CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

Question : 23

Total: 27

Prove that in a common-emitter amplifier, the output and input differ in phase by 180∘. In a transistor, the change of base current by 30µA produces change of 0.02V in the base-emitter voltage and a change of 4‌mA in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400.

Solution:

Proving the phase difference
Calculation of amplification factor
Calculation of load resistance
Input signal, Vi=∆IBri
Input signal, V0=−∆ICRL
Voltage amplification, AV=‌

● AV=−‌

∆IB

∆IC

×‌

● AV=−β× resistance gain.
Here negative sign indicates that output is 180∘ out of phase w.r.t. input signal.
β=‌

∆IC

∆IB

=‌

4×10−3

30×10−6

=‌

400

● ri=‌

∆VBE

∆IB

=‌

0.02

30×10−6

=‌

2×10−2

30×10−5

● ri=‌

×103Ω
AV=β‌

RL=‌

AV×ri

=‌

400×2×103×3

400×3

=2×103Ω

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