The mains at home change direction a hundred times every second. A phone battery can be charged only by a current in one direction; an alternating current would push charge in and pull it out again. Yet a phone plugged into the wall socket charges steadily. Somewhere between the socket and the battery, the alternating current has been made to flow one way.
The mains voltage
The mains in India is \(230\,\text{V}\) at \(50\,\text{Hz}\). It is a sine wave that repeats every \(1/50 = 20\,\text{ms}\), positive for 10 ms and negative for the next 10 ms. The 230 V is its rms value, the steady voltage that would heat a resistor at the same rate. The unit on electricity explains this; here we use one fact. The peak of a sine wave is \(\sqrt{2}\) times its rms value, so the mains peaks at \(230 \times 1.414 = 325\,\text{V}\).
A transformer first steps the mains down to a safer voltage. Take a winding that gives \(12\,\text{V}\) rms. Its peak is \(12 \times 1.414 = 17.0\,\text{V}\). In the figures below, the transformer's winding is drawn simply as an AC source.
One diode
A circuit that turns alternating current into current in one direction is a rectifier. The simplest has one diode in series with the load resistor \(R_L\). In each positive half-cycle the diode is forward biased, and the current flows through \(R_L\). In each negative half-cycle the diode is reverse biased, and no current flows. The load sees only the positive halves. This is a half-wave rectifier.
A half-wave rectifier is fed from the \(50\,\text{Hz}\) mains. How many pulses of current pass through the load each second?
The half-wave chip of the figure above shows it. Each output pulse is a positive half of the sine wave, one per 20 ms, so the output frequency is 50 Hz. A silicon diode takes 0.7 V of each pulse, so the peak output is \(17.0 - 0.7 = 16.3\,\text{V}\). The current through the load always flows the same way. That is what makes it direct current: "direct" means one direction, not steady.
The average of such pulses is what a DC meter would read. For half sine waves with peak \(V_p\) and gaps between them, the average is \(V_p/\pi\). Here that is \(16.3/\pi = 5.2\,\text{V}\). The output is zero for half the time, which wastes half of every cycle.
The output of a half-wave rectifier rises and falls from zero to a peak fifty times a second. Why is it still called direct current?
Direct current means current in one direction, not current of constant size. The diode lets the current through the load flow only one way, during the positive half-cycles, and blocks the other way. So the output varies in size but never reverses, which makes it direct current.
Using both halves: the centre-tap rectifier
A full-wave rectifier uses the negative halves too. One way needs a transformer winding with a connection in its middle, a centre tap. The two ends of the winding are always opposite in sign relative to the tap. So the winding acts like two equal sources in series, as the centre-tap chip of the figure draws it. Each end feeds the load through its own diode.
In one half-cycle the upper end is positive, and its diode conducts. In the next half-cycle the lower end is positive, and the other diode conducts. Both diodes feed the same end of \(R_L\), so the current through the load is in the same direction both times. Every half-cycle gives a pulse, so there are 100 pulses a second.
If each half of the winding gives 12 V rms, each pulse peaks at \(17.0 - 0.7 = 16.3\,\text{V}\). The average of a full-wave output is \(2V_p/\pi\), twice the half-wave value, because the gaps are filled. There is a cost. When one diode is off, it has the output's 16.3 V on one side and the far end's 17.0 V on the other. It must withstand about 33 V in reverse. And the winding must be twice as long, with a centre tap.
Four diodes: the bridge rectifier
The bridge rectifier needs no centre tap. Four diodes form a diamond. The winding joins two opposite corners, and the load joins the other two. In each half-cycle, two diodes on opposite arms conduct in series with the load, and the other two are reverse biased. The pairs change over each half-cycle, but both pairs send the current the same way through \(R_L\).
The bridge uses the whole winding in each half-cycle, so a 12 V rms winding gives the full 17.0 V peak. But the current passes through two diodes at a time, so two drops are lost: the peak output is \(17.0 - 1.4 = 15.6\,\text{V}\). Each off diode need withstand only about the winding's peak, 17 V. Most power supplies, including phone chargers, use a bridge.
