Chapter 14 · Semiconductor Electronics: Materials, Devices and Simple Circuits

14.3 The p–n Junction, the Semiconductor Diode and Rectifiers

Where p-type meets n-type silicon, carriers diffuse, a thin depletion region forms and a barrier of about $0.7$ V builds up. The result is a diode — a one-way valve for current — and two or four of them turn the AC mains into the DC that every electronic device needs.

Every phone charger, laptop adapter and LED bulb plugged into the mains has the same problem to solve: the socket supplies alternating current, reversing direction $100$ times a second, but electronic circuits need a steady, one-way current. The device that solves it is the semiconductor diode, which lets current through in one direction and almost completely blocks it in the other. Diodes also protect circuits from batteries inserted the wrong way round, demodulate radio signals and, in special forms, emit light, detect it and generate electricity from sunlight.

A diode is simply a single crystal that is p-type on one side and n-type on the other. In this section we see what happens at the boundary — the p–n junction — when it forms: carriers diffuse across, leave behind a layer of fixed charge, and set up a barrier potential. We then apply a voltage in either direction (forward and reverse bias), study the diode's I–V characteristic, and use diodes to build half-wave and full-wave rectifiers, smoothing their output with a capacitor.

Formation of the p–n Junction

A p–n junction is made inside one crystal, for instance by diffusing or implanting acceptor atoms into the surface of an n-type wafer until that layer becomes p-type. (Pressing a p-type piece against an n-type piece does not work: the surfaces would be separated by gaps and oxide many atoms thick.) On the p side holes are plentiful; on the n side electrons are plentiful. Two processes now begin.

Diffusion. Carriers spread from where they are concentrated to where they are scarce: holes diffuse from p to n, electrons from n to p. This is the diffusion current, conventionally from p to n. Near the junction the diffusing carriers meet and recombine.

Uncovered ions and the junction field. When a hole leaves the p side, it leaves behind an acceptor ion with charge $-e$ that cannot move; when an electron leaves the n side, it leaves behind a donor ion with charge $+e$. So a thin layer of fixed negative charge builds up on the p side of the boundary and a layer of fixed positive charge on the n side. Together they make an electric field pointing from n to p, which pushes holes back towards p and electrons back towards n — it opposes further diffusion.

Drift and equilibrium. The same field sweeps any minority carriers that wander into it across the junction (electrons from p to n, holes from n to p) — the drift current, conventionally from n to p. As diffusion proceeds, the layer of charge grows, the field strengthens and drift increases, until the drift current exactly cancels the diffusion current. In equilibrium the net current is zero.

📘 Definition 14.8 — p–n junction and depletion region
A p–n junction is the boundary between a p-type and an n-type region of a single semiconductor crystal. The depletion region (space-charge region) is the thin layer on both sides of the boundary, typically $0.1$–$1\ \mu$m wide, from which mobile carriers have been removed by diffusion and recombination. It contains only fixed ions: negative acceptor ions on the p side and positive donor ions on the n side, with equal total charges.
📘 Definition 14.9 — Barrier potential
The charge of the depletion region creates a potential difference across it, the n side being at the higher potential. In equilibrium (no external voltage) this barrier potential $V_0$ is about $0.6$–$0.7$ V for silicon and $0.2$–$0.3$ V for germanium. A majority carrier needs an energy of at least $eV_0$ (about $0.7$ eV in silicon) to diffuse across the junction against it.
✏️ Example 1 — The field inside a depletion region
A silicon junction has a barrier potential of $0.70$ V across a depletion region $0.50\ \mu$m wide. (a) Estimate the average electric field in the region and give its direction. (b) Compare it with the field ($3\times10^6$ V/m) at which air breaks down. (c) How much energy, in eV and in J, does an electron from the n side need to cross to the p side?
(a) Field $E\approx\dfrac{V_0}{W}=\dfrac{0.70}{0.50\times10^{-6}}=1.4\times10^{6}$ V/m, pointing from the n side (positive ions) to the p side (negative ions).
(b) Compare $\dfrac{1.4\times10^6}{3\times10^6}\approx0.47$ — about half the field that makes sparks in air, set up by less than one volt, because the layer is so thin.
(c) Energy The electron must climb the barrier: $eV_0=0.70$ eV $=(0.70)(1.6\times10^{-19})\approx1.1\times10^{-19}$ J.
Note You cannot measure $V_0$ with a voltmeter across a diode: the contact potentials at the metal leads exactly cancel it, so no current flows round a closed circuit.

