Appearance
2.2 — Diodes, Rectifiers and Power Supplies
A wall socket gives 230 V alternating at 50 Hz. A phone charger's output is 5 V, dead steady. Everything that happens between those two facts is built from the single junction of Chapter 2.1, and this chapter follows the signal all the way through.
1. The diode as a component
A diode is one PN junction with a wire on each end. The p side is the anode, the n side is the cathode. The symbol is a triangle pointing at a bar, and the triangle points the way conventional current is allowed to flow — from anode to cathode. On the physical part, a painted band marks the cathode, matching the bar in the symbol.
The behaviour, from the diode equation of Chapter 2.1:
I = I_S\left(e^{V/nV_T}-1\right)
is a curve, not a straight line, and that non-linearity is the whole point. Three models of the same device, used at different levels of precision:
| Model | Forward | Reverse | Use it when |
|---|---|---|---|
| Ideal | short circuit | open circuit | first-pass reasoning |
| Constant drop | 0.7 V source | open circuit | almost all hand analysis |
| Exponential | Shockley equation | -I_S | precision, temperature work |
The constant-drop model is the working default. It says a conducting silicon diode drops 0.7 V and that is that. The justification is the steepness from Chapter 2.1: since current changes tenfold for every 60 mV, a diode carrying anything from 1 mA to 1 A sits between roughly 0.6 and 0.8 V. Treating it as 0.7 costs you a tenth of a volt at worst.
Schottky diodes, made from a metal-semiconductor contact instead of a PN junction, drop only 0.2 to 0.4 V and switch far faster because they store no minority carriers. Germanium drops about 0.3 V. LEDs drop between 1.8 V (red) and 3.4 V (blue), for a reason section 6 explains.
Reading the datasheet numbers that matter
- V_F — forward voltage at a stated current. Not a constant; always quoted with its test current.
- I_F — maximum continuous forward current.
- V_{RRM} — peak repetitive reverse voltage, the highest reverse voltage the part will survive. This is the number that kills designs. Section 3 shows a rectifier where the diode sees twice the peak supply voltage.
- t_{rr} — reverse recovery time. Irrelevant at 50 Hz, decisive at 100 kHz.
- I_R — reverse leakage, usually at 25 °C, and remember it doubles every 10 °C.
2. Half-wave rectification
The simplest thing you can do with a diode: put it in series with a load and feed it AC.
During the positive half-cycle the diode conducts and the load sees the input minus 0.7 V. During the negative half-cycle the diode blocks and the load sees nothing. The output is a series of positive humps with gaps between them.
V_{dc(avg)} = \frac{V_m}{\pi} = 0.318V_m, \qquad V_{rms} = \frac{V_m}{2}
Derivation of the average. Average over a full period, but the waveform is zero for half of it:
V_{avg} = \frac{1}{2\pi}\int_0^\pi V_m\sin\theta\,d\theta = \frac{V_m}{2\pi}\left[-\cos\theta\right]_0^\pi = \frac{V_m}{2\pi}(1+1) = \frac{V_m}{\pi}
The diode must withstand V_m in reverse during the blocked half-cycle — the peak inverse voltage or PIV.
Half-wave rectification wastes half the input and produces a lumpy output with a large 50 Hz ripple that is hard to smooth. It is used almost nowhere except in tiny, cheap, low-current supplies. It is worth knowing mainly as the thing full-wave rectification improves on.
3. Full-wave rectification
The bridge rectifier
Four diodes in a diamond. Trace the current for each half-cycle and the trick becomes obvious.
On the positive half-cycle, the two green diodes conduct and the red pair blocks. On the negative half-cycle the roles swap — but because the pair that conducts is on the opposite diagonal, the current through the load still flows the same direction. Both halves of the input are used, and both come out positive.
V_{dc(avg)} = \frac{2V_m}{\pi} = 0.637V_m, \qquad V_{rms} = \frac{V_m}{\sqrt2}
Twice the average of half-wave, and the ripple frequency is doubled to 100 Hz (or 120 Hz on a 60 Hz supply), which is much easier to filter because a capacitor's impedance falls with frequency.
