MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)
4th Edition
ISBN: 9781266368622
Author: NEAMEN
Publisher: MCG
Question
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Chapter 11, Problem 11.92P

a.

To determine

The differential-mode voltage gain Ad=vo3/vd and the common-mode voltage gain Acm=vo3/vd

a.

Expert Solution
Check Mark

Answer to Problem 11.92P

  Ad=176ACM=0.114

Explanation of Solution

Given:

The given circuit is,

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 11, Problem 11.92P , additional homework tip  1

  β=200,VBE(on)=0.7V,VA=80V

Calculation:

Consider the given figure,

  I1=V+VBEVR1I1=120.7(12)12×103I1=1.94mAIc4=I1=1.94mAIcs=Ic4=1.94mA

  Ic1=Ic2=Ic52Ic1=1.94mA2=0.97mAgm2=Ic2VTgm2=0.97×1030.026=37.3mA/V

The voltage v02 is,

  v02=V+Ic2RC=12(0.97×103)(8×103)=4.24V

  Ic3=IE3=vo2VBERE=4.240.73.3×103=1.07mA

  rπ3=βVTIc3=200(0.026)1.07×103=4.86kΩ

The input resistance of the CE amplifier is,

  Rin3=rπ3+(1+β)RE=4.86×103+(1+200)(3.3×103)=668kΩThegainisACE=βRc2rπ3+(1+β)RE=200(8×103)4.86×103+(1+200)(3.3×103)=1.197

  Ad1=12gm2(Rc||Rin3)Ad1=12gm2(RcRin3Rc+Rin3)Ad1=12(37.3×103)(8×103(668×103)8×103+(668×103))Ad1=147.43Ad=v03vd=Ad1ACE=(147.43)(1.197)Ad=176

Now calculate the output resistance,

  Ro=r05Ro=VAIc5=801.94×103Ro=41.2kΩ

Now calculate rπ2 ,

  rπ2=βVTIc2=200(0.026)0.97×103=5.36kΩnow,Acm1Acm1=gm2(Rc||Rin3)1+2(1+β)Rorπ2Acm1=(37.3×103)8×103(668×103)8×103+(668×103)1+2(1+200)(41.2×103)5.36×103Acm1=0.09539nowAcm,Acm=Acm1ACEAcm=(0.09539)(1.197)Acm=0.114

Hence,

  Ad=176ACM=0.114

b.

To determine

The output voltage v03 .

To compare: The obtained result with the ideal output.

b.

Expert Solution
Check Mark

Answer to Problem 11.92P

  v03=5.28sinωtV

The output voltage obtained for ideal case is greater than the actual result obtained.

Explanation of Solution

Given:

The given circuit is,

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 11, Problem 11.92P , additional homework tip  2

  β=200,VBE(on)=0.7V,VA=80V

  v1=2.015sinωtVandv2=1.985sinωtV

Calculation:

Consider the given voltage,

  v1=2.015sinωtVandv2=1.985sinωtV

The differential input voltage will be,

  vd=v1v2=2.015sinωt1.985sinωt=0.03sinωtVnowvcmis,vcm=v1+v22=2.015sinωt+1.985sinωt2vcm=2sinωtV

Now calculate output voltage,

  v03=Advd+Acmvcm=176(0.03sinωt)+0.114(2sinωt)=5.052sinωtV

Now calculate output voltage for ideal case,

Put Acm=0 for ideal case,

  v03=Advd+Acmvcm=176(0.03sinωt)+0(2sinωt)=5.28sinωtV

Hence,

  v03=5.28sinωtV

It is observed that both the values are slightly different. The output voltage obtained for ideal case is greater than the actual result obtained.

c.

To determine

The differential-mode and common-mode input resistances.

c.

Expert Solution
Check Mark

Answer to Problem 11.92P

  Rid=10.72kΩRicm=4.15MΩ

Explanation of Solution

Given:

The given circuit is,

  MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL), Chapter 11, Problem 11.92P , additional homework tip  3

  β=200,VBE(on)=0.7V,VA=80V

Calculation:

Let calculatethe differential-mode input resistances,

  Rid=rπ1+rπ2=2rπ2[rπ1=rπ2]=2(5.36×103)=10.72kΩ

  r02=VAIc2=800.97×103=82.5kΩ

The common-mode input resistances will be,

  2Ricm=[(1+β)2R0]||[(1+β)r02]2Ricm=[(1+200)2(41.2×103)]||[(1+200)(82.5×103)]2Ricm=16.6×106||16.6×1062Ricm=16.6×106×16.6×10616.6×106+16.6×1062Ricm=8.3Ricm=8.3/2Ricm=4.15MΩ

Hence,

  Rid=10.72kΩRicm=4.15MΩ

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Chapter 11 Solutions

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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