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 14, Problem 14.61P

a.

To determine

Maximum value of tolerance of resistor ‘ x’for given CMRRdB .

a.

Expert Solution
Check Mark

Answer to Problem 14.61P

Maximum value of tolerance of resistor is xmax=0.474%

Explanation of Solution

Given:

The given difference amplifier circuit is,

  Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.61P , additional homework tip  1

Tolerance of each resistor is ±x% .

Minimum CMRRdB is 50dB .

Calculation:

Circuit with voltage and resistance notation is,

  Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.61P , additional homework tip  2

Tolerance of each resistor is ±x% .

  CMRRdB=20log10(CMRR) CMRR dB20=log10(CMRR)CMRR=10( CMRR dB 20 )CMRR=10( 50 10 )CMRR=316.228

KCL at VY node,

  VYR4+VYv l2R3=0R3VY+R4( V Y v l2 )R4R3=0R3VY+R4(VYv l2)=0(R3+R4)VYR4vl2=0VY=R4R3+R4vl2

As op-amp is ideal, so VX=VY

  VX=R4R3+R4vl2.......(1)

KCL at VX node;

  VXv l1R1+VXv0R2=0R2( V X v l1 )+R1( V X v 0 )R1R2=0R2(VXv l1)+R1(VXv0)=0(R1+R2)VXR2vl1R1v0=0

Now put VX from (1)

  (R1+R2)( R 4 R 3 + R 4 v l2)R2vl1R1v0=0R1v0=(R1+R2)( R 4 R 3 + R 4 )vl2R2vl1v0=( R 1 + R 2 R 1 )( R 4 R 3 + R 4 )vl2R2R1vl1v0=(1+ R 2 R 1 )( R 4 R 3 + R 4 )vl2R2R1vl1

If vd is differential mode input voltage and vcm is common-mode input voltage, then,

  vl1=vcmvd2vl2=vcm+vd2v0=(1+ R 2 R 1 )( R 4 R 3 + R 4 )(v cm+ v d 2)R2R1(v cm v d 2).....(2)

For vd=0 and Acm=v0vcm equation (2) results,

  v0=(1+ R 2 R 1 )( R 4 R 3 + R 4 )(v cm+02)R2R1(v cm02)v0=[(1+ R 2 R 1 )( R 4 R 3 + R 4 ) R 2 R 1]vcmv0v cm=(1+ R 2 R 1 )( R 4 R 3 + R 4 )R2R1Acm=(1+ R 2 R 1 )( R 4 R 3 + R 4 )R2R1

For vcm=0 and Ad=v0vd equation (2) results,

  v0=(1+ R 2 R 1 )( R 4 R 3 + R 4 )(0+ v d 2)R2R1(0 v d 2)v0=12[(1+ R 2 R 1 )( R 4 R 3 + R 4 )+ R 2 R 1]vdv0vd=12[(1+ R 2 R 1 )( R 4 R 3 + R 4 )+ R 2 R 1]Ad=12[(1+ R 2 R 1 )( R 4 R 3 + R 4 )+ R 2 R 1]

Now,

  CMRR=| A d A cm|CMRR=12[( 1+ R 2 R 1 )( R 4 R 3 + R 4 )+ R 2 R 1 ]( 1+ R 2 R 1 )( R 4 R 3 + R 4 ) R 2 R 1 CMRR=12[ R 4 R 3 ( 1+ R 2 R 1 ) 1 ( 1+ R 4 R 3 )+ R 2 R 1 ] R 4 R 3 ( 1+ R 2 R 1 )1 ( 1+ R 4 R 3 ) R 2 R 1 .........(3)

For minimum CMRR maximize the denominator, that is R4R3 will be maximum and R2R1 will be minimum.

  ( R 4 R 3 )max=( 1+x% 1x%)( R 4 R 3 )( R 4 R 3 )max=( 1+ x 100 1 x 100 )( 50k 10k)( R 4 R 3 )max=( 100+x 100x)( 50 10)( R 4 R 3 )max=5( 100+x 100x)

and

  ( R 2 R 1 )min=( 1+x% 1x%)( R 2 R 1 )( R 2 R 1 )min=( 1+ x 100 1 x 100 )( 50k 10k)( R 2 R 1 )min=( 100+x 100x)( 50 10)( R 2 R 1 )min=5( 100+x 100x)

Therefore,

   CMRR= 1 2 [ 5( 100+x 100x )( 1+5( 100x 100+x ) ) 1 ( 1+5( 100+x 100x ) ) +5( 100x 100+x ) ] 5( 100+x 100x )( 1+5( 100x 100+x ) ) 1 ( 1+5( 100+x 100x ) ) 5( 100x 100+x )

   2CMRR= 5[ ( 100+x 100x )( 1+5( 100x 100+x ) ) 1 ( 1+5( 100+x 100x ) ) +( 100x 100+x ) ] 5[ ( 100+x 100x )( 1+5( 100x 100+x ) ) 1 ( 1+5( 100+x 100x ) ) ( 100x 100+x ) ]

   2CMRR= [ ( 100+x 100x )( 6004x 100+x ) 1 ( 600+4x 100x ) +( 100x 100+x ) ] [ ( 100+x 100x )( 6004x 100+x ) 1 ( 600+4x 100x ) ( 100x 100+x ) ]

   2CMRR= [ ( 6004x 600+4x )+( 100x 100+x ) ] [ ( 6004x 600+4x )( 100x 100+x ) ]

   2CMRR= 1200008 x 2 400x

As x is very small. So neglect x2

  2CMRR=120000400xx=120000400×2CMRR.....(4)x=120000400×2×316.228xmax=0.474%

b.

To determine

Maximum value of tolerance of resistor ‘ x’ for given CMRRdB .

b.

Expert Solution
Check Mark

Answer to Problem 14.61P

Maximum value of tolerance of resistor is xmax=0.0267%

Explanation of Solution

Given:

The given difference amplifier circuit is;

  Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.61P , additional homework tip  3

Tolerance of each resistor is ±x%

Minimum CMRRdB is 75dB .

Calculation:

Minimum CMRRdB is 50dB .

  CMRRdB=20log10(CMRR) CMRR dB20=log10(CMRR)CMRR=10( CMRR dB 20 )CMRR=10( 75 10 )CMRR=5623.41

Using equation (4)

  x=120000400×2CMRR

Put the value of CMRR in above equation

  x=120000400×2×5623.41x=1200004498728xmax=0.0267%

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Chapter 14 Solutions

Microelectronics: Circuit Analysis and Design

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