Learn › Topic 11

Diodes, Rectifiers, Capacitor Filters and the Zener Regulator

8 min read · bitGateLab Team

Almost all electronic devices run on direct current, but the power grid delivers alternating current. Getting from one to the other starts with a very simple component: the diode. In this topic you'll see how to model a diode for quick calculations, how a half-wave rectifier and a full-wave bridge rectifier are built, how a capacitor smooths the result and how much ripple remains, and how a Zener diode holds a fixed voltage. By the end you will be able to calculate a basic power supply step by step.

The diode and its models

A diode has two terminals, the anode and the cathode, and lets current flow mainly in one direction. With the anode more positive than the cathode it is forward biased and conducts; the other way around it is reverse biased and only a negligible leakage current flows. Two simple models are used for circuit calculations:

Ideal model: when forward biased it is a closed switch (0 V drop), and when reverse biased an open switch. It is useful for understanding how a circuit works.
Constant-voltage-drop model: when forward biased it has a fixed drop V_D, about 0.7 V for silicon diodes, and when reverse biased it is an open circuit. It gives results close to the real ones when the voltages in the circuit are a few volts.

The actual curve is exponential: the drop grows slightly with current, but for most power supply calculations the 0.7 V model is good enough.

Example 1. A 5 V source drives a forward-biased silicon diode in series with 1 kΩ.

Ideal model: I = 5 V/1 kΩ = 5 mA.
Constant drop: I = (5 − 0.7) V/1 kΩ = 4.3 mA.

The difference is 14 %. With a 1.5 V source the difference would be much larger, and there the ideal model is no longer good enough.

Half-wave rectifier

A single diode in series between the transformer and the load lets the positive half-cycles through and blocks the negative ones. The output is a series of "humps" separated by intervals at zero, with the same frequency as the mains.

Output peak: V_p,out = V_p − 0.7 V.
Average value without a filter: V_p,out/π.
Peak inverse voltage the diode must withstand (PIV): V_p without a filter, and up to about 2·V_p if there is a filter capacitor, because the capacitor stays charged while the secondary reaches its negative peak.

Full-wave bridge rectifier

The bridge uses four diodes. Two of them conduct during the positive half-cycle and the other two during the negative one, but both half-cycles drive current through the load in the same direction. The result is that the negative half-cycles are "flipped" upward.

At any moment two diodes are conducting in series, so 2 · 0.7 = 1.4 V is lost: V_p,out = V_p − 1.4 V.
The output frequency is twice the mains frequency: with a 60 Hz input, the output pulses at 120 Hz.
Average value without a filter: 2·V_p,out/π, twice that of the half-wave rectifier.
Each diode has to withstand a reverse voltage of approximately V_p.

Two cycles of the input sinusoid (top), the output of the half-wave rectifier (middle) and that of the bridge (bottom), drawn with ideal diodes. With the diode drops, the output peaks end up 0.7 V or 1.4 V below the input peak.

Example 2. A transformer delivers 12 V RMS at 60 Hz on its secondary.

Secondary peak: V_p = 12 · √2 = 16.97 V.
Half-wave: output peak 16.97 − 0.7 = 16.27 V; average 16.27/π = 5.18 V; frequency 60 Hz.
Bridge: output peak 16.97 − 1.4 = 15.57 V; average 2 · 15.57/π = 9.91 V; frequency 120 Hz.

Capacitor filter and ripple voltage

A capacitor in parallel with the load charges up to the peak on each hump and, while the rectified voltage falls, supplies current to the load and discharges a little. The output is almost constant, with a small oscillation called ripple. If the ripple is small compared with the peak, it can be estimated like this:

V_r(pp) ≈ I_load / (f_r · C)

where f_r is the frequency of the humps: 60 Hz for a half-wave rectifier and 120 Hz for a bridge (with 60 Hz mains). The DC output voltage ends up at about V_p,out − V_r/2.

Example 3. The bridge from the previous example supplies a 100 mA load with a 2200 µF capacitor.

