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MHT-CET Physics · Semiconductor Devices

Diode Circuits and Rectifiers

Solve a diode circuit by deciding which diodes are forward biased (a short, or a 0.7 V drop for silicon) and which reverse (an open switch); a rectifier uses the same one-way action to turn a.c. into pulsating d.c.

Why this matters

18 PYQs, one HARD. Two shapes: the current in a circuit of ideal or silicon diodes with resistors — find the conducting branches first — and the half-wave and full-wave rectifier: output frequency, efficiency, and the order rectifier → filter → regulator.

Concept 1 of 2: Circuits With Diodes

Before any arithmetic, decide each diode's state from the battery's polarity. A forward-biased ideal diode is a plain wire; a reverse-biased one is a gap, so its whole branch drops out. What is left is an ordinary resistor circuit.

Definition

  • Ideal diode: forward = short circuit, reverse = open circuit. Silicon diode: forward drop 0.7 V.
  • A diode's forward resistance, if given, simply adds to its branch.
  • Between two points: a diode in one of two parallel branches makes RforwardR_{\text{forward}} the parallel value and RreverseR_{\text{reverse}} the other branch alone.
  • Current from a higher to a lower voltage through a forward diode: I=V1−V2RI = \dfrac{V_1 - V_2}{R}.

Silicon diode in series

I=V−0.7RI = \frac{V - 0.7}{R}

Worked example

A 12 V battery drives current through a forward-biased diode and a 3 kΩ3\,\text{k}\Omega resistor. Current if the diode is ideal, and if it is silicon?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q48Moderate

Example 1 · Semiconductor Devices · Diode Circuits and Rectifiers

For the diagram shown, the resistance between points A and B when the ideal diode D is forward biased is R1R_1 and that when reverse biased is R2R_2. The ratio R1R2\frac{R_1}{R_2} is

Including a reverse-biased branch

A reverse-biased ideal diode carries NO current, so everything in series with it vanishes from the circuit. Adding its resistor in parallel is the most common wrong answer.

Concept 2 of 2: Half-Wave and Full-Wave Rectifiers

A single diode passes only one half of each cycle: half-wave output, pulsing at the supply frequency. Two diodes with a centre-tapped transformer (or a bridge) pass both halves the same way round, so the pulses come twice as often and the efficiency doubles.

Definition

  • Half-wave: output frequency = input frequency; maximum efficiency 40.6%.
  • Full-wave: pulsating d.c. at TWICE the input frequency; maximum efficiency 81.2% — so x=2yx = 2y.
  • Centre-tap: each diode sees half the secondary, Vs2\dfrac{V_s}{2}.
  • A diode alone gives pulsating d.c., not steady d.c.; the order for steady d.c. is rectifier → filter → regulator.

Output frequency

fhalf=f,ffull=2ff_{\text{half}} = f, \qquad f_{\text{full}} = 2f

Worked example

A 60 Hz supply feeds a half-wave and a full-wave rectifier. Output ripple frequencies?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q25Easy

Example 2 · Semiconductor Devices · Diode Circuits and Rectifiers

If the frequency of the input voltage is 50 Hz, applied to (a) half wave rectifier and (b) full wave rectifier. The output frequency in both cases is respectively

Full-wave keeps the input frequency

Each half-cycle becomes a pulse, so a 50 Hz input gives 100 pulses a second. Answering 50 Hz for both rectifiers is the planted option.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (2)

Test yourself on Semiconductor Devices

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.