The Semiconductor Diode: Forward and Reverse Bias

A p–n junction with metal contacts on its two ends is a semiconductor diode. Its circuit symbol is an arrow (the p side, or anode) pointing into a bar (the n side, or cathode); the arrow shows the direction in which conventional current flows easily. An external voltage $V$ across the diode is called a bias.

Forward bias — p side connected to the positive terminal. The applied voltage opposes the barrier, so the barrier falls to $V_0-V$ and the depletion region becomes narrower. Many more majority carriers can now diffuse across: holes are injected into the n side and electrons into the p side, where they are minority carriers. The current is tiny until $V$ approaches $V_0$, then rises very steeply — roughly exponentially — reaching milliamperes or amperes.

Reverse bias — p side connected to the negative terminal. The applied voltage adds to the barrier, which rises to $V_0+V$, and the depletion region widens. Diffusion of majority carriers is cut off almost entirely. Only the minority carriers, swept across by the junction field, carry a current — and since their number is fixed by temperature, not by $V$, this reverse current is tiny (nanoamperes to microamperes in silicon) and almost independent of the voltage. At a large enough reverse voltage, the breakdown voltage, the reverse current suddenly rises steeply; an ordinary diode is damaged if it is driven far into breakdown.

📐 Result 14.5 — The I–V characteristic of a junction diode

Forward bias: the current is negligible below the knee (threshold) voltage — about $0.7$ V for silicon, $0.2$–$0.3$ V for germanium — and above it rises steeply, so the voltage across a conducting diode stays close to the knee voltage over a wide range of currents.

Reverse bias: a very small reverse current (μA or less) flows, almost independent of $V$, until breakdown.

The dynamic resistance at an operating point is $$r_d=\frac{\Delta V}{\Delta I},$$ the ratio of a small change in voltage to the resulting change in current (the inverse slope of the I–V curve). It is a few ohms in forward bias and many megohms in reverse bias. A diode is non-ohmic: $V/I$ is not constant (Section 3.2).

Working model for circuits: a forward-biased silicon diode is replaced by a constant drop of $0.7$ V (an ideal diode: a drop of $0$); a reverse-biased diode by an open switch.

✏️ Example 2 — Dynamic resistance from measurements
A student measures a silicon diode (illustrative data). In forward bias the current is $10$ mA at $0.70$ V and $30$ mA at $0.80$ V. In reverse bias it is $1.0\ \mu$A at $10$ V and still $1.0\ \mu$A at $20$ V. (a) Find the forward dynamic resistance between the two points. (b) Find the ratio $V/I$ in reverse bias at $10$ V. (c) Is the diode ohmic?
(a) Forward $r_d=\dfrac{\Delta V}{\Delta I}=\dfrac{0.80-0.70}{(30-10)\times10^{-3}}=\dfrac{0.10}{0.020}=5.0\ \Omega$.
(b) Reverse $\dfrac{V}{I}=\dfrac{10}{1.0\times10^{-6}}=1.0\times10^{7}\ \Omega=10$ M$\Omega$. Between $10$ and $20$ V the current does not change at all, so the reverse dynamic resistance is even larger (effectively infinite here).
(c) Ohmic? No: $V/I$ is $70\ \Omega$ at $0.70$ V, $27\ \Omega$ at $0.80$ V and $10$ M$\Omega$ in reverse — it depends on $V$ and on the direction.
Interactive 1 — The p–n Junction under Bias

Top left: the junction, with holes (open) on the p side, electrons (filled) on the n side and the fixed ions ($-$ and $+$) in the depletion region. Below it: the potential across the junction — the barrier the majority carriers must climb. Right: the diode's I–V characteristic (forward and reverse halves on different scales, as is usual), with the operating point. Drag the bias slider from reverse to forward, or press Play to sweep it. (Illustrative silicon diode model: the width follows $W\propto\sqrt{V_0-V}$, the forward current $I=I_s(e^{eV/\eta k_BT}-1)$, breakdown at $30$ V — not data for a particular diode.)

