Question

In a common emitter amplifier, using output resistance of \(5000 \Omega\) and input resistance of \(2000 \Omega\), if the peak value of input signal voltage is \(10 \mathrm{mV}\) and \(\beta=50\) then what is the peak value of output voltage? (A) \(2.5 \times 10^{-4}\) Volt (B) \(5 \times 10^{-6}\) Volt (C) \(1.25\) Volt (D) 125 Volt

Short answer

The peak value of the output voltage is (C) \(1.25V\).

Step-by-step solution

Determine the given values

In this exercise, we are given the following values: Output resistance, \(R_L = 5000\Omega\) Input resistance, \(R_{in} = 2000\Omega\) Peak value of input signal voltage, \(V_{in(peak)} = 10mV = 10\times 10^{-3}V\) Current gain of the transistor, \(\beta = 50\)

Calculate the voltage gain (A_v)

Using the formula for the voltage gain of a common emitter amplifier, we can calculate the voltage gain (A_v) as follows: \[A_v = -\frac{\beta(R_L)}{R_{in}}\] Plug in the values: \[A_v = -\frac{50\times 5000\Omega}{2000\Omega}\] Now, compute A_v: \[A_v = -\frac{250000}{2000}\] \[A_v = -125\]

Calculate the peak value of the output voltage (V_out(peak))

To find the peak value of the output voltage, we simply multiply the voltage gain by the input peak voltage: \(V_{out(peak)} = A_v \times V_{in(peak)}\) Plug in the values: \(V_{out(peak)} = (-125) \times (10\times 10^{-3}V)\) Now, compute \(V_{out(peak)}\): \(V_{out(peak)} = -1.25V\) Since we are asked for the peak value, we can take the magnitude of the output voltage, i.e., the peak value of the output voltage is \(1.25V\). The answer is (C) 1.25 Volt.