The transistor in the shown figure has the following values : VBE-0.7 V, af =0.98.oftionc) Insert an AC input signal Vin = 0.2 sin ot volt, in series with VEE.Determine the output voltage .Q52N2222AR1R26.Vcc-12 V1k26002VEE6 V5.

The transistor in the shown figure has the following values : VBE=0.7 V, af-0.98. c) Insert an AC input signal Vin = u.2 sin ot volt, in series with VEE. Determine the output voltage. Q5 2N2222A R1 R2 1kQ 6002 Vcc VEE -12 V 5.

The transistor in the shown figure has the following values : VBE=0.7 V, af-0.98. c) Insert an AC input signal Vin = u.2 sin ot volt, in series with VEE. Determine the output voltage. Q5 2N2222A R1 R2 1kQ 6002 Vcc VEE -12 V 5.

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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BJT (Bipolar Junction Transistor)

A BJT is a semiconductor device with 3 terminals that consist of a p-n junction diode that can amplify the signal or current. It is a current-controlled device. The three BJT terminals are base, collector, and emitter. The BJT works with the charge carrier, the two type of carriers in the BJT are holes and electrons, the charge carriers flow by the process of recombination. The signal to be amplifier is generally given as the input at the base terminal and an amplified form of it is available at the collector terminal.

Question

The transistor in the shown figure has the following values : VBE-0.7 V, af =0.98.
of
tion
c) Insert an AC input signal Vin = 0.2 sin ot volt, in series with VEE.
Determine the output voltage .
Q5
2N2222A
R1
R2
6.
Vcc
-12 V
1k2
6002
VEE
6 V
5.

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