Rectifiers and filters

The charger on your desk

A phone runs on direct current, but the socket in the wall gives alternating current. Inside every charger, a few diodes and a capacitor turn one into the other. The diode lets the current go one way only, and that is exactly what we need.

The half-wave rectifier

The simplest circuit has a transformer, one diode and a load resistor $R_L$. The transformer gives $v = V_m \sin \omega t$.

• In the positive half cycle, the diode is forward biased and conducts. The load gets the voltage.
• In the negative half cycle, the diode is reverse biased. No current flows, and the load gets no voltage.

The output is a train of positive humps, separated by gaps. Its mean value is not zero. We call it the direct (dc) part. For an ideal diode:

To see this, average one hump over the whole cycle. The mean of $\sin \omega t$ over a half cycle is $2/\pi$, and half of the cycle is empty. So the mean is $\dfrac{1}{\pi}$. For the rms value, the mean of $\sin^2 \omega t$ over a half cycle is $\tfrac{1}{2}$, and half of the cycle is empty, so the mean square is $V_m^2/4$.

The output has an alternating part too. The ripple factor measures it:

A ripple factor of $1.21$ is bad: the alternating part is larger than the direct part. The efficiency is the ratio of the direct power in the load to the power taken from the transformer:

Two more facts. The ripple frequency is the frequency of the humps, which is $f$ for a half-wave rectifier. The peak inverse voltage (PIV) is the largest reverse voltage across the diode, which is $V_m$ here, because the transformer voltage is all across the diode when the diode is off.

The full-wave rectifier

A full-wave rectifier uses both halves of each cycle. There are two circuits.

• The centre-tap rectifier. The transformer secondary has a tap at its middle. Two diodes join the two ends to the load. Each half of the secondary gives a voltage of peak $V_m$. One diode conducts in each half cycle. The PIV is $2V_m$.
• The bridge rectifier. Four diodes make a square. In one half cycle two of them conduct, and in the other half cycle the other two conduct. The current through the load always has the same direction. The PIV is $V_m$.

For an ideal full-wave rectifier, with $V_m$ the peak of the output:

The ripple frequency is $2f$. A bridge needs no centre tap, but two diodes conduct together, so a silicon bridge loses $2 \times 0.7 = 1.4$ V. The peak of its output is $V_m - 1.4$ V.

A worked example. A transformer gives $9.0$ V rms to a half-wave rectifier with an ideal diode. The load is $220\ \Omega$. Find $V_{dc}$, $I_{dc}$ and the efficiency.

1. The peak. $V_m = \sqrt{2} \times 9.0\ \text{V} = 12.73$ V.
2. The direct voltage. $V_{dc} = \dfrac{V_m}{\pi} = \dfrac{12.73\ \text{V}}{\pi} = 4.05$ V.
3. The direct current. $I_{dc} = \dfrac{V_{dc}}{R_L} = \dfrac{4.05\ \text{V}}{220\ \Omega} = 18.4$ mA.
4. The efficiency. $V_{rms} = \dfrac{V_m}{2} = 6.36$ V, so $\eta = \dfrac{V_{dc}^2}{V_{rms}^2} = \dfrac{(4.05\ \text{V})^2}{(6.36\ \text{V})^2} = 40.6\ \%$.

Smoothing with a capacitor

We do not want the humps. We want a steady voltage. We join a capacitor $C$ across the load. It is a filter.

• When the rectified voltage rises above the capacitor voltage, the diodes conduct and charge the capacitor to the peak $V_p$.
• When the rectified voltage falls, the diodes switch off. The capacitor then supplies the load by itself, and its voltage falls slowly.
• The next peak charges it again.

