In the circuit below: R 1 = 8.2 k, R 2 = 1.6 k, V CC = 10 V, R C = 1 k, R E = 30
ID: 2075199 • Letter: I
Question
In the circuit below: R1 = 8.2 k, R2 = 1.6 k, VCC = 10 V, RC = 1 k, RE = 300 , and dc = 150.
Note: The transistor is silicon.
A) Calculate the values of base voltage VB and emitter voltage VE. Round the final answer to two decimal places.
Base voltage VB = V
Emitter voltage VE = V
B) Calculate the value of collector current IC. Round the final answer to two decimal places. mA
C) Calculate the values of collector voltage VC, collector-emitter voltage VCE, and cutoff collector-emitter voltage VCE(off). Round the final answers to two decimal places.
Collector voltage VC = V
Collector-emitter voltage VCE = V
Cutoff collector-emitter voltage VCE(off) = V.
/77Explanation / Answer
Given
R1 = 8.2 k, R2 = 1.6 k, VCC = 10 V, RC = 1 k, RE = 300 , and dc = 150.
Note: The transistor is silicon.
A) Base voltage VB = V = IR = 1 A* 8.2 = 8.2 V
( Assumed I =1 A)
Emitter voltage VB = V = IR = 1 A* 1.6 = 1.6 V
( Assumed I =1 A)
b) IC = VCC / RCC = 10/ 1000 = 1/100 = 0.01 A
c)
Collector voltage VC = V = 0.01 * 8.2 = 0.082
Collector-emitter voltage VCE = V = 0.01 * 1.6 = 0.016
Cutoff collector-emitter voltage VCE(off) = V. = total = 0.098