Chopper Control Strategies and Load Current

The averages first, which need no waveform at all:

Set up the two intervals. With the switch on, the load sees \(V_s\) and the current tends towards \((V_s-E)/R\); with it off, the freewheeling diode clamps the terminal to zero and the current tends towards \(-E/R\):

Both targets are far outside the actual operating range — the current never gets near 80 A or \(-30\) A, because each interval is only a fiftieth of a time constant. That is exactly why the exponentials look straight.

Impose periodicity — the current at the end of the off interval must equal the current at the start of the on interval:

Two linear equations, solved simultaneously:

Matching the average found from the load equation, which confirms both. Note that the current never reverses — the minimum is 22.25 A, comfortably above zero — so this is continuous conduction.

Now the approximation. Treating each exponential as a straight line, the on-interval voltage across the inductor is \(V_s-E-I_oR = 220-60-50 = 110\) V acting for \(DT_s\):

Identical to three significant figures. The exact solution took two simultaneous equations; the approximation took one line, and the error is under 0.1%.

Why the approximation is so good. The error depends on how much of a time constant each interval occupies:

Every converter in Sets 18 to 24 switches at 100 kHz or more with load time constants in milliseconds, so \(T_s/\tau\) is under 0.01 and the straight-line model is exact for every practical purpose. This is why the rest of Part 3 never writes an exponential.

Note the ripple's dependence on duty ratio. The \(D(1-D)\) term peaks at \(D = 0.5\):

So the inductor must be sized at \(D = 0.5\) whatever the nominal operating point — unless the duty ratio can be guaranteed never to pass through it, which for a variable-speed drive it cannot.

PWM — fix the frequency, vary the on-time. The ripple follows \(D(1-D)\), so it is symmetric about the middle:

A worst case of 5.50 A that is known in advance and occurs at a known duty ratio. Everything about the filter, the EMI spectrum and the device losses can be designed once.

Frequency control — fix the on-time, vary the period. Now \(f_s = D/t_{on}\), so the frequency changes with the operating point:

A three-to-one variation in both frequency and ripple. The worst-case ripple of 8.25 A is twice the PWM figure at the same duty ratio, because the frequency has halved just when the ripple formula needed it most.

Hysteresis control — fix the ripple, let the frequency fall out. Switching whenever the current crosses a band of \(\Delta I = 4\) A:

The three side by side:

Why PWM won. Fixing the frequency is what makes everything else designable:

The last row matters most for a controller. A fixed-frequency converter has a well-defined transfer function that a compensator can be designed against; a variable-frequency one does not.

Where hysteresis still wins. Its one structural advantage is that it controls the current directly:

There is no loop to tune, no possibility of overshoot, and the response to a step demand is instantaneous. That is why hysteresis control survives in motor drives with fast torque demands and in current-source applications — and why almost everything else uses fixed-frequency PWM.

At \(D = 0.8\):

At \(D = 0.4\):

Torque unchanged at 60 N·m, because the current is unchanged. Speed varies with duty ratio — and unlike the cosine law of Set 12, the relationship is exactly linear, which makes the controller's job considerably easier.

Now the source side, which is where the real difference lies. The chopper draws current from the link only while the switch is on:

Compare with Set 12's phase-controlled drive. There, a drive running slower at constant torque drew exactly the same RMS supply current and merely wasted more of it as reactive power:

That last row is the point of Part 3. The rectifier feeding this chopper runs at \(\alpha = 0\) at every speed, so its displacement factor stays at unity and its power factor stays at 0.9 — the chopper's duty ratio is invisible to the supply.

Why the coupling was broken. In a phase-controlled converter the same devices set both the output voltage and the supply current's phase. In a chopper drive the two jobs are separated:

The DC link capacitor between them is what allows the two to be independent. It absorbs the difference between a 100 Hz rectified input and a kilohertz-chopped output, so neither stage needs to know what the other is doing.

The cost. Nothing is free:

Two extra percent of loss to double the power factor at low speed, and Set 6 already showed that the self-commutating device costs less than the thyristor's commutation circuit at any frequency above 12.5 Hz. The trade was decided decades ago.

Motoring first, for reference, at 100 rad/s:

Braking reverses the current, so the machine becomes the source and the 220 V link becomes the load. The machine terminal voltage is now:

Note the sign: the resistance always opposes the current, so with the current now flowing out of the machine, the terminal voltage is below the EMF rather than above it.

The chopper must boost 130 V up to 220 V. That is a Class B (step-up) chopper, and the boost relation of Set 1 applies:

During \(D\) the switch short-circuits the machine through the armature inductance, building current; during \((1-D)\) that current is forced into the link through a diode, at 220 V. The armature inductance is doing the boosting.

The power returned:

Which class does what:

This drive needs Class C — motoring and regenerative braking in one direction of rotation. That is one leg of a bridge: two switches with anti-parallel diodes, the same half-bridge that Set 7 designed the gate drive for.

The comparison that matters. Set 14 needed a second complete converter, or a field reversal taking a second, to reach quadrant II:

And because both switches are in the same leg, the transition between motoring and braking is simply a change of which one is modulating — no dead time of milliseconds, no changeover logic, no circulating current reactor. The dual converter of Set 16 is replaced by a single half bridge.

Bipolar switching. Both diagonal pairs are driven together, so the output alternates between \(+220\) and \(-220\) V:

Note that \(D = 0.5\) gives zero output, so the full \(-220\) to \(+220\) V range maps onto \(D\) from 0 to 1. All four quadrants come for free.

The ripple, from the inductor voltage:

Unipolar switching. One leg holds while the other modulates, so the output alternates between \(+220\) and 0:

And unipolar has a second advantage. If both legs are modulated with phase-shifted carriers, the output waveform changes state twice per switching period:

The devices still switch at 2 kHz — so the switching loss is unchanged — but the load sees 4 kHz. That is a genuinely free halving of the ripple, and it is the same trick that Set 18's interleaved converter uses.

Everything together:

Look at the second-to-last row. Bipolar switching applies the full \(\pm220\) V even when the demanded output is zero — so the ripple is worst at zero output, exactly where a servo holding position needs it least. Unipolar's steps shrink with the demand.

Why bipolar survives at all. Two reasons:

And (ii) the current can reverse freely at any duty ratio, so the converter never enters discontinuous conduction and the small-signal gain never changes. For a four-quadrant servo that has to cross zero torque smoothly, those two properties can be worth the extra ripple — which is the same argument the circulating-current dual converter of Set 16 made, at a fraction of the cost.

Choose the frequency from the worst-case ripple, which occurs at \(D = 0.5\):

Note that the machine's own inductance is doing the filtering — no external reactor is needed, unlike the phase-controlled drive of Set 13 which required 50 mH. Switching 30 times faster is worth a factor of six in inductance.

Device selection. The link is 220 V and the peak current is the rated current plus half the ripple:

An IGBT at 3 kHz — comfortably inside the crossover of Set 7, where the IGBT beat the MOSFET below 4.2 kHz.

Losses at \(D = 0.8\), using \(V_{CE(sat)} = 2.0\) V, \(V_F = 1.5\) V and \(E_{sw} = 1.5\) mJ:

Efficiency:

And note the loss breakdown: 91% of it is conduction, only 9% switching. At 3 kHz the frequency is essentially free, which is why there was no reason to go lower and accept a larger ripple.

The critical current, where conduction becomes discontinuous:

A factor of ten better, with a sixth of the inductance and no external reactor at all. The phase-controlled drive spent much of its operating range in discontinuous conduction with a changing loop gain; this one is continuous down to 4.6% of rated torque and stays linear throughout.

The design, assembled: