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.

Key concepts

Voltage Gain

In a common emitter amplifier, the concept of voltage gain is crucial for understanding how the amplifier increases the input signal's strength. The voltage gain, denoted as \(A_v\), is the ratio of the output voltage to the input voltage. In the context of this amplifier, the formula to compute the voltage gain is given by: \[A_v = -\frac{\beta(R_L)}{R_{in}}\] where \(\beta\) is the current gain, \(R_L\) is the load resistor or output resistance, and \(R_{in}\) is the input resistance. By substituting the provided values:

• \(\beta = 50\)
• \(R_L = 5000\Omega\)
• \(R_{in} = 2000\Omega\)

we find that the voltage gain \(A_v = -125\). This negative sign indicates a phase inversion, meaning the output voltage waveform is 180 degrees out of phase with the input waveform.

Input Resistance

Input resistance in amplifiers is an important factor because it determines how much of the input signal voltage actually makes it into the amplifier for processing. The input resistance, \(R_{in}\), is the resistance that the input signal "sees" when it enters the amplifier. For a common emitter amplifier, having a higher input resistance is generally beneficial because it allows for more signal voltage to be applied across the amplifier.In our case, the input resistance is given as \(R_{in} = 2000\Omega\). This means that the input signal voltage experiences a resistance equivalent to 2000 Ohms before getting amplified. This factor is crucial as it impacts how efficiently the input signal can drive the amplifier.

Output Resistance

Output resistance, or load resistance, is another key parameter in amplifiers like the common emitter. It plays a role in determining how well the amplifier can drive the load connected to it. The output resistance is represented as \(R_L\), which in our exercise is \(5000\Omega\). This value indicates the resistance faced by the amplified signal on its way out of the amplifier.The output resistance affects both the voltage gain and the power delivered to the load. In the case of this common emitter amplifier, a lower output resistance would mean more power can be delivered to a connected load, but it could potentially lower the voltage gain depending on the configuration of the amplifier.

Current Gain

The current gain, symbolized by \(\beta\) and also known as the transistor's beta, measures how effectively an amplifier can increase the input current. It is defined as the ratio of the collector current (I_C) to the base current (I_B). In this problem, the current gain is given as \(\beta = 50\).This means that the collector current is 50 times that of the base current. Current gain is a vital aspect of semiconductor performance in common emitter configurations because it affects how much current the amplifier can provide to its load, contributing to the overall amplification process. For these devices, a higher current gain typically indicates a more efficient amplification of the input current, thereby ensuring more effective signal processing.