Two diodes are always in series with the load, so the output peak is V_m - 1.4 V, not V_m - 0.7. On a 5 V output that 1.4 V is a serious tax, which is why low-voltage supplies often use Schottky bridges at 0.3 V each.
PIV per diode is V_m, the same as half-wave, which is a genuine advantage of the bridge.
The centre-tapped alternative
Two diodes and a transformer with a centre tap does the same job. Each diode handles one half-cycle, feeding from opposite ends of the winding. Only one diode is in series with the load at a time, so the drop is 0.7 V rather than 1.4 V — which mattered enormously in valve-era and germanium designs.
The cost: PIV is 2V_m, because the non-conducting diode sees the full winding voltage plus the output. Miss that and you fit a 100 V diode where 200 V was needed and it fails on the first cycle.
| Bridge | Centre-tapped | |
|---|---|---|
| Diodes | 4 | 2 |
| Drop | 1.4 V | 0.7 V |
| PIV | V_m | 2V_m |
| Transformer | plain | needs centre tap |
Modern practice is the bridge, because diodes are cheap and copper is not.
4. Smoothing — turning humps into DC
The rectified output is positive but far from steady. Put a large capacitor across the load and the behaviour changes completely.
When the rectified voltage is rising, the diodes conduct and the capacitor charges to the peak. When the rectified voltage falls below the capacitor's voltage, the diodes stop conducting — they are reverse biased — and the capacitor supplies the load on its own, discharging slowly through it.
The output is therefore the peak voltage with a saw-tooth ripple on top. The peak-to-peak ripple follows directly from the capacitor's defining equation, i = C\,dv/dt, rearranged:
\Delta V = \frac{I_{load}\,\Delta t}{C}
The discharge time \Delta t is approximately one ripple period, 1/(2f) for full-wave rectification of a mains at frequency f, so:
\boxed{V_{ripple} \approx \frac{I_{load}}{2fC}}
Worked example. A 12 V, 1 A supply from a full-wave rectifier on 50 Hz mains, with $C = 4700\ \mu$F:
V_{ripple} = \frac{1}{2\times50\times4700\times10^{-6}} = \frac{1}{0.47} = 2.13\ \text{V peak-to-peak}
Two volts of ripple on a 12 V rail is 18% — audible as hum in an audio circuit and unusable for anything digital. To get it under 0.5 V you would need about 20,000 µF, which is a physically large and expensive component.
This is why the capacitor is never the last stage. Section 5's regulator takes the ripply output and produces a clean one, and it does so far more cheaply than a bigger capacitor would.
The surge current problem nobody expects
At switch-on the capacitor is empty, so it looks like a short circuit (Chapter 1.5). The peak charging current is limited only by the transformer's winding resistance and the diodes, and it can be tens of amps for a few milliseconds. That surge is what blows the fuse on a large amplifier when you switch it on, and the standard fixes are a slow-blow fuse, a series NTC thermistor whose resistance is high when cold and drops as it warms, or a relay that shorts out a limiting resistor after a second.
Even in steady operation, the diodes conduct only during a narrow spike at each peak — perhaps 15% of the cycle — so the peak diode current is several times the average load current. A 1 A supply may need diodes rated at 4 A. This also makes the current drawn from the mains distinctly non-sinusoidal, which is a power quality problem at scale and the reason large equipment now requires power factor correction circuitry.
5. The Zener regulator
A Zener diode is designed to be run in reverse breakdown, where its voltage is nearly constant regardless of current. Chapter 2.1 explained the two breakdown mechanisms; the device exploits either, and the name is used for both.
The simplest regulator: a series resistor from the unregulated supply, with the Zener across the load, cathode to the positive side.
R_S = \frac{V_{in} - V_Z}{I_Z + I_L}
The series resistor must be chosen so that the Zener still gets enough current at the worst case, which is the lowest input voltage combined with the highest load current.
Worked example. Unregulated input varies from 15 to 20 V. Wanted: 9 V at up to 50 mA. Zener minimum operating current 5 mA.
