Answered: O +5 V Ic= aplE Тв Ic=Bplg Vec 100 μΑ…

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

Calculate the values of the currents and base-emitter voltage in the circuit in
if (a) the value of the voltage source is changed to 10 V and (b) the transistor’s
common-emitter current gain is increased to 50

O +5 V
Ic= aplE
Тв
Ic=Bplg
Vec
100 μΑ
100 μΑ
Ig = (Bp + 1)/
6 -9 V
(a)
(b)

Expert Solution

Step 1

(a)given that

$b a s e c u r r e n t I_{B} = 100 μ A c o l l e c t o r v o l t a g e V_{C} = 10 V s a t u r a t i o n c u r r e n t I_{S} = 10^{- 16} A f o r w a r d c o m m o n b a s e c u r r e n t g a i n α_{F} = 0.95$

forward common emitter current gain $β_{F} = \frac{α_{F}}{1 - α_{F}}$

$β_{F} = \frac{0.95}{1 - 0.95} = 19$

here NPN transistor in active region as $V_{B E} > 0 a n d V_{B C} < 0$

$I_{c} = β_{F} I_{B} = 19 \times 100 \times 10^{- 6} I_{c} = 1.9 m A$

now we will find out emitter current $I_{E}$ and $I_{B} = 100 μ A$ from fig (b)

$I_{E} = (β_{F} + 1) I_{B} = 20 \times 100 \times 10^{- 6} I_{E} = 2 m A$

base emitter voltage $V_{B E}$ as we know

$V_{B E} = V_{T} \ln (\frac{I_{C}}{I_{S}}) here V_{T} is terminal voltage at room temperature is 0.025 V and putting all value in equation V_{B E} = 0.025 \times \ln (\frac{1.9 \times 10^{- 3}}{10^{- 16}}) V_{B E} = 0.764 V$

base collector voltage $V_{B C} = V_{B} - V_{C}$

$V_{B C} = V_{B E} - V_{C} = 0.764 - 5 (V_{B E} = V_{B}) V_{B C} = - 4.24 V$

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