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From the given circuit, find: the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz  the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz  Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms

From the given circuit, find: the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz  the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz  Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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From the given circuit, find:

  • the upper cutoff frequency of the system. a) 8.6 MHz, b) 10.77 MHz, c) 690.15 kHz, d) 801 kHz 
  • the cutoff frequency due to CE a) 222 Hz, b) 327 Hz, c) 489 Hz, d) 577 Hz 
  • Input Impedance Zi a) 2 kohms, b) 1.32 kohms, c) 3.45 kohms, d) 2.22 kohms
Amplifier's Frequency Response Vcc 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 RC CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce-1 pF Cwi = 6 pF Cwo -8 pF Beta = 100 Сс Cce Rg Cw RL V; R2 Cw* Cbe CE Rg Determine the upper cutoff frequency, f2, of the system. а. 8.6 MHz b. 10.77 MHz с. 690.15 kHz d. 801 kHz +
Transcribed Image Text:Amplifier's Frequency Response Vcc 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 RC CE = 20 uF Cbc = 4 pF Cbe = 36 pF Cce-1 pF Cwi = 6 pF Cwo -8 pF Beta = 100 Сс Cce Rg Cw RL V; R2 Cw* Cbe CE Rg Determine the upper cutoff frequency, f2, of the system. а. 8.6 MHz b. 10.77 MHz с. 690.15 kHz d. 801 kHz +
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