD =0 2 VD 1

-2Vm


wt

Downloaded from www.jayaram.com.np Vs Single Phase Full Wave Bridge Rectifier Vm wt

2

D1

+ Vp -

D2

D3

+

R

V1 -

D4

-Vm VL Vm o

wt

2

Vs

o

2

-Vm

During positive half cycle of input voltage, the power is supplied to the load through diode D1 and D2 during negative half cycle the diode D3 and D4 conduct.

Q. A single phase full wave center tapped transformer has a purely resistive load of R. Determine: a) Efficiency b) Wave-form Factor c) Ripple Factor d) TUF e) PIVof diode D f) CF of input current. VL Solution: Average Output Voltage: Vm 1 π Vdc = ∫ VL dωt

π

=

0

2Vm

π

2

= 0.6366Vm

Average Load Current: 0.6366Vm V I dc = dc = R R (0.6366Vm ) 2 Pdc = Vdc .I dc = R RMS value of output voltage: Vrms

⎡1 =⎢ ⎣π

⎡1 =⎢ ⎣π ⎡V =⎢ ⎣π

1

⎤2 V d ω t ∫0 ⎥ ⎦ π

2 L

1

⎤2 ω V tdt sin ∫0 ⎥⎦

2 m

wt

o

π

2 m

∫

π

0

2

⎛ 1 − cos ωt ⎞ ⎤ ⎟dωt ⎥ ⎜ 2 ⎠ ⎦ ⎝

1 2

1

π V 2 ⎡⎡ sin 2ωt ⎤ ⎤ 2 = m ⎢ ⎢ωt − ⎥ π ⎢⎣ ⎣ 2 ⎥⎦ 0 ⎥⎦


Downloaded from www.jayaram.com.np 1 2

⎡V ⎤ = ⎢ .π ⎥ ⎣ 2π ⎦ V = m = 0.707Vm 2 2 m

RMS Value of Load Current: I rms =

Vrms 0.707Vm = R R

Pac = Vrms I rms = a) η =

(0.707Vm ) 2 R

Pdc (0.6366Vm ) .R = = 81% Pac (R.0.707Vm )2 2

b) FF =

0.707Vm Vrms = = 1.11 0.6366Vm Vdc

c) RF = FF 2 − 1 = 0.482 d ) Vs = I s2 = Is =

Vm 2

2 .R I rms 2 .R I rms

2

Vm

=

2 .R. 2 0.5Vm = R

=

Vm 2R

(0.6366Vm ) × 2 × R Pdc = = 57.32% 2Vs I s 2 × R × Vm × 0.5 × Vm 2

TUF =

e) PIV of diode = 2Vm f) CF =

I s ( peak ) Vm 2 R = = 2 I rms R.Vm

Q. A single phase full wave bridge rectifier has a purely resistive load of with the supply voltage 220V and supply frequency 50Hz. Then calculate: a) Efficiency b) Ripple Factor c) TUF d) PIV of diode e) CF Solution: R = 5Ω

Vm = 2 × 220V = 311V

Vdc = 0.6366Vm

Vrms = 0.707Vm

I dc =

0.6366Vm R

I rms =

0.707Vm R



a) η =

Pdc (0.6366Vm ) .R = = 81% 2 Pac R.(0.707Vm ) 2

b) FF =

Vrms = 1.11 Vdc

RF = FF 2 − 1 = 0.482 c) TUF =

0.6366Vm Pdc = = 81% Vs I s R.Vm (0.707Vm )

d) PIV = Vm e) CF =

I s ( peak ) Vm . 2 .R = = 2 Is R.Vm

Single Phase Full Wave Rectifier with RL Load: With resistive load, the load current is identical in shape to the output voltage. In practice, most loads are inductive to a certain extent and the load current depends on the value of load resistance R and load inductance L. This is shown in the figure. i

L

D1

+ D3 R -

D2

i

D4

L

-

E

I

a) Circuit Diagram

2

L

i Max L L

i min L

/2

2

b) Wave Form

Fig: Full-wave Bridge Rectifier with R-L Load If Vs = Vm sin ωt = 2Vs is the input voltage, the load current i L can be found from: di L + Ri L + E = 2Vs sin ωt dt which has the solution of the form, L

⎛ R⎞

iL =

⎜ − ⎟t 2Vs E sin (ωt − θ ) + A1e ⎝ L ⎠ − 2 R

1

2 2 where, z = ⎡ R 2 + (ωL ) ⎤ ⎢⎣ ⎥⎦ ⎛ ωL ⎞ θ = tan −1 ⎜ ⎟ ⎝ R ⎠ Case 1: Continuous Load Current iL = I1 At ωt = π ,

⎛ 2Vs sin θ E A1 = ⎜ I 1 + − ⎜ R z ⎝

⎞ ⎛⎜⎝ RL ⎞⎟⎠.⎛⎜⎝ ωπ ⎞⎟⎠ ⎟e ⎟ ⎠


Downloaded from www.jayaram.com.np ⎛ ⎞ 2Vs 2Vs E sin (ωt − θ ) + ⎜⎜ I 1 + − sin θ ⎟⎟ 2 2 R ⎝ ⎠ At ωt = 0, i L (ωt = 0 ) = I 1 IL =

⎛ R ⎞π ⎜− ⎟

2Vs 1 + e ⎝ L ⎠ω E − sin θ I1 = ⎛ R ⎞π 2 R ⎜− ⎟ 1 − e ⎝ L ⎠ω After substitution, we get ⎡ ⎛ R⎞ ⎤ ⎜ − ⎟t 2Vs ⎢ 2 E ⎝ L⎠ ⎥ iL = sin (ωt − θ ) + R e − . sin θ R π ⎛ ⎞ ⎛ ⎞ ⎥ R 2 ⎢ ⎜ − ⎟⎜ ⎟ 1 − e ⎝ L ⎠⎝ ω ⎠ ⎣⎢ ⎦⎥

0 ≤ ωt ≤ π ; i L ≥ 0

V Vm E o i

π

wt

L

α

β

Wave Form Case 2: Discontinuous Load Current Load Current Flows only during the period α ≤ ωt ≤ β ⎛ E ⎝ Vm

α = sin −1 ⎜⎜

⎞ ⎟⎟ ⎠

At ωt = α ; i L (ωt ) = 0 ⎡E ⎤ ⎛⎜⎝ RL ⎞⎟⎠ ⎛⎜⎝ ωπ ⎞⎟⎠ 2Vs A1 = ⎢ − sin (α − θ )⎥ e z ⎢⎣ R ⎥⎦ iL =

⎡E ⎤ RL ⎛⎜⎝ αω− β ⎞⎟⎠ E 2Vs 2Vs sin (ωt − θ ) + ⎢ − sin (α − θ )⎥ e − =0 z z R ⎢⎣ R ⎥⎦

β can be determined from this transidental equation sin x − e − x = 0 by an iterative method of solution. rms diode current 1

⎡ 1 β ⎤2 I r = ⎢ ∫ i L2 dωt ⎥ ⎣ 2π α ⎦ Average diode current 1 β Id = i L (dωt ) 2π ∫α


Downloaded from www.jayaram.com.np Multiphase Star Rectifier: In single phase full wave rectifier, the output contains the harmonics and the frequency of fundamental component is 2 times the source frequency (2 f ) .

In case of multi phase rectifier, the fundamental frequency of the harmonics is also increased

and is 9 times the source frequency (9 f ) . This rectifier is also known as star rectifier. D1 iL D2

1 2

Va

+ a

N

3 4

D3 D4

R

VL -

Circuit Diagram

For simplicity, let us assume three phase star rectifier. Vr = Vm sin ωt 2π ⎞ ⎛ V y = Vm sin (ωt − 120°) = Vm sin ⎜ ωt − ⎟v 3 ⎠ ⎝ 2π ⎞ ⎛ Vb = Vm sin (ωt + 120°) = Vm sin ⎜ ωt + ⎟ 3 ⎠ ⎝



Y

-120

-60

R

30

At ωt = 30°

120 150 180

240 270 300

360

At ωt = 150° V Vr = m ↓ 2 Vm Vy = ↑ 2 Vb = −Vm

Vm ↑ 2 V y = −Vm Vr =

Vb =

60

B

Vm ↓ 2

So, Vr is most positive phase voltage and D1 conducts from 30° to 150°. At ωt = 150° , V y is most positive phase voltage and D2 conducts from 150° to 270°. At ωt = 270° Vr = −Vm Vm ↓ 2 V Vb = m ↑ 2 So, Vb is most positive phase voltage and D3 conducts from 270° to 30° next cycle. Hence output Vy =

consists of 3 phase voltages V R ,V y ,Vb . Each lasting for a period of 120°. Q. A 3φ star rectifier has a purely resistive load with RΩ . Determine a) efficiency b) foam factor c) Ripple factor d) TUF e) PIV of each voltage f) Peak current through a diode. If the rectifier deliver I dc = 30A. At an output of Vdc = 140V . Solution: Average Output Voltage

Vdc =

5π

1 6 π Vm sin ωtdωt ⎛ 2π ⎞ ∫6 ⎜ ⎟ ⎝ 3 ⎠ 5π

1 ⎡Vm cos ωt ⎤ 6 = ⎥π 2π ⎢⎣ ω ⎦ 6 3 = 0.827Vm


Downloaded from www.jayaram.com.np I dc =

Vrms

Vdc 0.827Vm = R R

⎡ ⎢ 1 =⎢ ⎢ 2π ⎢⎣ 3

1

⎤2 5π ⎥ 2 ∫π66 (Vm sin ωt ) dωt ⎥ ⎥ ⎥⎦

= 0.84068Vm I rms =

0.84068Vm R

Pdc = Vdc × I dc =

(0.827Vm )2

Pac = Vrms × I rms =

R

(0.84068Vm ) 2 R

a) η =

Pdc (0.8272) = 96.77% = Pac (0.84068)2

b) F =

Vrms 0.84068Vm = = 1.0165 0.827Vm Vdc

2

c) RF = FF 2 − 1 = d) Vs =

(1.0165)2 − 1 = 0.1824

Vm 2

r 3I s2 R = I rms R

Is =

I rms 3

=

0.4855Vm R

(0.827Vm ) × 2 P = 0.6643 TUF = dc = 3Vs I s 3 × Vm × (0.4855Vm ) 2

e) PIV = 3Vm 3 φ bridge rectifier: i

L

D1

VB Nr

Vy

D3

D5

R

R

+ RVL -

D4

D6

D2

Fig : 3φ Bridge Rectifier

This is a full wave rectifier. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line to line voltage will conduct. Phase Voltages: Downloaded from www.jayaram.com.np /- 37

Downloaded from www.jayaram.com.np Vr = Vm sin ωt

V y = Vm sin (ωt − 120°)

Vb = Vm sin (ωt + 120°) Vry = Vr − V y = Vm sin ωt − Vm sin (ωt − 120°) ⎛ ωt − ωt + 120° ⎞ ⎛ ωt + ωt − 120° ⎞ = Vm .2 sin ⎜ ⎟. cos⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ 3 = 2Vm . cos(ωt − 60°) 2 = 3Vm cos(90° − (30 + ωt )) = 3Vm sin (ωt + 30°)

V yb = 3Vm sin (ωt − 90°) Vby = 3Vm sin (ωt − 200°) V6y

Vry

VD

Vrb

Vyb

Vyr

Vbr

Vby

wt -30

30

60

90

120

150

180

210

240

270

300

330

360

π

Average Output Voltage, Vdc =

=

2

1

⎛ π ⎞ π∫ ⎜ ⎟6 ⎝3⎠ π 1 2

π

∫π

6

Vry dωt

π⎞ ⎛ 3Vm sin ⎜ ωt + ⎟dωt 6⎠ ⎝

3 = 1.6539Vm 1

Vrms

⎡ ⎤2 ⎢1 π 2 π⎞ ⎥ ⎛ = ⎢ ∫π2Vm sin 2 ⎜ ωt + ⎟dωt ⎥ 6⎠ ⎝ ⎢π 6 ⎥ ⎢⎣ 3 ⎥⎦ = 1.6554Vm

Q 1. A 3 φ bridge rectifier has a purely resistive load of R. Determine a) efficiency b) foam factor c) Ripple factor d) TUF e) PIV of each diode f) Peak current through a diode. The rectifier deliver I dc = 60 A at an output voltage of Vdc = 280.7V and the source frequency is 60 Hz . Solution: Vdc = 1.6539Vm , Vrms = 1.6554Vm V V I dc = dc , I rms = rms R R 2 (1.6539Vm ) Pac = Vdc I dc = R -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 38

Downloaded from www.jayaram.com.np (1.6554Vm ) 2 Pac = Vrms I rms = R 2 (1.6539Vm ) .R P a) φ = dc = = 99.8% Pac (1.6554Vm )2 .R V 1.6554Vm b) FF = rms = = 1.009 Vdc 1.6539Vm

R

L

c) RF = FF 2 − 1 = 0.042 d) rms voltage of transformer secondary voltage, Vs =

Vm

-

2

I a = rms value of diode current 2 3I a2 R = I rms R I I a = rms 3 2 I s R = 2 I a2 R

I s = 2I a = e) TUF = =

2 3

I rms

Pdc 3Vs I s

(1.6539Vm )2 ×

2 × 3 .R

R × 3 × Vm × 2 × 1.6554Vm

f) PIV = 3Vm = 3 × 169.72 = 293.9V 3Vm 3 × 169.72 = = 62.9 A R 4.67 V I dc = dc R V R = dc = 4.67Ω I dc

g) Peak diode Current =

3φ bridge rectifier with RL Load: The output voltage becomes

2π 3 3 When Vry = line to line rms voltage , the lead current io can be found from, Vry = 2 × Vry sin ωt

for

π

≤ ωt ≤

di + Rio + E = 2Vry sin ωt which has the solution of the form dt ⎛R⎞ −⎜ ⎟ t 2Vry E sin (ωt − θ ) + A1e ⎝ ω ⎠ − − (1) io = z R L

(

Where, load impedance, z = R 2 + ω 2 L2

)

1 2

⎛ ωL ⎞ And load impedance angle, θ = tan −1 ⎜ ⎟ ⎝ R ⎠

A1 in the equation can be determined from the condition at ωt =

π 3

, io = I o


E

Downloaded from www.jayaram.com.np ⎛ R ⎞⎛ π ⎞ ⎡ 2Vry E ⎛π ⎞⎤ ⎜⎝ c ⎟⎠ ⎜⎝ 3 ω ⎟⎠ sin ⎜ − θ ⎟⎥ e A1 = ⎢ I o + − R z ⎝3 ⎠⎥⎦ ⎣⎢ After substitution, ⎛ R ⎞⎛ π ⎞ ⎡ 2Vry Vry E ⎛π ⎞⎤ ⎜⎝ c ⎟⎠⎜⎝ 3 −t ⎟⎠ E sin (ωt − θ ) + ⎢ I o + − sin ⎜ − θ ⎟⎥ e − io = 2 z R z R ⎝3 ⎠⎥⎦ ⎢⎣ 2π ⎞ π⎞ ⎛ ⎛ i o ⎜ ωt = ⎟ = io ⎜ ωt = ⎟ = I o 3 ⎠ 3⎠ ⎝ ⎝ ⎛ R ⎞⎛ π

