Consider the differential amplifier shown in Figure 9.24(a). Assume thateach resistor is 50 ( 1 ± x ) k Ω . (a) Determine the worst case common-modegain A C M = v O / v C M , where v C M = v 1 = v 2 . (b) Evaluate A C M andCMRR(dB) for x = 0.01 , 0.02 , and 0.05.

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Answer 1

Textbook Question

Chapter 9, Problem 9.62P

Consider the differential amplifier shown in Figure 9.24(a). Assume thateach resistor is $50 ( 1 ± x ) k Ω$ . (a) Determine the worst case common-modegain $A C M = v O / v C M$ , where $v C M = v 1 = v 2$ . (b) Evaluate $A C M$ andCMRR(dB) for $x = 0.01 , 0.02 ,$ and 0.05.

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

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Answer 2

Textbook Question

Chapter 9, Problem 9.62P

Consider the differential amplifier shown in Figure 9.24(a). Assume thateach resistor is $50 ( 1 ± x ) k Ω$ . (a) Determine the worst case common-modegain $A C M = v O / v C M$ , where $v C M = v 1 = v 2$ . (b) Evaluate $A C M$ andCMRR(dB) for $x = 0.01 , 0.02 ,$ and 0.05.

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

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Answer 3

Textbook Question

Chapter 9, Problem 9.62P

Consider the differential amplifier shown in Figure 9.24(a). Assume thateach resistor is $50 ( 1 ± x ) k Ω$ . (a) Determine the worst case common-modegain $A C M = v O / v C M$ , where $v C M = v 1 = v 2$ . (b) Evaluate $A C M$ andCMRR(dB) for $x = 0.01 , 0.02 ,$ and 0.05.

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

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Answer 4

Textbook Question

Chapter 9, Problem 9.62P

Consider the differential amplifier shown in Figure 9.24(a). Assume thateach resistor is $50 ( 1 ± x ) k Ω$ . (a) Determine the worst case common-modegain $A C M = v O / v C M$ , where $v C M = v 1 = v 2$ . (b) Evaluate $A C M$ andCMRR(dB) for $x = 0.01 , 0.02 ,$ and 0.05.

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Answer 8

(a)

Expert Solution

To determine

The value of the worst case common mode gain.

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The value of the worst case common mode gain.

Answer 13

Answer to Problem 9.62P

The value of the worst case common mode gain is $2x1+x$ .

Answer 14

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 1

Mark the voltages and redraw the circuit.

The required diagram is shown in Figure 2

Microelectronics: Circuit Analysis and Design, Chapter 9, Problem 9.62P , additional homework tip 2

The expression for the common mode voltage gain is given by,

$ACM=vOvCM$

The expression for the value of the voltage $v1$ is given by,

$v1=v2$

The expression for the voltage $v2$ by voltage division rule is given by,

$v2=(R4R3+R4)vI2$

Apply KCL at the inverting terminal.

$v I1−v1R1=v1−vOR2v I1R2−v1R1−v1R1R1R2=−vOR2vO=−vI1( R 2 R 1 )+v1(1+ R 2 R 1 )$

Substitute $v2$ for $v1$ in the above equation.

$vO=−vI1(R2R1)+v2(1+R2R1)$

Substitute $(R4R3+R4)vI2$ for $v2$ in the above equation.

$vO=−vI1(R2R1)+(( R 4 R 3 + R 4 )vI2)(1+R2R1)$ …… (1)

The expression for the common node input voltage is given by,

$vCM=vI1$

The expression for the common node input voltage is given by,

$vCM=vI2$

Substitute $vCM$ for $vI1$ and $vCM$ for $vI2$ in the above equation.

$vO=−vCM( R 2 R 1 )+(( R 4 R 3 + R 4 )v CM)(1+ R 2 R 1 )vOv CM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]$

Substitute $ACM$ for $vOvCM$ in the above equation.

$ACM=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )− R 2 R 1]=R4R3+R4(1− R 2 R 3 R 1 R 4 )$ …… (2)

In the above equation the worst case $ACM$ is obtained when $R4R3+R4$ is maximum and $1−R2R3R1R4$ is minimum. The second case arises when $R2$ and $R3R4$ is minimum and $R1$ is maximum. The third case arises when the value of $R1, R4$ are minimum and $R2$ , $R3$ is minimum.

