Set 6 — Forced Commutation Circuit Design
A thyristor conducting DC has no natural way to stop. The gate cannot help, the supply never reverses, and the device will conduct until something external drives its anode current below the holding value and then keeps it reverse-biased long enough for the stored charge to recombine. Building that something is forced commutation, and it dominated power electronics for thirty years.
Two numbers govern every circuit in this set. The device turn-off time \(t_q\) is a property of the silicon — typically 10 to 30 µs — and the circuit turn-off time \(t_c\) is what the commutation circuit actually provides. Every design here reduces to making \(t_c\) comfortably exceed \(t_q\), usually by a factor of two, and then discovering what that costs. Problem 6 adds up the cost.
The governing inequality, with a safety factor of about two:
\[ t_c \ge 2\,t_q \]\(t_q\) comes from the data sheet; \(t_c\) comes from the circuit you build.
A capacitor discharged by a constant load current gives a linear voltage ramp, so the reverse-bias interval is:
\[ t_c = \frac{CV_s}{I_l} \]Impulse commutation — Problems 1 and 5.
A capacitor discharged through a resistance gives an exponential, and the device is reverse-biased until the voltage crosses zero:
\[ t_c = 0.693\,RC \]Complementary commutation — Problem 4.
A resonant LC pair reverses the capacitor and can deliver a current pulse larger than the load current:
\[ I_p = V_s\sqrt{\frac{C}{L}}, \qquad T = 2\pi\sqrt{LC} \]Require \(I_p > I_l\), usually by 1.5 to 2 times — Problem 2.
An underdamped series RLC self-commutates when its current returns to zero:
\[ \omega_0 = \frac{1}{\sqrt{LC}}, \quad \alpha = \frac{R}{2L}, \quad \omega_d = \omega_0\sqrt{1-\zeta^2}, \quad t_{cond} = \frac{\pi}{\omega_d} \]Commutation costs energy every cycle, because the capacitor is charged one way and then the other:
\[ P_{comm} = C V_s^2 f_s \]Linear in frequency, and it is the term that killed the technique — Problems 5 and 6.
A thyristor with a turn-off time of \(t_q = 20\ \mu\text{s}\) carries a constant load current of 50 A from a 200 V DC supply. Commutation is by an auxiliary capacitor charged to \(V_s\), which is switched across the main thyristor to reverse-bias it. The circuit repeats at 500 Hz. Find the capacitance for a safety factor of two, the resulting circuit turn-off time, the stored energy, and the commutation power.
What the capacitor actually does. At the commutating instant the charged capacitor is thrown across the conducting thyristor, applying \(-V_s\) to it. The load current, held constant by its own inductance, then discharges the capacitor at a fixed rate, so the device voltage ramps linearly back toward zero:
The device is reverse-biased for as long as that ramp stays negative, which is what gives the circuit turn-off time its name.
Set \(t_c\) and solve for \(C\):
Confirm the achieved margin:
Note the direction of the dependence on load current. Doubling \(I_l\) halves \(t_c\), so the capacitor must be sized for the maximum load current the converter will ever commutate — and a converter that commutates reliably on a light load may fail on an overload, which is exactly when it must not.
Stored energy:
And the power that implies. That energy is cycled once per commutation:
One hundred watts, spent solely on turning a switch off, at only 500 Hz. Nothing in this figure depends on the load power — a lightly loaded converter pays exactly the same, so the light-load efficiency of a force-commutated circuit is dreadful. Problem 5 works this through on a real chopper and Problem 6 draws the conclusion.
The same 200 V, 50 A circuit with \(t_q = 20\ \mu\text{s}\) is to use resonant pulse commutation, in which an \(LC\) pair rings to produce a current pulse that opposes the load current. Design for a peak resonant current 1.5 times the load current and a circuit turn-off time of \(2t_q\). Find \(C\), \(L\), the peak resonant current, and the resonant period.
Why a peak larger than the load current is required. The resonant pulse must not merely reduce the main thyristor's current but drive it through zero and hold it there. If \(I_p \le I_l\) the device current dips but never reaches zero, and commutation simply fails:
The factor 1.5 also buys the interval over which the difference \(I_p - I_l\) flows into the capacitor, which is what generates the reverse bias.