− (2 )

⎞

⎛ 2π ⎞ ⎛π ⎞ ⎜ − ⎟⎜ ω ⎟ 2Vry sin ⎜ − θ ⎟ − sin ⎜ − θ ⎟e ⎝ L ⎠ ⎝ 3 ⎠ E ⎝ 3 ⎠ ⎝3 ⎠ Io = − − (3) For I o ≥ 0 ⎛ R ⎞⎛ π ⎞ −⎜ ⎟ ⎜ ω ⎟ R 1 − e ⎝ L ⎠⎝ 3 ⎠ After substitution, ⎡ ⎛ 2π ⎞ ⎛π ⎞⎤ sin (ωt − θ ) + sin ⎜ − θ ⎟ − sin ⎜ − θ ⎟ ⎥ ⎢ 2Vm ⎝ 3 ⎠ ⎝3 ⎠ ⎥ − E For π ≤ ωt ≤ 2π & i ≥ 0 ⎢ io = o ⎛ R ⎞⎛ π ⎞ −⎜ ⎟ ⎜ ω −t ⎟ z ⎢ 3 3 ⎥ R ⎝ L ⎠⎝ 3 ⎠ 1− e ⎥ ⎢ ⎦ ⎣ Rectifier Circuit Design: The design of rectifier involves determining the ratings of diodes like PIV, peak current, etc. The output of rectifier contains harmonics and filters can be used to smooth out DC output voltage of the rectifier and these are known as DC Filter. Due to rectification action, the input current of the rectifier contains harmonics also. An AC Filter is used to filter out some of the harmonics from the supply system. Q. A 3φ bridge rectifier supplies a highly inductive load such that the average load current l dc = 60 A and ripple contain is negligible. Determine the ratings of the diodes if the line to neutral voltage of star connected supply 120V at 60Hz. Solution: Average diode current =

1 2π

⎡ 1 RMS diode current = ⎢ ⎣ 2π

PIV =

5π 6

∫π

I dc dωt =

6

5π 6

∫π

6

5π

1 × 60 × ∫π 6 dωt = 20 A 2π 6

1

⎤2 I dc2 dωt ⎥ = 34.64 A ⎦

3 × Vm = 3 × 2 × 120 = 294V

Output Voltage with LC Filter: Figure

If Vdc is less than Vm current i L will begin to flow at ωt = α which is given by:


Downloaded from www.jayaram.com.np ⎛V ⎞ Vm sin α = Vdc which gives α = sin −1 ⎜⎜ dc ⎟⎟ ⎝ Vm ⎠ The output current i L is given by di Lc L = Vm sin ωt − Vdc dt which can be solved for i L V V 1 ωt ( iL = Vm sin ωt − Vdc )dωt = m (cos α − cos ωt ) − dc (ωt − α ) forωt ≥ α ∫ LC α ωLc ωLc



Chapter – 4 Thyristor Characteristics A cathod p

A

J1

n

J2

p

G R

G gate

J3

n

G cathod

Fig : Thyristor Symbol and Thyristor Junction

A thyrisor is 4 layered semi-conductor device of PNPN structure with 3 PN Junction. It has three terminals. When anode voltage is made positive with respect to cathode, the junction J 1 and J 2 are forward bias, the junction J 2 is reverse bias and only a small leakage current flows

from anode to cathode. The thyristor is then said to be in forward blocking or off state condition and leakage current is known as off state current I d . If anode to cathode voltage Vak is increased to sufficiently large value, the reverse bias junction J 2 will break. This is known as avalancy breakdown and the corresponding voltage is called forward break down voltage (VBD ) . Since the other junction J 1 and J o already forward bias there will be free movement of carriers across all the junctions resulting in a large forward anode current. The device will then be in a conducting state or on state. The anode current must be more than a value known as latching current I L . The latching current I L is the minimum anode current required to maintain the thyristor in on state immediately after a thyristor has been turned on and the gate signal has been removed. Once a thyristor conducts, it behaves liek a conducting diode and there is no control over the device. However, if the anode current is induced below a level known as holding current I h , a depletion region is developed around the junction J 2 due to reduce no. of carriers and thyristor will be in blocking state.

IL

Forward voltage drop gate troggered

IH Forward break down voltage VAk

VBR

Forward leavage current Reverse leavage current

Fig : v-i characteristics


Downloaded from www.jayaram.com.np When the cathode voltage is positive with respect to anode, the junction J 2 is forward biased but the junction J 1 and J 3 are reverse biased. The thyristor will be in reverse blocking state an a reverse leakage current would flow through the device. In practice, the forward voltage is maintain below VBO and the thyristor is turn only by applying a positive voltage between its gate and cathode. Two transistor model of thyristor: A IT

P

A IA= IT

n

n

I2

p

I3

J2 p

G IG

θ1

α1

IB1= IC2

IG1

θ1 P G θ2

α2 IG

IB2

Ik

Ik K

a) Basic Structure

θ2

b) Equivalent Circuit

K

Fig : Two transistor model of Thyristor

A thyristor can be consider as two complementary transistor one PNP transistor θ 1 and the other NPN transistor θ 2 as shown in the figure. The collector current is I c = αI E + I CBO where, I CBO = leakage amount of collect or base junction

α = common base current gain For transistor θ1, I c1 = α 1 I A + I CBO1

− (i )

where, α 1 = current gain for θ 1 I CBO1 = leakage current of collector base junction For transistor θ2, I c2 = α 1 I k + I CBO2

− (ii )

where, α 2 = current gain for θ 2 I CBO2 = leakage current gain for θ 2 Combining equ(i) and (ii), we have: I A = I c1 + I c2 = α 1 I A + I CBO1 + α 2 I k + I CBO2 − (3) But, I k = I A + I G

− (4) Downloaded from www.jayaram.com.np /- 43

Downloaded from www.jayaram.com.np Now, I A = α 1 I A + I CBO1 + α 2 I A + α 2 I G + I CBO2 IA =

α 2 I G + I CBO + I CBO 1 − (α 1 + α 2 ) 1

− (5)

2

[1 − (α 1 + α 2 )]I A = I CBO

1

+ I CBO2 + α 2 I G

If the gate current I G is suddenly increased, this will immediately increase anode current I A , which would further increase α 1 and α 2 . The increase in the value of α 1 and α 2 would furhter inrease I A I. Therefore, there is a regenerative of positive feed back effect. If α 1 + α 2 tends to infinity, the denominator of equation (5) approaches zero, resulting in a large value of anode current I A and the thyristor will turn on with a small gate current. Thryristor turn on: 1. Thermal 2. Light 3. High Voltage 4.

dv dt

5. Gate Current A thyristor a turn on by increasing the anode current. This can be done in one of the following ways: 1. Thermal : If the temperature of thyristor is high, there will be increse in the number of electro-hole pair, which would increase the leakage current and cause α 1 and α 2 to increase due to regenerative action.

α 1 + α 2 may tend to be unity and the thyristor may be turn on. This type of turn on may cause thermal runaway and is normally avoided. 2. Light : If light is allowed to strike the junction of thyrsitor, the electron hole pair will increase and the thyristor may be turn on. 3. High Voltage : If forward anode to cathode voltage is greater than the forward breakdown voltage V BO sufficient, leakage current will flow to initiate regenerative turn on. This type of turn on may be destructive and should be avoided. 4.

dv dt i j2 =

dVi2 di j2 d d a j2 = C j2 V j 2 = C j2 + V j2 dt dt dt dt

( )

(

)

If the rate of rise of anode to cathode voltage is high, the charging current of capacitive junction may be sufficient enough to turn the thyristor, a high value of charging current may damage the thyristor and device must be protected against high

dv . dt

5. Gate Current: If a thyristor is forward bias, the injection of gate current by applying positive voltage between the gate and cathode terminals would turn on the thyristor. As the gate current is increased, the forward blocking voltage is decreased as shown in the figure.


Downloaded from www.jayaram.com.np iT IG1 > IG2 > IG3 IL IA

IG2 IG1

V3 V2

O

IG=O

V1 VB0

VAk

V1 > V2 > V3

Fig : Effect of Gate Current on Forward Blocking Voltage

di Protection : dt Ls

i

T1

Im

+

R2

Dm

load

C2 -

A thyristor requires a minimum time to spread the current conduction uniformly through the junction. If the rate of rise of anode current is very fast compared to the spreading velocity of turn on process, a localize “hot spot” heating will occur due to high current density and the device may fail as a result of excessive temperature. di . Under steady state operation Dm The practical devices must be protected against high dt di can be very high conduct when the thyristor T1 is off. If T1 is free when Dm is still conducting dt di and limited only by the stray inductance of the circuit. In practice is limited by adding a series dt inductor Ls as shown in the figure. di = V s Ls dt dv Protection : dt VA k

+

Vs 0.632Vs

A

S1 Cs

Vs O

t=τ

Rs

t

T1

-


C

k


If switch s1 is close at t = 0, a step voltage will be applied across thyristor T1 and

dv may be dt

dv can be limited by connecting capacitor C s . When T1 is turn on, the dt discharge current of capacitor is limited by resistor Rs . enough to turn on the device.

dv 0.632Vs = dt τ dv 0.632Vs or , = dt Rs C s where, τ = Rs C s Rs =

Vs I TD

− 9i

where, I TD = discharge current

+ Ds Vs

R2 T1

R1 -

Cs

It is possible to use more resistor for I RD =

dv and discharging current such that: dt

Vs R1 + R2

Types of Thyristor: Depending upon the physical construction, turn on and turn off behaviour, thyristor can be broadly classified into following categories: 1. Phase Control Thyristor (SCR) 2. Fast Switching Thyristor (SCR) 3. Gate Turn off Thyristor (GTO) 4. Bidirectional Triode Thyristor (TRIAC) 5. Reverse Conducting Thyristor (RCT) 6. Static Induction Thyristor (SITH) 7. Light Activated Silicon-Controlled Rectifier (LASCR) 8. FET Controlled Thyristor (FET-CTH) 9. MOS Controlled Thyristor (MCT)


Downloaded from www.jayaram.com.np Phase Control Thyristor: This type of thyristor generally operate at the line frequency and is turn off by natural communication. The turn off time is of the order 500-100 µs . This is most suited for low speed switching application and is also known as converter thyristor. The modern thyristor use an amplifying gate where auxiliary thyristor TA is gated by a gate signal and then the amplified output of the TA is applied as a gate signal to the main thyristor Tm . A

IG

TA

Tm

K

Fig : Amplifying Gate Thyristor Fast Switching Thyristor: These are used in high speed switching application with forced communication (example chopper, inverter). They may have fast turn off time. Generally in a range 5-15 µs depending on the voltage range. Gate Turn-off Thyristor: A GTO line an SCR can be turn on by applying a positive gate signal. However it can be turn off by negative gate signal. It eliminates the requirement of equipment. Bi-directional Triod Thyristor: MT 2

MT 1

TA MT 2

TRIAC Sumbol

MT 2

Equivalent of TRIAC

A TRIAC can conduct in both directions and is normally used in AC phase control. It can be considered as two SCR connected in anti-parallel with a common gate connection as shown in the figure. FET Controlled Thyristor:

M1

T1

G

R


Downloaded from www.jayaram.com.np A FET controlled thyristor combines a MOSFET and thyristor in parallel as shown in the figure. If a sufficient voltage is applied to the gate of MOSFET typically 3r, a triggering current for the thyristor generated internally. MOS Controlled Thyristor: + ve → N − channel − ve → P − channel

G

pnp

(off-FET) (n-channel)

on FET(p-channel) off

npn

A

on

G

a) Equivalent Circuit

k

It is basically a thyristor with two MOFET built into the get structure. One MOSFET is used for turning on the MCT and other for turning off the device with the application of negative voltage pulse, on FET gets turn on and off FET is off. With on FET on, current begins to flow from anode A through ON FET and then as the base current and emitter current of NPN transistor and then to the cathode K. This turn on NPN transistor as a result collector current begins to flow in NPN transistor as OFF FET is off. This collector current of NPN Transistor acts as the base current of PNP transistor. Subsequently, PNP transistor is also turn on, MCT turn on. For turning off, OFF FET is energise by positive voltage pulse at the gate. With the application of positive voltage off FET is turn on and ON FET is turn off. After off FET is turn on emitter base terminal of PNP transistor are short circuited by off FET. So anode current begins to flow through off FET and therefore base current of PNP transistor begins to decrease. Further, the collector current of PNP transistor that forms a base current of NPN transistor also become to decrease. As a result MCT turn off. Thyristor Turn Off: v Vm i + V

VAK

R1

O

π

2π

wt

i Vm RL wt

O α

Fig: Line Commutated Thyristor Circuit


Downloaded from www.jayaram.com.np A thyristor which is on the ON state can be turned off by reducing the forward current to a level below the holding current I H . Series Operation of Thyristor: For high voltage application, two or more transistors can be connected in series to provide the voltage rating. However, due to production spread the characteristics of thyristor of the same type are not identical. ON-state T2 T1

off-state V1 V2 O

V

Is

Fig: Off State Characteristics of Two Thyristor

Series Operation of Thyristor: Consider n thyristor connected in series as shown in the figure:



String current

R VGm String voltage

I 1 = I − I bmn & I 2 = I − I rms where, I = total string current Voltage across SCR1 = Vbm = I1 R Voltage across (n+1) SCR = (n − 1)I 2 R Now, Vs = I 1 R + (n − 1)I 2 R = Vbm + (n − 1)(I − I bmx )R = Vbm + (n − 1)[I1 − (I bmx − I bmn )]R = Vbm + (n − 1)I1 R − (n − 1)R∆I b where, ∆I b = I bmx − I bmn RI 1 = Vbm As, Vs = nVbmn − (n − 1) × ∆I b R=

nVbm − Vs (n − 1)∆I b

Parallel Operation of Thyristor: When thyristor are connected in parallel, the load current is not shared equally due to the difference in their characteristics. If a thyristor carries more current then that of others, it’s power dissipation increases there by increasing the junction temperature and decreasing the internal resistance. This inturn will increase it’s current sharing and may damage the thyristor. This thermal runaway may be avoided by having a common heat sink so that all unit operate at same temperature. A small resistance may be connected in series with each thyristor to force equal current sharing but there will be considerable power loss in series resistance. R1 I1

IT

T1

R2 I2

T2

Fig : Current Sharing Thyristor

Thyristor Firing Circuit: In thyristor converter, different potential existed various terminals. The power circuit is subjected to a high voltage usually greater than 100V and gate circuit is held at low voltage typically -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 50

Downloaded from www.jayaram.com.np 12V to 30V and isolation circuit is required between individual thyristor and it’s gate pulse generating circuit. The isolation can be accomplish by either pulse trf or opto couples. ILED

+Vc 0

R1 +

A T1

V1

D1

G

k

R optocoupler Photoser

An optocoupler could be a phototransistor or photoSCR as shown in the figure. A short pulse to the input of ILED D1 turns on the photo SCR T1 and power thyristor TL is trigger. G

V1

R1

gate voltage

N2

N1

θ1 D1 C1

A simple isolation arrangement with pulse transformer is shown in the figure. When a pulse of adequate voltage is applied to the base of switching transistor θ1, the transistor saturate and DC voltage Vcc appears across the transformer primary, inducing a pulse voltage on the transformer secondarily which is applied between the thyristor gate and cathode terminal. When pulse is removed from base of the transfer turn off and voltage of opposite polarity is induced across the primary and the freewheeling diode Dm conducts. Vs R

Uni-junction Transistor (UJT)

RB2 E +

C -

B1

+

R B1 VB1 -

Fig : Circuit

Negative resistance region

VE

IE

B2

VE Vp

Saturation Region Vp

Peak Point Valley Point

Vv

VE(Set)

T

2T

T

2T

Vv VB1

IE IE (VA) 0

Ip

Iv

Fig : Characteristics


Fig : Waveform

Downloaded from www.jayaram.com.np UJT is commonly used for generating triggering signals for SCR. A basic UJT triggering circuit is shown in the figure. UJT has three terminals called emitters E, B1 and B2. When DC supply voltage Vs is applied, the capacitor C is charged through resistance R. Since the emitter circuit of UJT is in the open state, the time constant of charging circuit is T1 = RC . When the emitter voltage V E which is same as the capacitor voltage Vc reaches the peak voltage V p , the UJT turns on and the capacitor C will discharge through RB1 at a rate determine by time constant

τ 2 = RB1 XC . When the emitter voltage V E decays to the valley point VV , the emitter ceases to conduct, the UJT turns off and the charging cycle is repeated. The waveform of emitter and triggering voltage are shown in the figure C. The period of oscillation is approximately given by: T=

⎛ 1 ⎞ 1 ⎟⎟ ≈ RC ln⎜⎜ f ⎝1 −η ⎠

where η is intrinsic stand off ratio.

