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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Chapter 12, Problem 12.37P

Consider the series-shunt feedback circuit in Figure P 12.37 , with transistor parameters: h F E = 120 , V B E ( on ) = 0.7 V , and V A = . (a) Determine the small-signal parameters for Q 1 , Q 2 , and Q 3 . Using nodal analysis, determine: (b) the small-signal voltage gain A v f = v o / v i , (c) the input resistance R i f , and ( d ) the output resistance R o f

Chapter 12, Problem 12.37P, Consider the series-shunt feedback circuit in Figure P12.37 , with transistor parameters:
Figure P12.37

(a)

Expert Solution
Check Mark
To determine

The small signal parameter for Q1,Q2 and Q3 .

Answer to Problem 12.37P

The trans-conductance of the first transistor Q1 is 32.81mA/V , transistor Q2 is 19.12mA/V and for transistor Q3 is 78.08mA/V .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.37P , additional homework tip  1

Calculation:

The value of the Thevenin voltage is given by,

  VTH=(75kΩ400kΩ+75kΩ)(10V)=1.579V

The expression for the equivalent resistance is given by,

  RTH=(400kΩ)(75kΩ)(400kΩ)+(75kΩ)=63.2kΩ

The expression for the current IBQ1 is given by,

  IBQ1=VTHVBEQ11RTH+(hFE+1)RE1

Substitute 1.579V for VTH , 0.7V for VBEQ1 , 120 for hFE , 63.2kΩ for RTH and 0.5kΩ for RE1 in the above equation.

  IBQ1=1.579V0.7V63.2kΩ+(120+1)0.5kΩ=0.007106kΩ

The expression to determine the value of the collector current ICQ1 is given by,

  ICQ1=hFEIBQ1

Substitute 120 for hFE and 0.007106mA for IBQ1 in the above equation.

  ICQ1=(120)0.007106mA=0.853mA

The expression for the collector voltage VC1 is given by,

  VC1=10VICQ1(8.8kΩ)

Substitute 0.853mA for ICQ1 in the above equation.

  VC1=10V0.853mA(8.8kΩ)=2.49V

The expression for the value of the current IC2 is given by,

  IC2=VC1VBE23.6kΩ

Substitute 0.7V for VBE2 and 2.49V for VCE1 in the above equation.

  IC2=2.49V0.7V3.6kΩ=0.497mA

The expression for the collector voltage VC2 is given by,

  VC2=10VIC2(13kΩ)

Substitute 0.497mA for IC2 in the above equation.

  VC2=10V(0.497mA)(13kΩ)=3.54V

The expression for the value of the current IC3 is given by,

  IC3=VC2VBE31.4kΩ

Substitute 0.7V for VBE3 and 3.54V for VC2 in the above equation.

  IC3=3.54V0.7V1.4kΩ=2.03mA

The expression for the small signal input resistance is given by,

  rπ1=hFEVTICQ1

Substitute 120 for hFE , 0.026 for VT and 0.853mA for ICQ1 in the above equation.

  rπ1=(120)(0.026V)0.853mA=3.66kΩ

The expression for the trans-conductance of the first transistor is given by,

  gm1=ICQ1VT

Substitute 0.853mA for ICQ1 and 0.026V for VT in the above equation.

  gm1=0.853mA0.026V=32.81mA/V

The expression for the small signal input resistance is given by,

  rπ2=hFEVTIC2

Substitute 120 for hFE , 0.026 for VT and 0.497mA for IC2 in the above equation.

  rπ2=(120)(0.026V)0.497mA=6.28kΩ

The expression for the trans-conductance of the first transistor is given by,

  gm2=IC2VT

Substitute 0.497mA for IC2 and 0.026V for VT in the above equation.

  gm2=0.497mA0.026V=19.12mA/V

The expression for the small signal input resistance is given by,

  rπ3=hFEVTIC3

Substitute 120 for hFE , 0.026 for VT and 2.03mA for IC3 in the above equation.

  rπ3=(120)(0.026V)2.03mA=1.54kΩ

The expression for the trans-conductance of the first transistor is given by,

  gm3=IC3VT

Substitute 2.03mA for IC3 and 0.026V for VT in the above equation.

  gm3=2.03mA0.026V=78.08mA/V

Conclusion:

Therefore, the trans-conductance of the first transistor Q1 is 32.81mA/V , transistor Q2 is 19.12mA/V and for transistor Q3 is 78.08mA/V .

(b)

Expert Solution
Check Mark
To determine

The value of the small signal voltage gain Avf .

Answer to Problem 12.37P

The value of small signal voltage gain is 20.7.

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.37P , additional homework tip  2

Calculation:

Mark the values and draw the small signal equivalent circuit.

The required diagram is shown in Figure 2

  Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.37P , additional homework tip  3

The expression for the input voltage Vi is given by,

  Vi=Vπ1+Vε1Vε1=ViVπ1

Apply KCL at the node 1

  Vπ1rπ1+gm1Vπ1=Vε1R]E1Vε1VoRF

Substitute ViVπ1 for Vε1 in the above equation.

  Vπ1rπ1+gm1Vπ1=(ViVπ1)R]E1(ViVπ1)VoRFVπ1(1rπ1+gm1)=(ViVπ1)(1RE1+1RF)VoRF

Substitute 3.66kΩ for rπ1 . 32.81mA/V for gm1 , 0.5kΩ for RE1 and 10kΩ for RF in the above equation.