Ripple with the bridge: V_r = 0.1/(120 · 0.0022) = 0.38 V peak to peak.
Approximate DC voltage: 15.57 − 0.38/2 = 15.38 V.
With a half-wave rectifier and the same capacitor the ripple would be twice as large: 0.1/(60 · 0.0022) = 0.76 V.

Reverse design: for a bridge that delivers 500 mA with at most 1 V of ripple, C ≥ 0.5/(120 · 1) = 4167 µF; you choose the next standard value up, 4700 µF.

The capacitor recharges in brief pulses near each peak, so the diodes carry peak currents much larger than the average load current. Keep this in mind when choosing the diodes and the transformer.

The Zener diode as a regulator

A Zener diode is designed to operate in reverse bias: when it reaches its breakdown voltage V_Z, it conducts as much as necessary to keep that voltage almost constant. With a series resistor R_S and the load in parallel with the Zener you get a simple regulator. The resistor takes up the voltage difference, and the current through it splits between the Zener and the load:

R_S = (V_in − V_Z) / (I_L + I_Z)

The Zener needs a minimum current I_Z to regulate well. When the load draws less, the Zener takes the leftover current, so it must withstand the worst-case power: no load.

Example 4. You want 5.1 V for a 20 mA load from 12 V, leaving at least 5 mA for the Zener.

R_S = (12 − 5.1)/(0.020 + 0.005) = 6.9/0.025 = 276 Ω. You choose 270 Ω.
Current in R_S: 6.9/270 = 25.6 mA. That leaves 25.6 − 20 = 5.6 mA for the Zener.
Power in R_S: 6.9 · 0.0256 = 0.18 W; a 1/2 W resistor gives some margin.
Worst case with no load: the Zener takes the full 25.6 mA and dissipates 5.1 · 0.0256 = 0.13 W.
If the input dropped to 10 V, the current through R_S would be 4.9/270 = 18.1 mA, less than the 20 mA the load needs: the regulator would stop regulating. The input ripple must keep V_in above that limit.

Common mistakes

Forgetting that in the bridge the current passes through two diodes, for a total drop of 1.4 V.
Using the mains frequency instead of 120 Hz when calculating the ripple of a bridge.
Confusing the transformer's RMS value with its peak value: a 12 V secondary gives a peak of almost 17 V.
Choosing a filter capacitor with a voltage rating barely equal to the peak; leave some margin.
Calculating the Zener's power only with the load connected. The worst case is no load.

Try it in the simulator

Build the complete power supply in stages and look at each waveform on the virtual oscilloscope.

Place a sine-wave source with 16.97 V amplitude and 60 Hz to stand in for the transformer secondary.
Connect a diode and a 150 Ω load resistor, and look at the half-wave output.
Replace the diode with a bridge of four diodes and compare the waveform and its frequency.
Add a 2200 µF capacitor in parallel with the load and measure the peak-to-peak ripple.
Change the capacitor or the load and check that the ripple follows the formula I/(f_r·C).
Add a series resistor and a 5.1 V Zener with its load, and watch the ripple practically disappear.

Exercises

A 9 V RMS secondary feeds a half-wave rectifier with a silicon diode. What is the output peak?

Show answer
V_p = 9 · √2 = 12.73 V. Output peak = 12.73 − 0.7 = 12.03 V.
With the same secondary and a bridge, what are the output peak and the frequency of the humps with 60 Hz mains?

Show answer
12.73 − 1.4 = 11.33 V, with a frequency of 120 Hz.
A bridge supplies 50 mA with a 1000 µF capacitor from 60 Hz mains. Estimate the ripple.

Show answer
V_r = 0.05/(120 · 0.001) = 0.42 V peak to peak.
Design R_S for a 9.1 V Zener regulator fed from 15 V, with a 30 mA load and a minimum of 10 mA in the Zener.

Show answer
R_S = (15 − 9.1)/(0.030 + 0.010) = 5.9/0.04 = 147.5 Ω. With the standard value of 150 Ω, 5.9/150 = 39.3 mA flows and the Zener is left with 9.3 mA, enough if its actual minimum is lower. If you need to guarantee the 10 mA, use the next standard value down, 130 Ω: 45.4 mA flows and the Zener is left with 15.4 mA.