η = 0.51 to 0.82 UJT will turn on if Vs − I p R > V p R

Vs − VV IV

− (ii )

The width of triggering pulse is t g = RB1 .C RB2 =

10 4 ηV s

− (iii )

V p ≈ ηVs + VD (= 0.5V )

- (iv)

Q. Design triggering circuit of figure between the parameter of UJT are Vs = 30V , η = 0.51 V, I p = 10 µA, Vv = 3.5V , I V = 10mA. . The frequency of oscillation is f = 60 Hz and width of triggering pulse is t g = 50 µs . Solution: 1 1 T= = = 16.67ms f 60 V p = ηVs + 0.5 = 0.51 × 30 + 0.5 = 15.8V Let C = 0.5µF


R< R>

Vs − V p Ip

Downloaded from www.jayaram.com.np =

30 − 15.8 = 1.42mΩ 10

Vs − VV 30 − 3.5 = = 2.65 KΩ IV 10

⎛ 1 ⎞ ⎟⎟ T = RC ln⎜⎜ ⎝1 −η ⎠ 1 ⎛ 1 ⎞ = R × 0.5 ln⎜ ⎟ 60 ⎝ 1 − 0.5 ⎠ R = 46.7 KΩ which falls within limiting value we have t g = RB1 .C 50 = 100Ω 0.5 10 4 = = 654Ω ηV s

R B1 = R B2

Programmable Unijunction Transistors (PUT): +V3 Rv

R1

anode Gate

anod VA

Cathod

a) Symbol

+

+

+ Rk

R2

VG

-

VRk -

b) Circuit

PUT is a small thyristor shown in figure (a). PUT can be used as relaxation oscillator as shown in figure (b).The gate voltage VG is maintained from the supply by the resistor divided by R1 and R2 and determine the peak voltage V p in case of UJT V p is fixed for a device by DC supply voltage. But V p of PUT can be varied by varying resistor dividor R1 and R2 . If anode V A is less than ate voltage VG , the device will remain in its off state. If V A exceeds gaet voltage VG by one diode forward voltage Vd , the peak is reached and the device turns on. R2 .Vs R1 + R2 Intrinsic Ratio Vp R2 = η= Vs R1 + R2 Period of Oscillation Vp =

⎛ Vs ⎞ ⎛ ⎞ ⎛ 1 ⎞ ⎟ = RC ln⎜1 + R2 ⎟ ⎟⎟ = RC ln⎜ T = RC ln⎜⎜ ⎜ ⎜ ⎟ R1 ⎟⎠ ⎝1 −η ⎠ ⎝ ⎝ Vs − V p ⎠ The gate current I G at valley point is given by V I G = (1 − η ) s RG Downloaded from www.jayaram.com.np /- 53

Downloaded from www.jayaram.com.np where, RG = R1 = R1 =

R1 R2 R1 + R2 RG

η RG 1 −η

Q. Design triggering circuit of figure (i) the parameter of PUT are Vs = 30V , I G = 1mA frequency of oscillation f = 60Hz, pulse width is t g = 50µs . Peak triggering voltage is VRK = 10V . Solution: 1 1 = = 16.67ms f 60 V p = VRk = 10V T=

Let,

C = 0.5µF t g = Rk .C 50 = 100Ω 0.5 V p 10 1 = = η= Vs 30 3 RK =

⎛ 1 ⎞ ⎟⎟ T = RC ln⎜⎜ ⎝1 −η ⎠ ⎞ ⎛ ⎜ 1 1 ⎟⎟ ⎜ = R × 0.5 ln 60 ⎜ 1⎟ ⎜1− ⎟ 3⎠ ⎝ R = 82.2 KΩ V I G = (1 − η ) s R2 ⎛ 1 ⎞ 30 R 2 = ⎜1 − ⎟ = 20 KΩ ⎝ 3⎠ 1



Chapter - 5 Thyristor Commutation Technique Introduction: A thyristor is normally switched on by applying a pulse of gate signal. When a thyristor is in conduction mode it’s voltage drop is small. Once the thyristor is turn on, and the output requirement are satisfied, it is normally necessary to turn it off. Commutation is the process of turning off a thyristor and it normally causes transfer of current flow to other parts of the circuit. A commutation circuit normally uses additional component to accomplish the turn off. There are many techniques to commutate a thyristor. However, these can be broadly classified into two types: 1. Natural Commutation 2. Forced Commutation Natural Commutation:

+

Vs= Vmsinwt

+ Vo -

io R

V0 i 0= V0R

-

a) Circuit Diagram 2π

wt

b) Wave-Form Fig : Thyristor with natural commutation

If the source voltage is AC, the thyristor current goes through a natural zero and a reverse voltage appears across the thyristor. The device is then automatically turn off due to the natural behaviour of source voltage. This is known as natural commutation of line commutation . Forced Commutation: In some thyristor circuit, the input voltage is DC and the forward current of thyristor is forced to zero by an additional circuitry called commutation circuit to turn off the thyristor. This technique is called as forced commutation. The forced commutation of thyristor can be achieved by seven ways and can be classified as: 1) Self Commutation 2) Impulse Commutation 3) Resonant Pulse Commutation 4) Complementary Commutation 5) External Pulse Commutation 6) Load Side Commutation 7) Line Side Commutation


Downloaded from www.jayaram.com.np Self Commutation: i

+

+

T1 Vs

-

V2

L

+ Vc -

C

a) Circuit Diagram In this type of commutation, a thyristor is turn off due to the natural characteristics of the circuit. When the thyristor T1 is switch on, the capacitor charging current is given by: di 1 + idT + Vc (t = 0) dt c ∫ With initial conditions Vc (t = 0) = 0 ; (t = 0) V s = V L + Vc = L

The charging current is C sin ω m t L

i (t ) = Vs

And the capacitor voltage as : Vc (t ) = Vs (1 − cos ω m t ) where ω m =

1 LC

i(E) Vs

C /2

O

π

O

w 2t2

ωmto = π π to = ωm t o = π LC

O

π t0

b) Waveform

At time t = t o = π LC charging current becomes zero and thyristor t1 is switch off itself, t o is commutation time. This method of turning off a thyristor is called self commutation and the thyristor t1 is said to be self commutated. The figure shows a typical circuit where the capacitor is an initial voltage of − Vo when the thyristor t1 is fire the current that will flow through a circuit is given by:


Downloaded from www.jayaram.com.np T1

L

di 1 + idt + Vc (t = 0) = 0 dt C ∫

i

With Vc (t = ) = Vo , i (t = 0) = 0 i (t ) = Vo

L + Vc -

C

a) Circuit

C sin ω m t L

Capacitor Voltage, Vc (t ) = −Vo cos ω m t i(t)

O

w mt

π

V0

O

π/L

π

-V0

After time t = t o = π LC , the current becomes zero and the capacitor voltage is reverse to Vo , t r is called reversing time. Q. A thyristor circuit is shown in the figure. If thyristor T1 is switched on at t=0, determine the conduction time of thyristor T1 and the capacitor voltage after T1 is turned off. The circuit parameters are L = 10µH , C = 50µF & Vs = 200V . The inductor carries an initial current of I m = 250 A . T1 i

Im Vs

V0

+ -

C

Solution: di 1 + idt + Vc (t = 0) dt C ∫ With initial condition i (t = 0) = I m & Vc (t = 0) = Vo = Vs By solving above equation, i (t ) = I m cos ω m t 1 where ω m = LC 1 L sin ω m t + Vs Vc (t ) = ∫ idt + Vc (t = 0 ) = I m C C Vs = L


Downloaded from www.jayaram.com.np i (t)

Vc (t)

Im

O

π/2

w mt

Conduction time of thyristor T1 =

π/2

π 2

ωm =

π LC 2

Capacitor Voltage after T1 is turned off = I m

w mt

= 35.12 µs

L + Vs C

Impulse Commutation: Im T3 Vs

T1

+

Dm T2

Fig : Impulse Commutated Circuit

It is assumed that the capacitor is initially charge to voltage of − Vo . Let us assume that thyristor T1 is initially conducting and carrying a load current of I m . When the auxillary thyristor T2 is fire, thyristor T1 is reverse bias by the capacitor voltage and T1 is turned off. The current through T1 and cease to flow and the capacitor would carry the load current .The capacitor will discharge from − Vo to 0 and then charge to the DC input voltage Vs . When the capaitor current falls to zero and thyristor T2 turns off. The charge reversal of capacitor from Vo (= Vs ) to − Vo is then done by firing thyristor T3 . Thyristor T3 is self commutated. Time required for capacitor to discharge from − Vo to zero is called circuit turn off time (toff ) . Vo =

I m t off 1 toff I dt = m C ∫0 C

t off =

Vo C Im

Q. An impulse commutated thyristor circuit is shown in the figure. Determine the available turn off time of the circuit if Vs = 210V , R = 10Ω, C = 5µF & Vo = Vs .


Downloaded from www.jayaram.com.np +

-

Solution:

V0=Vs

1 Vs = Vc + Ri = ∫ idt + Vc (t = 0) + Ri v C With initial condition, Vc (t = 0) = −Vo = −Vs

+

Vc (t ) =

i

R

Vs T2 -

t

2Vs − RC e R Capacitor Voltage i (t ) =

T1

t − ⎛ 1 RC ⎜ ( ) idt V t 0 V 1 2 e + = = − c s⎜ C∫ ⎝

⎞ ⎟ ⎟ ⎠

At t = t off

Vc (t = t off ) = 0

t off ⎛ ⎞ − ⎜ Vs 1 − 2e RC ⎟ = 0 ⎜ ⎟ ⎝ ⎠ 2 t off = RC ln = 347 µs

Line-side Commutation: T1

Im

L T3

C

Vs

Dm

Lr T2

In this type of commutation, the discharging and recharging of capacitor are not accomplished through the load and the commutation circuit can be tested without connecting the load. When the thyristor T2 is fired, the capacitor is charged to 2Vs and T2 is self commutated. Thyristor T3 is fired to reverse the voltage of capacitor to − 2Vs and T3 is also self commutated. Assuming that thyristor T1 is conducting and carries a load current of I m thyristor T2 is fired to turn off T1 . The inductor L carries the load current I m and equivalent circuit during the commutation period is shown in the figure. L +

+

i (t) -

Im Vs

Vs

+ -

Vc(t)

-

di 1 + idt + Vc (t = 0) dt C ∫ With initial condition i (t = 0 ) = I m and Vc (t = 0) − 2Vs , Capacitor current Vs = L


Downloaded from www.jayaram.com.np i (t ) = I m cos ω m t + 3Vs

C sin ω m t L

− (1)

And capacitor voltage, L sin ω m t − 3Vs cos ω m t + Vs − (ii ) Vc (t ) = I m C 1 where, ω m = LC The circuit turn off time is obtained from the condition Vc (t = t off ) = 0 of equation (2). Commutation Circuit Design and Commutation Capacitor: The design of commutation circuit require determining the value of capacitor C and inductor L. In selecting a commutation capacitor, the specification of peak, rms and average current and peak to peak voltage must be satisfied.



