Diode Lab

Quick experiments

• Find where 0.7 V comes from. On the I-V tab, sweep the forward voltage from 0. Almost nothing happens until about 0.5 V, then current climbs very steeply. There is no actual threshold — 0.7 V is just where the exponential has become vertical enough to treat as fixed.
• Compare half-wave and full-wave ripple. Set a smoothing capacitor and switch between half-wave and bridge with the same load. The bridge ripple is roughly half, because the capacitor is topped up twice per cycle instead of once. Same part, twice the performance.
• Watch the bridge hand over. Step through a cycle on the full-wave tab. D1 and D2 light up on the positive half, D3 and D4 on the negative — and the load current flows the same way through both. That reversal is the whole trick of a bridge.
• Pay for the bridge's second diode. Compare the peak output of the bridge against the centre-tapped circuit at the same input. The bridge sits about 0.7 V lower, because its current passes through two diodes rather than one.
• Load it down and watch ripple grow. Reduce the load resistance with the capacitor fixed. The capacitor discharges further between peaks and the ripple deepens. Ripple is proportional to load current — which is why a supply that looks clean unloaded can be unusable under load.

Formula reference

Shockley diode equation

$$I = I_S\left(e^{\frac{V}{nV_T}} - 1\right)$$

The exponential behind the 0.7 V rule of thumb.

Rectified peak output

$$V_{peak} = V_{in(peak)} - n_d V_F$$

n_d is 1 for half-wave and centre-tapped, 2 for a bridge.

Ripple voltage

$$V_{ripple} \approx \frac{I_{load}}{f_{ripple} \, C}$$

More load or less capacitance means more ripple.

Ripple frequency

$$f_{half} = f_{in} \qquad f_{full} = 2 f_{in}$$

Full-wave refills twice per cycle, halving the ripple.

Average of an unsmoothed output

$$V_{avg(half)} = \frac{V_p}{\pi} \qquad V_{avg(full)} = \frac{2V_p}{\pi}$$

0.318 and 0.637 of the peak.

Peak inverse voltage

$$\text{PIV}_{bridge} = V_p \qquad \text{PIV}_{CT} = 2V_p$$

The centre-tapped circuit stresses its diodes twice as hard.

Common mistakes

• ✗Treating the forward drop as an exact constant.

  It rises with current and falls with temperature — roughly −2 mV per °C. At a few milliamps a silicon diode may drop 0.6 V; at several amps it can exceed 1 V. Use 0.7 V for estimates, and the datasheet curve when the margin is tight.

• ✗Forgetting that a bridge costs two diode drops.

  Current always passes through two diodes in series, so you lose about 1.4 V, not 0.7 V. On a low-voltage supply that can be a tenth of the output. A Schottky bridge drops around 0.3 V per diode if it matters.

• ✗Sizing the capacitor's voltage rating from the DC output.

  The reservoir capacitor charges to the peak, which is √2 times the RMS input, minus the diode drops. A 12 V RMS transformer gives about 16 V at the capacitor — a 16 V part is already marginal and a 25 V one is the sensible choice.

• ✗Ignoring peak inverse voltage in a centre-tapped rectifier.

  Each diode sees twice the peak secondary voltage when reverse-biased, not once. A part chosen on the bridge's PIV will fail in a centre-tapped circuit at the same voltage.

• ✗Assuming a bigger capacitor is always the fix for ripple.

  It works, but the charging current comes in ever-narrower, taller spikes, which stresses the diodes and the transformer and worsens power factor. Moving from half-wave to full-wave halves the ripple with no such penalty.

Frequently asked questions