Source Inductance and Overlap Angle

What has to happen during commutation. The load current must transfer from one thyristor pair to the other, which means the source current must change from \(+30\) A to \(-30\) A — a change of \(2I_{dc}\) through \(L_s\):

The supply voltage supplies those volt-seconds, and while it is doing so it is short-circuited through the two conducting pairs — so the load gets nothing.

Integrating the supply over the overlap gives the standard result:

Eleven degrees is 0.63 ms — not negligible against a 10 ms half cycle. And note what \(\mu\) depends on: the current and the inductance, not the power.

The voltage drop. Averaging the lost area over the half period:

Verify with the symmetric form, which averages the supply over the actual conduction pattern rather than subtracting a correction:

Two independent routes agreeing. The second form is the more fundamental — it says the output is the mean of the voltage at the start and end of commutation — and the first is that result rearranged into a drop.

Read the loss of control range. The 12 V drop is 6.7% of the ideal output at this angle, but it is a fixed voltage, so its proportional effect grows as the output is reduced:

A drive at low speed loses nearly a quarter of its output to commutation. And since the drop is proportional to current, it also means the machine's speed sags with load in a way the cosine law does not predict — which is Problem 2.

The drop is linear in current, which is the key structural observation:

So the converter behaves exactly as an ideal source \((2V_m/\pi)\cos\alpha\) in series with 0.40 Ω. Every question about regulation reduces to that Thevenin equivalent.

Sweep the load:

The drop is exactly linear; the overlap angle is not, because the cosine is nonlinear. Doubling the current from 30 to 60 A doubles the drop but raises \(\mu\) by only 81%.

Load regulation from no load to 60 A:

Poor for a supply and significant for a drive: the machine's speed will sag by that much between no load and full load at fixed firing angle, on top of the sag from armature resistance. A speed controller has to correct for both.

Now the question that catches people out: where does the power go? Twelve volts at 30 A looks like 360 W, but the source inductance is lossless:

So the 360 W is not dissipated anywhere. It is never drawn from the supply in the first place. The commutation process makes the converter draw less real power and more reactive power — the fundamental current shifts further behind the voltage, so \(V_sI_{s1}\cos\phi_1\) falls by exactly 360 W.

Confirm through the displacement factor. The extra shift is approximately half the overlap angle:

Against \(\cos30^\circ = 0.866\) without overlap — a 6.2% reduction, matching the 6.7% voltage drop to within the accuracy of the approximation.

This is the resolution: \(R_c\) is a bookkeeping device that correctly predicts the voltage while being physically a phase shift, not a dissipation. Calling it a resistance is convenient and calling it lossy is wrong.

Two practical consequences:

The three-phase overlap relation has the same form, with the line peak as the commutating voltage:

The voltage drop carries a factor of three, because there are six commutations per cycle rather than two:

Compare the two converters:

The three-phase converter is far less affected in relative terms — but that is mostly because its inductance is ten times smaller. The structural advantage is that the drop is compared against an output nearly three times larger.

Read the scaling properly. With the same source inductance the three-phase converter would have a larger absolute drop, since the coefficient is \(3/\pi\) rather than \(2/\pi\):

More commutations per cycle means more lost volt-seconds. What saves the three-phase converter is the denominator: its output is 2.7 times larger, so the same absolute drop is a much smaller fraction. That is the correct way to state the advantage.

Where the inductance comes from. It is rarely a component; it is the supply itself:

A transformer of 5% impedance feeding this converter would contribute roughly this much inductance. So the overlap angle is a property of how stiff the supply is, and a converter that behaves well on a strong grid can regulate poorly on a generator or at the end of a long cable — a common and confusing field problem.

What a notch is. During overlap two phases are short-circuited through the total commutating inductance. The instantaneous commutating voltage is divided across that inductance in proportion to where it sits — so the voltage at any point in the network collapses by the fraction of inductance upstream of it:

The full notch depth is the instantaneous line voltage at the commutation instant:

Note that this grows with firing angle: at \(\alpha = 90^\circ\) the full line peak is being short-circuited, which is why notching is worst at low output.

With the line reactor fitted, only a quarter of the inductance is upstream:

The notch area, which is the other half of the standard. It is the volt-seconds absorbed by the upstream inductance during the two commutations per cycle affecting each line:

Both limits satisfied with margin. Note that the area depends only on the upstream inductance and the current — not on the firing angle, and not on the reactor. The reactor helps the depth, not the area.

Now remove the line reactor:

A 50% notch, two and a half times the limit. Every other load on that busbar sees the supply voltage collapse by half for 2.4° of every 60 — enough to disrupt zero-crossing detectors, misfire other thyristor equipment, and cause audible noise in transformers.

The counter-intuitive part. Adding the line reactor increases the total commutating inductance, which makes the converter's own regulation and overlap worse:

The converter is made slightly worse so that the network is made much better. That is the whole purpose of a line reactor, and it explains why one is fitted to almost every large drive even though nothing in the converter's own performance calls for it.

The extinction angle now has two claimants. Set 14 treated \(\gamma = 180^\circ-\alpha\); with overlap the commutation itself consumes \(\mu\) of it:

Which is convenient, because \(\alpha+\mu\) is exactly the quantity the overlap equation gives directly.

Apply the overlap relation with \(\gamma_{min} = 5.4^\circ\) at 100 A:

There is also an absolute limit, independent of any turn-off time. The overlap equation requires \(\cos(\alpha+\mu) \ge -1\), since a cosine cannot go below it:

Beyond 168.1° the supply simply cannot deliver the volt-seconds needed to transfer 100 A before its polarity reverses. Commutation does not merely become risky — it becomes impossible.

Now raise the current to 300 A:

The ceiling has dropped by 8.4°. A converter operating safely at 167° with 100 A will commutation-fail at the same firing angle if the current rises to 300 — and a regenerating drive's current rises exactly when the braking demand increases.

The failure sequence, and why it is self-worsening:

Note the second row: a voltage dip is doubly bad, because it both increases the overlap and reduces the margin available. That is why a network fault elsewhere on the system is the classic cause of commutation failure in an HVDC inverter — and why those terminals monitor \(\gamma\) continuously and retard the firing angle the moment it narrows.

What failure costs. The outgoing thyristor never regains its blocking ability, so:

The DC-side inductor, which was there to smooth the current, now maintains the fault current at whatever it was. There is no natural zero crossing to clear it, so the protection must act — either a fast fuse or, in HVDC, retarding the firing to bring the converter briefly back into rectification.

The overlap angle:

Output voltage:

The displacement factor, both ways:

Overlap has cost 2.4% of power factor. Small here, but it scales with current: at 450 A the overlap would be about 18° and the penalty three times larger.

Verify the power balance, which is the check that confirms the lossless claim of Problem 2:

Against 0.844 from the factor product — agreement to 0.1%, which is as close as the \(\mu/2\) approximation allows. The real power drawn from the supply equals the real power delivered to the load; no energy has disappeared into \(R_c\).

Assemble the full accounting:

Read the last two rows together. The apparent power is unchanged because the RMS line current is set by the DC current alone; the real power fell; therefore the reactive power rose by exactly the difference. That is the whole physical content of the commutation drop.

What a controller must do about it. The drop is a load-dependent, firing-angle-independent error:

So an open-loop drive will run slower under load than its firing angle suggests, by 13.5 V here on top of the armature \(I_aR_a\) drop. A closed-loop speed controller corrects it automatically; an open-loop one needs an \(IR\)-compensation term with \(R_a+R_c\) in place of \(R_a\).