Chapter - 6 Controlled Rectifier Introduction: A diode rectifier provides a fixed output voltage only. In order to obtain controlled output voltage, phase controlled thyristor are used instead of diode. The output voltage can be varied by controlling the decay or fixing angle of thyristor. Since, these rectifiers convert from AC to DC, these controlled rectifiers are also called AC to DC converter and are used extensively in industrial applications specially variable speed drive ranging from fractional h.p. (horse power) to mega watt power level. The phase controlled converter can be classified into two types depending on the input supply. 1. Single Phase Converter 2. Three Phase Converter Each type can be sub-divided into 1. Semi Converter 2. Full Converter 3. Dual Converter

Semi Converter is the one quadrant converter and it has one polarity of voltage and current. Full Converter is a two quadrant converter and the polarity of its output voltage can be either positive or negative. However, the output current of full converter has only one polarity. Dual Converter can operate in four quadrant and both the output voltage and current can be either positive or negative. Vo

Vo

Vo Io

Io

Io

Fig : Semi Converter

Fig : Full Converter

Fig : Dual Converter

Principle of Phase Controlled Converter Operation:


Downloaded from www.jayaram.com.np T1

Vs

iv

Vm

+ Vs

Vp

o

Vo

α

π

2π

α

π

2π

α

π

2π

α

π

2π

V0

a) Circuit Diagram

Vm

i0

wt

V0/R

wt

VT1

o

wt

-Vm

b) Wave Form

During positive half of input voltage the thyristor anode is positive with respect to its cathode and the thyristor is said to be forward bias. When the thyristor T1 is fired at ωt = α , thyristor T1 conduct and input voltage appears across the load. When input voltage starts to be negative at ωt = π the thyristor anode is negative with respect to its cathode and thyristor T1 is said to be reverse bias and it is turned off. The average output voltage can be formed from Vm 1 π (1 + cos α ) Vdc = V td t sin ω ω = m 2π ∫α 2π V α = 0, Vdc = m When, π α = π , Vdc = 0 RMS Output Voltage is given by: Vrms

1 2

V ⎡ 1 ⎤ = ⎢ ∫ (Vm sin ωt ) dωt ⎥ = m 2 ⎣ 2π α ⎦ π

2

1

⎡1 ⎛ sin 2α ⎞⎤ 2 ⎢ π ⎜ π − α + 2 ⎟⎥ ⎠⎦ ⎣ ⎝

Q. If the converter of above figure has a purely resistive load of R and the delay angle is α = determine a) Rectification Efficiency e) PIV of thyristor T1 Solution:

α=

b) Foam Factor c) Ripple Factor d) TUF

π

2 -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 62

π 2

,

Downloaded from www.jayaram.com.np Vdc = I dc =

Vm

π

(1 + cos α ) = 0.1592Vm

Vdc 0.1592Vm = R R 1

V ⎡1 ⎛ sin 2α ⎞⎤ 2 Vrms = m ⎢ ⎜ π − α + ⎟ = 0.3536Vm 2 ⎣π ⎝ 2 ⎠⎥⎦ 0.3536Vm V I rms = rms = R R (0.1592Vm )2 Pdc = Vdc × I dc = R (0.3536Vm )2 Pac = Vrms .I rms = R

(0.1592Vm ) .R P a) Rectification Efficiency, η = dc = = 20.27% 2 Pac R(0.3536Vm ) 2

b) Foam Factor, FF =

Vrms 0.3536Vm = = 2.21 0.1592Vm Vdc

c) RF = FF 2 − 1 = 1.983 d) Vs =

Vs

0.3536Vm .I s = R 2

Vm

(0.1592Vm ) 2 R Pdc = = 0.1014 I s I s RVm × (0.3536Vm )

o

2

TUF =

e) PIV = Vm

π+α α

π

2π

v0

α

Single Phase Semi Converter:

π π+α

2π

wt

IT1, ID2

T1 Vp

io T3

Vs

Vo L T4

a) Circuit

R

+

T2

E

α

π

2π

α

π

2π

wt

IT2, ID1 Ia

IDma

wt

IA

i0 Ia wt

b) Wave Form

Fig shows a single phase semi converter with highly inductive lead such that current is as same continuous and ripple free. Downloaded from www.jayaram.com.np /- 63

Downloaded from www.jayaram.com.np During a positive half cycle, thyristor T1 is forward bias when T1 is fired at ωt = α the load is connected to the input supply through T1 and T2 . During the period α ≤ ωt ≤ π . During the period from α ≤ ωt ≤ π + α , the input voltage is negative and freewheeling diode Dm is forward bias. During negative half cycle of input voltage thyristor T2 is forward bias and firing of T2 at

ωt = π + α will reverse bias Dm . The load is connected to the supply through T2 and D1 . Average Vdc =

output voltage is

2 2π

π

∫α

Vm sin ωtdωt =

Vm

π

(1 + cos α ) 1

V ⎡1 ⎛ sin 2α ⎞⎤ 2 = m ⎢ ⎜π − α + ⎟ 2 ⎠⎥⎦ 2 ⎣π ⎝ Single Phase Semi Converter with R-L Load: In practice a load has finite inductance. The load current depends on the values of load resistor R, load inductor L and battery voltage E as shown in the figure above. A converter operation can be divided into two mode: mode 1 and mode 2.

Mode 1: This mode is valid for 0 ≤ ωt ≥ a during which the freewheeling diode Dm conducts. di + Ri L + E = 0 − (1) dt Initial conduction, i L (ωt = 0 ) = I Lo L

I L1 = I Lo e

−

R L .t

− E⎛ − ⎜⎜1 − e L.t R⎝

R

⎞ ⎟ for i L ≥ 0 ⎟ ⎠

At ωt = α , load current becomes I Lo e

− (2 )

⎛ − R ⎞⎛ α ⎞ ⎜ ⎟⎜ ⎟ ⎝ L ⎠⎝ ω ⎠

⎛ R ⎞⎛ α ⎞ −⎜ ⎟ ⎜ ⎟ ⎤ E⎡ − ⎢1 − e ⎝ L ⎠ ⎝ ω ⎠ ⎥ R ⎢⎣ ⎥⎦

for i L1 ≥ 0

Mode 2: This mode is valid for α ≤ ωt ≤ π while thyristor T1 conducts if Vo = − 2Vo sin ωt is the input voltage the load current I 1, 2 during mode 2 can be found as LdiI 2 + RI L 2 + E = 2Vs sin ωt , whose solution is of the form,

I L2 =

(

)

2Vs / Z sin (ωt − θ ) + A1e

[

z = R 2 + (ωL )

⎛R⎞ −⎜ ⎟.t ⎝I ⎠

−

E R

For I L2 ≥ 0

− (5) where,

]

1 2 −2

⎛ ωL ⎞ ⎟ ⎝ R ⎠

θ = tan −1 ⎜

A1 can be determined from the initial condition at ωt = α i L2 = i L1 ; α

⎡ ⎤ ⎛⎜ ⎞⎟ ⎛⎜ ⎞⎟ 2 E A1 = ⎢ I L1 + − .Vs sin (α − θ )⎥ e ⎝ L ⎠ ⎝ ω ⎠ R z ⎣ ⎦ R

− (6 )

With substitution of A1 , -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 64

I L2

Downloaded from www.jayaram.com.np R α ⎤ ⎛⎜⎝ L ⎞⎟⎠⎛⎜⎝ ω −t ⎞⎟⎠ 2 2 E ⎡ E = .Vs sin (ωt − θ ) − + ⎢ I L1 + − .Vs sin (α − θ )⎥ e z R ⎣ R z ⎦

for I L2 ≥ 0

− (7 )

Applying initial condition i L2 (ωt = π ) = I Lo ; we get : I Lo

⎛ R ⎞ (α −π ) ⎧⎪ 2 ⎫⎪ ⎜ ⎟ = ⎨ .Vs sin (π − θ ) − sin (α − θ )e ⎝ L ⎠ ω ⎬ ⎪⎩ z ⎪⎭

⎡ 1 ms current of thyristor, I r = ⎢ ⎣ 2π

T

∫α I

Average current of thyristor, I av =

2

1 2π

⎤ L 2 dω t ⎥ ⎦ T

∫α i

L2

1− e

⎛ R ⎞π ⎜− ⎟ ⎝ L ⎠ω

−

E R

− (8) R

for I Lo ≥ 0

1 2

dωt

Single Phase Full Converter: For shown a single phase full wave converter with highly inductive load so that the load current is continue and ripple free. Vs T1, T2

T 3,T 4

Vo

T1 Vp

α

T3

π

π+α

wt

2π

R Vo

Vs L T4

T2 E wt

Io=Ia

wt

During the positive half cycle thyristor T1 and T2 are forward biased and io = ia when these two thyristor are fired simultaneously at ωt = α , the load is connected to the input supply through T1 and T2 . Due to inductive load thyristor T1 and T2 will continue to conduct beyound ωt = π even those the input voltage is already negative. Single Phase Dual Converter: Lr Lr /2

T1

T3

+ Vs -

/2

+

T12

i0

V01 V0

+ Vs -

V02 -

T4

T42

T2

a) Circuit Diagram


T13

T11


Vm

π1

α1

π=α 1

wt

2π

Vo1

Vm sinwt Converter 1 o/p π

α1

π=α 1

wt

2π

Vm sinwt Vo2

-Vm sinwt Converter 2 o/p

Vm sinwt

If two of these full converters are connected back to back as shown in the figure above both output voltage and load current flow can be reversed. The system will provide four quadrant operation and is called dual converter. The delay angles are controlled such that one converter operates as rectifier and the other as inverter. If α 1 and α 2 are delay angles of converter 1 and converter 2 respectively, the corresponding output voltages are VDC1 and VDC 2 . V DC1 =

2Vm

π

cos α 1

&

VDC 2 =

2Vm

π

cos α 2

V DC1 = −VDC 2 cos α 1 = − cos α 2 = cos(π − α 2 )

α1 = π − α 2 α 2 = π − α1 Since the instances output voltages of two converter are out of phase there will be instantaneous voltage difference. This will result in circulating current between the two converters. The circuiting current will not flow through the load and is normally limited by a circulating current reactor Lr as shown in the figure. 1 ωt ir = θ r d (ωt ) ωLr ∫2π −α1 =

1 ωLr

∫ π α (V ωt

2 −

1

o1

)

+ Vo2 dωt


Downloaded from www.jayaram.com.np ωt 1 ⎡ = − + V sin ω td ω t m ∫2π −α1 − Vm sin ωtdωt ⎤⎥⎦ ωLr ⎢⎣ ∫2π −α1 ωt

ir =

2Vm (cos ωt − cos α 1 ) ωLr

Q. A single phase dual converter is operated from 120 V, 60Hz supply and the load resistance is R = 10Ω , the circulating inductance is Lr = 40mH . Delay angles are α 1 = 60° & α 2 = 120° .

Calcuate the peak circulating current and peak current of converter 1. Solution: At ωt = 2π I r (max ) =

2Vm (cos 2π − cos 60°) = 11.25 A ωLr

Vm = 16.97 A R Peak Current of Converter1 = 16.97 + 11.25 = 28.22 A

Peak Load Current =

Three Phase Half Wave Converter: T1 1 R 3 2 T2

T3

a) Circuit Diagram

The circuit functions in a manner such that only one thyristor is conducting at any given instant, the one which is connected to the phase voltage having highest instantaneous positive value. Here no any thyristor can be trigger below phase angle of 30° because it remains reverse bias by other conducting phases. The firing angle α for a particular thyristor connected in a particular phase is therefore measure from 30° w.r.t. corresponding phase voltage. T1 can conduct from ωt = 30° to150° as phase 1 is most positive in the period 30° to 150°. Similarly T2 can conduct from ωt = 150°to 270° and T3 from ωt = 270° to 30° of next cycle. With the resistive load, there are two modes of conduction:

1. Continuous Conduction Mode (α ≤ 30°)

2. Discontinuous Conduction Mode (α > 30°) When the firing angle α is taken in between 0° to 30° from the cross over point, the load current is continuous. This is due to the fact that the maximum value of conduction angle of thyristor is 120°. If α is more than 30°, conduction angle will be less than 120° and hence output voltage and current becomes discontinuous.


Downloaded from www.jayaram.com.np V2

V1

o

30

60

120

150 180

V3

240

230

wt

360

α=300 Continuous conduction)

5π

1 α+ 6 V sin ωtdωt Average load voltage for continuous conduction mode is Vdc = 2π ∫α + π6 m 3 3 3Vm cos α = 2π 1 π Average load voltage for discontinuous conduction mode is Vdc = V sin ωtdωt π 2π ∫6 +α m 3 3V ⎡ ⎛π ⎞⎤ = m ⎢1 + cos⎜ + α ⎟⎥ 2π ⎣ ⎝6 ⎠⎦ Rms Output Voltage are 1

Vrms

⎛1 ⎞2 3 cos 2α ⎟⎟ = 3Vm ⎜⎜ + ⎝ 6 8π ⎠

Vrms

⎡ρ α 1 ⎛π ⎞⎤ 2 = 3Vm ⎢ − + sin ⎜ + 2α ⎟⎥ ⎝3 ⎠⎦ ⎣ 24 4π 8π

for continuous conduction 1

for discontinuous conduction

Design of Converter Circuits: The design of converter circuit requires determining the rating of thyristors and diodes. The thyristors and diodes are specified by average current, r.m.s. current, peak current and PIV in the case of controlled rectifier, the current rating of device depend on the delay angle α . The output of converter contains harmonics that depend on delay angle and filters must be design to remove the harmonics.

Q. A 3φ half wave converter is operated from 3φ star connected 208V, 60Hz supply and the load resistance R = 10Ω . If it is required to obtain an average output of 50% maximum possible output voltage then calculate a) Delay angle α b) Rms and Average output current c) Average and rms thyristor current d) Rectification efficiency e) TUF f) Input Power Factor (IPF) Solution: 208V, 60Hz R=10 Ω


Downloaded from www.jayaram.com.np Phase Voltage (Vs ) =

208 3

V

Vm = 2Vs = 169.83V

3 3Vm 3 3 × 169.89 = 2π 2π

Vdc (max ) =

Vdc = 0.5 × Vdc (max ) = 70.23V I dc =

Vdc 70.23 = = 7.023 A R 10

Vdc =

3 3Vm cos α 2π

3 3Vm cos α 2π α = 59.9 > 30° (so it is working in discontinuous conduction mode ) 70.23V =

So, we have: Vdc =

3Vm 2π

⎡ ⎛π ⎞⎤ ⎢1 + ω ⎜ 6 + α ⎟⎥ ⎝ ⎠⎦ ⎣

α = 67.70 1

Vrms

⎡5 α 1 ⎛π ⎞⎤ 2 = 3Vm ⎢ − + sin ⎜ + 2α ⎟⎥ = 97.74 A ⎝3 ⎠⎦ ⎣ 24 4π 8π

I rms =

Vrms 94.74 = = 9.474 A R 10

c) Average current of thyristor, I A = Rms current of thyristor, I R =

I rms 3

I dc 7.023 = = 2.34 A 3 3

= 5.47 A

70.23 × 7.023 = 54.95% 94.74 × 9.474 P 70.23 × 7.023 e) TUF = dc = = 0.25 208 3Vs I s 3× × 5.47 3

d) η =

2 .R = 9.47 2 × 10 = 896.81W f) Output Power, Po = I rms

Input Power , Pi = 3Vs I s = 1970.84 Input P.F. =

896.81 = 0.455(lag ) 1970.84



Chapter – 8 DC Choppers DC Choppers: DC Choppers converts fixed DC input voltage to variable DC output voltage. It can step up or step down DC voltage. The main principal is to chop a constant DC voltage for a particular time interval and by changing the chopping time, output DC voltage can be controlled. Generally, power transistors are used as switch for chopper operation. Principal of Step-Down Operation: Vo Vs Chopper +

+

sw

VH

io t1

t

+ Vo -

Vs

t2

io

R

Is=Vs/R

-

a) Circuit Diagram t1

t2 t T

b) Wave Form

a) Fig : Step down chopper with resistive load When switch s w is closed for a time t1 , the input voltage Vs appears across the load. If the switch s w remains off for a time t 2 , the voltage across the load is zero. The chopper switch can be implemented by using 1. Power BJT 2. Power MOSFET 3. Power GTO 4. Forced Commutated Thyristor The average output voltage is given by: Va =

1 t 1 Vo dt = ∫ T 0 T

∫

t1

0

Vo dt =

Vs V .t1 = t l ∫ Vs = s t1 = kVs T T

t1 = duty cycle T T = chopping period

where, k =

f = chopping frequency Rms Output Voltage is


Downloaded from www.jayaram.com.np 1 2

⎤ ⎡ 1 kT Vo = ⎢ ∫ Vs2 dt ⎥ = kVs 0 ⎦ ⎣T Assuming a loss less chopper, the input power to the chopper is same as the output power and is given by