Size the capacitor from the turn-off requirement exactly as in Problem 1 — after the resonant reversal, the capacitor discharges into the constant load current:
Then \(L\) follows from the peak current requirement. The resonant peak of an \(LC\) discharge is set by the characteristic impedance:
Verify both simultaneously:
The resonant period, which sets how long the commutation sequence occupies:
Half of that — 84 µs — is the reversal itself. Add the 40 µs of turn-off time and the commutation occupies well over 100 µs of every cycle. At 1 kHz that is 12% of the period; at 5 kHz it is 60% and the converter has almost no time left to deliver power. This, rather than loss, is the first hard ceiling on switching frequency.
Compare the two schemes:
| Aspect | Impulse (Problem 1) | Resonant pulse |
|---|---|---|
| Components | \(C\) + auxiliary thyristor | \(C\), \(L\) + auxiliary thyristor |
| Peak device current | \(I_l\) | \(I_l+I_p = 125\) A |
| Commutation duration | 40 µs | > 120 µs |
| Tolerance of load variation | poor | better — \(I_p\) is fixed |
The resonant scheme is more robust because its commutating current is set by \(V_s\) and \(Z_0\) rather than by the load — but it costs an inductor, a longer sequence, and a peak device current of 125 A to switch a 50 A load.
A thyristor switches a series \(RLC\) circuit across a 100 V DC supply, with \(R = 10\ \Omega\), \(L = 5\) mH and \(C = 20\ \mu\text{F}\), the capacitor initially uncharged. Find the damping factor, the conduction time, the peak current, and the capacitor voltage when conduction ceases.
Establish that the circuit oscillates at all. Self-commutation requires the current to return naturally to zero, which happens only if the circuit is underdamped:
Underdamped, so the current is a decaying sinusoid and there will be a zero crossing. Had \(\zeta \ge 1\) the current would decay asymptotically without ever reaching zero, the thyristor would conduct indefinitely, and the class would not work at all.
The damped frequency and the conduction time:
The device conducts for exactly half a cycle of the damped oscillation, regardless of the supply voltage — a property that makes this class useful as a fixed-width current pulse generator, and useless as a variable-duty controller.
The current waveform:
The peak, which occurs before the quarter-cycle point because the exponential is already pulling the amplitude down:
Against 6.67 A for an undamped circuit — the resistance has taken 37% of the peak. Note also that the peak occurs at 416 µs, not at the 524 µs quarter-period; the waveform is skewed toward its start.
Capacitor voltage at the moment of commutation, which is what determines whether the thyristor is actually reverse-biased:
The capacitor overshoots to 135 V, i.e. 35 V above the supply. That excess is what reverse-biases the thyristor once the current tries to reverse, so it is the source of the circuit turn-off time — and it exists only because the circuit is underdamped. Lower \(\zeta\) gives more overshoot and more reverse bias; at \(\zeta\to1\) the overshoot vanishes along with the commutation.
In a complementary (Class C) circuit, two thyristors share a commutating capacitor and each turns the other off. The supply is 200 V, each load resistance is \(R = 10\ \Omega\), and \(t_q = 15\ \mu\text{s}\). Find the capacitance for \(t_c = 2t_q\), the achieved \(t_c\) with a preferred value, the peak current in the incoming thyristor, and the maximum switching frequency the recharge time permits.
The mechanism. While \(T_1\) conducts, the capacitor charges to \(V_s\) with the polarity that will oppose it. Firing \(T_2\) connects that charged capacitor directly across \(T_1\), driving its anode to \(-V_s\). The capacitor then discharges through the load resistance — an exponential, not a ramp, because the discharge path is resistive:
The zero crossing occurs when the exponential has covered half the swing, which is the familiar 0.693 time constants:
The incoming thyristor gets a rough moment. When \(T_2\) fires it must carry its own load current and the capacitor's discharge current, and at the first instant the capacitor looks like a \(V_s\) source:
Twice the steady load current, every commutation. The auxiliary path is also where the \(di/dt\) limit of Set 4 bites hardest, since the capacitor has no inductance to slow it down.
The hidden constraint: recharge. Before \(T_2\) can be commutated in its turn, the capacitor must fully reverse its charge through the other load resistance. That takes several time constants:
Because each half cycle must contain a full recharge. Fire faster than this and the capacitor starts a commutation only partly charged, so \(t_c\) falls below \(t_q\) and the circuit fails — typically as a shoot-through, at the worst possible moment.
