Answered: +10 V 3.6 k2 330 2 6.8 k2 420 2 Figure…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Find the Q-point for the circuit using
the graphical load-line approach. Use the characteristics in P5.26, assuming that the graph is a
plot of i_C vs. v_EC rather than i_C vs. v_{C E} .

+10 V
3.6 k2
330 2
6.8 k2
420 2
Figure P5.94

10 mA
Ig = 100 µA
= 80 µA
I = 60 uA
5 mA
1, - 40 μΑ
1, - 20 μΑ
OV
10 V
15 V
VCE
Figure P5.26
Collector current

Expert Solution

Step 1

Given

$\begin{array}{rcl} R_{1} & = & 6.8 kΩ \\ R_{2} & = & 3.6 kΩ \\ R_{C} & = & 330 kΩ \\ R_{E} & = & 420 kΩ \\ V_{C C} & = & 10 V \end{array}$

The expression for equivalent voltage is given by

$\begin{array}{rcl} V_{E Q} & = & \frac{R_{1}}{R_{1} + R_{2}} V_{C C} \\ = & \frac{6.8}{6.8 + 3.6} \times 10 \\ = & 6.54 V \end{array}$

Step 2

The equivalent resistance is given by

$\begin{array}{rcl} R_{E Q} & = & R_{1} ∥ R_{2} \\ = & \frac{6.8 \times 3.6}{6.8 + 3.6} \\ = & 2.354 kΩ \end{array}$

From the load line characterstics ghraph,

Collector-emitter voltage $V_{C E} = 5 V$

Froward common-emitter current gain is given by

$\begin{array}{rcl} β_{F} & = & \frac{I_{C}}{I_{B}} \\ = & \frac{5 \times 10^{- 3}}{60 \times 10^{- 6}} \\ = & 83 \end{array}$

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