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 6.1EP

The circuit parameters for the circuit in Figure 6.3 are V C C = 3.3 V , V B B = 0.850 V , R B = 180 , and R C = 15 . The transistor parameters are β = 120 and V B E (on) = 0.7 V . (a) Determine the Q−point values I C Q and V C E Q . (b) Find the small−signal hybrid− π parameters g m and r π . (c) Calculate the small−signal voltage gain. (Ans. (a) I C Q =0 .1mA , V C E Q = 1.8 V ; (b) g m = 3.846 mA/V , r π = 31.2 ; (c) A υ = 8.52 ).

(a)

Expert Solution
Check Mark
To determine

The quiescent collector current ICQ and the Q -point value VCEQ for the given transistor.

Answer to Problem 6.1EP

The quiescent collector current ICQ is 0.1 mA and Q -point VCEQ is 1.8 V .

Explanation of Solution

Given:

The circuit for common emitter is shown in Figure 1.

  Microelectronics: Circuit Analysis and Design, Chapter 6, Problem 6.1EP

The circuit parameters for the transistor circuit shown in Figure 1 are as follows:

  VCC=3.3 VRC=15 kΩRB=180 kΩVBE(on)=0.7 VVBB=0.85 V

The value of current gain β is 120 .

Concept used:

The expression for quiescent collector current is written below:

  ICQ=βIBQ ...... (1)

The expression for quiescent value VCEQ is written below.

  VCEQ=VCCICQRC ...... (2)

Calculation:

From DC analysis the ac voltage source is reduced to zero and the equation can be written as,

  IBQ=VBBVB(on)RB ...... (3)

Substitute 0.85 for VBB , 0.7 for VB(on) and 180×103 for RB in equation (3).

  IBQ=0.850.7180× 103=11200 mA

Substitute 120 for β and 11200 for IBQ in equation (1).

  ICQ=12011200 mA=110 mA=0.1 mA

Therefore, the quiescent collector current ICQ is 0.1 mA .

Substitute 3.3 for VCC , 0.1×103 for ICQ and 15×103 for RC in equation (2).

  VCEQ=3.3(0.1× 10 3)(15× 103)=3.31.5=1.8 V

Therefore, the Q -point VCEQ is 1.8 V .

Conclusion:

Thus, the quiescent collector current ICQ is 0.1 mA and Q -point VCEQ is 1.8 V .

(b)

Expert Solution
Check Mark
To determine

The transconductance gm and the diffusion resistance rπ for small signal analysis.

Answer to Problem 6.1EP

The transconductance gm is 3.846 mA/V and the diffusion resistance rπ is 31.2 kΩ .

Explanation of Solution

Concept used:

The expression for transconductance gm is written below.

  gm=ICQVT ...... (4)

Here, VT is thermal voltage and its value is 26 mV .

The expression for diffusion resistance is written below.

  rπ=βVTICQ ...... (5)

Calculation:

Substitute 0.1×103 for ICQ and 26×103 for VT in equation (4).

  gm=0.1× 10 326× 10 3=3.846 mA/V

Therefore, the transconductance gm is 3.846 mA/V .

Substitute 120 for β , 26×103 for VT and 0.1×103 for ICQ in equation (5).

  rπ=120( 26× 10 3 )0.1× 10 3=31.2×103 Ω

Therefore, the diffusion resistance rπ is 31.2 kΩ .

Conclusion:

Thus, the transconductance gm is 3.846 mA/V and the diffusion resistance rπ is 31.2 kΩ .

(c)

Expert Solution
Check Mark
To determine

The value of small signal voltage gain Av .

Answer to Problem 6.1EP

The value of small signal voltage gain Av is 8.52 .

Explanation of Solution

Concept used:

The expression for small signal voltage gain Av is written below.

  Av=(gmRC)(rπrπ+RB) ...... (6)

Calculation:

Substitute 3.846 for gm , 15 for RC , 31.2 for rπ and 180 for RB in equation (6).

  Av=[(3.846)(15)]( 31.2 31.2+180)8.52

Therefore, the voltage gain Av is 8.52 .

Conclusion:

Thus, the value of small signal voltage gain Av is 8.52 .

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Chapter 6 Solutions

Microelectronics: Circuit Analysis and Design

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