Note the trap in the design. The turn-off time and the recharge time both scale with \(RC\):
So there is no value of \(C\) that gives a longer turn-off time without proportionally lowering the maximum frequency. A device with a shorter \(t_q\) is the only escape — which is the whole argument of Problem 6 in miniature.
An impulse-commutated chopper supplies a 40 A load from 220 V at \(D = 0.5\) and \(f_s = 400\) Hz. The main thyristor has \(t_q = 25\ \mu\text{s}\) and the commutating inductor is sized for a peak reversal current of twice the load current. Find the commutating components, the commutation loss as a fraction of output, and what happens at 2 kHz.
The capacitor, from the turn-off requirement at the full load current:
The inductor, from the reversal current requirement:
Now the commutation loss. The capacitor is charged to \(+V_s\) and then to \(-V_s\) each cycle, so the energy cycled is twice \(\tfrac12CV_s^2\):
Put it against the useful output:
A 4.4% efficiency penalty before a single conduction loss is counted, at a switching frequency of only 400 Hz — a frequency at which the chopper's output filter must be enormous and the acoustic noise is squarely in the audible band.
Raise the frequency to fix the filter, and watch what happens:
| \(f_s\) | \(P_{comm}\) | Share of 4.4 kW output | Commutation time used |
|---|---|---|---|
| 400 Hz | 194 W | 4.4% | 2.2% |
| 1 kHz | 484 W | 11.0% | 5.5% |
| 2 kHz | 968 W | 22.0% | 11.0% |
At 2 kHz nearly a quarter of the delivered power is being spent on turning the switch off. And this is a term that cannot be engineered away: it is \(CV_s^2f_s\), and \(C\) is fixed by \(t_q\) and the load current.
The only lever is \(t_q\). Since \(C \propto t_q\), the commutation loss is directly proportional to the device's turn-off time:
A device with \(t_q = 2.5\ \mu\text{s}\) instead of 25 µs would cut this tenfold. A device with no \(t_q\) at all — one that turns off from its gate — removes the term entirely, along with the capacitor, the inductor, the auxiliary thyristor and its own gate drive.
The chopper of Problem 5 — 220 V, 40 A, \(D = 0.5\) — is to be rebuilt with an IGBT. Compare:
- SCR: \(v_T = 1.5+0.005\,i\), plus the commutation loss \(CV_s^2f_s\) with \(C = 10\ \mu\text{F}\);
- IGBT: \(V_{CE(sat)} = 2.0\) V, \(E_{sw} = 4\) mJ per cycle.
Find each total loss at 400 Hz and 2 kHz, and the break-even frequency.
Conduction, both devices. Each conducts 40 A for half the period:
The thyristor wins on conduction — it always does, because its four-layer structure gives a lower drop than a transistor at the same rating. That is the entire case in its favour.
Switching, both devices. Here the structures are completely different:
A factor of 121 in energy per switching event. The IGBT dissipates a few millijoules crossing between states; the thyristor circuit must cycle half a joule of capacitor energy to achieve the same thing.
Totals at the two frequencies:
| Conduction | Switching @ 400 Hz | Total @ 400 Hz | Total @ 2 kHz | |
|---|---|---|---|---|
| SCR + commutation | 34 W | 194 W | 228 W | 1002 W |
| IGBT | 40 W | 1.6 W | 41.6 W | 48 W |
At 2 kHz the thyristor chopper dissipates twenty-one times as much. And the IGBT achieved this while losing the conduction comparison.
Find where the two are equal:
Twelve and a half hertz. Above that frequency — which is to say, in every converter anyone would build — the IGBT wins on total loss alone, before counting the capacitor, the inductor, the auxiliary thyristor and its gate drive that the thyristor version also needs.