—
applied $V$—
barrier height—
depletion width $W/W_0$—
current $I$—
🧭 Problem-solving strategy — circuits containing diodes
  1. For each diode, decide whether it is forward biased (the circuit tries to push current from p to n, i.e. along the arrow) or reverse biased. If unsure, guess, solve, and check that the current in each "on" diode really flows along its arrow.
  2. Replace each forward-biased diode by a drop of $0.7$ V (Si), $0.3$ V (Ge) or $0$ (ideal), as the question says; replace each reverse-biased diode by a break in the circuit.
  3. Solve the remaining circuit with Ohm's law and Kirchhoff's rules (Section 3.4).
  4. For a graphical solution, draw the load line $I=(E-V)/R$ on the I–V characteristic: it joins $(V=E,\ I=0)$ to $(V=0,\ I=E/R)$, and the operating point is where it crosses the diode curve.
✏️ Example 3 — A diode and a resistor in series
A $5.0$ V battery, a $100\ \Omega$ resistor and a silicon diode are connected in series, with the diode's p side towards the positive terminal. (a) Find the current, using the $0.7$ V model. (b) Find the power dissipated in the diode. (c) What current flows if the diode is reversed?
(a) Forward The diode is forward biased: $I=\dfrac{E-0.7}{R}=\dfrac{5.0-0.7}{100}=0.043$ A $=43$ mA.
(b) Power $P=VI=(0.7)(0.043)\approx0.030$ W $=30$ mW.
(c) Reversed Reverse biased: only the reverse current of a few μA or less flows — practically zero. The full $5.0$ V appears across the diode.
Check The load line through $(5.0\ \text{V},0)$ and $(0,\ 50\ \text{mA})$ crosses the model curve in the lab below at about $0.71$ V and $43$ mA. ✓
Check: the operating point on a load line (illustrative Si diode model)
diode voltage $V_D$—
current $I$—
0.7 V model gives—
✏️ Example 4 — Two branches, one diode the wrong way round
A $10$ V battery feeds two branches connected in parallel across it. Branch 1 contains a silicon diode $D_1$, forward biased, in series with $1.0$ k$\Omega$. Branch 2 contains a silicon diode $D_2$, reverse biased, in series with $2.0$ k$\Omega$. Find the current drawn from the battery (a) treating the diodes as ideal, (b) with the $0.7$ V model.
Branch 2 $D_2$ is reverse biased: an open switch. No current flows in branch 2, whatever its resistor.
(a) Ideal $I=\dfrac{10}{1000}=10$ mA.
(b) 0.7 V model $I=\dfrac{10-0.7}{1000}=9.3$ mA.
Exam tip Spot the reverse-biased diodes first — they delete whole branches and make the circuit simple.

The Diode as a Rectifier

A rectifier converts alternating voltage into a voltage of one polarity. It relies on the diode conducting in forward bias only. The AC is usually first stepped down by a transformer (Section 7.3); we write the secondary voltage as $v=v_m\sin\omega t$ with peak value $v_m$ (often written $V_m$) and frequency $f$, and recall from Section 7.1 that a sinusoid has $V_\text{rms}=V_m/\sqrt2$.

Half-wave rectifier. One diode in series with the load resistor $R_L$ across the transformer secondary. During the half-cycle in which the diode's p side is positive, it conducts and the output across $R_L$ follows the input; during the other half-cycle it is reverse biased and the output is zero. The output is a series of positive half-sine pulses, one per cycle, so it repeats at the supply frequency $f$. The reverse voltage across the diode reaches $V_m$ in the blocked half-cycle, so the diode's breakdown voltage must be well above $V_m$.

Full-wave rectifier (centre-tap). The transformer secondary has a centre tap, connected to one end of $R_L$; its two ends are connected through diodes $D_1$ and $D_2$ to the other end of $R_L$. The two ends are always in antiphase: in one half-cycle end $A$ is positive and $D_1$ conducts, in the next end $B$ is positive and $D_2$ conducts. Either way, current flows through $R_L$ in the same direction, so every half-cycle gives an output pulse: the output repeats at $2f$. Here $V_m$ is the peak voltage of each half of the secondary. (A bridge rectifier of four diodes does the same job without a centre tap and is what most chargers use.)

📐 Result 14.6 — Output of half-wave and full-wave rectifiers

For a sinusoidal input of peak $V_m$ and frequency $f$, with ideal diodes and a resistive load (uses integration — the average of $\sin\omega t$ over a half-cycle is $2/\pi$):

Average (DC) outputrms outputOutput pulse frequency
Half-wave$V_\text{dc}=\dfrac{V_m}{\pi}\approx0.318\,V_m$$\dfrac{V_m}{2}$$f$
Full-wave$V_\text{dc}=\dfrac{2V_m}{\pi}\approx0.637\,V_m$$\dfrac{V_m}{\sqrt2}$$2f$

The average load current is $I_\text{dc}=V_\text{dc}/R_L$. With real silicon diodes each conducting diode reduces the output peak by about $0.7$ V.

