Diode Circuits and Applications Questions and Answers

Answer: D

The circuit is a full wave bridge rectifier with ideal diodes. Both half cycles of the input appear at the load with the same polarity, so the average (dc) output voltage is

\[ \begin{aligned} V_{dc} = \frac{2V_m}{\pi} \\ &= \frac{2 \times 250}{\pi} = 159.2\ \text{V} \end{aligned} \]

The value for a half wave rectifier would be \(V_m/\pi = 79.6\) V. The options 0.159, 1.59 and 15.9 V are 1000, 100 and 10 times too small, which would come from \(V_m\) taken as 0.25, 2.5 or 25 V. The answer is 159 V, option D.

Answer: A

The correct matching is (i) - Q; (ii) - S; (iii) - P; (iv) - R.

• The PIN diode, with its wide intrinsic layer and low capacitance, is used as a switch and attenuator in radio-frequency and microwave devices (Q).
• The tunnel diode has a region of negative differential resistance, which lets it sustain oscillations (S).
• The Zener diode holds a nearly constant voltage in reverse breakdown, the basis of voltage regulation (P).
• The photo diode turns incident light into current, for optoelectronic detection (R).

The other options give the Zener diode a role other than regulation, or the tunnel diode a role other than oscillation, so they fail on at least one pair.

Answer: A

The output is a square wave between 0 V and +6 V (A).

This is a positive clamper. The diode runs from ground (anode) to the output (cathode), so it conducts whenever the output tries to go below 0 V. During the -3 V half of the input the diode conducts and the capacitor charges to 3 V, with its output side 3 V above its input side. Afterwards the diode stays off and the capacitor simply adds 3 V to the input:

\[ \begin{aligned} V_{0} &= V_{in} + 3\ \text{V} \\ -3\ \text{V} &\to 0\ \text{V}, \quad +3\ \text{V} \to +6\ \text{V} \end{aligned} \]

The shape of the input is preserved, so triangular pulses (B) are wrong. A wave between +3 V and +6 V (C) has the wrong shift and amplitude, and a wave between -3 V and -6 V (D) is what a clamper with the diode reversed would give, and with a halved swing.

Answer: A

The correct graph is (A): \(v_{o} = 0\) for negative \(v_{i}\) and \(v_{o} = v_{i}\) for positive \(v_{i}\).

The circuit is a precision (super) diode. For \(v_{i} \gt 0\) the op-amp output rises until the diode conducts; the feedback taken from \(v_{o}\) then forces \(v_{o} = v_{i}\), and the op-amp output simply sits \(v_{d}\) higher to cover the diode drop. For \(v_{i} \lt 0\) the op-amp output swings negative, the diode is reverse biased, and \(R_{L}\) pulls \(v_{o}\) to 0.

Graph (B), with a threshold at \(v_{d}\), and graph (C), with the exponential diode curve, describe an ordinary diode without the op-amp; the feedback removes the diode drop entirely. Graph (D) is a plain follower with no rectification.