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VBB 3V (b) RB Μ 27 ΚΩ Rc 390 Ω Bbc = 125 Vcc 8 V

VBB 3V (b) RB Μ 27 ΚΩ Rc 390 Ω Bbc = 125 Vcc 8 V

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
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why does the V(BE) is -0.7 and when input to saturation it change to positive 

VBB 3 V (b) RB ww 27 ΚΩ Rc 390 Ω Bpc=125 Vcc 8 V
Transcribed Image Text:VBB 3 V (b) RB ww 27 ΚΩ Rc 390 Ω Bpc=125 Vcc 8 V
b.) VBE = -0.7 V IB=VBB-VBE -2.3 V -3 V-(-0.7 V) 27 ΚΩ -85.2 μα RB 27 ΚΩ IC BDCIB = 125(-85.2 µA) = -10.7 mA VCE = Vcc - IcRc= -8 V - (-10.7 mA) (390 2) = -3.83 V VCB = Vc - VB= (-3.83 V) + 0.7 V = -3.13 V VBE = -0.7 V, VCE= -3.83 V, VCB= -3.13 V IC(SAT)= Vcc= Rc 8V 390 Ω = 20.5 mA IB 3 V-0.7 V VBB-VBE = = 85.2 μA RB 27 ΚΩ Ic = BDCIB = 125(85.2 µA) = 10.7 mA Ic<IC(SAT) = Therefore, the transistor is not saturated.
Transcribed Image Text:b.) VBE = -0.7 V IB=VBB-VBE -2.3 V -3 V-(-0.7 V) 27 ΚΩ -85.2 μα RB 27 ΚΩ IC BDCIB = 125(-85.2 µA) = -10.7 mA VCE = Vcc - IcRc= -8 V - (-10.7 mA) (390 2) = -3.83 V VCB = Vc - VB= (-3.83 V) + 0.7 V = -3.13 V VBE = -0.7 V, VCE= -3.83 V, VCB= -3.13 V IC(SAT)= Vcc= Rc 8V 390 Ω = 20.5 mA IB 3 V-0.7 V VBB-VBE = = 85.2 μA RB 27 ΚΩ Ic = BDCIB = 125(85.2 µA) = 10.7 mA Ic<IC(SAT) = Therefore, the transistor is not saturated.
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