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 6, Problem 6.45P

Consider the circuit in Figure P6.45. The transistor parameters are β = 120 and V A = . Repeat parts (a)−(d) of Problem 6.44.

Chapter 6, Problem 6.45P, Consider the circuit in Figure P6.45. The transistor parameters are =120 and VA= . Repeat parts
Figure P6.45

a.

Expert Solution
Check Mark
To determine

The value of the ICQ and VCEQ .

Answer to Problem 6.45P

  ICQ=2.09mAVCEQ=3.69V

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  1

Calculation:

Let the BJT be the single node, then by applying Kirchhoff’s current law, the quiescent emitter current IEQ , the quiescent base current IBQ and the quiescent collector current ICQ are expressed as:

  IEQ=IBQ+ICQ ……………………….(1)

In Common-Emitter mode:

The relation between the quiescent collector current is ICQ and the quiescent base current is IBQ .

  ICQ=βIBQ...............(2)

For dc analysis,

All the capacitors act as an open circuit and all the ac sources act as short circuits.

DC analysis:

Redrawing the circuit by open circuit, the capacitors, and short circuit the ac sources.

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  2

Evaluating the Thevenin resistance RTH by using voltage division rule:

  RTH=R1R2=R1×R2R1+R2RTH=(10×1010+10)kΩRTH=5kΩ

Evaluating the Thevenin voltage, VTH :

  VTH=10R1+R2R2=(1010kΩ+10kΩ)×10kΩ=5V

Drawing the modified circuit:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  3

Applying the Kirchhoff s voltage law in the base-emitter loop,

  IBQRTHVBE(on)IEQRE+10VTH=0IBQRTHVBE(on)(1+β)IBQRE+10VTH=0IBQ=10VTHVBE(on)RTH+(1+β)RE

Substituting 5kΩforRTH,-5VforVTH,120forβ,2kΩforRE,0.7VforVBE(on)

  IBQ=[1050.7(5×103)+(1+120)×(2×103)]=0.0174mA

From equation (2):

  ICQ=βIBQ

Substituting 120forβ,and0.174mAforIBQ.

  ICQ=120×0.0174mA=2.09mA

Therefore, the value of ICQ is 2.09mA .

From equation (1) and (2):

  ICQ=βIBQIEQ=IBQ+ICQ

Substituting ICQβforIBQ ,

  IEQ=ICQβ+ICQIEQ=(1+ββ)ICQ

Substitute 120forβ,and2.09mAforICQ.

  IEQ=(1+120120)×2.09mA=2.11mA

Drawing the modified circuit:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  4

Applying Kirchhoff s voltage law to the loop,

  ICQRCVCEQIEQRE+10=0VCEQ=10ICQRCIEQREVCEQ=10(RC+1+ββRE)×ICQ{sinceIEQ=(1+ββ)ICQ}

Substituting 120forβ,1kΩforRC,2kΩforREand2.09mAforICQ.

  VCEQ={10(1+1+120120×2)kΩ×2.09mA}=3.69V

Therefore, the value of VCEQ is 3.69V.

b.

Expert Solution
Check Mark
To determine

To plot: The dc and the ac load lines.

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  5

Calculation:

Applying Kirchhoff s voltage law around the collector-emitter loop in figure 1.

  ICRCVCEIERE+10=0VCE=10ICRCIEREVCE=10{RC+(1+ββ)RE}ICVCE=10{1×103+(1+120120)2×103}(IC)VCE=10(3.02kΩ)IC........................(3)

Substitute 0VforVCE in equation (3):

  0=10(3.02kΩ)ICIC=103.02×103A=3.31mA

Substitute 0AforIC in equation (3).

  VCE=10(3.02kΩ)(0)=10V

The required “Q” point co-ordinates are ICQ=2.09mA,VCEQ=3.69V .

Draw the DC load line:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  6

Drawing the AC equivalent circuit:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  7

Evaluating the ac load rc :

  rc=RERE=(2kΩ2kΩ)=1kΩ

The expression for the ac load line:

  IC=(ICQ+VCEQrc)VCErC................(4)

Substitute 0VforVCE in equation (4).

  IC=(ICQ+VCEQrc)(0)=(2.09+3.691)mA=5.78mA

Substitute 0VforIC in equation (4).

  0=(ICQ+VCEQrc)VCErCVCE=VCEQ+ICQrC=3.69+(2.09mA)(1kΩ)=5.78V

Drawing the ac load line:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  8

c.

Expert Solution
Check Mark
To determine

The small signal voltage gain.

Answer to Problem 6.45P

The small signal voltage gain.

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  9

Drawing the small signal analysis of the circuit by short circuiting the dual dc sources and the capacitors:

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  10

Evaluating the diffusion resistance,

  rπ=βVTICQrπ=120×0.026V2.09×103A=1.49×103Ω=1.49

Evaluating the small signal transistor output resistance r0 ,

  r0=VAICQr0=ICQ=

Evaluating the input resistance Rib of the transistor:

  Rib=rπ+(1+β)(r0||RE||RL)=rπ+(1+β)(RE||RL){sincer0=}={1.49+(1+120)×(2||2)}kΩRib=122.5kΩ................(5)

Calculate the input resistance Ri .

  Ri=R1||R2||Rib=(10||10||122.5)kΩ=(5||122.5)kΩ=4.8kΩ

The small-signal voltage gain is, Aν is

  Av=(1+β)(RERL)rπ+(1+β)(RERL)(R1R2RibR1R2Rib+RS)=121×(22)1.49+(121×(22))(4.84.8+5)=0.484

d.

Expert Solution
Check Mark
To determine

The output resistance.

Answer to Problem 6.45P

The output resistance is 32.5Ω .

Explanation of Solution

Given:

The circuit is given as

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.45P , additional homework tip  11

Calculation:

From equation 5,

  Rib=122.5kΩ

Thus, the input resistance Rib is 122.5kΩ :

Calculate the output resistance, R0 :

  R0=(rπ+(R1||R2||Rs)1+β)||RE||r0=(rπ+(R1||R2||Rs)1+β)||RE{sincer0=}={(1.49+(10||10||5)1+120)||2}×103Ω=32.5Ω

Therefore, the output resistance R0 is 32.5Ω .

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Chapter 6 Solutions

Microelectronics: Circuit Analysis and Design

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