Lecture 5

Lecture 5 PN Junction and MOS Electrostatics(II) PN JUNCTION IN THERMAL EQUILIBRIUM

Outline 1. 2. 3. 4.

Introduction Electrostatics of pn junction in thermal equilibrium The depletion approximation Contact potentials

Reading Assignment: Howe and Sodini, Chapter 3, Sections 3.3-3.6 6.102 Spring 2007

Lecture 5

1

1. Introduction • pn junction – p-region and n-region in intimate contact

Why is the p-n junction worth studying? It is present in virtually every semiconductor device! Example: CMOS cross-section

Understanding the pn junction is essential to understanding transistor operation

6.102 Spring 2007

Lecture 5

2

2. Electrostatics of p-n junction in equilibrium Focus on intrinsic region:

Doping distribution of an abrupt p-n junction

6.102 Spring 2007

Lecture 5

3

What is the carrier concentration distribution in thermal equilibrium? First think of the two sides separately:

Now bring the two sides together. What happens?

6.102 Spring 2007

Lecture 5

4

Resulting carrier concentration profile in thermal equilibrium:

• Far away from the metallurgical junction: nothing happens – Two quasi-neutral regions

• Around the metallurgical junction: diffusion of carriers must counter-balance drift – Space-charge region 6.102 Spring 2007

Lecture 5

5

On a linear scale:

Thermal equilibrium: balance between drift and diffusion

J n (x) = J

drift n

(x) + J

diff n

(x) = 0

J p (x) = J pdrift (x) + J pdiff (x) = 0 We can divide semiconductor into three regions • Two quasi-neutral n- and p-regions (QNR’s) • One space-charge region (SCR) Now, we want to know no(x), po(x), ρ(x), E(x) and φ(x). We need to solve Poisson’s equation using a simple but powerful approximation 6.102 Spring 2007

Lecture 5

6

3. The Depletion Approximation • Assume the QNR’s are perfectly charge neutral • Assume the SCR is depleted of carriers – depletion region

• Transition between SCR and QNR’s sharp at – -xpo and xno (must calculate where to place these)

x < −x po ;

ni2 po (x) = N a , no (x) = Na

−x po < x < 0;

po (x), no (x) xno ; 6.102 Spring 2007

po (x) x no

Lecture 5

8

Electric Field Integrate Poisson’s equation

E(x2 ) − E(x1 ) =

1

x2

ρ(x) dx ∫ εs x 1

x < − x po ; − x po < x < 0;

E(x) = 0 E(x) − E( −x po ) =

x

1

εs − x∫ po

−qN a dx ′

⎡ qN a ⎤ x − qN a = ⎢− x⎥ = x + x po ε ε ⎣ ⎦ −x po s s

(

qN d

0 < x < xno ;

E(x) =

x > xno ;

E(x) = 0

6.102 Spring 2007

εs

)

(x − xno )

Lecture 5

9

Electrostatic Potential (with φ=0 @ no=po=ni) kT no φ= • ln ni q

kT po φ=− • ln ni q

In QNRs, no and po are known ⇒ can determine φ in p-QNR: po=Na ⇒

φp = −

N kT • ln a q ni

kT Nd φ = • ln in n-QNR: no=Nd ⇒ n q ni

Built-in potential:

Nd Na kT φ B = φn − φ p = • ln q ni2

This expression is always correct in TE! We did not use depletion approximation. 6.102 Spring 2007

Lecture 5

10

To obtain φ(x) in between, integrate E(x) x2

φ(x2 ) − φ(x1) = − ∫ E( x ′ )dx ′ x1

x < −x po ; −x po < x < 0;

φ(x) = φp φ(x) − φ(−x po ) = −

x

∫

−

qNa

−x po

(

εs

(x′ + xpo )dx′

)

2 qNa x + x po 2ε s 2 qN φ(x) = φ p + a x + x po 2ε s

=

(

0 < x < xno ;

φ(x) = φn −

)

qNd 2 x − x ( no ) 2ε s

φ(x) = φn Almost done ….

x > xno ;

6.102 Spring 2007

Lecture 5

11

Still do not know xno and xpo ⇒ need two more equations

1. Require overall charge neutrality:

qN a x po = qN d x no

2. Require φ(x) to be continuous at x=0; φp +

qN a 2 qN d 2 x po = φn − x no 2ε s 2ε s

Two equations with two unknowns — obtain solution:

2ε s φ B N a x no = q(N a + N d )N d

2ε s φ B N d x po = q(N a + N d )N a

Now problem is completely solved!

6.102 Spring 2007

Lecture 5

12

Solution Summary Space Charge Density

Electrostatic Field

Electrostatic Potential

6.102 Spring 2007

Lecture 5

13

Other results: Width of the space charge region:

x do = x po + x no

2ε s φ B (N a + N d ) = qN a N d

Field at the metallurgical junction:

2qφ B N a N d Eo = ε s (N a + N d )

6.102 Spring 2007

Lecture 5

14

Three Special Cases • Symmetric junction: Na = Nd

x po = x no • Asymmetric junction: Na > Nd

x po < x no • Strongly asymmetric junction – p+n junction: Na >> Nd

2 εs φB x po