Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
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Textbook Question
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Chapter 11, Problem 11.82P

The BiCMOS circuit shown in Figure P11.82 is equivalent to a pnp bipolar transistor with an infinite input impedance. The bias current is I Q = 0.5 mA The MOS transistor parameters are V T P = 0.5 V , K p = 0.7 mA / V 2 , and λ = 0 , and the BIT parameters are β = 180 , V B E (  on  ) = 0.7 V , and V A = (a) Sketch the small-signal equivalent circuit. (b) Calculate the small-signal parameters for each transistor. (c) Determine the small-signal voltage gain A v = v o / v i for (i) R L = 10 k Ω and (ii) R L = 100 k Ω .

Chapter 11, Problem 11.82P, The BiCMOS circuit shown in Figure P11.82 is equivalent to a pnp bipolar transistor with an infinite

(a)

Expert Solution
Check Mark
To determine

To sketch: The small-signal equivalent circuit of the given BiCMOS circuit.

Answer to Problem 11.82P

The required sketch is shown in Figure 3.

Explanation of Solution

Given:

The given diagram is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.82P , additional homework tip  1

Calculation:

Mark the value and redraw the circuit.

The required diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.82P , additional homework tip  2

Figure 1

The small signal equivalent circuit is shown in Figure 2

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.82P , additional homework tip  3

Figure 2

Conclusion:

Therefore, the required sketch is shown in Figure 2

(b)

Expert Solution
Check Mark
To determine

The small signal parameters for each of the transistor.

Answer to Problem 11.82P

The value of small signal parameters are gm2=15.78mA/V , gm1=0.5014mA/V and rπ2=11.41kΩ .

Explanation of Solution

iven:

The given diagram is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.82P , additional homework tip  4

Calculation:

The value of the current through the resistance R1 is calculated as,

  IR1=0.7V8kΩ=0.0875mA

The expression for the resistance IQ is calculated as,

  IQ=IR1+IE2

Substitute 0.5mA for IQ and 0.0875mA for IR1 in the above equation.

  0.5mA=0.0875mA+IE2IE2=0.4125mA

The value of the drain current is calculated as,

  ID1=IR1+IB2=0.002279mA

The expression for the collector current is given by,

  IQ=ID+IC=IR1+IB+IC=IR1+IBβ+IC

Substitute 180 for β , 0.5mA for IQ and 0.0875mA for IR1 in the above equation.

  0.5mA=0.0875mA+IC180+ICIC=0.41022mA

The value of the base current is calculated as,

  ID1=IR1+IB2=0.0875mA+IC2β=0.0875mA+0.41022mA180=0.08978mA

The transconductance of the first transistor is given by,

  gm1=2knID1

Substitute 0.08978mA for ID1 and 0.7mA/V2 for kn in the above equation.

  gm1=2(0.7mA/V)(0.08978mA)=0.5014mA/V

The transconductance of the first transistor is given by,

  gm2=IC20.026V

Substitute 0.41022mA for IC2 in the above equation.

  gm2=0.41022mA0.026V=15.78mA/V

The value of the resistance rπ2 is calculated as,

  rπ2=βgm2=18015.78mA/V=11.41kΩ

Conclusion:

Therefore, the value of small signal parameters are gm2=15.78mA/V , gm1=0.5014mA/V and rπ2=11.41kΩ .

(c)

Expert Solution
Check Mark
To determine

The value of small signal voltage gain for the given values of load resistor.

Answer to Problem 11.82P

The value of the small signal voltage gain for RL=10kΩ is 0.99736 and for RL=100kΩ is 0.99936 .

Explanation of Solution

Given:

The given diagram is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.82P , additional homework tip  5

The given values of load resistor is RL=10kΩ and RL=100kΩ

Calculation:

The expression for the voltage VO is calculated as,

  Vπ=gm1Vsg(R1rπ2R1+rπ2)

The expression for the gate to source voltage is given by,

  Vsg=VOVi

The expression to determine the value of the output voltage is given by,

  VO=(gm1Vsg+gm2Vπ)RLVO=(gm1(VOVi)+gm2gm1(VOVi)(R1rπ2R1+rπ2))RLVO=(VOVi)(gm1+gm2gm1(R1rπ2R1+rπ2))RLVOVi=gm1[1+gm2(R1rπ2R1+rπ2)]RL1+gm1[1+gm2(R1rπ2R1+rπ2)]RL …. (1)

Substitute 8kΩ for R1 , 0.5014mA/V for gm1 , 15.78mA/V for gm2 , 11.41kΩ for rπ2 and 10kΩ for RL in the above equation.