Smoothing with a filter capacitor. The rectified output is one-way but very bumpy. A large capacitor connected across the load smooths it: on each rising pulse the capacitor charges up to almost the peak voltage; when the input falls, the diode becomes reverse biased (the capacitor holds its side at a higher potential) and the capacitor slowly discharges through $R_L$, keeping the output up until the next pulse arrives. The remaining small variation is the ripple. It is smaller when $C$ and $R_L$ are larger (slower discharge, i.e. a larger product $CR_L$) and in a full-wave rectifier, where the gaps between pulses are half as long. With a good filter the DC output is close to $V_m$ rather than to $V_m/\pi$ or $2V_m/\pi$.

✏️ Example 5 — A half-wave rectifier on a 12 V transformer
A transformer supplies $12$ V rms at $50$ Hz to a half-wave rectifier with a $1.0$ k$\Omega$ load. (a) Find the peak output voltage with an ideal diode and with a silicon diode. (b) For the ideal diode, find the DC (average) output voltage, the average load current and the rms output voltage. (c) What is the frequency of the output pulses?
(a) Peak $V_m=\sqrt2\times12\approx17.0$ V (ideal diode); with a silicon diode about $17.0-0.7=16.3$ V.
(b) DC and rms $V_\text{dc}=\dfrac{V_m}{\pi}=\dfrac{16.97}{\pi}\approx5.4$ V, so $I_\text{dc}=\dfrac{5.40}{1000}\approx5.4$ mA; $V_\text{rms}=\dfrac{V_m}{2}\approx8.5$ V.
(c) Frequency One pulse per input cycle: $50$ Hz.
Check Interactive 2 (half-wave, $V_m=17$ V, no capacitor, ideal diode) averages the output numerically over a cycle and gives $5.41$ V and $8.50$ V rms. ✓
✏️ Example 6 — A full-wave rectifier and the effect of a filter
A centre-tap full-wave rectifier is fed from $50$ Hz mains; each half of the secondary gives $9.0$ V rms. The load is $1.0$ k$\Omega$. (a) Find $V_m$, the DC output voltage and the average load current (ideal diodes). (b) What is the ripple frequency? (c) A large capacitor is now connected across the load. Roughly what DC voltage do you expect, and why?
(a) DC $V_m=9.0\sqrt2\approx12.7$ V; $V_\text{dc}=\dfrac{2V_m}{\pi}=\dfrac{2(12.73)}{\pi}\approx8.1$ V; $I_\text{dc}\approx8.1$ mA.
(b) Ripple Two pulses per input cycle: $2\times50=100$ Hz.
(c) Filter The capacitor charges to nearly the peak and discharges only a little between pulses, so the output is close to $V_m\approx12.7$ V (a little less, with a small $100$ Hz ripple; with silicon diodes subtract about $0.7$ V).
Interactive 2 — Rectifier Lab

Left: the circuit, with the conducting diode and the current path lit up as the input goes through its cycle. Right: the $50$ Hz input (top) and the voltage across the $1.0$ k$\Omega$ load (bottom), simulated step by step. Choose half-wave or full-wave, ideal or silicon diodes, and a filter capacitor. The readouts are computed by averaging the simulated output over a full cycle.

17
DC (average) output—
rms output—
ripple (peak to peak)—
ripple frequency—
⚠️ Common mistakes with junctions, diodes and rectifiers

1. "The depletion region is full of carriers." It is the one place that has almost none; it contains only fixed ions.

2. Getting the barrier direction wrong. The n side is at the higher potential (positive donor ions); the field in the depletion region points from n to p.

3. Forward bias "connects + to n". No: forward bias connects the positive terminal to the p side (the arrow of the symbol points along the easy current).

4. Thinking the reverse current grows with voltage. Below breakdown it is almost constant: it is limited by the supply of minority carriers, not by the voltage.

5. Mixing up average and rms. Half-wave: $V_\text{dc}=V_m/\pi$, $V_\text{rms}=V_m/2$. Full-wave: $2V_m/\pi$ and $V_m/\sqrt2$. And the ripple frequency of a full-wave rectifier is $2f$, not $f$.

