Consider the fix-biased common emitter amplifier with bypassed emitter degenerative resistor shown below. VCC RC 12V RB

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answerhappygod
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Consider the fix-biased common emitter amplifier with bypassed emitter degenerative resistor shown below. VCC RC 12V RB

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Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 1
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 1 (63.91 KiB) Viewed 33 times
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 2
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 2 (10.77 KiB) Viewed 33 times
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 3
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 3 (14.02 KiB) Viewed 33 times
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 4
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 4 (4.03 KiB) Viewed 33 times
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 5
Consider The Fix Biased Common Emitter Amplifier With Bypassed Emitter Degenerative Resistor Shown Below Vcc Rc 12v Rb 5 (4.38 KiB) Viewed 33 times
Consider the fix-biased common emitter amplifier with bypassed emitter degenerative resistor shown below. VCC RC 12V RB Cout VC Vout 500 Ohms Cin HE Rg RLERC + VE Vg Vin CE RE
(15 pts) 1. Design the amplifier by finding RB, RC, and RE, so that the amplifier is set to operate with the following specifications: Vc = 8V, VE = 2V, Ic = 1mA, ß = 100, VBE = 0.7V = = = = = 1
(30 pts) 2. Using the components value determined from #1 with Cin = 1uF, Cout = 1uF, and CE = 10uF, solve the low cut-off frequencies fcin, fce, and fcout of the amplifier using the Short Circuit Time Constant (SCTC) method.
Sketch/draw here the working small-signal equivalent circuit followed by the solution in finding fce.
Sketch/draw here the working small-signal equivalent circuit followed by the solution in finding fcout
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