  Vπ1(13.66kΩ+32.81mA/V)=(ViVπ1)(10.5kΩ+110k)Vo10kΩ35.18Vπ1=2.10Vi0.10Vo ........... (1)

Apply KCL at node 2.

  Vπ2=(gm1Vπ1)(RC1rπ2RC1+rπ2)

Substitute 6.28kΩ for rπ2 . 32.81mA/V for gm1 and 8.8kΩ for RC1 in the above equation.

  Vπ2=32.81mA/V(Vπ1)((8.8kΩ)(6.28kΩ)(8.8kΩ)+(6.28kΩ))=120.2Vπ1

Apply KCL at the node 3

  gm2Vπ2+Vπ3+VoRC2+Vπ3rπ3=0

Substitute 1.54kΩ for rπ3 . 19.12mA/V for gm2 and 13kΩ for RC2 in the above equation.

  (19.12mA/V)Vπ2+Vπ3(113kΩ+11.54kΩ)+Vo13kΩ=0(19.12mA/V)Vπ2+0.7263Vπ3+0.07692Vo=0

Substitute 120.2Vπ1 for Vπ2 in the above equation.

  (19.12mA/V)(120.2Vπ1)+0.7263Vπ3+0.07692Vo=0Vπ3=(316.42)Vπ1(0.1059)Vo ........... (2)

Apply KCL at the node 4

  Vπ3rπ3+gm3Vπ3=VoRE3+VoVε1RF

Substitute ViVπ1 for Vε1 in the above equation.

  Vπ3rπ3+gm2Vπ3=VoRE3+Vo(ViVπ1)RF

Substitute 1.4kΩ for RE3 , 1.54kΩ for rπ3 . 78.08mA/V for gm3 and 10kΩ for RF in the above equation.

  Vπ31.54kΩ+(78.08mA/V)Vπ3=Vo1.4kΩ+Vo(ViVπ1)10kΩ78.73Vπ3=(0.8143)Vo0.1Vi+0.1Vπ1 ........... (3)

Substitute (316.42)Vπ1(0.1059)Vo in Vπ3 in equation (2)

  78.73((316.42)Vπ1(0.1059)Vo)=(0.8143)Vo0.1Vi+0.1Vπ1Vπ1(2.49×105)Vπ1=9.152Vo0.1ViVπ1=(3.674×105)Vo(4.104×107)Vi

Substitute (3.674×105)Vo(4.104×107)Vi for Vπ1 in equation (1).

  35.18((3.674×105)Vo(4.104×107)Vi)=2.10Vi0.10VoVoVi=20.7

Conclusion:

Therefore, the value of small signal voltage gain is 20.7.

(c)

Expert Solution
Check Mark
To determine

The value of the resistance Rif .

Answer to Problem 12.37P

The value of the input resistance is 62.4kΩ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.37P , additional homework tip  4

Calculation:

The expression to determine the value of the current IRTH is given by,

  IRTH=ViRTH

Substitute 63.2kΩ for RTH in the above equation.

  IRTH=Vi63.2kΩ=1.582×105ViA

The value of Vπ1 from equation (1) is given by,

  35.18Vπ1=2.10Vi0.10VoVπ1=0.0596Vi(2.843×103)Vo

Substitute 20.7Vi for Vo in the above equation.

  Vπ1=0.0596Vi(2.843×103)20.7Vi=7×104Vi

The expression for the current Irπ1 is given by,

  Irπ1=Vπ1rπ1

Substitute 7×104Vi for Vπ1 and 3.66kΩ for rπ1 in the above equation.

  Irπ1=7×104Vi3.66kΩ=1.913×107Vi

The expression for the input current Ii is given by,

  Ii=IRTH+Irπ1

Substitute 1.582×105ViA for IRTH and 1.913×107Vi for Irπ1 in the above equation.

  Ii=1.582×105ViA+1.913×107ViViIi=62.4kΩ

Conclusion:

Therefore, the value of the input resistance is 62.4kΩ .

(d)

Expert Solution
Check Mark
To determine

The value of the resistance Rof .

Answer to Problem 12.37P

The value of the output resistance is 1.39Ω .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 12, Problem 12.37P , additional homework tip  5

Calculation:

Consider the equation (1).

  78.73Vπ3+Ix=(0.8143)Vo0.1(0)+0.1Vπ178.73Vπ3+Ix=(0.8143)Vo+0.1Vπ1 ........... (4)

Substitute 0 for Vi in equation (3).

  35.18Vπ1=2.10(0)0.10VoVπ1=(0.002843)Vo

The equation for Vπ2 is given by,

  Vπ2=120.2Vπ1

Substitute (0.002843)Vo for Vπ1 in the above equation.

  Vπ2=120.2((0.002843)Vo)=0.3417Vo

Consider the equation given,

  (19.12mA/V)Vπ2+0.7263Vπ3+0.07692Vo=0

Substitute 0.3417Vo for Vπ2 in the above equation.

  (19.12mA/V)(0.3417Vo)+0.7263Vπ3+0.07692Vo=0Vπ3=9.1Vo

Substitute 9.1Vo for Vπ3 and (0.002843)Vo for Vπ1 in equation (4).

  78.73(9.1Vo)+Ix=(0.8143)Vo+0.1((0.002843)Vo)VoIx=1.39×103kΩVoIx=1.39Ω

Conclusion:

Therefore, the value of the output resistance is 1.39Ω .

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Chapter 12 Solutions

Microelectronics: Circuit Analysis and Design

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