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 gain will beA- 9-9B- 99C- 990D- 9900

The correct option is C 990 Voltage gain = current gain × Resistance gain Now β =α/1 α=0.99/1 0.99=0.99/ 0.01=99 Again, voltage gian =99×10k Ω/1 k Ω=990 ...

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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 gain will be

9.9

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990

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9900

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Solution

The correct option is C
990

Voltage gain = current gain $\times$ Resistance gain

Now $β = \frac{α}{1 - α} = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$

Again, voltage gian
$= 99 \times \frac{10 k Ω}{1 k Ω} = 990$

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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 gain will be

The correct option is C 990 Voltage gain = current gain × Resistance gain Now β=α1−α=0.991−0.99=0.990.01=99 Again, voltage gian =99×10kΩ1kΩ=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 gain will be

The correct option is C
990

Voltage gain = current gain $\times$ Resistance gain

Now $β = \frac{α}{1 - α} = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$

Again, voltage gian
$= 99 \times \frac{10 k Ω}{1 k Ω} = 990$