If the load current $I$ is nearly constant, the capacitor loses a charge $I\,\Delta t$ in the time between peaks $\Delta t = 1/f_r$. The voltage falls by

This is the peak-to-peak ripple. The mean output is about $V_{dc} = V_p - \tfrac{\Delta V}{2}$. The ripple has the shape of a saw-tooth, with an rms value of $\dfrac{\Delta V}{2\sqrt{3}}$, so the ripple factor is

A larger $C$ or a larger $R_L$ (a lighter load) gives a smaller ripple. A full-wave circuit has twice the ripple frequency of a half-wave circuit, so it needs half the capacitance for the same ripple. Designers also use a coil in series, or a second capacitor, to smooth the output further.

A worked example. A bridge on $50$ Hz mains has a peak output of $12$ V. The load takes $80$ mA. A capacitor of $1000\ \mu$F is across the load. Find the ripple and the ripple factor.

1. The ripple frequency. $f_r = 2 \times 50\ \text{Hz} = 100$ Hz.
2. The ripple. $\Delta V = \dfrac{I}{f_r C} = \dfrac{0.080\ \text{A}}{100\ \text{Hz} \times 1000 \times 10^{-6}\ \text{F}} = 0.80$ V.
3. The mean output. $V_{dc} = 12\ \text{V} - 0.40\ \text{V} = 11.6$ V.
4. The ripple factor. $r = \dfrac{0.80\ \text{V}}{2\sqrt{3} \times 11.6\ \text{V}} = 0.0199$, which is $1.99\ \%$.

Make the capacitor as small as it can go. Then make it large. At what capacitance does the ripple look small to you?

Practise with new numbers

Try it in Python

The first program makes a sine wave of peak $17$ V and rectifies it with an ideal diode, in the half-wave way and in the full-wave way. It finds the mean, the rms value, the ripple factor and the efficiency by averaging over many samples.

The second program steps a bridge rectifier with a capacitor through time. The diodes conduct when the source is above the capacitor voltage. Otherwise the capacitor discharges through the load. It compares the ripple that it finds with the formula $I/(f_r C)$.

Common mistakes

• Using the rms voltage as the peak. The peak is $\sqrt{2}$ times the rms voltage of the transformer.
• Using $V_m/\pi$ for a full-wave rectifier. For a full-wave rectifier it is $2V_m/\pi$.
• Forgetting the diode drops of a bridge. Two diodes conduct together, so a silicon bridge takes off $1.4$ V.
• Using the mains frequency in the ripple formula of a full-wave rectifier. The ripple frequency is $2f$.
• Thinking that a capacitor filter makes direct current out of nothing. It stores charge and gives it back to the load between peaks.
• Mixing up the PIV of the bridge and of the centre-tap circuit. The bridge has $V_m$. The centre-tap circuit has $2V_m$.

Remember these

• Half-wave, ideal: $V_{dc} = V_m/\pi$, $V_{rms} = V_m/2$, $r = 1.21$, $\eta = 40.6\ \%$, ripple frequency $f$, PIV $= V_m$.
• Full-wave, ideal: $V_{dc} = 2V_m/\pi$, $V_{rms} = V_m/\sqrt{2}$, $r = 0.48$, $\eta = 81.2\ \%$, ripple frequency $2f$. PIV $= 2V_m$ for the centre-tap circuit and $V_m$ for the bridge.
• A silicon bridge loses $1.4$ V, a centre-tap rectifier loses $0.7$ V.
• A capacitor filter gives $\Delta V = \dfrac{I}{f_r C}$ and $V_{dc} \approx V_p - \dfrac{\Delta V}{2}$.

References

• NCERT Physics Class 12, Semiconductor Electronics, the section on the junction diode as a rectifier.
• OpenStax University Physics Volume 3, the chapter Condensed Matter Physics, on semiconductor devices.
• Horowitz and Hill, The Art of Electronics, third edition, Chapter 1, the sections on rectifiers and on smoothing.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A half-wave rectifier gets $30$ V rms from a transformer. What is the peak inverse voltage across its diode?

2. The mains frequency is $60$ Hz. What is the ripple frequency of a half-wave and of a full-wave rectifier?

3. A bridge rectifier uses silicon diodes. The transformer has a peak voltage of $20$ V. What is the peak output?

4. Why does a capacitor filter work better with a lighter load?