Capacitor-Filtered Rectifiers and Ripple

The circuit is now a peak detector. The capacitor charges to near \(V_m = 325.3\) V, and between peaks it alone supplies the load, discharging at a nearly constant current. The linear-discharge approximation is excellent because \(RC \gg T\):

The factor of two is the bridge's doing — the capacitor is topped up twice per cycle, so it discharges for only half as long as in a half-wave circuit.

But \(I_{dc}\) depends on \(V_{dc}\), which depends on the ripple, so the equations are coupled. Start from \(V_{dc} \approx V_m\) and iterate:

Compare with the unfiltered bridge of Set 9:

The output has risen by half, from one capacitor, because the circuit now holds the peak rather than averaging the waveform. That is also why the output voltage of a capacitor-input supply is sensitive to line voltage but almost independent of load — until the load gets large enough that the ripple becomes significant.

The ripple factor. For a roughly triangular ripple of peak-to-peak \(V_{r}\), the RMS of the AC component is \(V_r/2\sqrt3\):

Against 48.3% for the same bridge without the capacitor — a factor of nearly 16 for one component. Note the shape assumption: the discharge is exponential, not linear, but over a window this short the difference is under 1%.

Read the scaling before moving on. Every result here follows from one relation:

So doubling the capacitance halves the ripple; doubling the load current doubles it; and a 60 Hz supply gives 17% less ripple than a 50 Hz one for the same components. Equipment designed for both must be sized at 50 Hz.

When do the diodes conduct? Only while the rectified sine exceeds the capacitor voltage. Conduction begins when the rising sine reaches the minimum capacitor voltage:

The conduction window runs from there to roughly the peak:

Out of a 10 ms half period — so the diodes are off for 86% of the time. All the energy the load consumes over 10 ms must be delivered in 1.44 ms.

Now use charge, not Ohm's law. The charge removed by the load between pulses must be replaced during each pulse:

Approximating the pulse as a triangle of base \(t_{cond}\):

Twenty-one amps peak to deliver 1.54 A average — a ratio of 13.9. The diodes must be rated for this repetitive peak, not for the DC current.

The RMS current, for a triangular pulse of base \(t_{cond}\) repeating every \(T/2\):

A sine wave has a crest factor of 1.41. At 4.56 this current is heating the supply wiring, the transformer and the diodes three times harder than a sinusoid delivering the same average power.

The power factor is the consequence:

Compare Set 9's inductively loaded bridge at 0.90. The capacitor halved the power factor. And as in Set 9, the displacement is near unity — the pulse is centred on the voltage peak — so all of the shortfall is distortion.

The whole trade in one table:

The capacitance follows directly:

Exactly double, because the capacitor must ride out a full period rather than half. This is a stronger argument for the bridge than anything in Set 9: at these values the capacitor is one of the largest and most expensive components in the supply.

Now find what that costs. The tighter ripple means a shallower discharge, so the sine catches up much later:

The window has shrunk from 26° to 10° — and the charge to be delivered has gone up, because the load current is larger.

The peak current is the consequence:

Seventy-two amps peak, from a supply delivering 2 A. The diodes now need a repetitive peak rating of at least 100 A, and the crest factor has risen from 4.56 to 7.33 — purely from asking for less ripple.

The capacitor's own rating. Its RMS current is everything the diodes supply minus what goes straight to the load:

A 4000 µF capacitor rated for 9.6 A of ripple current at 100 Hz is a substantial and expensive part — and ripple current, not capacitance or voltage, is usually what sets its physical size. Electrolytic lifetime halves for every 10 °C of self-heating, so this figure decides how long the supply lasts.

The power factor has got worse too:

The trend, stated plainly:

Every improvement in output quality is paid for at the input, and the exchange rate is poor: an 85% reduction in ripple cost a 3.3× increase in peak current and a third of the power factor.

Why an inductor first changes everything. With the inductor in front, the diodes cannot deliver a spike — the inductor forbids a step change of current. Provided its current never reaches zero, the diodes conduct continuously and the input current returns to the square wave of Set 9.

The output reverts to the averaging value, because the circuit now averages the waveform rather than holding its peak:

Which is the first cost: 207 V instead of 323 V from the same supply. A choke-input supply needs a transformer with a 56% higher secondary voltage to reach the same output.