🚀 Beyond the board syllabus (JEE Main · NEET) — the Zener diode as a voltage regulator

Syllabus note: special-purpose diodes are no longer in the NCERT book or the CBSE syllabus, and they are not in JEE Advanced, but the Zener diode and its use as a voltage regulator are listed in the NEET and JEE Main syllabi. A Zener diode is a heavily doped junction designed to work in reverse breakdown. Heavy doping makes its depletion region very thin, so the field there is enormous ($\gt10^{7}$ V/m) and pulls electrons straight out of their bonds once the reverse voltage reaches the Zener voltage $V_Z$. Beyond that point the reverse current can change greatly while the voltage stays almost exactly $V_Z$ — and the diode is not damaged, provided its power rating is respected.

Voltage regulator. Connect the Zener diode, reverse biased, in parallel with the load $R_L$, and feed both from an unsteady DC supply $V_\text{in}$ through a series resistor $R_s$. As long as the diode is in breakdown, the load voltage is pinned at $V_Z$; any rise in $V_\text{in}$ only increases the current through $R_s$ and the Zener diode, and the extra voltage appears across $R_s$: $$I_s=\frac{V_\text{in}-V_Z}{R_s},\qquad I_L=\frac{V_Z}{R_L},\qquad I_Z=I_s-I_L.$$ Example. $V_Z=6.0$ V, $R_s=300\ \Omega$, $R_L=1.2$ k$\Omega$, $V_\text{in}=12$ V: $I_s=20$ mA, $I_L=5.0$ mA, $I_Z=15$ mA. If $V_\text{in}$ rises to $15$ V, $I_s=30$ mA and $I_Z=25$ mA, but the load still sees $6.0$ V. (The circuit works only while $I_Z\gt0$, i.e. $V_\text{in}$ must stay above $V_Z(1+R_s/R_L)=7.5$ V here.)

LEDs, photodiodes and solar cells — the other special-purpose junctions — are described in the extension box of Section 14.1.

