Analog Electronics - Study Mode

[#81] In a p + n junction diode under reverse bias, the magnitude of electric field is maximum at
Correct Answer

(C) The p + n junction

Explanation

Solution: Option A: The edge of the depletion region on the p-side At the edge of the depletion region on the p-side, the electric field is not maximum. The field strength gradually reduces as you move away from the junction towards the edges of the depletion region. Therefore, this option is incorrect . Option B: The edge of the depletion region on the n-side Similarly, at the edge of the depletion region on the n-side, the electric field is not at its peak. The field strength diminishes as you approach the edges of the depletion region. Thus, this option is incorrect . Option C: The p + n junction In a p + n junction diode under reverse bias, the electric field is maximum at the junction itself. This is because the maximum charge separation occurs at the junction, resulting in the highest electric field strength. Hence, this option is correct . Option D: The center of the depletion region on the n-side The center of the depletion region on the n-side does not experience the maximum electric field. As you move away from the junction, the electric field strength decreases, making this option incorrect . Conclusion: In a p + n junction diode under reverse bias, the magnitude of the electric field is maximum at the p + n junction (Option C).

[#82] To prevent a DC return between source and load, it is necessary to use
Correct Answer

(C) Capacitor between source and load

Explanation

Solution: Option A: Resistor between source and load A resistor between the source and load allows both AC and DC components of the signal to pass. It does not block the DC component, which is not suitable when the goal is to prevent DC return. Therefore, this option is incorrect . Option B: Inductor between source and load An inductor primarily opposes changes in current and allows DC to pass through while blocking high-frequency AC signals. Since it does not block DC, it cannot prevent a DC return. Hence, this option is incorrect . Option C: Capacitor between source and load A capacitor blocks DC components and allows AC signals to pass through due to its property of impedance, which is inversely proportional to frequency. This makes it the ideal component to prevent DC return between the source and load. Therefore, this option is correct . Option D: Either A or B Neither a resistor nor an inductor can effectively prevent DC return as they allow DC to pass through. Thus, this option is incorrect . Conclusion: To prevent a DC return between the source and load, it is necessary to use a capacitor (Option C).

[#83] A constant current signal across a parallel RLC circuits gives an output of 1.4v at the signal frequency of 3.89KHZ and 4.1KHZ. At the frequency of 4KHZ, the output voltage will be
Correct Answer

(B) 2 v

Explanation

Solution: A constant current signal across a parallel RLC circuit gives an output of 1.4v at the signal frequencies of 3.89kHz and 4.1kHz. This indicates that the circuit is operating near its resonant frequency. At resonance, the impedance of the parallel RLC circuit is maximum, leading to a maximum voltage across it for a given current. In this case, the output voltage is 1.4v at frequencies close to resonance. As we move further away from the resonant frequency, the impedance decreases, and consequently, the output voltage also decreases. Therefore, at 4kHz, which is closer to the resonant frequency than 3.89kHz or 4.1kHz, the output voltage will be **slightly higher** than 1.4v. Among the given options, Option B: 2v is the closest to this expected value.

[#84] If $$alpha $$ = 0.98, $${I_{{ ext{CO}}}} = 6mu { ext{A}}$$ xa0 and $${I_x08eta } = 100mu { ext{A}}$$ xa0 for a transistor, then the value of $${I_{ ext{C}}}$$ will be
Correct Answer

(D) 5.2 mA

Explanation

Solution: $$eqalign{
& {I_{ ext{C}}} = frac{{{I_{{ ext{CO}}}}}}{{1 - alpha }} + frac{alpha }{{1 - alpha }} imes {I_x08eta } cr
& = frac{6}{{1 - 0.98}} + frac{{0.98}}{{1 - 0.98}} imes 100 cr
& = 5.2,{ ext{mA}} cr} $$

[#85] Which of the following is not associated with a p-n junction
Correct Answer

(D) Channel length modulation

Explanation

Solution: Explanation of each option: Option A: Junction capacitance is associated with a p-n junction. This capacitance arises due to the separation of charge carriers across the junction, creating an electric field that contributes to capacitance. Option B: Charge storage capacitance is also associated with a p-n junction. It refers to the capacitance that stores charge at the junction when it is forward biased, as the charge carriers accumulate at the junction. Option C: Depletion capacitance is related to a p-n junction as well. It occurs due to the depletion region that forms at the junction when it is reverse biased, and this region acts like a capacitor. Option D: Channel length modulation is not associated with a p-n junction. This phenomenon is related to field-effect transistors (FETs), where the effective length of the channel is modulated by the drain-to-source voltage, and does not occur in p-n junctions. Conclusion: The correct answer is Option D: Channel length modulation because it is not associated with a p-n junction.