Pi =

1 T

∫

kT

0

1 T

Vo idt =

Vo2 Vs2 dt k = ∫0 R R 0

The effective input resistance seen by the source is Ri =

Vs V R = s = V k Ia k s R

The duty cycle k can be vary from 0 to 1 by varying t1 , T or f . Therefore, the output voltage can be varied from 0 to Vs by controlling k/ Two types of operation can be defined: a) Constant Frequency Operation: The chopping frequency f or chopping period 'T ' is kept constant and on time t1 is varied and this type of control is known pulse width modulation (PWM) control. b) Variable Frequency Operation: The chopping frequency ' f ' is varied either 'on' time t 1 or off time t 2 is kept constant. This is called frequency modulation. Q. A step down DC chopper has a resistive load of R = 10Ω and the input voltage Vs = 220V . When the chopper remains on its voltage and chopping frequency f = 1KHz . If the duty cycle is 50% determine (i) average output voltage VA ii) Rms output voltage Vo iii) chopper efficiency iv) effective input resistance Ri of the chopper v) on-time and off-time of the chopper. Vch Solution: Vs = 220V , R = 10Ω, Vch = 2V , k = 0.5, f = 1Kmz i) Va = k (Vs − Vch ) = 109V

Vo

Vs

ii) Vo = k (Vs − Vch ) = 154.15 1 kT Vo2 1 iii) Output Power, Po = ∫ dt = 0 T R f Input Power, Pi = =

iv) η =

1 T

∫

kT

0

Vs idt =

1 T

∫

kT

0

∫

kT

(Vs − Vch )2

0

R

k (Vs − Vch ) dt = = 2398kW R 2

1 V (V − Vch ) ⎛ V − Vch ⎞ Vs ⎜ s .kT ⎟dt = . s s T R ⎝ R ⎠

kVs (Vs − Vch ) = 2398kW R

Po 2376.2 = = 99.09% 2398 Pi

R 10 = = 20Ω k 0.5 vi) on-time t1 = kT = 0.5 × 1 = 0.5ms

v) Ri =

off-time t 2 = T − t1 = 0.5ms Downloaded from www.jayaram.com.np /- 71


Step Down Chopper with R-L Load:

Vo Vs

t1

t2

i

t

+

T

+ L Dm

Vo

Vs

Countineous Current

IV R I1

-

-

kt

(1-k)T

E

t T

Fig : Chopper with R-L Load I2

i1

Discountineous Current

i2

F

Fig : Wave Form

The operation of the chopper can be divided into two modes. During mode1, the chopper is switch on and current flows from supply to the load. During mode2, the chopper is switch off and the load current continuous to flow through freewheeling diode. The load current for mode1 can be found + di L Vs = Ri1 + L + E − (i ) dt Vs With initial current R i1 (t = 0 ) = I 1 gives the load current as

(Vs − E ) ⎛⎜

⎞ ⎟ − i1 (t ) = I 1e + 1 e − (2) ⎜ ⎟ R ⎝ ⎠ This mode is valid for 0 ≤ t ≤ t1 (= kT ) at the end of this mode, the load current becomes −t

R L

−t

I 1 (t = t1 = kT ) = I 2

R L

E

− (3)

The load current for mode 2 can be found from di2 +E − (4 ) dt With initial condition 0 = RiL + L

i2 (t = 0 ) = I 2

i2 (t ) = I 2 e

−t

R L

−t E⎛ ⎜1 − e L ⎜ R⎝

R

−

⎞ ⎟ ⎟ ⎠

− (5)

This mode is valid for 0 ≤ t ≤ t 2 (= (1 − k )T ) at the end of this mode, load current becomes i2 (t = t 2 ) = I 3 − (6)


Downloaded from www.jayaram.com.np Under steady state condition

−(7 )

I1 = I 3 Now from (2) I 2 = I 1e

− kT

R L

+

(Vs − E ) ⎛⎜

⎜1 − e ⎝

R

− kT

R L

⎞ ⎟ ⎟ ⎠

− (8)

From (5) I 3 = I1 = I 2 e

− (1− R )T

− (1− k )T E⎛ L − ⎜⎜1 − e R⎝

R L

R

⎞ ⎟ ⎟ ⎠

− (9)

Peak to peak ripple current is ∆I = I 2 − I 1 which after simplification becomes V 1− e ∆I = s R

− kT

R L

+e

−T

R L

−

−e

TR − (1− k ) L

TR L

1− e The condition for maximum ripple d (∆I ) gives =0 dk k = 0.5 The maximum peak to peak ripple current at (k = 0.5) is

∆I max =

Vs R tanh R 4 fL

For 4 FL >> R, tanh θ ≈ θ ∆I max =

&

Vs 4 FL

Q. Step down chopper is feeding RL load with Vs = 220V , R = 5Ω, L = 7.5mH , f = 1KHz , K = 0.5 and E = 0. Calculate: a) Minimum instantaneous load current ' I 1 ' b) Peak instantaneous load current ‘ I 2 ' c) Maximum peak to peak load ripple Solution: R R − KI − KT ⎞ V −E⎛ ⎜1 − e L ⎟ I 2 = I 1e L + s ⎟ R ⎜⎝ ⎠ R R − (1− K )T − kT ⎞ E⎛ L I1 = e − ⎜⎜1 − e L ⎟⎟ R⎝ ⎠ I2 E I1



− (i )

I 2 = 0.7165 I 1 + 12.473

− (ii )

I 1 = 0.7165 I 2

Solving these two equations, we get: I 1 = 18.37 A, I 2 = 25.63 A I ∆I = I 2 − I 1 = 7.26 A

Q. The step down chopper has a load resistance R = 0.25Ω , input voltage Vs = 550V and E = 0V . The average load current I A = 220 A in chopping frequency f = 250 Hz . Use the average output voltage. Calculate the load inductance ‘L’ which would limit the maximum load ripple current to 90% of I A . Solution: ∆i = 10% of I a = 0.1 × 200 = 20 A T=

1 = 0.004 sec f

Average output voltage, Va = kVs = RI a Voltage across inductor is di L = Vs − RI a = Vs − kVs = (1 − k )Vs dt V ∆i = (1 − k ) s .kT L ∆i.L = (1 − k )Vs .kT

∆i = 0, k = 0.5 dk Principle of Step-up operation: +

iL D1

+

Chopper

Vs

CL

Load

_

_ a) Step-up Arrangement Vo/Vs 6 i

5

I2 i2

i1

4 E

3

I1 t1

2

t2 T

b) Current Wave Form

t 1

0.2

0.4

0.6

0.8

c) Output Voltage


1.0

K

Downloaded from www.jayaram.com.np When a switch is closed for time t1 , the inductor current rises and energy is stored in the inductor L. If the switch is opened for time t 2 , the energy stored in the inductor is transfer to the load through diode D1 and the inductor current falls. When the chopper is turn, the voltage across the inductor is di V L = Vs = L dt And, this gives the peak to peak ripple current in the inductor as V .t ∆i = s 1 L The instantaneous output voltage is di Vo = V s + L dt ∆i Vo = V s + L t2 V .t = Vs + s 1 t2 ⎛ t ⎞ = Vs ⎜⎜1 + 1 ⎟⎟ ⎝ t2 ⎠ T = Vs . t2 T = Vs . T − t1 1 = Vs . t 1− 1 T Vo =

Vs 1− k

If a large capacitor C L is connected across the load as shown by dash line in figure, the output voltage will be continuous. The voltage across the load can be step up by varying the duty cycle k. This principle can be applied to transfer energy from one voltage source to another as shown in the figure. L +

i iL

L D1

+

i

+

iL

i

L

I2 D1

Vs

Chopper Vs

_ _

_

Mode 1

i2

H +

_

The inductor current for mode 1 is given by: di Vs = L dt And, it is expressed as V i1 (t ) = s t + I 1 where, I 1 is initial current. L During mode 1, current must rises and necessary condition Downloaded from www.jayaram.com.np /- 75

I1 t1

t2 t

Downloaded from www.jayaram.com.np di > 0 or Vs > 0 dt The current for mode 2 is given by di Vs = L 2 + E dt And is solved as ⎛V − E ⎞ i2 (t ) = ⎜ s ⎟t + I 2 ⎝ L ⎠ where, I 2 is initial current for mode 2. For stable system, the current must fall and condition is: di2 < 0 or Vs < E dt Performance Parameter: The power semi-conductor devices require a minimum time to turn on and turn off. Therefore, the duty cycle k can only be controlled between a minimum value K min & K max , thereby limiting the minimum and maximum value of output voltage, we have: ∆I max =

Vs 4 fL

Load ripple current depends inversely on chopping frequency f and frequency should be as high as possible to reduce load ripple current. Chopper Classification: Depending on the direction of current and voltage flow chopper can be classified into 5 types: 1. Class A Chopper 2. Class B Chopper 3. Class C Chopper 4. Class D Chopper 5. Class E Chopper Class A Chopper: νi

VL

VL

VL

IL iL Class A Chopper

vL

νL

-IL

o

Class B Chopper

IL

-IL

o

IL

Class C Chopper


Downloaded from www.jayaram.com.np vL

vL

VL

VL

iL

+IL

o

+IL

iL

-VL -VL

Class E Chopper

Class D Chopper

Class A Chopper: The load current flows into the load. Both the load voltage and load current are positive. This is a single quadrant chopper and is to be operated as rectifier. Its example is step-down chopper. Class B Chopper: i2

I2

i s iL +

-

R

I1 t

Vs

S1

VL

E VL

a) Circuit Diagram

kt

T

Load Current

Vs

The load current flows out of the load, the load voltage is positive but the load current is negative. This is also a single quadrant chopper but operates in second quadrant and said to be operated as inviter. When switch s1 is closed, the voltage E drives current through inductor L and load voltage V L becomes zero. The current I m which rises is described by:

di + Ri L + E dt With initial current i L (t = 0 ) = I 1 , gives 0=L

i L = I 1e

⎛R⎞ −⎜ ⎟ t ⎝L⎠

⎛R⎞ ⎛R⎞ −⎜ ⎟ t ⎞ −⎜ ⎟ ⎞ i ⎛⎜ E ⎛⎜ ⎝L⎠ ⎟ − 1− e − 1 − e ⎝ L ⎠ ⎟ for 0 ≤ t ≤ kT ⎜ ⎟ ⎜ ⎟ R⎝ ⎠ R⎝ ⎠

at t = t1 , i L = (t = t1 = kT ) = I 2 When switch is turned off, a magnitude of energy stored in inductor L is returned to the supply Vs . Vi diode D1 . The load current falls. The load current is described by, Vs = L

di + Ri L + E dt

An initial current of I 2 ; iL = I 2 e

⎛R⎞ −⎜ ⎟ t ⎝L⎠

+

R − t ⎞ Vs − E ⎛ ⎜1 − e L ⎟ ⎟ R ⎜⎝ ⎠

for 0 ≤ t ≤ t 2



Class C Chopper: VL

S1

D1 iL

Vs

L

R

+ S2

D2 -

a) Circuit Diagram

The load current is either positive or negative. The load voltage is always positive. This is known as two quadrant chopper. Class A and Class B chopper can be combined to form Class C chopper. S1 and D2 operate as Class A Chopper whereas S2 and D1 operates as Class B Chopper. Care must be taken to insure that two switches are not fire together; otherwise, the supply Vs will be short circuited. Class C Chopper can be operated either as a rectifier or as an invitor. Class D Chopper: D3

S1

L

E

Vs +

-

D2

a) Circuit Diagram

Class D Chopper can also operate as rectifier or as inverter. If S1 and S4 are turn on VL, iL become positive. If S1 and S4 are turn off, load current will be positive and continue to flow for a highly inductive load. Class E Chopper: VL

S1

VL

D1

S3

D3

Inverting VL +ve iL -ve

Rectifying VL +ve iL +ve iL

Vs L S2

D2

R V2

S4

D4

VL -ve iL -ve Rectifying

VL -ve iL +ve Inverting

a) Circuit Diagram b) Polarities

1st Quadrant Operation: S2 and S3 are kept off, S4 is kept on and S1 is operated when S1 is on, VL and iL are positive. When S1 is switched off, current free wheels through S4 and D2. 2nd Quadrant Operation: -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 78

Downloaded from www.jayaram.com.np S2 is operated while S1, S3, S4 are off when S2 on reverse current flows through R-L-E leand and D4 and inductor stores energy. With S2 off, energy stored by inductor is turned to supply through D1 and D4. 3rd Quadrant Operation: S1, S2, S3 are kept off and only S4 is operated. With S4 on the current freewheels through D2. With S4 turn off current is feed to the source through D2 and D3. V Input +

Vr dc chopper

Vc

+

Va

Vg

o

T

t

Vs Ve Vref

Control Vc

Vr

Amplifier -

-

o

a) Block Diagram

KT T

b) Control Signal

DC chopper can be used as switching mode regulator to convert DC voltage normally unregulated to a regulated DC output voltage. The regulation is normally achieved by pulse width modulation (PWM) at a fix frequency and the switching device is normally BJT or MOSFET. Control voltage Vc is obtained by comparing the output voltage with its desired value. Vc can be compared with saw tooth voltage to generate the PWM control signal for DC Choppers. Therefore, there are four basic topologies of switching regulator. 1. Buck Regulator 2. Boost Regulator 3. Buck-Boost Regulator 4. Cuk Regulator Buck Regulator:

Figures In a Buck regular, the output voltage Va is less than the output voltage Vs .This is like a step down chopper. The circuit operation can be divided into two modes. Mode 1 begins when θ 1 is switch on and mode 2 begins when transistor θ 2 is switch off. Assuming the inductor current rises linearly from I 1 to I 2 in time t1 V s − Va = L t1 =

L∆I Vs − I a

∆I t1

− (1) − (2 ) Downloaded from www.jayaram.com.np /- 79

Downloaded from www.jayaram.com.np & inductor current falls linearly from I 2 to I 1 in time t 2 , Va = L

di t2

− (3)

L∆I Va

t2 =

where, ∆I = I 2 − I 1 = peak to peak ripple current of inductor from (1) and (2), ∆I =

(Vs − Va )t1 L

=

Va t 2 L

V s − Va t 2 = Va t1

or ,

⎛ Vs ⎞ (1 − k )T ⎜⎜ − 1⎟⎟ = kT ⎝ Va ⎠ Vs 1− k 1 = 1+ = Va k k

Va = kVs

-(5)

Assuming a lossless circuit, Vs I s = Va I a = kVs I a , average input current − (6)

I s = kI a

The switching period T can be expressed as, I 1 ∆IL = t1 + t 2 = + L (From (2 ) and (4 )) f V s − Va Va

T= =

∆ILVs − (7 ) Va (Vs − Va )

which gives peak to peak ripple current as ∆I =

Va (Vs − Va ) fLV s

or , ∆I =

Vs k (1 − k ) fL

− (8)

− (9)

We can write inductor current i L as i L = iC + i D If we assume ∆i D is small and negligible, ∆i L = ∆iC which flows into for IC =