Exercises

I
Formation of the p–n junction. $e=1.6\times10^{-19}$ C.
  • 1 Why is a p–n junction not made by pressing a p-type piece against an n-type piece?
  • 2 Name the two processes that carry charge across a junction in equilibrium, and state the direction of each current.
  • 3 What is the sign of the fixed charge on each side of the depletion region, and where does it come from?
  • 4 A barrier potential of $0.70$ V exists across a depletion region $1.0\ \mu$m wide. Estimate the average field.
  • 5 What minimum energy, in eV and in joules, must a hole on the p side have to diffuse to the n side across a $0.70$ V barrier?
  • 6 Why is the depletion region so called?
  • 7 In an unbiased junction, which side is at the higher potential? Explain.
1 The surfaces could never be brought into atomic contact: roughness and oxide layers would leave gaps thousands of atoms wide. The junction must be a continuous single crystal.
2 Diffusion of majority carriers (holes p → n, electrons n → p; current from p to n) and drift of minority carriers under the junction field (current from n to p). In equilibrium they cancel.
3 p side: negative — acceptor ions left uncovered when holes diffused away (and electrons arrived and recombined). n side: positive — donor ions left behind by electrons that diffused away.
4 $E\approx\dfrac{0.70}{1.0\times10^{-6}}=7.0\times10^{5}$ V/m, from n to p.
5 $0.70$ eV $=(0.70)(1.6\times10^{-19})\approx1.1\times10^{-19}$ J.
6 It is depleted (emptied) of mobile carriers — electrons and holes have diffused out of it or recombined — leaving only fixed ions.
7 The n side, because it carries the positive donor ions of the depletion region; the field points from n to p, and potential falls along the field.
II
Bias and the I–V characteristic. Take $V_0=0.70$ V for silicon.
  • 1 To forward bias a diode, to which terminal of the battery is the p side connected? What happens to the barrier and to the depletion width?
  • 2 Find the height of the barrier of a silicon junction under (a) $0.50$ V forward bias, (b) $5.0$ V reverse bias.
  • 3 A diode's current rises from $5.0$ mA to $15$ mA when the voltage rises from $0.65$ V to $0.70$ V (illustrative data). Find its dynamic resistance in this range.
  • 4 The same diode passes $2.0\ \mu$A at a reverse bias of $20$ V. Find $V/I$.
  • 5 Explain why the reverse current hardly changes as the reverse voltage is increased (below breakdown).
  • 6 A $9.0$ V battery, a $1.0$ k$\Omega$ resistor and a forward-biased silicon diode are in series. Find the current.
  • 7 Two silicon diodes, both forward biased, are connected in series with a $470\ \Omega$ resistor across a $6.0$ V supply. Find the current.
  • 8 Sketch the I–V characteristic of a silicon diode, marking the knee voltage, the reverse saturation region and the breakdown voltage.
1 To the positive terminal. The barrier falls (to $V_0-V$) and the depletion region narrows.
2 (a) $0.70-0.50=0.20$ V; (b) $0.70+5.0=5.7$ V.
3 $r_d=\dfrac{0.05}{0.010}=5.0\ \Omega$.
4 $\dfrac{20}{2.0\times10^{-6}}=1.0\times10^{7}\ \Omega=10$ M$\Omega$.
5 The reverse current is carried by minority carriers that drift across the junction. Their number is set by thermal generation (temperature), not by the voltage; once the field sweeps all of them across, a larger voltage cannot increase the current.
6 $I=\dfrac{9.0-0.7}{1000}=8.3$ mA.
7 $I=\dfrac{6.0-2(0.7)}{470}=\dfrac{4.6}{470}\approx9.8$ mA.
8 Forward (first quadrant): nearly zero current up to about $0.7$ V, then a steep rise. Reverse (third quadrant, μA scale): a small, nearly constant current; at the breakdown voltage a sudden, nearly vertical increase.
III
Circuits with diodes.
  • 1 A $12$ V battery feeds two parallel branches: a forward-biased diode in series with $3.0$ k$\Omega$, and a reverse-biased diode in series with $1.0$ k$\Omega$. Treating the diodes as ideal, find the battery current.
  • 2 Repeat part 1 with silicon diodes ($0.7$ V model).
  • 3 A $1.5$ V cell drives current through a silicon diode and a $220\ \Omega$ resistor in series. Find the current.
  • 4 An alternating voltage of peak $5.0$ V is applied to a silicon diode in series with a resistor. For what fraction of each cycle does the diode conduct ($0.7$ V model)?
  • 5 A diode is in series with $50\ \Omega$ across a $3.0$ V battery. Find the two intercepts of the load line, and the current using the $0.7$ V model.
  • 6 Find the power dissipated in the diode of part 5.
1 The reverse-biased branch carries nothing: $I=\dfrac{12}{3000}=4.0$ mA.
2 $I=\dfrac{12-0.7}{3000}\approx3.8$ mA.
3 $I=\dfrac{1.5-0.7}{220}\approx3.6$ mA.
4 It conducts while $5.0\sin\theta\gt0.7$: from $\theta=\sin^{-1}0.14\approx8.0^\circ$ to $172^\circ$, i.e. $164^\circ$ out of $360^\circ$ — a fraction of about $0.46$.
5 $(V=3.0\ \text{V},\ I=0)$ and $(V=0,\ I=3.0/50=60\ \text{mA})$. $I=\dfrac{3.0-0.7}{50}=46$ mA.
6 $P=(0.7)(0.046)\approx0.032$ W $=32$ mW.
IV
Rectifiers. Assume ideal diodes unless told otherwise.
  • 1 The input to a rectifier is $50$ Hz AC. What is the frequency of the output pulses for (a) a half-wave and (b) a full-wave rectifier?
  • 2 A half-wave rectifier has an input of peak $20$ V. Find the DC (average) and rms output voltages.
  • 3 Repeat part 2 for a full-wave rectifier with $20$ V peak across each half of the secondary.
  • 4 The $230$ V rms mains is stepped down by a transformer with turns ratio $20:1$ and fed to a half-wave rectifier. Find the peak and DC output voltages.
  • 5 A centre-tap transformer gives $24$ V rms across its whole secondary. Find the DC output of the full-wave rectifier.
  • 6 Explain how a capacitor across the load reduces the ripple, and how the ripple changes if the load draws more current.
  • 7 Modelling. A small power supply uses a centre-tap full-wave rectifier with silicon diodes, $6.0$ V rms across each half of the secondary, and a large filter capacitor. Estimate its DC output voltage at light load.
  • 8 Find the average current in a $2.0$ k$\Omega$ load in part 3.
1 (a) $50$ Hz; (b) $100$ Hz.
2 $V_\text{dc}=\dfrac{20}{\pi}\approx6.4$ V; $V_\text{rms}=\dfrac{20}{2}=10$ V.
3 $V_\text{dc}=\dfrac{40}{\pi}\approx12.7$ V; $V_\text{rms}=\dfrac{20}{\sqrt2}\approx14.1$ V.
4 Secondary $\dfrac{230}{20}=11.5$ V rms, so $V_m=11.5\sqrt2\approx16.3$ V and $V_\text{dc}=\dfrac{V_m}{\pi}\approx5.2$ V.
5 Each half: $12$ V rms, $V_m=12\sqrt2\approx17.0$ V; $V_\text{dc}=\dfrac{2(16.97)}{\pi}\approx10.8$ V.
6 The capacitor charges to near the peak on each pulse and, while the diodes are off, discharges only slowly through the load, filling in the gaps. A larger load current discharges it faster between pulses, so the ripple grows (a larger $C$ is then needed).
7 $V_m=6.0\sqrt2\approx8.49$ V; one diode conducts at a time, so the capacitor charges to about $8.49-0.7\approx7.8$ V. At light load the DC output is close to this.
8 $I_\text{dc}=\dfrac{12.73}{2000}\approx6.4$ mA.
🎯 Exam corner — junctions, diodes and rectifiers BoardJEE MainNEET