Two numbers measure how good a rectifier is. The ripple factor compares the part of the output that still varies with the steady average: 1.21 for half-wave and 0.48 for full-wave. The efficiency is the fraction of the input power that reaches the load as steady DC. With ideal diodes it can be at most 40.5 % for half-wave and 81.1 % for full-wave. Full-wave is better on both counts.
A bridge rectifier with silicon diodes is fed from a 12 V rms, 50 Hz winding. Find the peak of the input, the peak of the output, the output frequency and the average output.
\(V_{rms} = 12\,\text{V}\), \(f = 50\,\text{Hz}\). Two silicon diodes conduct at a time.
Input peak \(= \sqrt{2} \times 12\). Output peak \(= 17.0 - 2 \times 0.7\). Average \(= 2V_p/\pi\).
Input peak \(= 17.0\,\text{V}\). Output peak \(= 15.6\,\text{V}\). Output frequency \(= 2 \times 50 = 100\,\text{Hz}\). Average \(= 2 \times 15.6/\pi = 9.9\,\text{V}\).
The load sees 100 pulses a second, each peaking at 15.6 V, and a DC meter across it reads about 9.9 V.
A bridge rectifier with silicon diodes is fed from a \(9\,\text{V}\) rms winding. What is the peak output voltage, in volts? Take \(\sqrt{2} = 1.414\).
In a working bridge rectifier, how many diodes conduct at any moment during a half-cycle?
A full-wave rectifier gives pulses of peak \(20\,\text{V}\). What does a DC voltmeter across the load read, in volts? Take \(\pi = 3.14\).
Smoothing the pulses
A phone needs a steady voltage, not pulses. The pulses are smoothed by a capacitor \(C\) across the load. A capacitor stores charge on two plates, and its voltage is proportional to that charge. Connected across a resistor, it discharges through it gradually. It loses most of its voltage in a time of about \(R_L C\).
Near each peak the diodes conduct and charge the capacitor almost to the peak voltage. As the input falls, the capacitor's voltage is higher than the input, so the diodes switch off. The capacitor alone now feeds the load and loses voltage slowly. At the next peak the diodes switch on again for a moment and top it up. The dips between the pulses are filled in, and what remains is a small ripple.
The last chip of the rectifier figure shows it with \(R_L = 100\,\Omega\) and \(C = 220\,\mu\text{F}\). Notice that the diodes now light up only for a short moment near each peak. The output falls from 15.6 V to 11.0 V between peaks, a ripple of about 4.5 V. It still repeats 100 times a second. The dashed curve is the output without the capacitor.
The rule for good smoothing is that \(R_L C\) must be much longer than the time between peaks. For a full-wave output that time is 10 ms. With \(1000\,\mu\text{F}\), \(R_L C = 100\,\text{ms}\), ten times longer, and the ripple is small. A larger capacitor, or a larger load resistance (a smaller load current), gives a smaller ripple. A full-wave rectifier also smooths more easily than a half-wave one, because its peaks come twice as often.
Which changes make the ripple of a filtered rectifier smaller?
The first thing the 230 V mains meets inside a phone charger is a small bridge of four diodes and a capacitor. Together they turn the mains into nearly steady DC at about 325 V, the peak of the mains. A switching circuit then chops this DC on and off tens of thousands of times a second. A tiny transformer steps the chopped voltage down, and a second diode and capacitor smooth it to about 5 V. The fast chopping lets the transformer be small, which is why a charger fits in a hand.
- Giving the output frequency of a full-wave rectifier as 50 Hz.Both halves of each cycle give a pulse, so the output repeats 100 times a second.
- Forgetting the second drop in a bridge.Two diodes conduct in series, so a bridge loses \(2 \times 0.7 = 1.4\,\text{V}\) from each peak.
- Using the rms value as the peak.A 12 V rms winding peaks at \(12\sqrt{2} = 17.0\,\text{V}\). Diode drops are subtracted from the peak.
- Thinking the filter capacitor needs the diodes to conduct all the time.With a filter, the diodes conduct only for a moment near each peak, to top the capacitor up.