Worst case for the Zener is V_{in} = 15 V and I_L = 50 mA:
R_S = \frac{15-9}{0.055} = 109\ \Omega
Choose the nearest standard value below, 100 Ω, giving a bit of margin.
Now check the other extreme, which is the one that destroys parts: V_{in} = 20 V with no load at all, so all the current goes through the Zener:
I_Z = \frac{20-9}{100} = 110\ \text{mA}, \qquad P_Z = 9\times0.11 = 0.99\ \text{W}
You need a 1 W Zener at least, probably 1.3 W for margin. A 400 mW part, the common cheap one, would fail. Always check the no-load case — it is where these circuits die.
Efficiency here is poor: the resistor burns (20-9)\times0.11 = 1.2 W to deliver 0.45 W. A Zener regulator is for reference voltages and small loads, not for power.
Real regulators
A three-terminal linear regulator (the 7805 family, and its low-dropout descendants) contains a voltage reference, an error amplifier and a series pass transistor. It compares the output with the reference and adjusts the transistor continuously to hold the output steady. That is a feedback loop, and Part 6 is entirely about how such loops behave.
The linear regulator's flaw is fixed by physics: it drops the excess voltage across a transistor, and P = (V_{in}-V_{out})I is dissipated as heat. Taking 12 V down to 5 V at 1 A wastes 7 W to deliver 5 W — 42% efficient. A switching regulator instead chops the input with a transistor and uses an inductor to convert voltage without dissipating the difference, reaching 90% and above. Chapter 9.5 builds one.
6. The rest of the diode family
The LED
A light-emitting diode is a forward-biased junction in which recombination releases a photon instead of heat. The photon's energy is set by the band gap:
E_{photon} = E_g = \frac{hc}{\lambda}
so the wavelength — the colour — is fixed by the semiconductor's band gap, and cannot be changed by the driving circuit. Red LEDs use gallium arsenide phosphide with a gap near 1.9 eV; blue uses gallium nitride at about 2.7 eV.
That relationship also explains the forward voltage. An LED cannot conduct until the applied voltage is roughly E_g/q, so a red LED needs about 1.8 V and a blue one about 3.2 V. You cannot light a blue LED from a 3 V coin cell reliably, and no resistor value will fix that.
Silicon makes a hopeless LED, because its band gap is indirect, meaning an electron cannot drop across it without also handing momentum to the lattice — a much less likely event, so the energy comes out as heat. Direct-gap materials such as gallium arsenide emit readily. This is also why silicon lasers were a research problem for decades.
LEDs are current devices, not voltage devices. The exponential curve means a small voltage change causes an enormous current change, so you never connect one straight to a supply. The series resistor is compulsory:
R = \frac{V_{supply}-V_F}{I_F}
For a red LED at 20 mA on 5 V: R = (5-1.8)/0.02 = 160\ \Omega, so use 180 Ω. The blue LED story explains the 2014 Nobel Prize. Red and green LEDs had existed since the 1960s. Blue needed gallium nitride, and nobody could grow crystals of it good enough to work until Akasaki, Amano and Nakamura managed it in the early 1990s.
Why blue mattered so much: a white LED is a blue one with a phosphor coating that converts part of the blue light to yellow, and blue plus yellow reads as white. Without blue there is no white, and without white there is no LED lighting. That is a Nobel Prize for a colour.
The photodiode
Run a junction in reverse bias and shine light on it. Photons with energy above the band gap break bonds, creating electron-hole pairs in the depletion region, which the field immediately sweeps out as current. The current is proportional to the light intensity over many orders of magnitude, which makes photodiodes excellent light meters and the receiving end of every fibre optic link (Chapter 8.5).
With no bias at all, the same device generates a voltage — that is a solar cell, which is a large-area photodiode operated in the fourth quadrant of its curve, delivering power instead of absorbing it.
