Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
Question
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Chapter 11, Problem 11.70P

(a)

To determine

The value of the output resistance.

(a)

Expert Solution
Check Mark

Answer to Problem 11.70P

The value of the output resistance RO=73.2kΩ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.70P , additional homework tip  1

Figure 1

Calculation:

The expression for the input current is given by,

  Iπ=VxVπ4rO2+gm2Vπ2+gmVπ4+VxrO4 ....................(1)

The expression to determine the value of the transconductance is given by,

  gm2Vπ2+VxVπ4rO2=Vπ4rπ4+rπ2VxVπ4rO2=Vπ4rπ4+rπ2gm2Vπ2

The expression for the relation for the voltage for hybrid pie parameter.

  Vπ4=Vπ2

Substitute Vπ4rπ4+rπ2gm2Vπ2 for VxVπ4rO2 in equation (1).

  Iπ=Vπ4rπ4||rπ2gm2Vπ2+gm2Vπ2+gmVπ4+VxrO4Iπ=Vπ4rπ4||rπ2+gmVπ4+VxrO4VxrO2=Vπ4[1rπ4||rπ2+1rO2+gm2] ....................(2)

The value of the collector current IC4 is calculated as,

  IC4=(β1+β)IQ2=(120120+0)1mA2=0.496mA

The value of the collector current IC2 is calculated as,

  IC2=(β1+β)IQ2(11+β)=(120120+0)1mA2(11+120)=0.0041mA

The expression to determine the expression for the diffusion resistance is given by,

  rπ2=βVTIC2

Substitute 0.026V for VT , 120 for β and 0.0041mA for IC2 in the above equation.

  rπ2=(120)(0.026V)0.0041mA=761kΩ

The value of the transconductance is calculated as,

  gm2=IC20.026V

Substitute 0.0041mA for IC2 in the above equation.

  gm2=0.0041mA0.026V=0.158mA

The expression to determine the value of the resistance rO2 is given by,

  rO2=VAIC2

Substitute 100V for VA and 0.0041mA for IC2 in the above equation.

  rO2=100V0.0041mA=24.4MΩ

The expression to determine the expression for the diffusion resistance is given by,

  rπ4=βVTIC4

Substitute 0.026V for VT , 120 for β and 0.496mA for IC4 in the above equation.

  rπ4=(120)(0.026V)0.496mA=6.29kΩ

The value of the transconductance is calculated as,

  gm4=IC40.026V

Substitute 0.496mA for IC4 in the above equation.

  gm4=0.496mA0.026V=19.08mA/V

The expression to determine the value of the resistance rO4 is given by,

  rO4=VAIC4

Substitute 100V for VA and 0.496mA for IC4 in the above equation.

  rO4=100V0.496mA=202kΩ

Substitute 24.4MΩ for rO2 , 202kΩ for rO4 , 6.29kΩ for rπ4 , 761kΩ for rπ2 and 0.158mA for gm2 in equation (2)

  Vx202kΩ=Vπ4[16.29||rπ2+124.4MΩ+0.158mA]Vx202kΩ=[16.29||761kΩ+124.4MΩ+0.158mA]Vπ4=[Vx(0.318)rO2]

The expression for the input current Ix is given by,

  Ix=VxrO1+VxrO4+Vπ4(gm4gm21rO2)

Substitute Vx0.318rO2 for Vπ4 in the above equation.

  Ix=VxrO1+VxrO4+Vx0.318rO2(gm4gm21rO2)

Substitute 24.4MΩ for rO2 , 202kΩ for rO4 , 19.08mA/V for gm4 and 0.158mA for gm2 in the above equation.

  Ix=VxrO1+VxrO4+Vx0.318rO2(19.08mA/V0.158π124.4MΩ)VxIx=135×10kΩ

The expression to determine the output resistance is given by,

  Ro2=VxIx

Substitute 135×10kΩ for VxIx in the above equation.

  Ro2=VxIx

The value of the quiescent current is given by,

  IC=IQ2=1mA2=0.5mA

The small signal output resistance is calculated as,

  ro6=βIC=800.5mA=160kΩ

The value of the output resistance is calculated as,

  RO=rO2||rO6=(135kΩ)(160kΩ)(135kΩ+160kΩ)=73.2kΩ

Conclusion:

Therefore, the value of the output resistance RO=73.2kΩ .

(b)

To determine

The value of the differential mode voltage gain.

(b)

Expert Solution
Check Mark

Answer to Problem 11.70P

The value of the differential voltage gain is 704 .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.70P , additional homework tip  2

Figure 1

Calculation:

The required small signal circuit is shown in Figure 2

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.70P , additional homework tip  3

Figure 2

The expression for the transconductance is given by,

  gmc=Δivd2

The expression for the differential voltage gain is given by,

  Ad=gmcRO=Δivd2RO

The expression for the change in current is given by,

  Δi=gm1Vπ1+gm3Vπ3 ....................(3)

Consider the value of Vπ3 is given by,

  Vπ1+Vπ3=vd2 ....................(4)

Substitute the values in the above equation.

  Vπ3=rπ3(gm1+1rπ1)Vπ1=6.29kΩ(120+1761kΩ)Vπ1=Vπ1

Substitute Vπ1 for Vπ3 in equation (4).

  Vπ1+Vπ1=vd2Vπ1=vd4

Substitute Vπ1 for Vπ3 in equation (3).

  Δi=gm1Vπ1+gm3Vπ1

Substitute 19.08mA/V for gm1 , vd4 for Vπ1 in the above equation.

  Δi=(19.08mA/V)vd4+(19.08mA/V)vd4=9.62(vd2)

Substitute 9.62(vd2) for Δi in the equation for transconductance.

  gmc=9.62(vd2)vd2=9.62

The value of gain is given by,

  Ad=(9.62)(73.2kΩ)=704

The expression for the input resistor is given by,

  Ri=rπ1+(1+β)rπ3=761kΩ+(1+120)(6.29kΩ)=1522kΩ

The value of expression for the differential input resistor is calculated as,

  Rid=2Ri=2(1522kΩ)=3.044MΩ

Conclusion:

Therefore, the value of the differential voltage gain is 704 .

(c)

To determine

The value of the differential mode input resistance Rid .

(c)

Expert Solution
Check Mark

Answer to Problem 11.70P

The value of Rid=3.044MΩ .

Explanation of Solution

Given:

The given diagram is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 11, Problem 11.70P , additional homework tip  4

Calculation:

The expression for the input resistor is given by,

  Ri=rπ1+(1+β)rπ3=761kΩ+(1+120)(6.29kΩ)=1522kΩ

The value of expression for the differential input resistor is calculated as,

  Rid=2Ri=2(1522kΩ)=3.044MΩ

Conclusion:

Therefore, the value of Rid=3.044MΩ .

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Chapter 11 Solutions

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