Semiconverters, Freewheeling and Discontinuous Conduction

The output. Conduction from the supply runs from \(\alpha\) to \(\pi\); beyond that the freewheeling diode clamps the load to zero rather than letting the output reverse:

Account for all three intervals. Each half cycle contains 135° of supply conduction and 45° of freewheeling, and every ampere has to be assigned to a device:

Check the averages sum correctly: two thyristors at 7.5 plus the freewheeling diode at 5.0 gives 20 A &checkmark. And note that all three device currents depend on \(\alpha\) — unlike the full converter, where they did not. As \(\alpha\) rises the freewheeling diode takes over more of the duty, so it must be rated for the largest firing angle used.

The supply carries current only during the 135° window:

The two factors and the power factor:

Against 0.636 for a full converter at the same firing angle. Note that the distortion factor has also improved — from 0.900 to 0.960 — because the gap in the current waveform makes it closer to a sinusoid, exactly as the 120° blocks of Set 11 did.

Where the improvement comes from. Both factors gained, for related reasons:

The freewheeling interval removes the tail of the current block rather than shifting the whole thing, so the fundamental's centroid moves by only half the firing angle. That is the entire mechanism — and the reason it works is the same reason the converter cannot invert.

Set up the periodic steady state. In continuous conduction the current at the start of one half cycle must equal the current at the start of the next, since successive half cycles are identical apart from sign:

The general solution over one half cycle is the forced sinusoid, the forced DC term and a transient:

Impose periodicity. Writing \(A = (V_m/Z)\sin(\alpha-\phi)\) and \(k = e^{-\pi/\tan\phi}\), and using \(\sin(\alpha+\pi-\phi) = -\sin(\alpha-\phi)\):

Conduction is continuous if and only if this is non-negative. If it comes out negative, the current would have to reverse to satisfy periodicity — which thyristors forbid — so the current instead reaches zero early and the assumption fails.

Evaluate at \(L = 10\) mH:

Substitute:

Negative, so 10 mH is not enough — the converter is discontinuous, narrowly. Note how close it is: a small change in \(L\), \(E\) or \(\alpha\) flips the mode, and with it the formula that applies.

Find the critical inductance by solving \(I_{min}(L) = 0\). Every term depends on \(L\) through \(Z\), \(\phi\) and \(k\), so this is transcendental and must be solved numerically:

So 10 mH misses by 4%. In practice one would specify 15 mH or more, because \(L_{crit}\) depends on the operating point — and the operating point is what a drive changes for a living.

How the boundary moves:

The second row is the one that matters for drives. A motor's back EMF rises with speed, so a lightly loaded machine running fast is in the worst position — large \(E\), small \(I_{dc}\), and therefore firmly discontinuous.

The current starts from zero. Because the previous pulse has already died, there is no initial current to carry over, and the transient constant is fixed by \(i(\alpha) = 0\):

Two transient terms now, one for each forced component. The \(E/R\) bracket is often dropped by mistake — it is what makes the back EMF suppress the current from the very first instant.

The parameters at 5 mH:

Solve \(i(\beta) = 0\) numerically for the extinction angle:

Consistent with Problem 2, which predicted this from the sign of \(I_{min}\) without solving for \(\beta\) at all. The idle interval is only 14°, so this is marginally discontinuous — and marginal cases are exactly where using the wrong formula does the most damage, because the answer still looks plausible.

The output voltage has two contributions. During conduction the load sees the supply; during the idle interval it sees its own back EMF, because no current flows and therefore no voltage appears across \(R\) or \(L\):

The second term is the one that gets forgotten. Omitting it would give 162.0 V — wrong by 5% — and would also break the current check below.

The average current, two ways. First from the voltage:

And independently, by integrating the current pulse itself:

The agreement confirms \(\beta\), \(V_{dc}\) and the idle-interval term together. In DCM this check is not a formality — it is the only thing distinguishing a correct solution from a plausible one.

Compare with the continuous-conduction formula:

Half again as much current as the cosine law predicts, from a converter that is only 14° short of continuous. The error in current is far larger than the error in voltage, because — as in Set 9, Problem 6 — the back EMF turns the converter into a difference amplifier.

The three outputs:

The semiconverter delivers half again as much as the full converter at the same firing angle — because it does not throw away the negative volt-seconds. The half-wave circuit delivers exactly half the semiconverter's output, since it uses only one half cycle.

Power factors, assembled from the two component factors:

Everything side by side:

Which quadrants each reaches. Since thyristors conduct one way, current is always positive; only the voltage sign can change:

Neither of the freewheeling circuits can produce a negative average output at any firing angle, because the freewheeling path clamps the output at zero the moment it tries. That is the whole limitation, and it comes from the same component that gives them their power factor.

Choosing between them:

The last row is Part 3's answer and it wins on every count: a diode bridge at \(PF = 0.9\) feeding a chopper, which does the varying at 20 kHz where no supply displacement is involved at all.

Start of charge. The converter must supply the EMF plus the drop across the internal resistance:

The power factor there:

End of charge, with the battery risen to 140 V:

The power factor improves as the battery charges — from 0.741 to 0.810 — because a higher terminal voltage needs less firing delay. A charger is therefore at its worst electrically when the battery is at its most depleted, which is when it will be run hardest.

Where the power goes, at the start of charge:

Charging efficiency at the battery terminals is \(2400/2600 = 92.3\%\), and the 200 W lost internally is unavoidable at this current — it is the battery's own resistance, not the converter's.

Why the semiconverter suits this job. Three reasons, all structural:

(ii) A battery charger never needs to return power to the supply, so quadrant IV is worthless; and (iii) the freewheeling path means an accidental loss of firing pulses leaves the current decaying safely to zero rather than the battery discharging back through the converter. Note the first figure: at this large firing angle the full converter would be at 0.230, so the semiconverter is worth more than a factor of three here.

Set up the boundary condition from Problem 2, but solve it for \(E\) rather than \(L\) — because in a drive the inductance is fixed and the EMF is what varies:

Note how close \(\phi\) is to 90° and how close \(k\) is to 1 — a highly inductive load barely decays between pulses, which is what makes continuous conduction possible at all.

Evaluate the sinusoidal term:

Set \(I_{min} = 0\) and solve for \(E\):

Translate into a current and a speed. At the boundary the converter is still just continuous, so the cosine law still applies:

So at this firing angle the drive stays continuous only while it is delivering more than 11.5 A — about 14 N·m of torque. Below that it goes discontinuous, and 11.5 A is a perfectly ordinary light load for a machine of this size.

What happens below the boundary. The output rises, so the speed rises with it, and the speed–torque characteristic changes shape:

The speed regulation becomes poor exactly where a user would expect it to be best — at light load. And the small-signal gain from firing angle to speed rises sharply, so a controller tuned in the continuous region can become oscillatory or unstable when the load falls away.

The remedies, in the order a designer would consider them:

A series reactor is the standard answer and is why converter-fed DC drives almost always have one. Beyond that: use a semiconverter, whose freewheeling path keeps current flowing longer; add a bleeder to guarantee a minimum load; or gain-schedule the controller against measured current so it retunes itself when the mode changes. Modern digital drives do the last, having first detected the mode by checking whether the current reaches zero.