A common emitter amplifier is designed with n-p-n transistor - - 0-99 - The input impedance is 1 k - and load is 10 k - The voltage will be -

The correct option is A 990 Voltage gain = β× resistance gain #160; β = α1 α = 0.99 (1 0.99) = 99 Resistance gain = 10 × 10^310^3 = 10 ...

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Standard XII

Physics

P-N Junction

A common emit...

Question

A common emitter amplifier is designed with n-p-n transistor $(α = 0.99) .$ The input impedance is $1 k Ω$ and load is $10 k Ω$ . The voltage will be :

9900

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99

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9.9

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990

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Solution

The correct option is A $990$
Voltage gain $= β \times$ resistance gain
$β = \frac{α}{1 - α} = \frac{0.99}{(1 - 0.99)} = 99$
Resistance gain $= \frac{10 \times 10^{3}}{10^{3}} = 10$
Voltage gain $= 99 \times 10 = 990$

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A common emitter amplifier is designed with n-p-n transistor (α=0.99). The input impedance is 1kΩ and load is 10kΩ. The voltage will be :

The correct option is A 990Voltage gain =β× resistance gain β=α1−α=0.99(1−0.99)=99Resistance gain =10×103103=10Voltage gain =99×10=990

A common emitter amplifier is designed with n-p-n transistor $(α = 0.99) .$ The input impedance is $1 k Ω$ and load is $10 k Ω$ . The voltage will be :

The correct option is A $990$
Voltage gain $= β \times$ resistance gain
$β = \frac{α}{1 - α} = \frac{0.99}{(1 - 0.99)} = 99$
Resistance gain $= \frac{10 \times 10^{3}}{10^{3}} = 10$
Voltage gain $= 99 \times 10 = 990$