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An NPN transistor is used in common emit...

An NPN transistor is used in common emitter configuration as an amplifier with `1k Omega` load resistance. Signal voltage of 10 mV is applied a across the base-emitter. This produces a 3 mA change in the collector current and `15 mu A` change in the base currect of the amplifier. The input resistance and voltage gain are :

A

`0.33k Omega, 1.5`

B

`0.67k Omega, 300`

C

`0.67k Omega, 200`

D

`0.33k Omega, 300`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will find the input resistance and the voltage gain of the NPN transistor in common emitter configuration. ### Step 1: Identify the given values - Load resistance, \( R_L = 1 \, k\Omega = 1000 \, \Omega \) - Signal voltage, \( V_{in} = 10 \, mV = 10 \times 10^{-3} \, V \) - Change in collector current, \( \Delta I_C = 3 \, mA = 3 \times 10^{-3} \, A \) - Change in base current, \( \Delta I_B = 15 \, \mu A = 15 \times 10^{-6} \, A \) ### Step 2: Calculate the input current The input current \( I_{in} \) is given as: \[ I_{in} = \Delta I_B = 15 \times 10^{-6} \, A \] ### Step 3: Calculate the input resistance The input resistance \( R_{in} \) can be calculated using the formula: \[ V_{in} = R_{in} \times I_{in} \] Rearranging this gives: \[ R_{in} = \frac{V_{in}}{I_{in}} \] Substituting the known values: \[ R_{in} = \frac{10 \times 10^{-3}}{15 \times 10^{-6}} = \frac{10}{15} \times 10^{3} \, \Omega = \frac{2}{3} \times 10^{3} \, \Omega \approx 0.67 \, k\Omega \] ### Step 4: Calculate the output voltage The output voltage \( V_{out} \) can be calculated using the load resistance and the change in collector current: \[ V_{out} = R_L \times \Delta I_C \] Substituting the known values: \[ V_{out} = 1000 \times 3 \times 10^{-3} = 3 \, V \] ### Step 5: Calculate the voltage gain The voltage gain \( A_v \) is given by the formula: \[ A_v = \frac{V_{out}}{V_{in}} \] Substituting the known values: \[ A_v = \frac{3}{10 \times 10^{-3}} = \frac{3}{0.01} = 300 \] ### Final Answers - Input Resistance, \( R_{in} \approx 0.67 \, k\Omega \) - Voltage Gain, \( A_v = 300 \)
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