Analog Electronics MCQ Questions & Answers | ECE

The correct answer is Option A: \( \frac{1}{11} \) kΩ

Explanation:

The input resistance of a current shunt negative feedback amplifier is given by the formula:

\[ R_{in(f)} = \frac{R_{in}}{1 + A\beta} \]

where,

\( R_{in} \) = Input resistance without feedback = 1 kΩ

\( A \) = Open-loop voltage gain = 50

\( \beta \) = Feedback factor = 0.2

Substituting the values:

\[ R_{in(f)} = \frac{1}{1 + (50 \times 0.2)} \]

\[ R_{in(f)} = \frac{1}{1 + 10} = \frac{1}{11} \text{ kΩ} \]

Thus, the input resistance of the current shunt negative feedback amplifier is \( \frac{1}{11} \) kΩ, which corresponds to Option A.

Option A: A unity gain non-inverting amplifier is correct because the most commonly used amplifier in sample & hold circuits is a unity gain non-inverting amplifier. This configuration provides a stable output with no amplification (gain = 1), which is ideal for sample & hold applications where the input signal is maintained without distortion.

Option B: A unity gain inverting amplifier is incorrect because inverting amplifiers flip the signal, meaning the output is inverted with respect to the input. This is not typically desired in sample & hold circuits, where the signal should remain in phase with the input.

Option C: An inverting amplifier with a gain of 10 is incorrect because amplifying the signal in this way would distort the original input, which is undesirable in sample & hold circuits where the purpose is to maintain the exact value of the sampled signal.

Option D: An inverting amplifier with a gain of 100 is incorrect because, like Option C, amplifying the signal with a high gain (such as 100) would lead to a distorted output, which is not suitable for sample & hold circuits.

Conclusion: The correct answer is Option A: A unity gain non-inverting amplifier because this configuration is commonly used in sample & hold circuits due to its ability to maintain the input signal without inversion or amplification.

Option A: Voltage shunt feedback is incorrect because voltage shunt feedback involves taking a portion of the output voltage and feeding it back in parallel with the input. This is not what an unbypassed resistor in a common emitter configuration does.

Option B: Current series feedback is incorrect because current series feedback involves taking a portion of the output current and feeding it back in series with the input. This is not characteristic of an unbypassed resistor in a common emitter amplifier.

Option C: Negative voltage feedback is correct because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback. This resistor, typically placed in the emitter leg of the transistor, causes a decrease in the gain by reducing the overall amplification and stabilizing the output, leading to better linearity and reduced distortion.

Option D: Positive current feedback is incorrect because positive current feedback would increase the overall gain and instability, which is not the role of an unbypassed resistor in a common emitter configuration.

Conclusion: The correct answer is Option C: Negative voltage feedback because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback, which stabilizes the gain and reduces distortion in the amplifier.

Option A: $$g_m r_o$$ is incorrect because this expression represents the voltage gain in a transistor, not the current gain. The term $$r_o$$ is the output resistance, and multiplying $$g_m$$ (transconductance) with $$r_o$$ gives the voltage gain, not the current gain.

Option B: $$\frac{g_m}{r_o}$$ is incorrect because this expression represents the inverse relationship between transconductance and output resistance, which is related to the voltage gain of the amplifier, not the current gain.

Option C: $$g_m r_\pi$$ is correct because this is the standard expression for the current gain of a BJT in terms of the transconductance ($$g_m$$) and the base-emitter resistance ($$r_\pi$$). The current gain is the product of these two parameters.

Option D: $$\frac{g_m}{r_\pi}$$ is incorrect because this expression does not correctly represent the current gain of a BJT. Instead, it gives a ratio that relates to the voltage gain and other characteristics of the transistor, but not the current gain.

Conclusion: The correct answer is Option C: $$g_m r_\pi$$ because the current gain of a BJT is given by the product of transconductance ($$g_m$$) and base-emitter resistance ($$r_\pi$$).

Option A: Transistor capacitances is correct because at high frequencies, the capacitances present in a bipolar transistor (such as base-emitter and base-collector junction capacitances) start to have a significant effect. These capacitances limit the current gain, causing a drop in the transistor's performance at high frequencies. The capacitive effects become more dominant as frequency increases, leading to a reduction in the current gain.

Option B: High current effects in the base is incorrect because high current effects in the base generally affect the transistor's thermal stability and the possibility of transistor saturation. While they can affect the performance, they are not the primary reason for the drop in current gain at high frequencies.

Option C: Parasitic inductive elements is incorrect because while parasitic inductive elements may have some impact on the high-frequency behavior of a transistor, they do not directly cause the current gain to drop. The primary limiting factor at high frequencies is the capacitance, not inductance.

Option D: The early effect is incorrect because the Early effect refers to the variation of the transistor's collector current due to changes in the collector-base voltage. While the Early effect can influence the transistor's operation, it is not the main cause for the reduction in current gain at high frequencies.

Conclusion: The correct answer is Option A: Transistor capacitances because at high frequencies, the transistor's junction capacitances play a significant role in limiting the current gain, leading to a reduction in performance.

Option A: Increase in both input & output impedances is incorrect because voltage series (series-shunt) feedback does not increase both input and output impedances. While it does increase the input impedance, it decreases the output impedance. Therefore, this option does not correctly describe the effect of series-shunt feedback.

Option B: Decreases in both input & output impedances is incorrect because voltage series feedback actually increases the input impedance and decreases the output impedance. Hence, this option is not applicable.

Option C: Increase in input impedance & decreases in output impedance is correct because when voltage series feedback (also known as series-shunt feedback) is applied, it increases the input impedance and decreases the output impedance. This makes the amplifier more stable and improves its performance in certain applications.

Option D: Decrease in input impedance & increase in output impedance is incorrect because voltage series feedback increases the input impedance, not decreases it, and it also decreases the output impedance, not increases it. Therefore, this option does not describe the behavior of voltage series feedback.

Conclusion: The correct answer is Option C: Increase in input impedance & decreases in output impedance because voltage series (series-shunt) feedback increases the input impedance while decreasing the output impedance, improving amplifier performance.