& Average capacitor current t1 t 2 T + = is 2 2 2

∆I 4 -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 80

Downloaded from www.jayaram.com.np & Peak to peak ripple voltage of capacitor T

1 ∆I ∆I .T ∆I ∆VC = ∫ 2 = dt = 0 4 8C 8 fC C

− (10 )

from (8) and (10) ∆V c =

Va (Vs − Va ) 8 LCf 2Vs

− (11)

∆VC =

Vs k (1 − k ) 8 LCf 2

− (12 )

Q. The buck regulator has an input voltage of Vs = 12V . The required average output voltage is Va = 5V & peak to peak ripple voltage is 20 mv. The switching frequency is 29 kHz. If the peak to peak ripple current of inductor is limited to 0.8A, determine (a) Duty Cycle K (b) Filter inductance L (c) Filter capacitor. Vs = 12V , Va = 5V , ∆Vc = 20mV , f = 25kHz

Solution: Va = kVs

a)

k=

5 = 0.4167 12

Va (Vs − Va ) fLV s

b) ∆I = L=

Va (Vs − Va ) ∆IfVs

L = 148.83µH c) ∆Vc = C=

∆I 8 fC

∆Ι = 200 µF 8 f∆Vc

Boost Regulator: Figures In boost regular the output voltage is greater than the input voltage. The circuit operation can be divided into two modes. Mode 1 begins when MOSFET M1 is switch on and mode 2 begins when MOSFET is switch off. Mode-1 ∆I Vs = L − (1) t1 or , t1 = L

∆I Vs

− (2)

Mode-2 Downloaded from www.jayaram.com.np /- 81

Downloaded from www.jayaram.com.np V s − Va = − L or , t 2 =

∆I t2

− (3)

∆I .L − (4) Va .Vs

Vs .t1 (Va − Vs ) = t 2 - (5) [(t1 = kT , t 2 = (1 − k ) / T )] L L

∆I =

Va =

Vs 1− k

Assuming loss less circuit, V s I s = Va I a ⎛ V ⎞ Vs I s = ⎜ s ⎟ I a ⎝1− k ⎠ I Is = a − (6 ) 1− k 1 Now, T = = t1 + t 2 f L∆I L.∆I = + Vs Va − V s ∆ILVa = Vs (Va − Vs ) V (V − Vs ) ∆I = s a − (7 ) fLVa ∆I =

Vs .k fL

- (8)

Peak to peak ripple voltage is 1 t1 ∆Vc = ∫ I C dt C 0 t 1 1 = ∫ I a dt C 0 I .t = a 1 C I (V − Vs ) = a a − (9 ) Va fC ∆VC =

I a .k fC

-

(10)

Q. A boost regular has an input voltage Vs = 5V , Average output voltage 15V , Average output load current I a = 0.5 A . The switching frequency f = 25kHz, L = 1500µH & C = 220 µC . Determine a) Duty cycle b) ripple current of inductor δi c) peak current of indcutor I s d) ripple voltage of filter capacitor. -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 82

Downloaded from www.jayaram.com.np Solution: V a) Va = s k Vs .k = 0.89 A b) ∆I = fL I c) I s = a = 1.5 A 1− k I k d) ∆VC = a fC I s = 2.5 A 0.69 ∆I I2 = Is + = 1.5 + = 1.945 A 2 2 I .k ∆Vc = a = 80.61mV fC

Figure

Buck Boost Regulator:

Figures It provides the output voltage which is less than or greater than the inut voltage. But the output voltage polarity is opposite to that of the input voltage. The circuit operation can be divided into two modes. Mode 1 begins when T1 is turn on and mode 2 when T1 is turn off. Mode 1: L.∆I Vs = − (1) t1 ∆I Vs

or, t1 = L.

− (2)

Mode 2: ∆I t2

− (3)

∆I Va

− (4)

Va = − L t2 = −L ∆I =

Vs .t1 Vt =− a 2 L L

Va = −

kVs 1− k

⎛ t = kT ⎞ ⎟⎟ - (5) ⎜⎜ 1 ⎝ t 2 = (1 − k )T ⎠

Assuming loss less circuit, Vs I s = −Va I a = Is =

I a .k 1− k

kVa 1− k

- (6) Downloaded from www.jayaram.com.np /- 83


Now, T =

∆IL ∆IL ∆IL(Va − Vs ) 1 = t1 + t 2 = − = f Vs Va V sV a V sV a - (7) fL(Va − Vs )

∆I =

Vs .k − (8) fL Peak to peak ripple voltage of capacitor is I aV s 1 t1 1 t1 1 ∆Vc = ∫ I C dt = ∫ I a .dt = I a.t1 = − (9 ) (Va − Vs ) fC C 0 C 0 C ∆I =

∆VC =

I a .k fC

- (10)

Q. The buck boost regulator has an input voltage Vs = 10V , duty cycle k = 0.5V and switching frequency is 25kHz. The inductance L = 150 µH and fileter capacitance C = 220µF . The average load current I a = 1.25µA . Determine a) average output voltage Va b) Peak to peak output ripple voltage ∆Vc c) Peak to peak ripple current of inductor. Solution: Vk a) Va = − s = −4V 1− k I k b) ∆Vc = a = 56.8mV fC Vk c) ∆I = s = 0.8 A fL

Figures

It provides an output voltage which is less than or greater than an input voltage but the output voltage polarity is opposite to that of input voltage. The circuit operation can be divided into two modes. Mode 1 begins when θ 1 is turn on and mode 2 begins when θ 1 is turn off. Mode-1 Vs =

L1 ∆I 1 t1

or , t1 =

L1 ∆I 1 Vs

Vs − Vc1 =

t2 = −

− (2 )

− L1 ∆I 1 t2

− (3)

∆I 1 .L1 Vs − Vc1

When ∆I 1 = I L12 − I L11 ∆I 1 =

− (1)

(

)

Vs .t1 − Vs − Vc1 t2 = L1 L1 -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 84

Downloaded from www.jayaram.com.np Vc1 =

Vs 1− k

− (5)

Now, Vc1 + Va = L2 t1 =

∆I 2. L2 Vc1 + Va

− (6 )

− (7 )

L 2 ∆I 2 t2

Va = − t2 =

∆I 2 t1

− ∆I 2 .L2 − (8) Va

(V

∆I 2 =

c1

)

+ V a t1 L2

=−

Va t 2 L2

Va − (9 ) k From (5) and (9), we get: Vc 2 = −

KVs 1− K

− (10 )

∆I 1 =

V s .K fL2

− (11)

∆I 2 =

KVs fL2

− (12)

Va = −

∆Vc1 =

I s (1 − k ) − (13) fc1

∆V c 2 =

KVs 8c 2 L2 f

2

Limitation of single stage conversion: 1. The four regulator use only one transistor employing one stage conversion due to current handling limitation of single transistor, the output power of these regulator is small, there is no isolation between input and output voltage which is highly desirable characteristics in most applications. For high power application, multi-stage conversion are used where a dc-voltage is converted to ac by inverter. The ac output is isolated by the transformer and converted to dc by rectifier. Chopper circuit design: 1. Identify the modes of operation for cheaper circuit 2. Determine the equivalent circuit for the various mode. 3. Determine current and voltage for the modes and their waveform. 4. Evaluate value of L and C that would satisfy design limit. 5. Determine voltage and current rating requirement of all components.



Chapter 10 DC Drives Basic Characteristics of DC motor: If, i f + La

Lf vf, Vf

Ra

va Va

Rf -

+ eg -

w

Td

Tc

B

Fig : Equivalent Circuit of Separately Excited DC Motor

When a separately excited DC motor is excited by a filed current of If and armature current of Ia flows in the armature circuit, the motor develops a back emf and a torque to balance the lead torque at a particular speed. The instantaneous field current if is described as: di f Lef = R f i f + L f dt The instantaneous armature current can be found from Va = Ra ia + La

dia + eg dt

The motor back emf is expressed as: e g = kVωi f The torque developed by motor is Td = k t i f ia The developed torque must be equal to load torque, dω Td = I + Bω + TL dt Where, ω = motor speed , rad / s Β = viscous friction cons tan t N − m / rad / s K v = Voltage Cons tan t V / A − rad / s K t = K V = torque cons tan t

La = armature circuit induc tan ce, H

L f = field circuit induc tan ce, H Ra = armature circuit resis tan ce, Ω R f = field circuit resis tan ce, Ω TL = Load torque, N − m


Downloaded from www.jayaram.com.np di f di Under steady state conditions, a =0 dt dt Vf = Rf I f E g = K v ωΙ f V a = R a I a + K v ωI f Td = K t I f I a = Bω + TL Developed power is Pd = Td .ω

Eg

w = constant applcoximately linear region If

Fig : Magnetization Characteristics

The speed of separately excited motor can be found from,

ω=

Va − I a Ra Va − I a Ra = Vf Kv I f Kv. Rf

The motor speed can be varied by: 1. Controlling armature voltage Va known as voltage control. 2. Controlling the field current, if , known as field control. 3. Torque demand, which corresponds to rated armature voltage, rated field current and rated armature current is known as base speed. Id, Pd Speed

Torque Td

Speed, w ia , if Ia

armature current

If

field current o

Speed, w Constant torque

Constant power

Fig : Characteristics of separately excited motor


Downloaded from www.jayaram.com.np For a speed less than base speed, armature current and field current are maintained constant to meet the torque demand and armature voltage Va is varied. For speed higher than base speed, armature voltage is maintained at rated value and field current is varied to control the speed. However, the power developed by motor remains constant. Ia = If + ia = if La, Ra

Lf, Rf

va Va

-

w

Td

Tc

B

E g = K v ωI f

Va = (Ra + R f )I a + E g = (Ra + R f )I a + K vωI f Td = K t I f I a = BωtI f

ω=

Va − (Ra + R f )I a Kv I f

Speed

Id, Pd Power Pd

Torque Td

armature current

Speed, w Constant torque

Constant power

Fig : Characteristics of DC Series Motor

The speed can be varied by controlling (1) armature voltage Va or (2) armature current Ia which is a measure of torque demand. Numericals: Q. A 15 HP, 220 V, 2200 r.p.m. separately excited d.c. motor controls a load requiring a torque of TL = 45 Nm at a speed of 1200 r.p.m. The field circuit resistance is Rf = 147 Ω and the armature circuit resistance is Ra = 0.25Ω and voltage constant of motor is K v = 0.7032 V / A rad/s. The field voltage -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 88

Downloaded from www.jayaram.com.np V f = 220V . The viscous friction and no load losses are negligible. The armature current may to amuse continuous and ripple free; determine: a) Back emf E g b) Required armature voltage Va c) Rated armature current of motor. Solution: R f = 147Ω,

Ra = 0.25Ω

V f = 220V

K v = K f = 0.7032 V / A rad / s

Td = TL = 45 Nm

ω = 2π × If =

Vf Rf

1220 = 125.66rad / s 60 =

220 = 1.497 A 147

a) E g = E v ωI f = 0.7032 × 125.66 × 1.497 = 132.28V b) Va = I a Ra + E g Td = K t I a I f Ia =

Td 45 = = 42.75 A KI f 0.7032 × 1.497

Va = 42.75 × 0.25 + 132.25 = 142.97V c) I h p = 74600 I rated =

15 × 746 = 50.87 A 220

Operating Modes: (i) Motoring Ia = If

Ia

Rf, If

Ra Ra, La La

Va

+ M

Eg Separately Excited Motor

Series motor


Downloaded from www.jayaram.com.np Back emf is less than supply voltage Va . Both armature and field current are +ve, the motor develops torque to meet the load demand. Operating Modes: ii) Regenerating Braking Ia

Ia

Ra Ra, La RF

La

Va

Rb

iF

+ Eg

+ Eg

M -

M -

The motor acts a generator and develop induce emf voltage E g . E g must be greater than the supply voltage V A . The armature current is –ve. iii) Dynamic Braking Ia

Ia

Ra Ra, La Rb

RF

La

+

Rb

iF

Eg

+

-

Eg -

Series motor

It is similar to regenerate braking except the supply voltage V A is replaced by braking resistance Rb . The kinetic energy of motor is dissipated in Rb . iv) Plugging Ia=If Rf, Lf

If

Rb

La

Va

RF iF

+ Eg

Vf

Va

+ Eg

M -

M -

Seperately excited moter


Downloaded from www.jayaram.com.np v) 4 Quadrants Speed

Ia

Ia

Ra, ta VaEg

Eg

M -

Forward Braking Forward Motering Torque Ia

Ia

Ra, La -

Va

+ M Eg -

+

Va

+

M

+ Eg Va < Eg -

Reverse Braking

Va > Eg Reverse Motoring

Fig : Condition for four quadrature

Single Phase Motor: If the armature circuit of dc motor is connected to the output of single phase control rectifier, the armature voltage can be vary by varying the delay angle of converter α a . A converter is also applied in field circuit to control the field current by varying delay angle . Depending on the type of single phase converter, single phase drive may be sub-divided into: 1. Single Phase Half-wave Converter Drive 2. Single Phase Semi-conductor Device 3. Single Phase Full-converter Drive 4. Single Phase Dual-converter Drive Single Phase Half-wave Converter Drive: Ia

is

Ia

if

+ id Va

Ia

La, Ra

Dm +

-

wt

is

Eg

M -

Vc LF, Rf

id

αa

π

αa

π

wt

Ia a) circuit


2π

wt

Downloaded from www.jayaram.com.np With a single phase half wave converter in the armature circuit, the average armature voltage is Va =

Vm (1 + cos α a ) 2π

for 0 ≤ α a ≤ π

Where, Vm = peak voltage of ac supply With a semi-converter in the field circuit average field current is Vf =

Vm

π

(1 + cos α )

for 0 ≤ α f ≤ π

f

Ia

Single Phase Semi-converter: wt

is Id if

π= αa

ia

αa

Vs

Vs

+ id

Ia

LF, Rf

Eg

M

wt

2π

id

La, Ra

is

π

-

αa

π= αa

π

wt

2π

With a single phase semiconductor in armature circuit, the average armature is V Va = m (1 + cos α a ) for 0 ≤ α a ≤ π

π

When, Vm = peak voltage of ac supply. With a semiconductor in filed circuit, average field voltage V f =

Vm

π

(1 + cos α ) for 0 ≤ α f

f

≤π .

if

ia Va

La, Ra

+ Eg

M

Vs

Ia

LF, Rf

-

o

wt is

a) Circuit

Ia

Is

αa

π π+ αa

Va

-Ia

b) Quadrature

c) Waveform


wt

Downloaded from www.jayaram.com.np Va = Vf =

2Vm

π 2Vm

π

cos α a ,

for 0 ≤ α a ≤ π where, Vm = peak voltage of supply

cos α f


Single Phase Dual Converter: converter-1 α a1

converter-2 α a2

if

Ia La, Ra Vs

Vs LF, Rf

If converter 1 operates with a delay angle of α a1 , the average armature voltage is Va =

2Vm

π

cos α a1

for 0 ≤ α a1 ≤ π

If converter 1 operates with a delay angle of α a 2 , the average armature voltage is Va =

