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 7, Problem 7.21P

For the circuit in Figure P7.21, the transistor parameters are β = 120 , V B E (on)=0 .7V , and V A = 50 V . (a) Design a bias−stable circuit such that I E Q =1 .5mA . (b) Using the results of part (a), find the small−signal midband voltage gain. (c) Determine the output resistance R o . (d) What is the lower 3dB corner frequency?

Chapter 7, Problem 7.21P, For the circuit in Figure P7.21, the transistor parameters are =120 , VBE(on)=0.7V , and VA=50V .
Figure P7.21

a.

Expert Solution
Check Mark
To determine

To design: The bias stable circuit for the given condition.

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.21P , additional homework tip  1

IEQ =1.5mA.

The parameters of the circuit is given as:

  β=120VBE( on)=0.7VVA=50V

The design needed for the bias stability is given as:

  RTH𑨀(1+β)RE

Recalling the general rule that the circuit is considered to be stable only, if:

  RTH=(0.1)(1+β)RE

Evaluating the Thevenin resistance RTH :

  RTH=(0.1)(1+β)RE

Substituting the known values in the equation:

  RTH=(0.1)(1+120)(4kΩ)=48.4

Evaluating the value of base current IBQ :

  IBQ=IEQ1+β

Substituting the known values in the above equation:

  IBQ=1.5mA1+120=0.01239mA

The expression for Thevenin s voltage VTH :

  VTH=IBQRTH+VBE(on)+IEQRE

Substituting the known values in the above equation:

  VTH=(0.01239mA)(48.4kΩ)+0.7V+(1.5mA)(4kΩ)=0.6+0.7+6=7.3V

The other expression for Thevenin s voltage VTH :

  VTH=1R1(RTH)(VCC)

Substituting the known values in the above equation:

  7.3=1R1(48.4kΩ)(12)R1=( 48.4kΩ)( 12)7.3=79.6kΩ

The value of resistance R2 is evaluated as:

  RTH=R1𑨀R2=( R 1 )( R 2 )R1+R2

Substituting the known values in the above equation:

  48.4kΩ=( 79.6kΩ)( R 2 )79.6kΩ+R23852.64×106+48.4×103R2=(79 .6×103)(R2)31.2×102R2=3852.64×106R2=3852 .64×10631 .2×103=123.5kΩ

Therefore, the designed values are:

  R1=79.6kΩR2=123.5kΩIBQ=0.01239mA

b.

Expert Solution
Check Mark
To determine

The small signal mid band voltage gain.

Answer to Problem 7.21P

The value of voltage gain Av is 0.991.

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.21P , additional homework tip  2

IEQ =1.5mA.

The parameters of the circuit is given as:

  β=120VBE( on)=0.7VVA=50V

Evaluating the value of collector current ICQ :

  ICQ=βIBQ

Substituting the known values:

  ICQ=(120)(0.01239mA)=1.49mA

Evaluating the value of rπ :

  rπ=(β)(VT)ICQ

Substituting the known values into the above equation:

  rπ=( 120)( 26× 10 3 )1.49× 10 3=2.094

Evaluating the value of output resistance r0 :

  r0=VAICQ

Substituting the known values into the above equation:

  r0=201.49× 10 3=33.56

Evaluating the value of voltage gain Av .

  Av=(1+β)(r0𑨀RE𑨀RL)rπ(1+β)(r0𑨀RE𑨀RL)

Substituting the known values into the above equation:

  Av=( 1+120)( 33.56kΩ𑨀4kΩ𑨀4kΩ)2.094× 103( 1+120)( 33.56kΩ𑨀4kΩ𑨀4kΩ)=( 121)( 33.56kΩ𑨀( ( 4kΩ )( 4kΩ ) 4kΩ+4kΩ ))2.094× 103+( 121)( 33.56kΩ𑨀( ( 4kΩ )( 4kΩ ) 4kΩ+4kΩ ))=( 121)( 33.56kΩ𑨀2kΩ)2.094× 103+( 121)( 33.56kΩ𑨀2kΩ)=( 121( ( 33.56kΩ𑨀2kΩ ) ( 33.56kΩ+2kΩ ) ))2.094× 103+( 121( ( 33.56kΩ𑨀2kΩ ) ( 33.56kΩ+2kΩ ) ))=( 121)( 1.89kΩ)2.094× 103+( 121)( 1.89kΩ)=228.69kΩ230.784kΩ=0.991

Therefore, the value of voltage gain Av is 0.991.

c.

Expert Solution
Check Mark
To determine

The value of the output resistance.

Answer to Problem 7.21P

The value of output resistance R0 is 17.23Ω .

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.21P , additional homework tip  3

IEQ =1.5mA.

The parameters of the circuit is given as:

  β=120VBE( on)=0.7VVA=50V

Evaluating the output resistance R0 :

  R0=RE𑨀r0𑨀rπ1+β

Substituting the known values into the above equation:

  R0=(4kΩ)𑨀(33.56kΩ)𑨀2.094kΩ1+120=(4kΩ)𑨀(33.56kΩ)𑨀2.094kΩ121=(4kΩ)𑨀(33.56kΩ)𑨀(17.31)=( 4kΩ)( 33.56kΩ)( 4kΩ)+( 33.56kΩ)𑨀(17.31)=(3.574kΩ)𑨀(17.31)=( 3.574kΩ)( 17.31)( 3.574kΩ)+( 17.31)=17.23Ω

Therefore, the value of output resistance R0 is 17.23Ω .

d.

Expert Solution
Check Mark
To determine

The lower 3dB corner frequency for the given circuit.

Answer to Problem 7.21P

The value of lower corner frequency is 19.81Hz.

Explanation of Solution

Given:

The circuit is given as:

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.21P , additional homework tip  4

IEQ =1.5mA.

The parameters of the circuit is given as:

  β=120VBE( on)=0.7VVA=50V

Evaluating the value of lower 3dB corner frequency:

  fL=12π(R0+RL)CC2

Substituting the known values to the above equation:

  fL=12π( 17.23+4× 10 3 )( 2× 10 6 )=19.81Hz

Hence, the value of lower corner frequency is 19.81Hz.

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Chapter 7 Solutions

Microelectronics: Circuit Analysis and Design

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