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 :
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
AI Generated Solution
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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