What the loss comparison still leaves out:
| Factor | SCR chopper | IGBT chopper |
|---|---|---|
| Power components | 2 thyristors, C, L, 2 diodes | 1 IGBT, 1 diode |
| Peak device current | \(3I_l\) during commutation | \(I_l\) |
| Practical \(f_s\) | < 1 kHz — audible | 10–20 kHz — silent |
| Behaviour on overload | commutation fails → shoot-through | gate turns it off |
| Response to a short | fuse only | desaturation detect, < 10 µs |
The last row may matter more than all the watts. A force-commutated circuit becomes less able to commutate exactly as the current rises, so its failure mode under fault is to stop turning off — the one thing it exists to do.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Design inequality | \(t_c \ge 2t_q\) | Circuit must beat the device |
| Impulse turn-off time | \(t_c = \dfrac{CV_s}{I_l}\) | Constant-current discharge — Problems 1, 5 |
| Commutating capacitor | \(C = \dfrac{t_cI_l}{V_s}\) | Size at maximum load current |
| Resistive turn-off time | \(t_c = 0.693\,RC\) | Complementary class — Problem 4 |
| Recharge time | \(t_{rc} \approx 5RC = 7.2\,t_c\) | Sets \(f_{s,max}\) — Problem 4 |
| Resonant peak current | \(I_p = V_s\sqrt{C/L} = V_s/Z_0\) | Need \(I_p > I_l\) — Problem 2 |
| Characteristic impedance | \(Z_0 = \sqrt{L/C}\) | Fixes \(L\) once \(C\) is known |
| Resonant period | \(T = 2\pi\sqrt{LC}\) | Occupies part of every cycle |
| Damping factor | \(\zeta = \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\) | Must be < 1 to self-commutate |
| Damped frequency | \(\omega_d = \omega_0\sqrt{1-\zeta^2}\) | Problem 3 |
| Conduction time | \(t_{cond} = \pi/\omega_d\) | Half a damped cycle |
| Peak RLC current | \(i_{pk} = \dfrac{V_s}{\omega_dL}e^{-\alpha t}\sin\omega_dt\) | Peak at \(\tan\omega_dt = \omega_d/\alpha\) |
| Capacitor overshoot | \(v_C = V_s\left(1+e^{-\alpha\pi/\omega_d}\right)\) | Source of the reverse bias |
| Commutation loss | \(P_{comm} = CV_s^2f_s\) | Independent of load — Problems 5, 6 |
| Break-even frequency | \(f_s = \dfrac{\Delta P_{cond}}{CV_s^2-E_{sw}}\) | 12.5 Hz here — Problem 6 |
Common Mistakes
Designing \(t_c = t_q\). The data-sheet turn-off time is a typical value at a stated temperature and \(dv/dt\); a factor of two is standard and even that is not generous — Problem 1.
Sizing the commutating capacitor at the nominal load current. \(t_c\) is inversely proportional to \(I_l\), so a circuit that commutates at rated load will fail on overload — exactly when it must not — Problem 1.
Using a resonant peak current smaller than the load current. The pulse must drive the device current through zero, not merely dent it — Problem 2.
Choosing \(L\) before \(C\). The capacitor is fixed by the turn-off requirement; the inductor then follows from \(Z_0\). Reversing the order leaves \(t_c\) to chance — Problem 2.
Forgetting to check that a self-commutating circuit is underdamped. At \(\zeta \ge 1\) the current never reaches zero and the thyristor never turns off — Problem 3.
Taking the RLC current peak at the quarter period. Damping shifts it earlier: 416 against 524 µs here — Problem 3.
Ignoring the capacitor recharge time. It is 7.2 times \(t_c\) in the complementary class and sets the maximum switching frequency — Problem 4.
Sizing the incoming thyristor for the load current alone. It carries the load plus the full capacitor discharge — twice the steady current here — Problem 4.
Treating commutation loss as a small correction. It is \(CV_s^2f_s\), independent of load, and reached 22% of output at 2 kHz — Problem 5.
Comparing devices on conduction loss alone. The thyristor wins that comparison and loses the total by a factor of twenty-one at 2 kHz — Problem 6.
Six circuits, all solving the same problem, and all solving it expensively. Whether the reverse bias came from a capacitor discharged by the load, an \(LC\) ring, an underdamped load or a complementary pair, the requirement never changed: \(t_c \ge 2t_q\), paid for in stored energy, recycled every cycle at \(CV_s^2f_s\). Problem 6 put a number on the alternative, and the number was 12.5 Hz.
So the rest of this book assumes a device that turns off from its gate. That assumption is not free either — it moves the whole problem into the gate circuit. The charge that must be moved to switch a MOSFET or IGBT, the resistance through which it moves, the Miller plateau where nothing appears to happen, the false turn-on that a neighbouring device's \(dv/dt\) can induce, and the tail current that no gate can hurry: these are the new constraints, and they are what decide whether a converter switches at 20 kHz or fails at 2.
Next: Set 7 — MOSFET and IGBT Losses and Gate-Drive Design, where gate charge is turned into transition times, transition times into switching energy, and the gate resistor becomes the single knob that trades loss against \(dv/dt\), EMI and shoot-through.