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CBSE Class 12-science Answered

Q. 15
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Asked by nigam3176 | 20 Apr, 2019, 04:08: PM
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Voltage gain for common-emitter amplifier = begin mathsize 12px style negative fraction numerator R subscript L over denominator r subscript E plus R subscript e end fraction end style ........................(1)
where RL is the load resistance, rE is the emitter resitance of transistor and Re is the resistance connected
in the circuit between emitter of transitor and ground.
 
Initially, Re is zero, hence voltage gain  = ( 5000/rE ) = 200  ...............(2)
 
from eqn.(2), we get rE = 25Ω . 
 
when Re = 2 kΩ, voltage gain = 5000/( 25 + 2000 ) ≈ 2.5
Answered by Thiyagarajan K | 21 Apr, 2019, 02:49: PM

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CBSE 12-science - Physics
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