2Vm

π

cos α a 2

for 0 ≤ α a 2 ≤ π

where α a 2 = π − α a1

With a full converter in the field circuit, the average field voltage is Vf =

2Vm

π

cos α f


Q. The speed of the separately excited motor is a controlled by 1φ semi-converter. The field current which is also controlled by a semi-converter is to the maximum possible value. The ac supply to the armature 1φ , 208V , 60 Hz. The armature resistance Ra = 0.25Ω , R f = 147Ω the motor voltage constant K v = 0.7032V / A − rad / s load torque TL = 45 Nm at 1000 rpm. The viscous friction and no load losses are negligible. The inductance of the armature and field circuit are sufficient enough to make the armatural field current, continuous and ripple free. Determine a) Field current i f b) Delay angle of converter in the armature circuit α a c) Input power factor of armature circuit converter Solution: Vs = 208V , Vm = 2 × Vs = 294.16V

Ra = 0.25Ω, R f = 147Ω K v = K t = 0.7032V / A rad / s Td = TL = 45 Nm

ω = 2π ×

1000 = 104.72 rad / s 60 Downloaded from www.jayaram.com.np /- 93

Downloaded from www.jayaram.com.np a) V f =

Vm

(1 + cos α )

π When, α f = 0 Vf =

If =

f

2Vm

π Vf Rf

2 × 294.16

=

=

π

= 187.27V

187.27 = 1.274 A 147

b) α a = ? Va =

Vm

π

(1 + cos α a )

Td = k t I a I f Ia =

Td 45 = = 50.20 A k t I f 0.7032 × 1.274

E g = Ra I a + E g = 0.25 × 50.23 + 93.82 = 106.38V 294.16

106.38 =

π

(1 + cos α a )

α a = 82.2° c) The output power, Po = Va I a = 106.38 × 50.23 = 5343.5W If the losses in armature converter are neglected, the power from supply is Pa = Po = 5343.5W The rms input current of the armature converter is ⎡ 2 I sa = ⎢ ⎣ 2π

π

∫α

a

1

1 2

⎛π −αa ⎞2 ⎤ I a2 dωt ⎥ = I a ⎜ ⎟ = 37.03 A ⎦ ⎝ π ⎠

Input volt-amp rating VI = Vs I a = 208 × 37.03 = 7702.24W Pa 5343.5 = = 0.694(lagging ) VI 7702.24 Closed Loop Control of DC drives: The speed of dc motor changes with the load torque to maintain a constant speed, the armature (and or field) voltage should be varied by continuously by varying the delay angle of ac to dc converter or duty cycle of chopper. A close loop control system has the advantages of improve accuracy fast dynamic response and reduce effects of load disturbances and system non linearity.

Input Power Factor, PF =

Power supply T2 Vr + -

Ve

Speed Converter

Vc

Va Converter

dc mo tor

w

Seed sesing

Fig : Block Diagram of Closed Loop Converter Field DC motor drive


Downloaded from www.jayaram.com.np If the speed of the motor decreases due to the application of additional load torque, the speed error Ve increases. The speed controller responses with increase controlled signal Vc , change the delay angle or duty cycle of converter and increase the armature voltage of the motor and increase armature voltage develops more torque to restore the motor speed to the original value. Open Loop Transfer Function: If + Lm, Rm + Vs -

Vf

Conerter of gain K2

-

w

Td

T2 B

Fig : Converter Fed Separately Excited dc motor

The motor speed is adjusted by setting reference voltage Vr . Assuming a linear power conver of

gain K 2 . The armature of the motor is Va = k 2Vr − (i )

Assuming filed current I a remains constant during any transient disturbances, the system equation are: eg = K v I f ω

− (2)

V a = R m i a + Lm

dia + K V I f ω − (3) dt

Td = K t I f ia

− (4 )

dω + Bω + TL − (5) dt Taking Laplace transform of above equation: Td = J

Va (s ) = K 2V1 (s )

− (6)

Va (s ) = Rm I a ( s ) + Lm sI a (s ) + K V I f ω (s ) Td (s ) = K t I f I a (s ) = sJω (s ) + TL (s ) From (7), the armature current is Va (s ) − K v I f ω (s ) I a (s ) = (sLm + Rm ) T=

V a ( s ) − k v I f ω (s )

Where, Ta =

Rm (sTa + 1)

− (7 )

− (8)

− (9)

− (10)

Lm = time constant of motor armature circuit Ra

From (8), motor speed is w(s ) =

Td (s ) − TL (s ) (SI + B )

−(11) Downloaded from www.jayaram.com.np /- 95

Downloaded from www.jayaram.com.np =

Td (s ) − TL (s ) B(sTm + 1)

− (12)

Where, Tm = J / B = mechanical time constant of motor Two possible disturbances are control voltage Vr and load torque TL . The response due to a step change in the reference voltage is obtained by setting TL to zero. T L(s) Ia (s)

Kt If

T d(s)

1 B(STm+1)

w(s)

+ -Kv If

Vr(s)

1 Rm(STa-1)

K2

Fig : Open Loop Block Diagram of Separetly Excited d.c. motor

ω (s )

V r (s )

=

K 2 K r I f / ( R m .B )

s 2 (ToTm ) + s (Ta + Tm ) + 1(K v I f

) / ( R .B ) 2

− (13)

m

The response due to a change in load torque TL can be obtained by setting Vr to zero.

ω (s )

TL (s )

=

− (1 B )(sTa + 1)

s (Ta I m ) + s (Ta + Tm ) + 1 + (kV I f 2

) /(R .B ) 2

− (14)

m

Using Final value theorem steady-stage relationship of change in speed k2 kv I f ∆ω = ∆V r − (15) 2 Rm B + (K v I f ) ∆ω =

Rm

Rm B + (K V I f

)

2

∆T1

− (16)

Close Loop Transfer Function: TL(s)

Vrs

Vc

K2

Va

la

Kv If

+ - Eg(s)

-

Kf If K1


w(s)

Downloaded from www.jayaram.com.np To change open loop arrangement into closed loop system a speed senser connected to the output shaft. The closed loop step response due to a change in reference voltage can be setting Vr to zero. K 2 K r i f / Rm B W (s ) = 2 − (1) Vr (s ) S (Taτ m ) + S (Ta + τ m ) + 1 + (K v I f )2 + K 1 K 2 K v I f / (Rm B )

[

]

The response due to change in load torque I L can be obtained by setting Vr to zero.

(− 1 / B )(Sτ a + 1) W (s ) = 2 − (2) Vr (s ) S (Taτ m ) + S (Ta + τ m ) + 1 + (K v I f )2 + K 1 K 2 K v I f / (Rm B )

[

]

Using final value theorem the steady change in speed due to step change in control voltage ∆I L can be found from (1) and (2) by substituting S=0. K 2 KV I f ∆W = ∆V r − (3) 2 Rm B + (K v I f ) + K 1 K 2 K v I f ∆W =

Rm B + (K v I f

Rm

)

2

+ K1 K 2 K v I f

∆TL

− (4)

Phase Locked Loop ( PLL) Control: Phase detector

Low pas filter

Converter k2

DC mortor

Speed encoder

In PLL control system, the motor speed is converted to a digital pulse train by using speed encoder the output of encoder acts as a speed signal of frequency to the phase detector compares a reference pulse train f r with the feed back frequency f o and provides a pulse with modulated output voltage Vc to a continuous dc level Vc which varies the output of power converter and in turn the motor speed. When the motor runs at same speed as the reference pulse train the low frequencies could be synchronized (or blocked) with a phase difference. The output of the phase detector would be constant voltage proportional to the phase difference and motor speed would be main trained at first value. Any disturbances contributing to the speed change would result in phase differences in the output of phase detector could respond immediately to vary the speed of motor. In such direction and magnitude as to retain the locking of reference and feedback frequencies. Micro-computer control of DC Drives:



Line synchronizin g ckt Determination of thyristor of be fired M Delay angle generator

If Rf Lf

Time and Logic

Current controller Pulse amplifier Current comparator AD

Motor Current

AD

Spe ed Signal

Speed controller

Speed comparato r

Speed reference (ωr) start/stop command

A micro computer control reduces the size and cost of hardware electronics improving reliability and control performance. This control scheme is implemented in software and it flexible to change the control strategy to meet difference performance characteristics. A micro computer control system can also perform various desirable functions: ON/OFF of main power supply, start/stop of drive, speed control, current control, etc. The speed signal is fed into micro-computer using A/D converter to limit armature current of motor inner current control loop is used the line synchronizing circuit is required to synchronize the generation of firing pulses with a supply line frequency. The pulse amplifier provides the necessary isolation and produces gate pulses of required magnitude and duration.



Chapter 11 Properties of Devices & Circuit Introduction: Due to the reverse recovery process of power devices and switching action in the presence of circuit inductance voltage transient occurring converter circuit, short circuit fault conditions may exist resulting in excessive current flow through the devices. In practice power devices are protected from: 1. Thermal runaway by heat sink dv di 2. High and by shutter circuit dt dt 3. Reverse recovery transient 4. Supply and load side transient 5. Fault condition by fuse Cooling & heat sink: Due to on state and switching loses, heat is generated within the power devices. This heat must be transfer from the devices to a cooling medium to maintain the operating junction temperature within the specified values. Although heat transfer can be accomplished by conduction, convection and radiation, natural or forced air, convection cooling is commonly used in industrial application. Heat must be flow from the device to the case and then to the heat sink in cooling medium. If Pa is the average power loss in the device, the electrical analog of a device which is mounted on a heat sink is shown in the figure: T1

Ric

Tc

R cs

PA

Ts

RSA TA

The junction temperature of a device is given by: T j = PA (R JC + RCS + RSA )

Where, R JC = thermal resistance from junction to case RCS = thermal resistance from case to sink c/w RSA = thermal resistance from sink to ambient c/w T A = ambient temperature

In high power application, the devices are more effectively cooled by liquids, normally oil or water. Snubber Circuits: Downloaded from www.jayaram.com.np /- 99


C T1 R1

D1

R1

C

R D

R

R 1>R

D

a) Polarized

T1

C R 1>R

T2

Ls

R

b) Reverse Polarized

dv within the dt maximum allowable rating. The snubber could be polarized orc)unpolarized. Unpolarized The forward polarized dv and snubber is suitable when thyristor is connected with anti-parallel diode; R limit the forward dt R1 limit the discharge current of capacitor when the device is turn on. R 1>R

L

s An RC-snubber is normally connected across the semiconductor device to limit

dv where R1 limits do change current of capacitor. dt When a pair of thyristor is connected in inverse parallel, the snubber must be effective in either direction. A reverse polarized snubber limits reverse

Design of snubber circuit: Rs

Cs

Rs

L

L

+

+ T1

RL

-

t − ⎛ i = I ⎜⎜1 − e T ⎝

Where, I =

RL

-

When switch is closed

Vs = (Rs + RL )i + L

Vs

di which gives dt

⎞ ⎟ ⎟ ⎠ Vs Rs + RL

&

T=

L Rs + RL

Now, t − Vs R + RL − Tt Vs − Tt di 1 = Ie T . = . s .e = e dt T Rs + R L R L -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 100

Downloaded from www.jayaram.com.np Vs ⎛ di ⎞ at t = 0 ⎜ ⎟ = ⎝ dt ⎠ max L

L=

Vs ⎛ di ⎞ ⎜ ⎟ ⎝ dt ⎠ max

- (1)

Voltage across SCR, a = Rs .i dv a di = Rs . dt dt ⎛ dv a ⎞ ⎛ di ⎞ ⎜ ⎟ = Rs ⎜ ⎟ ⎝ dt ⎠ max ⎝ dt ⎠ max V = Rs . s L L ⎛ dv ⎞ − (3) Rs = ⎜ a ⎟ Vs ⎝ dt ⎠ max Rs = 2 ρ

− (2)

L R cs where, ρ = damping ratio = cs 2 L 2

⎛ 2ρ ⎞ ⎟⎟ .L C s = ⎜⎜ ⎝ Rs ⎠

− (4)

When SCR is turned on capacitor Cs will discharge a maximum current of

Vs and total current Rs

⎛V V ⎞ through thyristor will be ⎜⎜ s + s ⎟⎟ . It should be less than peak current rating of thyristor. ⎝ Rs R L ⎠

Q. Following are the specification of a thyristors operating from a peak supply of 500V and ⎛ di ⎞ ⎛ dv ⎞ repetitive peak current I p = 250 A⎜ ⎟ = 60 A / sec, ⎜ ⎟ = 200V / sec . Design the snubber ⎝ dt ⎠ max ⎝ dt ⎠ max

circuit, if the minimum load resistance is 20 Ω . Take ρ = 0.65 . Take a factor of safety for the given data. Solution: For a factor of safety 2, permitted values are: Ip =

250 = 125 2

60 ⎛ di ⎞ = 30 A / sec ⎜ ⎟ = 2 ⎝ dt ⎠ max

200 ⎛ dv ⎞ = 100V sec µs ⎜ ⎟ = 2 ⎝ dt ⎠ max

Now, L=

Vs 500 = = 16.67 µH 30 ⎛ di ⎞ ⎜ ⎟ ⎝ dt ⎠ max Downloaded from www.jayaram.com.np /- 101

Downloaded from www.jayaram.com.np Rs =

16.67 × 100 L ⎛ dv ⎞ = 3.33Ω ⎜ ⎟ = 500 Vs ⎝ dt ⎠ max

When thyristor is turned on, the peak current through thyristor is Vs Vs 500 500 + = + = 175.15 A Rs R L 3.33 20

This peak current is more than permissible peak current of 125A. The value of Rs must be increased. Take Rs = 7Ω , peak current through thyristor is

500 500 + = 96.42 A , which is also less than 7 20

permissible peak current. Also, Rs =

L=

Cs

L ⎛ dv ⎞ ⎜ ⎟ Vs ⎝ dt ⎠ max

Rs .Vs 7 × 500 = = 35µH 100 ⎛ dv ⎞ ⎜ ⎟ ⎝ dt ⎠ max

2 2 ( 2ρ ) ⎛ 2 × 0.65 ⎞ = .L = × 35 = 1.20 µF

⎜ ⎝

Rs

⎟ ⎠

7

Take C s = 1µF R=7-2

C=1uF 15aH

+ 20Ω

Vs -

Reverse Recovery Transients: Due to the reverse recovery time t rr and reverse current I R and amount of energy is trapped in the circuit inductance and as a result transient voltage appears across the device. In addition to dv protect, the snubber limits the peak transient voltage across the device. The equivalent circuit for dt a circuit arrangement is shown in the figure. IR +

L C R

Divice under recovery

-

The energy stored in the inductor L which is transferred to the snubber capacitance C is dissipated mostly in snubber resistor. Supply and Load side transient: -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 102

Downloaded from www.jayaram.com.np + Vp

S1

Lm

C Np

+ C

Vo

Ns

R

R -

a) Circuit Diagram

C Vo

R -

L

I0

i +

-

b) Equivalent Circuit during turn off

L O A D

c) Equivalent Circuit due to load disconnection

A transformer is normally connected to the input side of converter, under steady state condition and amount of energy is stored in the magnetizing inductance Lm of the transformer and switching off the supply provides a transient voltage to the input of converter, a capacitor may be connected across the primary or secondary of the transformer to limit the transient voltage. When load is disconnected the transient voltage are produced due to the energy stored in inductance. Voltage Protection by Selenium Diodes and Metal Oxide Varistors: i

_

Clamping Vz Voltage v

o

v

+

b) Symbol a) V-I characteristics

The selenium diode may be used for protection against transient over voltage. The characteristics of selenium diode are shown in the figure. Normally, the operating point lies before knee of the characteristics curve and draws very small current from the circuit. However, when an over voltage appears, the knee point is crossed and the reverse current flow through the selenium increases suddenly, thereby, limiting the transient voltage. Varistors are non-linear variable impedance device consisting of metal oxide particles. As the applied voltage is increased, the film become conductive an the current is increased. The current is expressed as I = KV α where, K = constant and V = applied voltage. The value of α varies between 30° and 40°. Current Protection: The power converter may develop short circuit of faults and the fault current must be clear quickly. Fast acting fuses are normally used to protect the semi-conductor device. As the fault current increases, fuse opens and clears a fault current in few milliseconds. Fusing:



T1

T3 L

ac supply

F1

F3

F2

F2

T4

R

T2

Fig : Controlled Rectifier

The semi-conductor device may be protected by carefully choosing the location of fuse. When the fault current rises, the fuse temperature also rises and fuse melts and arcs are developed across the fuse. Due to the arc, the impedance of fuse as increased thereby reducing current. Crowbar Protection Circuit: F + L

Dm Vs

Tc R θ1

_

Fig : Protection by crowbar

Transistors can be protected by a crowbar circuit. A crowbar is used for protecting circuits or equipment under fault condition where the amount of energy involved is to high and the normal protection circuit cannot be used. A crowbar consists of thyristor with voltage or current sensitive firing circuit. The crowbar thyristor is placed across the converter circuit to be protected. If the fault condition are sensed and crowbar thyristor Tc is fired. A virtual short circuit is created and fuse link F1 is blown thereby relieving the converter circuit from over current.

