Single-Phase Controlled Converters and Firing Angle

The conduction window. The thyristor blocks until fired at \(\alpha\), then conducts until the supply current falls to zero — which, for a resistive load, is exactly at 180°:

The RMS needs the integral of \(\sin^2\), which is where the \(\sin2\alpha\) term originates:

With \(\alpha = \pi/4\) and \(\sin90^\circ = 1\):

Currents and power. For a resistive load both follow directly:

The power follows the RMS, never the average — the Set 1 lesson. Using 3.54 A would give 313 W, an error of a factor of three.

The power factor, which for a purely resistive load has a particularly simple form:

A neat result: with a resistive load the power factor is just the fraction of the supply RMS that reaches the load. No reactive component exists anywhere, and yet the power factor is 0.674 — because the current is neither sinusoidal nor in phase.

Compare with full conduction:

Note how insensitive the output is near \(\alpha = 0\): 45° of delay costs only 9% of the power, because the sine is small near the zero crossing. Control is coarse at low angles and fine near 90° — a nonlinearity every phase-control system has to live with.

The cosine law. With a large inductance the current never breaks, so each thyristor pair conducts for a full 180° starting at \(\alpha\) — which means the output includes a negative excursion:

Half the uncontrolled output at 60°. And unlike the resistive case, the cosine law is exact and linear in \(\cos\alpha\) — which is why controlled drives are always designed around inductive loads.

Thyristor currents, unchanged from the diode bridge because each device still conducts for half of every cycle:

Independent of \(\alpha\) — the firing angle moves when each device conducts, not for how long. So the devices are rated once, for the maximum load current, regardless of the operating point.

The supply current is the same square wave, delayed by \(\alpha\):

Exactly as in Set 9 — the shape has not changed at all, so the distortion factor is identical. What has changed is where the square wave sits relative to the voltage.

The displacement factor is the new term. The current block is delayed by \(\alpha\), so its fundamental lags the voltage by \(\alpha\):

The reactive power the converter demands from the supply:

Nearly two kVAr for every kilowatt delivered. And note where it comes from: there is no inductor on the AC side and no capacitor anywhere. The reactive power is created purely by the firing delay, and it must be supplied by the network.

How the power factor varies:

At \(\alpha = 90^\circ\) the converter delivers zero average power while still drawing 30 A of RMS current — pure reactive loading, the worst possible operating point. Beyond it, the average voltage reverses and power flows the other way, which is Problem 4.

The freewheeling path changes the output. When the supply voltage tries to go negative, the two diodes conduct together and clamp the load to zero rather than letting the output reverse. So the negative excursion of Problem 2 never happens:

Fifty per cent more output than the full converter at the same firing angle — because it has stopped throwing away the negative volt-seconds. Exactly the freewheeling-diode result of Set 9, Problem 4, now under phase control.

The supply current is no longer a full square wave. During the freewheeling interval the load current circulates through the two diodes and the supply carries nothing:

Less RMS current for more delivered power — the two changes that together transform the power factor.

The fundamental and the displacement. The current block is now centred differently, so it lags by only half the firing angle:

This is the key structural difference: \(\cos(\alpha/2)\) instead of \(\cos\alpha\). At 60° that is 0.866 against 0.500 — the reactive demand is far smaller because only half the delay is seen by the fundamental.

Assemble the power factor:

The comparison, at the same firing angle:

Better on every measure but one — and that one is decisive for a drive. The freewheeling path that improves the power factor is precisely what prevents the output from ever going negative, so a semiconverter can never return power to the supply.

Choosing between them. The question is whether regeneration is required:

Which way does the current flow? A thyristor conducts in one direction only, so \(I_a\) stays positive whatever happens. For power to flow out of the machine, the converter voltage must become negative:

Note that the \(I_aR_a\) term still adds — the resistance always opposes the current, so it always requires driving voltage regardless of the direction of power flow. The converter must be 130 V negative, not 150.

