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Asked: 2023-07-22T06:09:02-04:00 In:

Consider the common-emitter amplifier, under the following conditions: Rsig= 5kΩ, RB1= 33 kΩ ,RB2= 22 kΩ, RE= 3.9 kΩ, RC= 4.7 kΩ, RL= 5.6 kΩ, VCC= 5 V. The dc emitter current can be shown to be IE=0.3 mA, at which β = 120. If CC1= CC2= 1 μF and CE=20 μF, use the method of short-circuit time constants to find fL

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Consider the common-emitter amplifier, under the following conditions: Rsig= 5kΩ, RB1= 33 kΩ ,RB2= 22 kΩ, RE= 3.9 kΩ, RC= 4.7 kΩ, RL= 5.6 kΩ, VCC= 5 V. The dc emitter current can be shown to be IE=0.3 mA, at which β = 120. If CC1= CC2= 1 μF and CE=20 μF, use the method of short-circuit time constants to find fLConsider the common-emitter amplifier, under the following conditions: Rsig= 5kΩ, RB1= 33 kΩ ,RB2= 22 kΩ, RE= 3.9 kΩ, RC= 4.7 kΩ, RL= 5.6 kΩ, VCC= 5 V. The dc emitter current can be shown to be IE=0.3 mA, at which β = 120. If CC1= CC2= 1 μF and CE=20 μF, use the method of short-circuit time constants to find fL

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