Answered: Vcc +18V Rc C3 3.3 КQ R1 Vout 47 k2 C…

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

If the circuit shown in Figure #1 had a 10 kΩ resistor connected as a load resistor, RL and an input signal is applied to the base of the transistor through the coupling capacitor C1, with an amplitude of 22 mVpp at a frequency of 1 kHz. Determine the new voltage gain AV and the output voltage VOUT. Is the output signal clipped?

Vcc
+18V
Rc
C3
3.3 КQ
R1
Vout
47 k2
10 uF
Vin
Bac = Bpc = 75
%3D
10 uF
R2
12 kN
C2
RE
1 k.
10 uF
Figure #1

Expert Solution

Step 1

DC analysis of the given circuit

Step 2

The capacitors are open in the DC circuit.

Applying thevenin theorem and calculating thevenin voltage

$V_{t h} = \frac{18 \times R_{2}}{R_{1} + R_{2}} V_{t h} = \frac{18 \times 12 K}{47 K + 12 K} V_{t h} = 3.6610 V$

Step 3

Calculating thevenin resistance

$R_{t h} = R_{1} ∥ R_{2} R_{t h} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}} R_{t h} = \frac{47 k \times 12 k}{(47 k + 12 k)} R_{t h} = \frac{47 k \times 12 k}{(47 k + 12 k)} R_{t h} = 9.56 Ω$

Step 4

Applying KVL at the input

$- V_{t h} + I_{B} R_{t h} + V_{B E} + I_{E} R_{E} = 0$

Hence, thyristor is in active region.

$I_{c} = β I_{B} I_{c} = α I_{E}$

Step 5

Rewrite the equation

$\begin{array}{rcl} - V_{t h} + \frac{I_{c}}{β} R_{t h} + V_{B E} + I_{c} R_{E} & = & 0 \\ I_{c} & = & \frac{V_{t h} - V_{B E}}{\frac{R_{t h}}{β} + R_{E}} \\ I_{c} & = & \frac{3.6610 - 0.7}{\frac{9.56}{75} + 1 k} \\ I_{c} & = & \frac{3.6610 - 0.7}{\frac{9.56}{75} + 1 k} \\ I_{c} & = & \frac{2.961}{\frac{9.56 k}{75} + 1 k} \\ I_{c} & = & 2.625 mA \end{array}$

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