The critical inductance is the value at which the ripple current just reaches zero. For a full-wave rectifier, where the dominant ripple is the second harmonic:

A 110 mH choke carrying 2 A continuously — that is a substantial iron- cored component, weighing on the order of a kilogram. Note also that \(L_{crit} \propto R\), so a supply that must regulate down to no load needs an infinite choke. Real designs add a bleeder resistor to define a minimum load.

What the input current looks like now:

The choke-input filter is better on every electrical measure and worse on the two that usually decide: it is heavy, and it delivers a lower voltage. That is why it dominated valve-era power supplies, where the transformer was large anyway, and disappeared from consumer electronics.

The regulation difference is worth noting too. A capacitor-input supply's output sags as the load grows, because the ripple grows; a choke-input supply's output is nearly constant above the critical current:

So the choke gives better regulation as well — a third advantage it cannot cash in, because a switching regulator downstream does not care about any of them.

An empty capacitor is a short circuit. At the first instant it holds 0 V, so the entire supply peak appears across the source resistance:

Six hundred and fifty amps in a supply rated at 2 A — a factor of 325. And this happens every time the equipment is switched on, not as a fault.

The energy involved:

Of which an equal amount is dissipated in the resistance during charging — a standard result independent of \(R_s\). So 212 J passes through the diodes in a few milliseconds.

Judge it as an \(I^2t\), exactly as in Set 4. The current decays exponentially with time constant \(R_sC\):

Against a 200 A²s diode rating — the bridge fails on the first switch-on, by a factor of two.

Add an NTC thermistor of \(10\ \Omega\) cold in series, giving \(10.5\ \Omega\) total:

A factor of 21 below the rating — comfortable. Note how the \(I^2t\) scales: it is proportional to \(1/R_s\), so a 21× increase in resistance gives a 21× reduction in delivered energy, while the peak current falls only 21× as well. Both improve together, unlike almost every other trade in this set.

What it costs in normal running. An NTC's resistance falls as it self-heats, typically to about \(0.1\ \Omega\), but the RMS input current from Problem 3 is 9.77 A:

Nine and a half watts, permanently, on a 646 W supply — 1.5% of the output, dissipated in a component whose only job lasted ten milliseconds. Note also that it is the RMS input current that matters, and Problem 3's poor crest factor has made that 4.9 times the DC current.

The better answer, above a few hundred watts:

The relay closes a few hundred milliseconds after start-up, shorting out the limiter once the capacitor is charged. It also fixes the NTC's other weakness: after a brief power interruption the thermistor is still hot and still low-resistance, so it offers no protection at all on the restart — the exact moment the capacitor may be partly discharged.

Extract the fundamental from the power factor. Since the displacement factor is close to unity for a symmetric pulse centred on the peak, the power factor is essentially the distortion factor:

So of 9.77 A of RMS current, only 2.80 A does any work. The other 9.36 A of harmonic content circulates, heats everything it passes through, and delivers nothing.

Why the low-order harmonics are so large. A narrow pulse has a broad spectrum: its harmonic amplitudes stay near the fundamental until the harmonic period becomes comparable with the pulse width. With a 10° pulse, that means harmonics up to about the 18th are barely attenuated:

This is the opposite of the square wave of Set 9, whose harmonics fell as \(1/n\) from the start. Narrowing the conduction window flattens the spectrum.

Estimate the low orders at roughly 0.85, 0.6 and 0.4 of the fundamental for the 3rd, 5th and 7th — typical measured ratios for a 10° conduction angle:

Every low-order limit is exceeded, several by 50%. And the failure is structural, not marginal — no choice of capacitor value fixes it, since Problem 3 showed that a smaller capacitor improves the harmonics only by making the ripple unacceptable.

Why the standard exists. Third harmonics are zero-sequence: in a three-phase four-wire installation they do not cancel in the neutral but add:

So a building full of these supplies can carry more current in its neutral than in any phase — in a conductor traditionally sized smaller than the phases, and protected by no breaker. Overheated neutrals in office buildings were what prompted the limits in the first place.

What actually fixes it. The options, in order of effectiveness:

Active power-factor correction puts a boost converter between the bridge and the reservoir capacitor and controls its input current to follow the rectified sine. It is why nearly every mains-powered supply above 75 W contains a boost stage that appears, at first sight, to serve no purpose — and it is the subject of Set 24.

Where the threshold falls. The Class A limits are absolute currents, so a smaller supply passes them trivially:

Which is roughly why the regulation applies above 75 W with the tighter Class D limits — those are proportional to power rather than absolute, and a capacitor-input rectifier fails them at essentially any rating.