The adjoining figure shows a common emitter transistor amplifier which uses a silicon transistor. If the quiescent emitter current is 1 mA what is the base biasing voltage?

Asked by Topperlearning User | 10th Jul, 2014, 02:32: PM

Expert Answer:

Because of the emitter current the voltage drop across the 1kΩ resistor connected to the emitter is 1 V.

[1 mA x 1 kΩ = (1/1000) A x 1000 Ω = 1 V].

The voltage drop across the base-emitter junction of the silicon transistor is 0.7 V.

Therefore, the base voltage under no signal (quiescent) condition is 1 V + 0.7 V = 1.7 V.


Answered by  | 10th Jul, 2014, 04:32: PM