How a Phone Charger Turns AC Into DC

Mains electricity is AC, yet a phone battery needs DC. The part that makes the conversion possible is the p-n junction diode — a one-way valve for current, made by joining p-type and n-type semiconductor.

This part covers how a p-n junction forms, the diode's I-V characteristics, and its use as a rectifier.

When p-type and n-type regions are joined, electrons diffuse into the p-side and holes into the n-side, leaving a thin layer of fixed ions with no free carriers — the depletion region — whose electric field sets up a potential barrier that stops further diffusion.

Formation step by step:

- Diffusion — electrons move from the n-side, where they are plentiful, to the p-side, and holes move the other way
- Fixed ions left behind — positive donor ions on the n-side, negative acceptor ions on the p-side
- Electric field — points from the n-side to the p-side and drives a drift current opposite to diffusion
- Equilibrium — diffusion and drift currents balance, so no net current flows

Worked example. A silicon junction has a barrier potential of about $0.7$ V across a depletion region $5\times 10^{-7}$ m wide. The average field is

$$E=\frac{V}{d}=\frac{0.7}{5\times 10^{-7}}=1.4\times 10^6{\text{ V m}}^{-1}$$

and an electron needs $0.7$ eV $=1.12\times 10^{-19}$ J to cross it.

An everyday example. Opening a door between a crowded hall and an empty room — people spread into the empty room until those already inside start pushing back, and the flow stops.

The substance. The depletion region is not empty of atoms — only of free carriers; its fixed ions create the barrier.

**In forward bias the external voltage lowers the barrier, and above a threshold of about $0.7$ V for silicon the current rises steeply; in reverse bias the barrier grows and only a tiny current flows, until breakdown at a large reverse voltage.

Forward bias — p-side to the positive terminal:

- The depletion region narrows and the barrier falls
- Current is negligible below the threshold voltage**, about $0.7$ V for silicon and $0.2$ V for germanium
- Beyond it, current rises steeply, reaching milliamperes

Reverse bias — p-side to the negative terminal:

- The depletion region widens
- A tiny, nearly constant current of minority carriers flows, in microamperes
- At the breakdown voltage, the current rises sharply

Dynamic resistance. From the curve, $r_d=\frac{\Delta V}{\Delta I}$.

Worked example. Raising the forward voltage from $0.7$ V to $0.8$ V increases the current from $10$ mA to $20$ mA:

$$r_d=\frac{0.8-0.7}{(20-10)\times 10^{-3}}=\frac{0.1}{0.01}=10\Omega$$

In reverse bias, a change from $-10$ V to $-20$ V that changes the current by only $0.01\mu$A gives $r_d=\frac{10}{10^{-8}}=10^9\Omega$.

An everyday example. The valve in a bicycle tyre lets air in but not out — a diode does the same for current.

The substance. A diode is non-ohmic — its resistance depends on the voltage, so no single resistance value describes it.

A single diode conducts only in the half-cycles when it is forward biased, giving a half-wave rectifier, while two diodes fed by a centre-tap transformer conduct in alternate half-cycles and send current the same way through the load all the time — a full-wave rectifier.

Half-wave rectifier:

- A transformer's secondary drives one diode in series with the load
- Positive half-cycles pass; negative half-cycles are blocked
- The output pulses at the same frequency as the input

Full-wave rectifier:

- A centre-tap transformer feeds two diodes
- One diode conducts in each half-cycle
- The output pulses at twice the input frequency

Filter. A capacitor across the load charges at each peak and discharges slowly between them, smoothing the output.

Worked example. A $230$ V, $50$ Hz supply is stepped down to $12$ V RMS:

$$V_{peak}=12\sqrt{2}\approx 17\text{ V}$$

With a $0.7$ V diode drop, the output peak is about $16.3$ V. The half-wave output pulses $50$ times a second and the full-wave output $100$ times.

An everyday example. Every phone charger and laptop adapter contains rectifier diodes that turn mains AC into the DC a battery needs.

The substance. Full-wave output is easier to smooth — pulses arrive twice as often, so the capacitor has less time to discharge.