6 is added to the circuit in parallel with the series combination of R. 1 and R. 4.
Thomas L. Floyd. Electronics Fundamentals, 6e. Electric Circuit Fundamentals,
6e.
Chapter 6 Series –Parallel Circuits A simple series-parallel circuit
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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R4 is added to the circuit in series with R1.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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R5 is added to the circuit in series with R2.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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R6 is added to the circuit in parallel with the series combination of R1 and R4.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Illustration of voltage relationships.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Balanced (Calibrated) Wheatstone bridge when the shows 0V and 0 A Between A and B
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
•Used in Sensitive Measurement Instruments (Strain, Temperature, Etc.) •Converts Change in Transducer Resistance (RX) to a Voltage Output (VA-B) •Voltage Output Sent to Amplifier, Conversion and Readout Circuits •RV is Used to Calibrate VAB to 0V at Known Rx Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Wheatstone Bridge
VOUT = 0V (Calibrated) When: •R1/R3 = R2/R4 Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Using a Transducer
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Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
VA = (1K Ohm/1.9K Ohm) 12V = 6.32V VB = 6V VA – B = 6.32V – 6V = 320mV
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Example of a more complex Wheatstone Load Cell
Calibration is done in a downstream Circuit
Loading Effect
RL in parallel with R2 Creates: •More Current Through R1 •More Voltage Drop Across R1 •Lower voltage at VOUT Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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The effect of a load resistor.
RL 10 Times Larger Than R2
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VOUT Unloaded =(10K Ohms /14.4K Ohms) 5V = 3.5V Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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VOUT Unloaded =(10K Ohms /14.4K Ohms) 5V = 3.5V
VOUT = (5K Ohms/9.7K Ohms) 5V = 2.6V
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
VOUT = (9.1K Ohms/13.8K Ohms) 5V = 3.4V
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Maximum Power Transfer – Impedance Matching
Load
Maximum power is transferred to the load when RL = RS.
Maximum Power Transfer: Common Application
Output of Audio Amplifier Impedance Matched with Speaker
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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In this case, Maximum power is transferred to the load when RL = 75 Ohms
Load
75 Ohms
•The Source “Sees” a 75 Ohm Load •The Load “Sees” a 75 Ohm Source
Maximum Power Transfer RL = 0 Ohms VRL = 0V IRL = 0A PRL = 0W RL = 25 Ohms VRL = 2.5V IRL = 100mA PRL = 250mW RL = 75 Ohms VRL = 5V IRL = 67mA PRL = 335mW
Curve showing that the load power is maximum when RL = RS Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
RL = 125 Ohms VRL = 6.25V IRL = 50mA PRL = 313mW Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
The general form of a Thevenin equivalent circuit is a voltage source in series with a resistance.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Example of the simplification of a circuit by Thevenin’s theorem.
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Superposition – Determining Current from Two Sources
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
RT = 150 Ohms IT = 66.7mA IR2(Vs2) = 33.4mA
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3. Replace VS1 With a Short 4. Calculate IR2 Created From VS2
Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
RT = 150 Ohms IT = 33.3mA IR2 (Vs2) = 16.7mA
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5. Calculate IR2 as Algebraic Sum of Previous Calculations
IR2(Total) = |33.4mA -16.7mA| = 16.7mA Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e
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Troubleshooting 1 Should be 4.22V
Fault – R2 Open
Troubleshooting 2a Should be 4.22V
Fault – R2 Open
Troubleshooting 2b
Fault – R2 Open
Troubleshooting 3
Correct Value
Should be 13.8V
Fault – R7 Shorted
