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JEE Mains Physics · Semiconductor Electronics

Diode Circuits and Rectifiers

Decide first which way each diode is biased; then a conducting diode is a wire (or a fixed small drop) and a blocking one is a break, and the rest is an ordinary circuit.

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Why this matters

Twenty-two PYQs, seventeen of them multiple choice, and four from 2026. Ten are resistor networks with ideal diodes, or diodes with a stated forward resistance, where the answer turns on which diodes conduct. Six give each diode a fixed voltage drop, including LEDs. Six are about rectifiers, filters and the output waveform. Nearly every one comes with a circuit figure, and the first step is always the same: read the battery's polarity and each diode's direction.

Concept 1 of 3: Ideal diodes in resistor networks

An ideal diode is a one-way switch. Pointing from higher potential to lower, it is closed and acts as a plain wire. Pointing the other way, it is open, and its whole branch carries no current. So a diode network is solved in two passes: first decide each diode's state from the battery, then redraw the circuit without the open branches and solve it with series and parallel rules.

Definition

  • Ideal diode: forward biased, zero resistance; reverse biased, infinite resistance (an open branch).
  • A battery's long plate is its positive terminal. Mark the higher-potential end before looking at any diode.
  • A diode conducts when its arrow points from the higher potential towards the lower one.
  • Delete every branch that holds a reverse-biased diode, then combine the rest.
  • A diode with a forward resistance rfr_f adds rfr_f in series in its own branch.
  • A conducting ideal diode connected straight across a resistor shorts that resistor.
  • Two diodes in parallel facing opposite ways (antiparallel): one of them always conducts, whichever way the current goes.

A diode as a switch

forward: RD=rf  (0 if ideal),reverse: RD→∞\text{forward: } R_D = r_f\ \ (0 \text{ if ideal}), \qquad \text{reverse: } R_D \to \infty

Worked example

A 12 V battery drives three branches in parallel: a 4 Ω resistor with a diode whose arrow points from the positive side to the negative side, a 12 Ω resistor with a diode pointing the opposite way, and a plain 6 Ω resistor. The diodes are ideal. Find the current from the battery.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q17Moderate

Example 1 · Semiconductor Electronics · Diode Circuits and Rectifiers

The value of net resistance of the network as shown in the given figure is:

A blocked branch is gone, resistor and all

A reverse-biased ideal diode carries no current, so the resistor in series with it carries none either. Leaving that resistor in the parallel combination gives a wrong, smaller resistance.

Read the battery before the diodes

The long plate is positive. Reading the battery the wrong way round flips every diode's state at once and gives an answer that is usually among the options.

Forward resistance goes in series

When a diode has a forward resistance r_f, add it to the resistor in its own branch before combining branches. It is not a separate parallel path.
Drill 9 more on ideal diodes in resistor networks

Concept 2 of 3: Diodes with a fixed forward voltage drop

A real diode needs a small voltage across it before it conducts, and once it conducts that voltage hardly changes. So treat a conducting diode as a small fixed battery opposing the current: subtract its drop from the supply, and divide what is left by the resistance. Below its cut-in voltage it does not conduct at all.

Definition

  • Usual drops: about 0.7 V for silicon and 0.3 V for germanium, unless the question states a cut-in value.
  • In a series loop: I=V−∑VD∑RI = \dfrac{V - \sum V_D}{\sum R}. The drops of all conducting diodes are subtracted.
  • If the supply is below the cut-in voltage, the diode is off and the current is zero.
  • Identical diodes in parallel share the current equally.
  • An LED also has a fixed drop. From its power rating at a given current, VLED=P/IV_{\text{LED}} = P/I; the series resistor is then RS=(V−VLED)/IR_S = (V - V_{\text{LED}})/I.
  • An LED in series with a Zener diode lights only when the supply exceeds VLED+VZV_{\text{LED}} + V_Z.

Current with fixed diode drops

I=V−∑VD∑R,VLED=PI,RS=V−VLEDII = \frac{V - \sum V_D}{\sum R}, \qquad V_{\text{LED}} = \frac{P}{I}, \qquad R_S = \frac{V - V_{\text{LED}}}{I}

Worked example

A 9 V supply drives a silicon diode (drop 0.7 V), a red LED (drop 2.0 V) and a 630 Ω resistor, all in series and forward biased. Find the current and the voltage across the resistor.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q105Moderate

Example 2 · Semiconductor Electronics · Diode Circuits and Rectifiers

In the given circuit, the voltage across load resistance (RL)\left( R_{L} \right) is:

Subtract every conducting diode's drop

Two diodes in series take two drops out of the supply. Subtracting only one, or none, gives a current that is too large.

