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CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

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Question : 23 of 27
Marks: +1, -0
Prove that in a common-emitter amplifier, the output and input differ in phase by 180∘180^{\circ}. In a transistor, the change of base current by 30 μ A30\ \mu\ \mathrm{A} produces change of 0.02 V0.02\ \mathrm{V} in the base-emitter voltage and a change of 4 mA4\ \mathrm{mA} in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400400.
Solution:  
Proving the phase difference
Calculation of amplification factor
Calculation of load resistance
Input signal, Vi=ΔIBriV_i = \Delta I_B r_i
Input signal, V0=−ΔICRLV_0 = -\Delta I_C R_L
Voltage amplification, AV=  V0ViA_V = \;\frac{V_0}{V_i}
● AV=−  ΔIBΔIC×  riRLA_V = -\;\frac{\Delta I_B}{\Delta I_C} \times \;\frac{r_i}{R_L}
● AV=−β×A_V = -\beta \times resistance gain.
Here negative sign indicates that output is 180∘180^{\circ} out of phase w.r.t. input signal.
β=  ΔICΔIB=  4×10−330×10−6=  4003\beta = \;\frac{\Delta I_C}{\Delta I_B} = \;\frac{4 \times 10^{-3}}{30 \times 10^{-6}} = \;\frac{400}{3}
● ri=  ΔVBEΔIB=  0.0230×10−6=  2×10−230×10−5r_i = \;\frac{\Delta V_{BE}}{\Delta I_B} = \;\frac{0.02}{30 \times 10^{-6}} = \;\frac{2 \times 10^{-2}}{30 \times 10^{-5}}
● ri=  23×103Ωr_i = \;\frac{2}{3} \times 10^{3} \Omega
AV=β  RLriA_V = \beta \;\frac{R_{L}}{r_i}
RL=  AV×riβR_L = \;\frac{A_V \times r_i}{\beta} =  400×2×103×3400×3=2×103Ω= \;\frac{400 \times 2 \times 10^{3} \times 3}{400 \times 3} = 2 \times 10^{3} \Omega
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