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Electronics Circuit: BJT biasing .A Voltage Divider bias NPN BJT (Beta = 80) circuit is connected to a 16VDC supply with the following components R1= 62 KOhms, R2= 9.1 KOhms, Rc= 3.9 KOhms, and Re= 0.68 KOhms. Determine the Emitter Voltage (Ve). a. 1.16 Volts b. 1.86 Volts c. 9.33 Volts d. 8.17 Volts

Electronics Circuit: BJT biasing .A Voltage Divider bias NPN BJT (Beta = 80) circuit is connected to a 16VDC supply with the following components R1= 62 KOhms, R2= 9.1 KOhms, Rc= 3.9 KOhms, and Re= 0.68 KOhms. Determine the Emitter Voltage (Ve). a. 1.16 Volts b. 1.86 Volts c. 9.33 Volts d. 8.17 Volts

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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Electronics Circuit: BJT biasing .A Voltage Divider bias NPN BJT (Beta = 80) circuit is connected to a 16VDC supply with the following components R1= 62 KOhms, R2= 9.1 KOhms, Rc= 3.9 KOhms, and Re= 0.68 KOhms. Determine the Emitter Voltage (Ve). a. 1.16 Volts b. 1.86 Volts c. 9.33 Volts d. 8.17 Volts
Transcribed Image Text:Electronics Circuit: BJT biasing .A Voltage Divider bias NPN BJT (Beta = 80) circuit is connected to a 16VDC supply with the following components R1= 62 KOhms, R2= 9.1 KOhms, Rc= 3.9 KOhms, and Re= 0.68 KOhms. Determine the Emitter Voltage (Ve). a. 1.16 Volts b. 1.86 Volts c. 9.33 Volts d. 8.17 Volts
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