The firing angle follows from the cosine law:

Past 90°, which is the boundary between rectification and inversion. Note that nothing about the hardware changed — the same four thyristors, the same connections, the same supply. Only the gate timing.

The power returned:

The machine generates 6 kW, the armature resistance eats 0.8 kW, and 5.2 kW reaches the supply. This is regenerative braking: the vehicle's kinetic energy is returned to the network rather than burned in a resistor.

Why \(\alpha\) cannot reach 180°. The outgoing thyristor needs to stay reverse-biased for longer than its turn-off time. The angle remaining after commutation is the extinction angle:

Add the commutation overlap \(\mu\) from source inductance — typically 5 to 10° — and a safety margin, and the practical limit is around \(\alpha_{max} \approx 160^\circ\). The 128.9° required here is comfortable.

What happens if that limit is exceeded is worth stating plainly, because it is the characteristic failure of line-commutated inverters:

And it is self-worsening: a momentary dip in supply voltage increases the overlap angle \(\mu\), reducing \(\gamma\) at the very moment the margin was already tight. This is why inverter-mode operation is always given a generous margin and a fast-acting protection.

The load parameters, as in Set 9:

Do not assume the cosine law. It holds only if the current is still flowing when the next pair is fired, i.e. if the conduction period reaches 180°. The current after firing at \(\alpha\) is:

The second term is the transient that forces \(i(\alpha) = 0\) — the current must start from zero because the previous pair has stopped conducting.

Find the extinction angle by setting that expression to zero:

With \(\alpha = 60^\circ\) and \(\phi = 32.14^\circ\), so \(\sin(\alpha-\phi) = \sin27.86^\circ = 0.467\), solving numerically:

The current reaches zero 28° before the next pair fires, and during that gap the load is disconnected from the supply entirely.

The actual output voltage, averaging over the conduction window only:

Verify by integrating the current directly, which is the check that makes the answer trustworthy:

Two routes — the voltage average divided by \(R\), and the direct current integral — agreeing exactly. They must, because in steady state the inductor carries no average voltage, so \(V_{dc} = I_{dc}R\) whatever the conduction mode.

Compare with the continuous-conduction formula:

The cosine law understates the output by a quarter, and in the same direction as Set 1's DCM buck: discontinuous conduction always gives a higher output than the continuous formula predicts, because the negative excursion is cut short.

How to avoid it. Three remedies, in order of practicality:

(ii) add a freewheeling diode, which converts the circuit to the semiconverter of Problem 3 and removes the negative excursion altogether; or (iii) accept it and use the correct DCM analysis, which is what a drive controller's lookup table effectively does. Note that a DC machine has a back EMF as well, which makes discontinuous conduction more likely at light load — and that is exactly when a speed controller's gain changes underneath it.

At \(\alpha = 30^\circ\):

Speed, torque and power follow from the machine equations:

Check: \(P_{dc} = (179.3)(25) = 4.48\) kW, minus \(I_a^2R_a = 313\) W of copper loss, gives 4.17 kW &checkmark. Note the torque depends only on the current, and the current is held constant — so the firing angle controls speed at constant torque.

The power factor at that point:

Now slow the machine to 500 rpm at the same 37.5 N·m, so the current stays at 25 A:

And the power factor collapses:

Look at the last column. The supply current is unchanged at 25 A while the delivered power has more than halved — so the apparent power is the same 5.75 kVA in both cases. Slowing the drive does nothing whatever to reduce its loading on the network.

Why this is the drive's defining weakness. A phase-controlled drive at constant torque draws constant RMS current at every speed:

So as the speed falls, the real power falls and the reactive power rises to fill the gap. At standstill with rated torque — a crane holding a load — \(\alpha \approx 87^\circ\), the power factor is near 0.05, and the converter draws almost pure reactive current while delivering essentially no mechanical power at all.

What replaced it:

The third row is the practical answer and the subject of Part 3: rectify once at full conduction, where the power factor is 0.9, and do the varying with a chopper on the DC side where no supply displacement is involved at all.