E1. Board · 5 marks Explain, with a diagram, how the depletion region and the barrier potential are formed at a p–n junction. Draw the circuit for studying the I–V characteristics of a diode in forward and reverse bias, sketch the characteristics, and explain their shape.

Formation Holes diffuse p → n and electrons n → p because of the concentration difference. They leave uncovered acceptor ions ($-$) on the p side and donor ions ($+$) on the n side: a thin region with no mobile carriers (depletion region). Its charge sets up a field from n to p and a potential difference $V_0$ (n side higher) that opposes further diffusion. Equilibrium: diffusion current = drift current.
Circuits Forward: battery (through a rheostat) with + to p, milliammeter in series, voltmeter across the diode. Reverse: battery reversed, a microammeter in series, a higher-voltage supply.
Shape Forward: negligible current below the knee voltage ($\approx0.7$ V for Si), then a steep rise as the lowered barrier lets majority carriers diffuse. Reverse: a tiny, almost constant current of minority carriers (μA) up to breakdown, where it rises sharply.
Exam tip Label the axes with different scales (mA/V forward, μA/V reverse) and mark the knee and breakdown voltages — each is a mark.

E2. Board · 3 marks Draw the circuit diagram of a full-wave rectifier using a centre-tap transformer and two diodes. Explain its working and draw the input and output waveforms. If the input frequency is $50$ Hz, what is the frequency of the output?

Circuit Ends $A$ and $B$ of the secondary go through diodes $D_1$, $D_2$ to one end of $R_L$; the centre tap goes to the other end.
Working When $A$ is positive with respect to the centre tap, $D_1$ is forward biased and $D_2$ reverse biased: current flows through $R_L$. In the next half-cycle $B$ is positive, $D_2$ conducts and $D_1$ is off — current through $R_L$ in the same direction. Output: a train of positive half-sine pulses with no gaps.
Frequency Two pulses per input cycle: $100$ Hz.

E3. JEE Main A $3.0$ V cell, a $100\ \Omega$ resistor and two silicon diodes $D_1$ and $D_2$ are all connected in series. $D_1$ points along the direction in which the cell drives current; $D_2$ points the opposite way. The current in the circuit is (A) zero (B) $16$ mA (C) $23$ mA (D) $30$ mA.

Concept $D_2$ is reverse biased and in series with everything else: it blocks the whole circuit (apart from a negligible reverse current).
Answer (A). Exam tip: (C) $=(3.0-0.7)/100$ is the answer if you forget $D_2$; (B) $=(3.0-1.4)/100$ treats both as forward biased.

E4. NEET When a p–n junction diode is forward biased, (A) the depletion region widens and the barrier rises (B) the depletion region narrows and the barrier falls (C) the depletion width is unchanged but the barrier falls (D) the depletion region narrows but the barrier rises.

Concept The applied voltage opposes the built-in potential: the barrier becomes $V_0-V$, and less uncovered charge is needed, so the depletion region narrows.
Answer (B).

E5. JEE MainNEET (Beyond NCERT: Zener regulator.) A Zener diode with $V_Z=6.0$ V is connected in parallel with a $1.2$ k$\Omega$ load, and the combination is fed from a $12$ V supply through a $300\ \Omega$ series resistor. The current through the Zener diode is (A) $5.0$ mA (B) $15$ mA (C) $20$ mA (D) $25$ mA.