Fault Current with AC source and DC source: The fault current and fuse clearing time will be dependent on time constant of fault circuit. The fuse manufacture specify the current-time characteristics for ac circuit and there is no equivalent curve for dc circuit for a circuit operating from dc-voltage, the voltage rating of fuse should be typically 1.5 times the equivalent ac rms voltage. The fuse protection of dc circuit require more careful design then that of ac circuit.



Chapter 7 Power Transistors Introduction: Power transistor have controlled turn ON and turn OFF characteristics. The power transistors can be classified broadly into four categories: i) By-polar junction transistor (BJT) ii) Metal-oxide semi-conductor filed effect transistor (MOSFET) iii) Static-induction transistor (SIT) iv) Insulated gate by-polar transistor (IGBT) i) BJT: co llector

co llector C Ic

C Ic

n

n IB p

IB

p

B

Base

Base

n

B

n

IE

IE

E

Fig : a) NPN transistor

emitter

emitter

Ic

RC

IB VC F1 VCE1

+ IB RB VB

E

Fig : b) NPN transistor

+ _

VCE _ VBE

IE

Fig a) Circuit Diagram IC

VCE Active region

Cut off IB3

Active region

Saturation

VCC

IB2 IB1 IB=0

VCE

VCE

Fig c) Output Characterisitcs

Fig d) Transfer Characterisitcs

There are three operating region of transistor circuit active and saturation. In circuit off region, the transistor is off or the base current is not enough to turn it ON. In active region, the transistor acts amplifier by a gain. In the saturation region the base current is sufficiently high so that the collection emitter voltage VCE is low and the transistor acts as switch. Downloaded from www.jayaram.com.np /- 105

Downloaded from www.jayaram.com.np The equation neglecting the current is − (1)

Ie = Ic + IB Current gain,

β=

Ic IB

− (2 )

− (3) where, I CEO = collector to emitter leakage current

I C = βI B + I CEO

From (1) and (3), we have

I E = I B (1 + β ) + I CEO − (4) ≈ I B (1 + β )

⎛ ⎛ β + 1⎞ 1⎞ ⎟⎟ − (5) I I E ≈ I C ⎜⎜1 + ⎟⎟ = I C ⎜⎜ ⎝ β⎠ ⎝ β ⎠ The collector current can be expanded as: − (6)

I C ≈ αI E Where, α = Or, β =

β

− (7 )

1+ β

α 1−α

− (8)

Let us consider the circuit of fig 1(a) where transistor is operated as switch. IB =

V B − V BE − (9 ) RB

− (10)

VC = VCE = VCC − I e Rc = VCC − β I B RC = VCC −

β RC RB

VCE = VCB + VBE

(VB − VBE ) − (11)

VCB = VCE − VBE The maximum collector current in active region which can be obtained by setting VCB = 0 and VBE = VCE is I CM =

VCC − VCE VCC − V BE = RC RC

− (12 )

And the corresponding value of base current I BM =

I CM

β

− (13)

The transistor saturation may be defined as the point above which any increase in base current does not increase collector current significantly. In the saturation, the collector current remains almost constant I CsS =

VCC − VCE (sat ) − (14 ) RC


Downloaded from www.jayaram.com.np I And the corresponding base current I BS = CS − (15)

β

Normally, the circuit is designed so that I B is higher than I BS . Overdrive factor, ODF =

IB I BS

Forced β = β f =

− (16 ) I CS − (17 ) IB

The total power less in two junctions is PT = VBE I B + VCE I C

− (18)

Q. R Ic RB + V _ B

+ _

VBE

VCC

IE

The bipolar transistor shown in the figure is specified to have β in the range 8 to 40. The load resistance is RC = 11Ω . The dc supply voltage VCC = 200V and the input voltage to the base circuit is N B = 10V . If VCE ( sat ) = 1V & VBE (sat ) = 1.5V . Find a) Value of R3 that results in saturation with overdrive factor (ODF) of 5. b) Force β F c) Power loss Pt in the transistor. Solution: RC = 11Ω VCC = 200V VB = 10V ODF = 5 VCE ( sat ) = 1V VBE ( sat ) = 1.5V β min = 8 β max = 40 a) R B = ? VB − V BE ( sat ) RB = IB VCC − VCE ( sat ) 200 − 1 I CS = = = 18.1A RC 11 I 18.1 I BS = CS = = 2.26 A 8 β min I ODF = B I BS I B = ODF × I BS = 5 × 2.26 = 11.31A 10 − 1.5 RB = = 0.7514Ω 1.31 I 18.1 = 1 .6 b) β s = CS = I B 11.31 PT = VBE I B + VCE I C = 1.5 × 11.31 + 1 × 18.1 = 35.07W Downloaded from www.jayaram.com.np /- 107

Downloaded from www.jayaram.com.np Switching Characteristics: VB V1

t KT

(1-K)T

-V2

VB VB

V1

IB

t

t KT -IB

-V2

VB Ic s 0.9c s 0.1Ic s td

tn

t s t f to

t

Fig : Switching times of bipolar transistor

When the input voltage is reverse from V1 to − V2 and the base current is also changed to − I B 2 the collector current does not change for a time t s called the storage time. Ts is required to remove the saturating charge from the base. Due to the change in polarity of V B from V1 to − V2 , the reverse current − I B2 helps to discharge base current .The turn on time ToN is the sum of delay time t d and rise time t r . ToN = t d + t r Q. The wave form of the transistor switch is shown in the figure. The parameters are VCC = 250V , VBE ( sat ) = 3V , I B = 8 A. VCE ( sat ) = 2V , I CS = 100 A, Td = 0.5µs, Tr = 1µs, Ts = 5µs, T f = 3 and f s = 10KHz. The duty cycle is 50%, the collector to emitter leakage current is I ceo = 3mA . Determine the power loss due to collector current a) During turn on (ton = td + tr ) b) During conduction period (tn) c) During turn off (Toff = ts + tf) d) During off time to e) Total average power loss (PT)


Downloaded from www.jayaram.com.np VCE

VCC

VCC(Sat) t

o t on

t off

IC Ics

ICCo t iB

td

tr

tr

ts

tf

to

IBs

KT

(1-K)T t

o T=1/ fc

iB

VBE(Sat)

Solution: 1 T= = 100µs fs o K = 0.5, t r = 1µs, t d = 0.5µs, t s = 5µs, t f = 3µs KT = t d + t r + t n = 50µs t n = 50 − 1 − 0.5 = 48.5µs (1 − K )T = t s + t f + t o = 50µs t o = 50 − 3 − 5 = 42 µs a) Due to delay time, 0 ≤ t ≤ td ic (t ) = I CEO VCE (t ) = VCC Instantaneous power due to collector current, PC (t ) = iC (t )VCE (t ) = I CEO .VCC = 5 × 10 −3 × 250 = 0.75W Average power loss during delay time is 1 td Pd = ∫ PC (t )dt = f s × 0.75 × t d = 3.75mW T 0 During rise time, ic (t ) =

(I CS + I CEO )t tr

+ I CEO

⎛ VCC − VCE (sat ) ⎞ ⎟⎟ + VCC VCE (t ) = −⎜⎜ tr ⎝ ⎠


Downloaded from www.jayaram.com.np PC (T ) = I C (T )VCE (T ) Average Power during rise time, Pr =

1 T

∫

tr

0

PC (t )dt =

1 T

∫

tr

0

⎞ I CS .t ⎛ VCC ⎜⎜ − .t + VCC ⎟⎟dt = 42.33W tr ⎝ tr ⎠

Total power loss during turn on time Pon = Pd + Pr = 3.75mW + 42.32W = 42.33W Total power PT = 274.05W Switching Limits: i) Secondary Breakdown: Secondary (second) break down is caused by localize thermal runaway, resulting from high current concentration. The current concentration may be caused by defects in transistor structure. ii) Forward Bias Safe Operating Area (FBSOA): FBSOA indicates I C − VCE limits of X’sistor during turn off and for reliable operation, the transistor must not be subjected to greater power dissipation, then that shown by FBSOA curve. iii) Reverse Bias Safe Operating Area (RBSOA): The manufacture provide I C − VCE I limit during reverse bias turn off, as RBSOA iv) Power Derating: The ambient temperature and thermal resistance must be consider when interpreting the rating of device. The manufacturer shows the derating curve for thermal derating and second breakdown derating. Base Drive Control: Ton can be reduced by allowing base current picking during turn Ton. and Toff and be reduce by reducing base current and allowing base current picking turning turn off. IB IB1 IBS t

-IB2

FIg : Base Drive Current Waveform

The commonly used technique for optimizing the base drive of transistor are: 1. Turn on Control 2. Turn off Control 3. Proportional Control 4. Anti-saturation Control -By Manoj Basnet (Ass. Lecturer, Eastern College of Engineering) /- 110

Downloaded from www.jayaram.com.np Turn on Control: VB

IC

C V1

Rc

R1 t1

t2

+ t

Vcc

R2 VB

IE

_ -V2

When an input voltage is turned on, the base current is limited by R1 and initial value of base current V − V BE V − V BE and final value of base current is I R1 = 1 is I BO = 1 R1 R1 + R2 The capacitor C charges up to infinity VC ≅ V1 .

R1 R1 + R2

Turn off Control: If the input voltage is changed to –V2, during turn off, the capacitor voltage VCE is added to V2 as a reverse voltage across the transistor. There will be base current picking during turn off. Proportional Base Control: If the collector current changes due to change in load demand, the base drive current is changed in proportional to collector current. Anti saturation Control: If the transistor is driven hard, the storage time which is proportional to the base current increases and switching speed is reduced. The storage time can be reduced by operating the transistor in sub-saturation rather than hard saturation. Power MOSFET: A power MOSFET is a voltage controlled device. The switching speed is very high. Switching times are of the order of nanosecond. MOSFET are of two types: 1) Diplation MOSFET 2) Enhancement MOSFET ID

D

D

RD

RD

G

G

+

VDD

VGS _

+

VDD

VGS _

a) n-channel

b) p-channel



D

D

RD

G

RD

G S

+

+

VDD

VGS _

VDD

S

VGS _

a) n-channel

b) p-channel

Fig : Enhancement MOSFET ID

VP

VGS

o

VP

o

VGS

-ID

b) p-channel

a) n-channel

a) Depletion MOSFET Vr VGS

o

VGS

Vr

-ID

a) n-channel

b) p-channel

b) Enhancement MOSFET

Lin ear reg ion

Saturation region

VGS2 VGS1 VGS = VG

VDS

Fig : Output Characteristics VG

Switching Characteristics: V1

t

O VGS V1 VGSo VT t d(on)

tr

td(off) tf



t d (on ) = turn - on delay time t r = rise time t d (off ) = turn off delay time t f = fall time Gate Drive: RD G

VDD

+ RS

R1

VG

RG

_

Turn on delay time can be reduced by connecting RC circuit. SIT: D

G S

Symbol A SIT is a high power, high frequency device. A SIT has a short channel length, loop gate series resistance, loop gate source capacitance and small thermal resistance. It has low noise, low distortion and high audio frequency power capacity. The turn on and turn off times are very small. Typically 0.25 µs . The current rating of SIT can be output 300A, 1200V and switching speed can be as high as 100kHz. It is most suitable for high frequency applications. Example: audio, VHF, UHF and microwave amplifier. IGBT: C

B

E

Symbol Downloaded from www.jayaram.com.np /- 113


IGBT combines the advantes of BJT and MOSFET. IGBT has high input impedance like MOSFET and low ON-state conduction loss like BJT. IGBT is inherently fast then BJT. However, the switching speed of IJBT is inferior to that of MOSFET. The current rating of IGBT can be up to 400A, 1200V and switching frequency can be up to 20 kHz. IGBT’s are finding increasing application in medium power application such as DC and AC motor drive, power supplies. Series and Parallel Operation: Transistor may be operated in series to increase their voltage handling capability. It is very important that the series connected transistor are turn on and turn off. Simultaneously, the device should be match for gain, transconductance threshold voltage, on-state voltage, turn on time and turn off time. Even the gate or base drive characteristics should be identical. Voltage sharing n/w, similar to diode could be used. Transistor are connected in parallel if one device cannot handle load current demand for equal current sharing, the transistor should be matched for gain, transconductance, turn on time and turn off time. But, in practice, it is not always possible to meet this requirements, a reasonable amount of current sharing can be obtained by connecting resistor in series with emitter (or source) terminals as shown in figure. IT RC θ1

θ2

Vcc

IE2

IE1

RC2

RC1

Q. Two MOSFET are connected in parallel to carry a total current of I T = 20A. The drain to source voltage of MOSFET m1 is V DS1 = 2.5V1 and that of MOSFET m2 is V DS 2 = 3V . Determine the drain current of each transistor (MOSFET). IT = 20A D

D

M1

I T = I D1 + I D 2 − (i ) V DS1 + I D1 Rs1 = V Ds 2 + I D 2 .Rs 2

RD

M2 S

S

RS1

RS2

VDD

− (ii )


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