  VOVi=(0.5014mA/V)[1+(15.78mA/V)((8kΩ)(11.41kΩ)8kΩ+11.41kΩ)](10kΩ)1+0.5014mA/V[1+(15.78mA/V)((8kΩ)(11.41kΩ)8kΩ+11.41kΩ)](10kΩ)=0.99736

Substitute 8kΩ for R1 , 0.5014mA/V for gm1 , 15.78mA/V for gm2 , 11.41kΩ for rπ2 and 100kΩ for RL in equation (1).

  VOVi=(0.5014mA/V)[1+(15.78mA/V)((8kΩ)(11.41kΩ)8kΩ+11.41kΩ)](100kΩ)1+0.5014mA/V[1+(15.78mA/V)((8kΩ)(11.41kΩ)8kΩ+11.41kΩ)](100kΩ)=0.99936

Conclusion:

Therefore, the value of the small signal voltage gain for R1=10kΩ is 0.99736 and for R1=100kΩ is 0.99936 .

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Chapter 11 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 11 - Prob. 11.7EPCh. 11 - Prob. 11.4TYUCh. 11 - Prob. 11.5TYUCh. 11 - The parameters of the diff-amp shown in Figure...Ch. 11 - For the differential amplifier in Figure 11.20,...Ch. 11 - The parameters of the circuit shown in Figure...Ch. 11 - The circuit parameters of the diff-amp shown in...Ch. 11 - Consider the differential amplifier in Figure...Ch. 11 - The diff-amp in Figure 11.19 is biased at IQ=100A....Ch. 11 - Prob. 11.10TYUCh. 11 - The diff-amp circuit in Figure 11.30 is biased at...Ch. 11 - Prob. 11.11EPCh. 11 - Prob. 11.12EPCh. 11 - Prob. 11.11TYUCh. 11 - Prob. 11.12TYUCh. 11 - Redesign the circuit in Figure 11.30 using a...Ch. 11 - Prob. 11.14TYUCh. 11 - Prob. 11.15TYUCh. 11 - Prob. 11.16TYUCh. 11 - Prob. 11.17TYUCh. 11 - Consider the Darlington pair Q6 and Q7 in Figure...Ch. 11 - Prob. 11.14EPCh. 11 - Consider the Darlington pair and emitter-follower...Ch. 11 - Prob. 11.19TYUCh. 11 - Prob. 11.15EPCh. 11 - Consider the simple bipolar op-amp circuit in...Ch. 11 - Prob. 11.17EPCh. 11 - Define differential-mode and common-mode input...Ch. 11 - Prob. 2RQCh. 11 - From the dc transfer characteristics,...Ch. 11 - What is meant by matched transistors and why are...Ch. 11 - Prob. 5RQCh. 11 - Explain how a common-mode output signal is...Ch. 11 - Define the common-mode rejection ratio, CMRR. What...Ch. 11 - What design criteria will yield a large value of...Ch. 11 - Prob. 9RQCh. 11 - Define differential-mode and common-mode input...Ch. 11 - Sketch the de transfer characteristics of a MOSFET...Ch. 11 - Sketch and describe the advantages of a MOSFET...Ch. 11 - Prob. 13RQCh. 11 - Prob. 14RQCh. 11 - Describe the loading effects of connecting a...Ch. 11 - Prob. 16RQCh. 11 - Prob. 17RQCh. 11 - Prob. 18RQCh. 11 - (a) A differential-amplifier has a...Ch. 11 - Prob. 11.2PCh. 11 - Consider the differential amplifier shown in...Ch. 11 - Prob. 11.4PCh. 11 - Prob. D11.5PCh. 11 - The diff-amp in Figure 11.3 of the text has...Ch. 11 - The diff-amp configuration shown in Figure P11.7...Ch. 11 - Consider the circuit in Figure P11.8, with...Ch. 11 - The transistor parameters for the circuit in...Ch. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - The circuit and transistor parameters for the...Ch. 11 - Prob. 11.13PCh. 11 - Consider the differential amplifier shown in...Ch. 11 - Consider the circuit in Figure P11.15. The...Ch. 11 - Prob. 11.16PCh. 11 - Prob. 11.17PCh. 11 - For the diff-amp in Figure 11.2, determine the...Ch. 11 - Prob. 11.19PCh. 11 - Prob. D11.20PCh. 11 - Prob. 11.21PCh. 11 - The circuit parameters of the diff-amp shown in...Ch. 11 - Consider the circuit in Figure P11.23. Assume the...Ch. 11 - Prob. 11.24PCh. 11 - Consider the small-signal equivalent circuit of...Ch. 11 - Prob. D11.26PCh. 11 - Prob. 11.27PCh. 11 - A diff-amp is biased with a constant-current...Ch. 11 - The transistor parameters for the circuit shown in...Ch. 11 - Prob. D11.30PCh. 11 - For the differential amplifier in Figure P 11.31...Ch. 11 - Prob. 11.32PCh. 11 - Prob. 11.33PCh. 11 - Prob. 11.34PCh. 11 - Prob. 11.35PCh. 11 - Prob. 11.36PCh. 11 - Consider the normalized de transfer...Ch. 11 - Prob. 11.38PCh. 11 - Consider the circuit shown in Figure P 11.39 . The...Ch. 11 - Prob. 11.40PCh. 11 - Prob. 11.41PCh. 11 - Prob. 11.42PCh. 11 - Prob. 11.43PCh. 11 - Prob. D11.44PCh. 11 - Prob. D11.45PCh. 11 - Prob. 11.46PCh. 11 - Consider the circuit shown in Figure P 11.47 ....Ch. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Consider the MOSFET diff-amp with the...Ch. 11 - Consider the bridge circuit and diff-amp described...Ch. 11 - Prob. D11.53PCh. 11 - Prob. 11.54PCh. 11 - Prob. 11.55PCh. 11 - Consider the JFET diff-amp shown in Figure P11.56....Ch. 11 - Prob. 11.57PCh. 11 - Prob. 11.58PCh. 11 - Prob. D11.59PCh. 11 - The differential amplifier shown in Figure P 11.60...Ch. 11 - Prob. 11.61PCh. 11 - Consider the diff-amp shown in Figure P 11.62 ....Ch. 11 - Prob. 11.63PCh. 11 - The differential amplifier in Figure P11.64 has a...Ch. 11 - Prob. 11.65PCh. 11 - Consider the diff-amp with active load in Figure...Ch. 11 - The diff-amp in Figure P 11.67 has a...Ch. 11 - Consider the diff-amp in Figure P11.68. The PMOS...Ch. 11 - Prob. 11.69PCh. 11 - Prob. 11.70PCh. 11 - Prob. D11.71PCh. 11 - Prob. D11.72PCh. 11 - An all-CMOS diff-amp, including the current source...Ch. 11 - Prob. D11.74PCh. 11 - Consider the fully cascoded diff-amp in Figure...Ch. 11 - Consider the diff-amp that was shown in Figure...Ch. 11 - Prob. 11.77PCh. 11 - Prob. 11.78PCh. 11 - Prob. 11.79PCh. 11 - Prob. 11.80PCh. 11 - Consider the BiCMOS diff-amp in Figure 11.44 ,...Ch. 11 - The BiCMOS circuit shown in Figure P11.82 is...Ch. 11 - Prob. 11.83PCh. 11 - Prob. 11.84PCh. 11 - For the circuit shown in Figure P11.85, determine...Ch. 11 - The output stage in the circuit shown in Figure P...Ch. 11 - Prob. 11.87PCh. 11 - Consider the circuit in Figure P11.88. The bias...Ch. 11 - Prob. 11.89PCh. 11 - Consider the multistage bipolar circuit in Figure...Ch. 11 - Prob. D11.91PCh. 11 - Prob. 11.92PCh. 11 - For the transistors in the circuit in Figure...Ch. 11 - Prob. 11.94PCh. 11 - Prob. 11.95PCh. 11 - Prob. 11.96PCh. 11 - Consider the diff-amp in Figure 11.55 . The...Ch. 11 - The transistor parameters for the circuit in...
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