Germanium and silicon differ

Silicon drops about 0.7 V and germanium about 0.3 V. A circuit with one of each loses 1.0 V, not 1.4 V or 0.6 V.

Below cut-in there is no current

A drop is not a resistance. If the supply is smaller than the cut-in voltage, the diode is off and the current is zero, not a small value from Ohm's law.
Drill 5 more on diodes with a fixed forward voltage drop

Concept 3 of 3: Rectifiers, filters and clipping

A rectifier uses a diode's one-way action on alternating voltage. One diode passes only one half of each cycle (half-wave); two or four diodes arranged right pass both halves the same way round (full-wave). The output is still bumpy, so a filter smooths it. A diode in series with a cell lets through only the part of the wave that rises above the cell's voltage.

Definition

  • Half-wave: one diode; one pulse per cycle, so the output repeats at the input frequency ff.
  • Full-wave (centre-tap with two diodes, or a bridge of four): two pulses per cycle, ripple frequency 2f2f. In a bridge, two diodes conduct at a time, so the peak output is Vm−2VDV_m - 2V_D.
  • The direction of the diode sets the sign: reversing it passes the negative half-cycles instead.
  • Filters: a capacitor across the load, or an inductor in series with it, smooths the pulsating output.
  • The full chain: a transformer steps the ac voltage up or down, the rectifier turns ac into pulsating dc, the filter smooths it, and a stabilizer (such as a Zener regulator) holds it constant.
  • Clipping: an ideal diode in series with a cell of emf EE conducts only while the input exceeds EE; the output then follows Vin−EV_{in} - E. Two diodes facing opposite ways across the output limit it to between −VD-V_D and +VD+V_D.

Output frequency and peak

fout=f (half-wave),2f (full-wave);Vpeak=Vm−VD per conducting diodef_{\text{out}} = f\ (\text{half-wave}), \quad 2f\ (\text{full-wave}); \qquad V_{\text{peak}} = V_m - V_D\ \text{per conducting diode}

Worked example

A bridge rectifier made of silicon diodes (0.7 V each) is fed with an ac voltage of peak 12 V at 50 Hz. Find the peak output voltage and the ripple frequency.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 23 · Q19Moderate

Example 3 · Semiconductor Electronics · Diode Circuits and Rectifiers

Statement-I: To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RLR_{L}. Statement-II: To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RLR_{L}. In the light of the above statements, choose the most appropriate answer from the options given below:

Capacitor across, inductor in series

Both smooth the output, but in different places. A capacitor goes in parallel with the load; an inductor goes in series with it. A capacitor in series would block the dc altogether.

Full-wave doubles the frequency

A full-wave rectifier gives two pulses per input cycle, so its ripple is at 2f, 100 Hz on 50 Hz mains. The half-wave output repeats at f.

A reversed diode passes the other half

Turning the diode round in a half-wave rectifier gives negative half-sines across the load, not positive ones. Check the diode's direction before choosing a waveform.
Drill 5 more on rectifiers, filters and clipping

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 (3)

  • Ideal diodes in resistor networks

    A diode as a switch

    forward: RD=rf  (0 if ideal),reverse: RD→∞\text{forward: } R_D = r_f\ \ (0 \text{ if ideal}), \qquad \text{reverse: } R_D \to \infty
  • Diodes with a fixed forward voltage drop

    Current with fixed diode drops

    I=V−∑VD∑R,VLED=PI,RS=V−VLEDII = \frac{V - \sum V_D}{\sum R}, \qquad V_{\text{LED}} = \frac{P}{I}, \qquad R_S = \frac{V - V_{\text{LED}}}{I}
  • Rectifiers, filters and clipping

    Output frequency and peak

    fout=f (half-wave),2f (full-wave);Vpeak=Vm−VD per conducting diodef_{\text{out}} = f\ (\text{half-wave}), \quad 2f\ (\text{full-wave}); \qquad V_{\text{peak}} = V_m - V_D\ \text{per conducting diode}

Watch out for (9)

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