Chapter 5: BJT AC Analysis
Two-Port Systems Approach This approach: • Reduces a circuit to a two-port system • Provides a “Thévenin look” at the output terminals • Makes it easier to determine the effects of a changing load With Vi set to 0 V: Z Th Z o R o
The voltage across the open terminals is: E Th A vNL Vi
where AvNL is the no-load voltage gain.
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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Effect of Load Impedance on Gain
This model can be applied to any current- or voltagecontrolled amplifier. Adding a load reduces the gain of the amplifier: Av
Vo RL A vNL Vi RL Ro
Ai A v
Zi RL
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Effect of Source Impedance on Gain The fraction of applied signal that reaches the input of the amplifier is: R i Vs Vi Ri Rs
The internal resistance of the signal source reduces the overall gain: A vs
Vo Ri A vNL Vs R i R s
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Combined Effects of RS and RL on Voltage Gain Effects of RL: Vo R L A vNL Vi RL Ro R Ai A v i RL
Av
Effects of RL and RS: Vo Ri RL A vNL Vs R i R s R L R o R Ri A is A vs s RL
A vs
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Cascaded Systems • • • • •
The output of one amplifier is the input to the next amplifier The overall voltage gain is determined by the product of gains of the individual stages The DC bias circuits are isolated from each other by the coupling capacitors The DC calculations are independent of the cascading The AC calculations for gain and impedance are interdependent
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R-C Coupled BJT Amplifiers This image cannot currently be display ed.
Input impedance, first stage: Z i R 1 || R 2 || re
Output impedance, second stage: Zo R C
Voltage gain: A v1
R C || R 1 || R 2 || re re
A V2
RC re
A v A v1 A v 2
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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R-C Coupled BJT Amplifiers Draw the ac equivalent circuit and calculate the voltage gain, input resistance and output resistance for the cascade BJT amplifier in above Figure. Let the parameters are:
VCC 20V , Q1 Q 2 200, V BE ( ON ) 0.7V , r0 R1 R3 15k, R2 R4 4.7 k, RC1 RC 2 2.2k, RE1 R E 2 1k Solution Dc Analysis
I BQ1 19.89A I CQ1 3.979mA 1re1 2 re 2 1.307 k
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
I BQ 2 19.89A I CQ 2 3.979mA
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R-C Coupled BJT Amplifiers Ac Analysis: The voltage gain at Q1 is:
AV 1 0.153(2.2k // 957.36) 102.06 The voltage gain at Q2 is
AV 2 0.153( 2.2k ) 336.6 The overall gain is then,
AV AV 1 AV 2 ( 102.06)( 336.6) 34,353 The input resistance is:
Ri R1 // R2 // Br e 957.36 The o/p resistance is:
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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Cascode Connection This example is a CE–CB combination. This arrangement provides high input impedance but a low voltage gain. The low voltage gain of the input stage reduces the Miller input capacitance, making this combination suitable for highfrequency applications.
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Cascode Connection
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Darlington Connection The Darlington circuit provides a very high current gain—the product of the individual current gains: bD = b1b2 The practical significance is that the circuit provides a very high input impedance.
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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DC Bias of Darlington Circuits This image cannot currently be display ed.
Base current: V VBE I B CC R B DR E Emitter current: I E ( D 1)I B D I B Emitter voltage: VE I E R E Base voltage: VB VE VBE
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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AC small signal of Darlington Circuits
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Feedback Pair This is a two-transistor circuit that operates like a Darlington pair, but it is not a Darlington pair. It has similar characteristics: • High current gain • Voltage gain near unity • Low output impedance • High input impedance The difference is that a Darlington uses a pair of like transistors, whereas the feedback-pair configuration uses complementary transistors.
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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DC Analysis of Feedback Pair
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AC Analysis of Feedback Pair
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Current Mirror Circuits Current mirror circuits provide constant current in integrated circuits.
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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Current Mirror Example Find I?
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Current Mirror More Configurations
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Current Source Circuits Constant-current sources can be built using FETs, BJTs, and combinations of these devices.
I IE
VZ VBE RE
IE IC more…
Electronic Devices and Circuit Theory, 10/e Robert L. Boylestad and Louis Nashelsky
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Current Source Circuits
VGS = 0V ID = IDSS = 10 mA
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