All parameters associated with the instrumentation amplifier in Figure 9.26are the same as given in Exercise Ex 9.8, except that resistor R 3 , which isconnected to the inverting terminal of A3, is R 3 = 30 k Ω ± 5 % . Determinethe maximum common-mode gain.

Question

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Answer 1

Textbook Question

Chapter 9, Problem 9.72P

All parameters associated with the instrumentation amplifier in Figure 9.26are the same as given in Exercise Ex 9.8, except that resistor $R 3$ , which isconnected to the inverting terminal of A3, is $R 3 = 30 k Ω ± 5 %$ . Determinethe maximum common-mode gain.

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

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Answer 2

Textbook Question

Chapter 9, Problem 9.72P

All parameters associated with the instrumentation amplifier in Figure 9.26are the same as given in Exercise Ex 9.8, except that resistor $R 3$ , which isconnected to the inverting terminal of A3, is $R 3 = 30 k Ω ± 5 %$ . Determinethe maximum common-mode gain.

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

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Answer 3

Textbook Question

Chapter 9, Problem 9.72P

All parameters associated with the instrumentation amplifier in Figure 9.26are the same as given in Exercise Ex 9.8, except that resistor $R 3$ , which isconnected to the inverting terminal of A3, is $R 3 = 30 k Ω ± 5 %$ . Determinethe maximum common-mode gain.

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

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Answer 4

Textbook Question

Chapter 9, Problem 9.72P

All parameters associated with the instrumentation amplifier in Figure 9.26are the same as given in Exercise Ex 9.8, except that resistor $R 3$ , which isconnected to the inverting terminal of A3, is $R 3 = 30 k Ω ± 5 %$ . Determinethe maximum common-mode gain.

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

Answer 8

Expert Solution & Answer

To determine

The value of maximum common mode gain.

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)

The general expression for the output equation is given by,

$vO=Acmvcm+Advd$

From above and from equation (3), the expression for the common mode gain is evaluated as,

$Acm=(−R4R 3x)+(1+R4R 3x)(R4R3+R4)$

Substitute $90 kΩ$ for $R4$ , $30 kΩ$ for $R3$ and $30 kΩ±5%$ for $R3x$ in the above equation.

$Acm=(− 90 kΩ 30 kΩ±5%)+(1+ 90 kΩ 30 kΩ±5%)( 90 kΩ 30 kΩ+90 kΩ)=14[3−( 90 30±5%)]$

From above the range of the common mode voltage gain is given by,

$14[3−( 90 30−5%)]≤Acm≤14[3−( 90 30+5%)]−0.0395≤Acm≤0.0357$

The maximum value of the voltage gain is given by,

$|Acm|max=0.0395$

Conclusion:

Therefore, themaximum value of the voltage gain is $0.0395$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The value of maximum common mode gain.

Answer 12

Answer to Problem 9.72P

The maximum value of the voltage gain is $0.0395$ .

Answer 13

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 1

Mark the voltages and current and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.72P , additional homework tip 2

The expression for the voltage $vb$ by voltage division rule is given by,

$vb=R4vO2R3+R4$

The expression for the current at the node $va$ is given by,

$va=vb$

Substitute $R4vO2R3+R4$ for $vb$ in the above equation.

$va=R4vO2R3+R4$

Apply KCL at the node $va$ .

$v O1−vaR 3x=va−vOR4vO=(− R 4 R 3x )vO1+(1+ R 4 R 3x )va$

Substitute $R4vO2R3+R4$ for $va$ in the above equation.

$vO=(−R4R 3x)vO1+(1+R4R 3x)R4vO2R3+R4$ ........ (1)

The expression for the value of the current $i1$ is given by,

$i1=vI1−vI2R1$

The expression for the voltage $vO1$ is given by,

$vO1−vI1=i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO1=vI1+(v I1−v I2R1)R2$

The expression for the voltage $vO2$ is given by,

$vO2=vI2−i1R2$

Substitute $vI1−vI2R1$ for $i1$ in the above equation.

$vO2=vI2−(v I1−v I2R1)R2$

Substitute $vI1+(v I1−v I2R1)R2$ for $vO1$ and $vI2−(v I1−v I2R1)R2$ for $vO2$ in equation (1)

$vO=(− R 4 R 3x )(v I1+( v I1 − v I2 R 1 )R2)+(1+ R 4 R 3x )R4R3+R4(v I2−( v I1 − v I2 R 1 )R2)=( v I1 − v I2 R 1 )R2[( − R 4 R 3x )−(1+ R 4 R 3x )(1+ R 4 R 3 + R 4 )]−vI1( R 4 R 3x )+vI2(1+ R 4 R 3x )( R 4 R 3 + R 4 )$ ........ (2)

The expression for the common mode voltage is given by,

$vcm=vI1+vI22$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

The expression for the voltage $vI2$ is given by,

$vI2=vcm+vd2$

The expression for the voltage $vI1$ is given by,

$vI1=vcm−vd2$

Substitute $vd$ for $vI2−vI1$ , $vcm+vd2$ for $vI2$ and $vcm−vd2$ for $vI1$ in equation (2).

$vO=[( v d R 1 ) R 2[ ( − R 4 R 3x )−( 1+ R 4 R 3x )( 1+ R 4 R 3 + R 4 )]−( v cm − v d 2 )( R 4 R 3x )+( v cm + v d 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]=[ v d[ ( R 2 R 1 + 1 2 )( R 4 R 3x )( R 2 R 1 + 1 2 )( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]+ v cm[ ( − R 4 R 3x )+( 1+ R 4 R 3x )( R 4 R 3 + R 4 )]]$ ........ (3)