For worst condition the value of the resistance $R1$ is given by,

$R1=50(1+x)$

For worst condition the value of the resistance $R2$ is given by,

$R2=50(1−x)$

For worst condition the value of the resistance $R3$ is given by,

$R3=50(1−x)$

For worst condition the value of the resistance $R4$ is given by,

$R4=50(1+x)$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$ACM=50( 1+x)50( 1−x)+50( 1+x)(1− ( 50( 1−x ) )( 50( 1−x ) ) ( 50( 1+x ) )( 50( 1+x ) ))=2x1+x$ …… (3)

Conclusion:

Therefore, the value of the worst case common mode gain is $2x1+x$ .

Answer 15

(b)

Expert Solution

To determine

The value of the CMRR and the common mode gain.

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The value of the CMRR and the common mode gain.

Answer 20

Answer to Problem 9.62P

The value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .

Answer 21

Explanation of Solution

Calculation:

The expression for the common mode gain is given by,

$vCM=v I2+v I12vI1=2vCM−vI2$

The expression for the differential voltage is given by,

$vd=vI2−vI1$

Substitute $2vCM−vI2$ for $vI1$ in the above equation.

$vd=vI2−2vCM+vI2vd=2vI2−2vCMvI2=vd2+vCM$

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

$vI1=vCM−vd2$

The expression for the common mode voltage is given by,

$vCM=vI2+vI12$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in the above equation.

$vCM= v d 2+v CM−v CM− v d 22=0 V$

Substitute $vCM−vd2$ for $vI1$ and $vd2+vCM$ for $vI2$ in equation (1).

$vO=−(vCM−vd2)(R2R1)+[( R 4 R 3+ R 4)( R 1+ R 2 R 1)](vd2+vCM)$

Substitute $0$ for $vCM$ in the above equation.

$vO=( v d 2)( R 2 R 1 )+[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )]( v d 2)=[( R 4 R 3 + R 4 )( R 1 + R 2 R 1 )+( R 2 R 1 )]vd2$

Substitute $50(1+x)$ for $R1$ , $50(1−x)$ for $R2$ , $50(1−x)$ for $R3$ and $50(1+x)$ for $R4$ in the above equation.

$vO=[( 50( 1+x ) 50( 1−x )+50( 1+x ))( 50( 1−x )+( 50( 1+x ) ) 50( 1−x ))+( 50( 1−x ) 50( 1+x ))]vd2vO=[21+x]vd2vOvd=[21+x]$

Substitute $Ad$ for $vOvd$ in the above equation.

$Ad=[11+x]$

The expression for the $CMRR$ is given by,

$CMRR(dB)=20log10|AdACM|$

Substitute for $11+x$ for $Ad$ and $2x1+x$ for $ACM$ in the above equation.

$CMRR(dB)=20log10| 1 1+x 2x 1+x|=20log10|12x|$ …… (4)

Substitute $0.01$ for $x$ in equation (3).

$ACM=2( 0.01)1+0.01=0.0198$

Substitute $0.01$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.01)|=33.979 dB$

Substitute $0.02$ for $x$ in equation (3).

$ACM=2( 0.02)1+0.02=0.0392$

Substitute $0.02$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.02)|=27.959 dB$

Substitute $0.05$ for $x$ in equation (3).

$ACM=2( 0.05)1+0.05=0.0952$

Substitute $0.05$ for $x$ in equation (4).

$CMRR(dB)=20log10|12( 0.05)|=20 dB$

Conclusion:

Therefore, the value of $ACM$ for $x$ equals to $0.01$ is $0.0198$ , for $x$ equal to $0.02$ is $0.0392$ and for $x$ equals to $0.05$ is $0.0952$ . The value of the CMRR for for $x$ equals to $0.01$ is $33.979 dB$ , for $x$ equal to $0.02$ is $27.959 dB$ and for $x$ equals to $0.05$ is $20 dB$ .