Concept Zener regulators are beyond the CBSE and JEE Advanced syllabi but in NEET and JEE Main. The load voltage is pinned at $V_Z$: $I_s=\dfrac{12-6.0}{300}=20$ mA, $I_L=\dfrac{6.0}{1200}=5.0$ mA.
Answer (B) $I_Z=20-5.0=15$ mA. Exam tip: always check $I_Z\gt0$ — otherwise the Zener is not in breakdown and does not regulate.

E6. Board · Case study A simple phone-charger power supply steps the mains ($50$ Hz) down with a transformer, rectifies it with a full-wave rectifier and smooths it with a capacitor. (i) (2 marks) If the peak of the rectified voltage is $10$ V, find the ripple frequency and the DC output without the capacitor (ideal diodes). (ii) (1 mark) What does the capacitor do to the DC output and to the ripple? (iii) (1 mark) In a bridge rectifier two diodes conduct at a time. With silicon diodes, by how much does this lower the output peak?

(i) Ripple frequency $2f=100$ Hz; $V_\text{dc}=\dfrac{2V_m}{\pi}=\dfrac{20}{\pi}\approx6.4$ V.
(ii) It raises the DC output towards the peak ($\approx10$ V) and greatly reduces the ripple, by charging at each peak and discharging slowly through the load between peaks.
(iii) Two drops of about $0.7$ V: about $1.4$ V. Exam tip: this loss is why low-voltage supplies use special diodes with smaller drops.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How large a filter capacitor does a simple mains power supply need — and why did that push the world's chargers to switch-mode designs?

Why it matters. Billions of chargers are made every year. Old "linear" adapters had a heavy transformer, a rectifier and a big capacitor; modern switch-mode chargers sold everywhere from Shenzhen to São Paulo are a fraction of the size. Part of the reason is the physics of ripple.

What to do.

  1. Use Interactive 2 (full-wave, ideal diodes, $V_m=17$ V, load $1.0$ k$\Omega$). Record the peak-to-peak ripple for $C=47$, $220$ and $1000\ \mu$F, and for the half-wave circuit with $220\ \mu$F.
  2. Engineers estimate the ripple with $\Delta V\approx\dfrac{I_\text{dc}}{f_rC}$, where $I_\text{dc}$ is the load current and $f_r$ the ripple frequency (an approximation beyond the syllabus, valid for small ripple). Compare it with your readings.
  3. A charger must supply $0.50$ A with at most $0.50$ V of ripple from a full-wave rectifier on $50$ Hz mains. Find the capacitance needed. Repeat for a half-wave rectifier.
  4. A switch-mode charger rectifies the mains, then chops the DC at about $100$ kHz before a tiny transformer and a second rectifier (full-wave). What capacitance does the same $0.50$ A, $0.50$ V specification need now?

Tools: Interactive 2, a calculator; optionally a capacitor catalogue to see how size grows with capacitance.

Hand in: Your table of ripple values with the formula's predictions, the two capacitances of step 3, the answer to step 4, and a paragraph explaining why higher frequency means smaller, lighter chargers.

Going further. Even a perfect filter cannot remove the effect of changes in the mains voltage or the load current. Find out how a voltage regulator (for example the Zener circuit in this section's extension box, or an integrated regulator chip) holds the output fixed.

Instructor note: using the simulated $I_\text{dc}$ ($16$–$17$ mA) the formula gives about $3.3$ V ($47\ \mu$F), $0.76$ V ($220\ \mu$F) and $0.17$ V ($1000\ \mu$F) for full-wave, and $1.5$ V for half-wave at $220\ \mu$F. The widget's simulated ripples are $2.7$, $0.69$, $0.16$ and $1.4$ V: close for small ripple, smaller for large ripple, because the formula assumes the capacitor discharges for the whole interval between peaks at constant current. Step 3: $C=I/(f_r\Delta V)=0.50/(100\times0.50)=10$ mF full-wave, $20$ mF half-wave — large, heavy electrolytic capacitors. Step 4: $f_r=200$ kHz gives $C=5\ \mu$F — thousands of times smaller.

🧪 Self-check quiz

0/5 correct

Q1. The depletion region of a p–n junction contains

Q2. In reverse bias, the current through a junction diode (below breakdown) is

Q3. A silicon diode in series with a $1.0$ k$\Omega$ resistor is forward biased by a $5.7$ V battery. The current is about

Q4. The input to a full-wave rectifier has a frequency of $50$ Hz. The fundamental frequency of the ripple in its output is

Q5. A half-wave rectifier with an ideal diode has an input of peak $10$ V. The average (DC) output voltage is about

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