Q1) For the circuit in Fig. 1, find Vz, Vz, and Vc for Rs = 100 k, 10 k, and 1 k. Let ß=100.+3 VRBmFigure 1mH1 ΚΩVcVE1 ΚΩ

Answered: Q1) For the circuit in Fig. 1, find V₁,…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Q1) For the circuit in Fig. 1, find Vz, Vz, and Vc for Rs = 100 k, 10 k, and 1 k. Let ß=100.
+3 V
RB
m
Figure 1
m
H
1 ΚΩ
Vc
VE
1 ΚΩ

Expert Solution

Step 1

The given transistor is in an emitter bias configuration. When $R_{B} = 100 k Ω$ . The required parameters are calculated as shown below:

$I_{B} = \frac{V_{C C} - V_{B E}}{R_{B} + (β + 1) R_{E}} I_{B} = \frac{3 - 0.7}{100 \times 10^{3} + (100 + 1) \times 1 \times 10^{3}} A I_{B} = 11.443 μ A$

The collector's current is,

$I_{C} = β I_{B} I_{C} = 100 \times 11.443 μ A I_{C} = 100 \times 11.443 μ A I_{C} = 1.143 m A$

The collector-emitter voltage is,

$V_{C E} = V_{C C} - I_{C} (R_{C} + R_{E}) V_{C E} = 3 - 1.143 \times 10^{- 3} (1 \times 10^{3} + 1 \times 10^{3}) V V_{C E} = 0.7114 V$

The collector voltage is,

$V_{C} = V_{C C} - I_{C} R_{C} V_{C} = 3 - 1.143 \times 10^{- 3} \times 1 \times 10^{3} V V_{C} = 1.857 V$

The emitter voltage is,

$V_{E} = V_{C} - V_{C E} V_{E} = 1.857 - 0.7114 V V_{E} = 1.1456 V$

The base voltage is,

$V_{B} = V_{B E} + V_{E} V_{B} = 0.7 + 1.1456 V V_{B} = 1.8456 V$

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