CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

Question : 23 of 27

Marks: +1, -0

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

30 µ {\A}

produces change of

0.02 {\V}

in the base-emitter voltage and a change of

4 {\mA}

in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is

{4 0 0}

Solution:

Proving the phase difference
Calculation of amplification factor
Calculation of load resistance
Input signal,

V_i=∆ I_B r_i

Input signal,

V_0=-∆ I_C R_L

Voltage amplification,

A_V=\;{V_0}/{V_i}

●

A_V=-\;{∆ I_B}/{∆ I_C} × \;{r_i}/{R_L}

●

A_V=-β ×

resistance gain.
Here negative sign indicates that output is

180^{∘}

out of phase w.r.t. input signal.

β=\;{∆ I_C}/{∆ I_B}=\;{4 × 10^{-3}}/{30 × 10^{-6}}=\;{400}/{3}

●

r_i=\;{∆ V_{B E}}/{∆ I_B}=\;{0.02}/{30 × 10^{-6}}=\;{2 × 10^{-2}}/{30 × 10^{-5}}

●

r_i=\;{2}/{3} × 10^3 Ω

A_V=β \;{R_{ {\L}}}/{r_i}

R_L=\;{A_V × r_i}/{β}

=\;{400 × 2 × 10^3 × 3}/{400 × 3}=2 × 10^3 Ω

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