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Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 kohm Cs = 10 uF Cc = 1 uF CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce 1 pF Cwi= 6 pF Cwo = 8 pF Beta = 100 Rs +I 2 V₁ + C. d. R₁ Vcc R₂ Cw₁ Cbc RC Сbe RE Cc Cce Calculate the cutoff frequency due to CE. a. 222 Hz b. 327 Hz 489 Hz 577 Hz RL Cwo 111 CE ●●● 4

Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 kohm Cs = 10 uF Cc = 1 uF CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce 1 pF Cwi= 6 pF Cwo = 8 pF Beta = 100 Rs +I 2 V₁ + C. d. R₁ Vcc R₂ Cw₁ Cbc RC Сbe RE Cc Cce Calculate the cutoff frequency due to CE. a. 222 Hz b. 327 Hz 489 Hz 577 Hz RL Cwo 111 CE ●●● 4

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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AMPLIFIER'S FREQUENCY RESPONSE 5

Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 kohm Cs = 10 uF Cc = 1 uF CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce 1 pF Cwi= 6 pF Cwo = 8 pF Beta = 100 Rs +I 2 V₁ + C. d. R₁ Vcc R₂ Cw₁ Cbc RC Сbe RE Cc Cce Calculate the cutoff frequency due to CE. a. 222 Hz b. 327 Hz 489 Hz 577 Hz RL Cwo 111 CE ●●● 4
Transcribed Image Text:Amplifier's Frequency Response Vcc=20V R1 = 40 kohm R2 = 10 kohm Rc = 4 kohm RE=2 kohm Rs = 1 kohm RL = 2.2 kohm Cs = 10 uF Cc = 1 uF CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce 1 pF Cwi= 6 pF Cwo = 8 pF Beta = 100 Rs +I 2 V₁ + C. d. R₁ Vcc R₂ Cw₁ Cbc RC Сbe RE Cc Cce Calculate the cutoff frequency due to CE. a. 222 Hz b. 327 Hz 489 Hz 577 Hz RL Cwo 111 CE ●●● 4
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