The Schottky diode
A metal-to-semiconductor junction rather than p-to-n. Forward drop 0.2 to 0.4 V, and — the important part — no minority carrier storage, so no reverse recovery delay. It switches in picoseconds. Every switching power supply and every high-frequency rectifier uses them. The cost is higher reverse leakage and lower reverse voltage ratings.
The varactor
A junction used deliberately as a voltage-controlled capacitor in reverse bias, as Chapter 2.1 described. It replaced the mechanical tuning capacitor in every radio and television made after about 1975, and it is what lets a phone's radio retune itself in microseconds.
7. Clippers and clampers — shaping a waveform
Two small circuits worth knowing because they appear constantly inside larger designs.
A clipper (or limiter) removes part of a waveform. Put a diode across the signal to ground, and any excursion beyond 0.7 V in the conducting direction is shorted out. Add a bias voltage in series with the diode and the clipping level moves to V_{bias}+0.7. Two diodes back to back clip symmetrically, producing a rough square wave from a sine — which is exactly how a guitar distortion pedal works, and why it is called clipping.
The same circuit is used defensively: a pair of diodes from an input pin to the supply rails clamps any spike to just outside the rails, protecting whatever is downstream. Every microcontroller input pin has this pair built in, and it is why a static discharge that would otherwise destroy the chip usually does not.
A clamper shifts a waveform's DC level without changing its shape, using a capacitor in series and a diode to ground. The capacitor charges to whatever offset is needed to keep the diode from conducting past its threshold, and thereafter the whole waveform rides on top of that offset. A voltage doubler is two clampers stacked one on the other.
Chain more of them together and you get a charge pump: a circuit that shuttles charge from one capacitor to the next in step with a clock, each stage handing its neighbour a little more voltage than it received. Because it needs only capacitors and switches, and no inductor at all, it fits on a chip. That is how a circuit produces a negative rail, or a voltage higher than its own supply, from a single positive input — and it is what let an old serial port put out ±12 V on a board that only had 5 V available.
The diode conducts one way and blocks the other, and that is all it can do. The next chapter adds a third terminal, and a small current at that terminal controls a large current between the other two. That is amplification, and it is the invention that created modern electronics.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Rectifiers and supplies
Half-wave:
V_{avg} = \frac{1}{2\pi}\int_0^\pi V_m\sin\theta\,d\theta = \frac{V_m}{\pi} = 0.318V_m, \qquad V_{rms} = \frac{V_m}{2}, \qquad \text{PIV} = V_m
Full-wave:
V_{avg} = \frac{1}{\pi}\int_0^\pi V_m\sin\theta\,d\theta = \frac{2V_m}{\pi} = 0.637V_m, \qquad V_{rms} = \frac{V_m}{\sqrt2}
PIV is V_m for a bridge and 2V_m for a centre-tapped pair.
Output peak after diode drops: V_m - 1.4 V (bridge), V_m - 0.7 V (centre-tapped).
Ripple
V_{ripple} \approx \frac{I_{load}}{2fC} \quad\text{(full-wave)}, \qquad \frac{I_{load}}{fC} \quad\text{(half-wave)}
Derivation: the capacitor supplies the load alone for approximately one ripple period. From i = C\,dv/dt, \Delta V = I\Delta t/C, and \Delta t = 1/(2f) for full-wave.
Ripple factor:
r = \frac{V_{ripple(rms)}}{V_{dc}} \approx \frac{1}{2\sqrt3 fCR_L}
Zener regulator
R_S = \frac{V_{in}-V_Z}{I_Z+I_L}
Size R_S from the worst case: minimum V_{in} with maximum I_L. Then check the Zener's dissipation at maximum V_{in} with no load, where I_Z = (V_{in}-V_Z)/R_S and P_Z = V_ZI_Z. That second check is the one that saves the part.
LED
R = \frac{V_{supply}-V_F}{I_F}, \qquad E_{photon}=E_g=\frac{hc}{\lambda} \;\Rightarrow\; \lambda = \frac{1240}{E_g\text{[eV]}}\ \text{nm}
A 2.0 eV gap gives 620 nm, which is red. A 2.7 eV gap gives 459 nm, which is blue.