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VCE) Vo=+Vcc +6 ↓ +5√ +4 +2 0 V(-) V₁ = +Vcc ↓ VC+) > V(-) +Vcc Vo=+5V ༡- 1 2 3 -2 V(+Y

VCE) Vo=+Vcc +6 ↓ +5√ +4 +2 0 V(-) V₁ = +Vcc ↓ VC+) > V(-) +Vcc Vo=+5V ༡- 1 2 3 -2 V(+Y

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1) For this Open Loop Amplifier - Sketch and Label  Vo (t) for each switch position:  

a) Switch   “S1” connected to  (B)  +2v 

VCE) Vo=+Vcc +6 ↓ +5√ +4 +2 0 V(-) V₁ = +Vcc ↓ VC+) > V(-) +Vcc Vo=+5V ༡- 1 2 3 -2 V(+Y <V(-) -VCC -4 Vo=-5V 5V -6 ↑ Vo-=- Vcc V(+) 4 tcmal 4 Vo=-Vcc
Transcribed Image Text:VCE) Vo=+Vcc +6 ↓ +5√ +4 +2 0 V(-) V₁ = +Vcc ↓ VC+) > V(-) +Vcc Vo=+5V ༡- 1 2 3 -2 V(+Y <V(-) -VCC -4 Vo=-5V 5V -6 ↑ Vo-=- Vcc V(+) 4 tcmal 4 Vo=-Vcc
B) V (-) 2 + Vcc (+5V) 4 5.OV U1 5 7 3/1 CA) SI V(+) 3 (B) 1 + 5.OV -Vcc (-5V) 6 t Vo
Transcribed Image Text:B) V (-) 2 + Vcc (+5V) 4 5.OV U1 5 7 3/1 CA) SI V(+) 3 (B) 1 + 5.OV -Vcc (-5V) 6 t Vo
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