CBSE Class 12-science Answered
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 | 10 Jul, 2014, 14:32: 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 | 10 Jul, 2014, 16:32: PM
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