Single-Phase Diode Rectifiers — R and RL Loads

Start from the peak. Every quantity in a rectifier problem is a fraction of \(V_m\):

Average output. The integral runs over the conducting half only, but is divided by the whole period:

RMS output. Square first, then average over the whole period — the half-cycle of zeros halves the mean square:

Compare with the full sine's \(V_m/\sqrt2 = 230\) V. Removing half the cycle divides the mean square by two, hence the RMS by \(\sqrt2\) — the same \(\sqrt{D}\) scaling as Set 1.

Form and ripple factor:

Both are pure numbers, independent of the supply voltage and the load — they are properties of the waveform shape. A ripple factor above unity means the AC content exceeds the DC content, so this is a very poor DC supply.

Rectification efficiency — the fraction of the delivered power that is genuinely DC:

The famous 40.5%, and note what it is not: no energy is lost here at all, since the diode is ideal and the load is a resistor that receives all 1322 W. The 59.5% is delivered as ripple rather than as DC — useful for a heater, useless for anything needing a steady voltage.

Transformer utilisation factor compares the DC output against the VA the transformer must be rated for:

A transformer must be rated at 3.5 times the DC power delivered. Worse, the secondary carries a DC component that saturates the core — which is the practical reason half-wave rectifiers are not used above a few watts, quite apart from the ripple.

Both half-cycles now reach the load, so the integral covers the whole period:

The RMS output equals the supply RMS exactly, because rectification does not change the magnitude of anything — it only folds the negative half up. Squaring is blind to sign, so the mean square is untouched.

Form and ripple factor:

The ripple factor falls from 1.211 to 0.483 — a factor of 2.5 — purely because the gaps have been filled. The ripple frequency also doubles to 100 Hz, which makes any subsequent filter smaller for a second, independent reason.

Rectification efficiency:

Exactly twice the half-wave figure, and equal to \(8/\pi^2\). This is the theoretical ceiling for an unfiltered full-wave rectifier.

Each diode carries half the job. With four diodes conducting in opposite pairs, each sees the load current for one half cycle:

Note the PIV: for a bridge it is \(V_m\), because the two conducting diodes clamp the blocked pair to the supply. For a centre-tapped rectifier it is \(2V_m\), and confusing the two doubles the device cost.

The comparison, side by side:

Four diodes instead of one, for double the voltage, half the ripple, twice the ripple frequency and no core saturation. Since Part 1 showed a diode costs a couple of watts and a transformer costs kilograms, this trade is never close.

The load impedance:

The current cannot stop at 180°. An inductor's current is continuous, so when the supply reverses, the stored energy keeps the diode conducting and drives current against the negative supply. The current is the sum of a steady-state sinusoid and a decaying transient:

The exponential term exists precisely because the current must start from zero at \(\theta = 0\), where the steady-state term alone would give \(-\sin\phi\).

The extinction angle is where that expression returns to zero:

This is transcendental — there is no closed form. Solving numerically with \(\phi = 0.561\) and \(\tan\phi = 0.6283\):

So the diode conducts for 212° instead of 180° — an extra 32°, all of it into the negative half cycle. Note that \(\beta\) depends only on \(\phi\), not on the supply voltage: a more inductive load extends conduction, a larger supply does not.

The average output voltage, integrating the supply over the whole conduction window including the negative part:

Why the average fell. With a resistive load the same circuit would give \(V_m/\pi = 103.5\) V. The inductance has cost 7.9 V, or 7.6%:

The extra 32° of conduction happens while the supply is negative, so it contributes negative volt-seconds to the average. The inductor is returning stored energy to the source. The load current is smoother than the resistive case — which is what the inductance was for — but the mean voltage is lower.

The RMS current, which is what actually rates the diode, must be found by integrating the current expression numerically:

Against an average of 9.56 A, a form factor of 1.51 — still peaky, because 20 mH is nowhere near enough to smooth a 50 Hz waveform. Genuine smoothing needs \(\omega L \gg R\), i.e. hundreds of millihenries here.

What the freewheeling diode does. The moment the supply tries to go negative, the output node would fall below zero — but that forward-biases the freewheeling diode, which clamps the load to 0 V and offers the inductor current a path that does not involve the source.

The inductor still gets to run its current down gradually; it just does so in a local loop rather than by pushing against the supply.

The output voltage is now the resistive-load waveform, whatever the inductance:

The result is independent of \(L\) entirely, which is the striking part. The inductance now affects only the smoothness of the current, not the average voltage — the two properties have been decoupled by one diode.

Device currents with large inductance. If \(\omega L \gg R\) the load current is essentially constant at 10.35 A, and the two diodes simply take turns:

Both diodes need the same rating, and it is set by the RMS, not the average — the Set 2 lesson reappearing. Note also that the freewheeling diode carries current for half of every cycle even though it delivers no power, so it needs a real heatsink.

What changed and what did not:

The last row is the mechanism behind all the others. Without the freewheeling path the inductor's stored energy has nowhere to go but back into the supply; with it, that energy circulates through the load and is delivered rather than returned.

The output side is simple:

The input current is a square wave. Since the load current is constant and the bridge simply reverses which pair of diodes carries it, the supply sees \(+20\) A for one half cycle and \(-20\) A for the next:

Its fundamental comes straight from the Fourier series of a square wave, whose fundamental amplitude is \(4/\pi\) times the level:

Distortion follows from the ratio of the two:

A pure number set by the waveform shape alone. Note also that \(48.3\%\) is numerically identical to the bridge's voltage ripple factor from Problem 2 — not a coincidence, since both measure the non-fundamental content of a square-edged waveform.

Displacement, and why it is unity here:

The current square wave is centred on the voltage sine, because diodes commutate at the zero crossings and have no way to delay. Everything degrading the power factor here is distortion, none of it is displacement — which is exactly the distinction that will separate diode rectifiers from the controlled converters of Set 12.

The power factor, and a check:

A power factor of 0.9 with a completely resistive-looking load, caused entirely by harmonics that carry no average power at all. The 5th and 7th harmonics circulate, heat the supply transformer and distort the voltage at the point of common coupling — and they are the reason standards exist for this.

The ideal answer first, so we can see what the drops cost:

Notice the sensitivity: the numerator is a small difference between two large numbers. A 1% error in \(V_{dc}\) is 2 V, which is 7.6% of the 27 V driving the current. Any approximation in the rectifier model is amplified enormously by the back EMF.

Now include the diodes. Two are always in the conduction path, and their drop depends on the current, which depends on the drop:

Iterating from the ideal value:

The ideal calculation overstated the current by 8.8%, from a rectifier drop of only 2.2 V out of 207 — a 1.1% error in voltage amplified eightfold. This is why a motor drive must never be analysed with an ideal rectifier model.

Diode currents and loss. Each conducts for half the cycle at the constant armature current:

Cross-check the loss against the voltage drop, which must give the same answer:

Two independent routes agreeing — per-device summation and terminal drop times current — confirms both the iteration and the current split.

Where the power goes:

The rectifier costs 1.1% and the armature resistance costs 12%. Which is worth noticing: the power electronics is nearly free compared with the machine it feeds, and the 27 V across \(R_a\) that sets the current is also 1.24 kW of pure heat.