Question

In a common emitter amplifier, output resistance is \(5000 \Omega\) and input resistance is \(1000 \Omega .\) If peak value of signal voltage is \(1 \mathrm{mV}\) and \(\beta=100\), then the peak value of output voltage is (A) \(0.1 \mathrm{~V}\) (B) \(0.3 \mathrm{~V}\) (C) \(0.2 \mathrm{~V}\) (D) \(0.5 \mathrm{~V}\)

Short answer

The peak value of output voltage is \(0.5 V\).

Step-by-step solution

Understand the Given Data

The provided data includes: - Output resistance, \(R_o = 5000 \Omega\), - Input resistance, \(R_i = 1000 \Omega\), - Peak value of signal voltage, \(V_s = 1mV\) = \(0.001V\), - Beta, \(\beta = 100\) (current gain).

Calculate Voltage Gain

With these values, you can calculate the voltage gain \(A_v\). Using the formula \(A_v=\beta \frac{R_o}{R_i}\), \(A_v=100 \left(\frac{5000}{1000}\right)\) which simplifies to \(A_v=500\).

Calculate Output Peak Voltage

Finally, the peak output voltage (\(V_o\)) can be determined by multiplying the voltage gain by the peak signal voltage. The formula is \(V_o= A_v \cdot V_s\). Substituting the values, we get \(V_o= 500 \cdot 0.001V = 0.5V\), indicating that option (D) \(0.5 V\) is the correct answer.

Key concepts

Output Resistance

In a common emitter amplifier, the output resistance is a crucial factor that influences the performance of the circuit. It is denoted as \( R_o \) and in our exercise, it is provided as \( 5000 \Omega \).
This resistance is important because it defines how much load the amplifier can drive. A high output resistance indicates that the amplifier cannot drive heavy loads directly. In our case, since \( R_o \) is relatively large, the amplifier will be more suitable for driving lighter loads.
Additionally, output resistance plays a role in determining the overall voltage gain and the efficiency of signal transfer. In a way, it's like a bottleneck that can limit the signal reaching the output if not managed properly. This resistance, when combined with the load resistance, will affect the total voltage the amplifier delivers to the load.

Input Resistance

The input resistance denoted as \( R_i \) is another significant parameter in a common emitter amplifier. For our exercise, this value is \( 1000 \Omega \). The input resistance determines how much of the input signal is actually picked up by the amplifier.
A higher input resistance is typically desirable because it allows the amplifier to absorb more of the incoming signal, affecting its sensitivity and subsequent amplification quality. It minimizes the loading effect on the input source, hence preserving the integrity of the input signal.
Think of input resistance as the gateway through which signals enter the amplifier. A low input resistance could lead to signal loss before amplification, negatively impacting the performance.

Voltage Gain

Voltage gain is the measure of how much an amplifier increases the input signal's voltage. For a common emitter amplifier, the voltage gain \( A_v \) can be calculated using \[ A_v = \beta \frac{R_o}{R_i} \]Here, \( \beta \) is the current gain, also known as the transistor's amplification factor, given as 100 in the exercise.
The calculation involves the given output and input resistances: \[ A_v = 100 \left(\frac{5000}{1000}\right) = 500 \]This means that the amplifier will increase the input voltage signal by 500 times.
A high voltage gain is typically desirable as it allows significant amplification of weak input signals, making them more usable at the output. However, it is essential to balance gain with stability to avoid distortion and quality loss.

Peak Output Voltage

The peak output voltage of an amplifier is the highest voltage level it can produce at its output for a given input signal.
In the context of our problem, it is calculated by multiplying the voltage gain \( A_v \) by the peak input voltage \( V_s \):\[ V_o = A_v \cdot V_s \]Given that \( V_s = 0.001 \, V \) and \( A_v = 500 \), the peak output voltage is \[ V_o = 500 \cdot 0.001 = 0.5 \, V \].
This result indicates how the amplifier, using its voltage gain, significantly boosts the low input signal, yielding a much higher voltage at the output.

• This higher voltage allows the output signal to effectively drive connected devices or further stages in an electronic system.
• Proper calculation ensures that the amplifier remains within its operating limits without causing distortion or damage to the circuit.