Class C Chopper (Two Quadrant Chopper)

A Class A and a Class B chopper sharing one half-bridge leg. Two switches, two diodes — and a motor that drives or brakes on demand, with nothing changing but the duty ratio.

Table of Contents:

Introduction — half a drive, twice over

A Class A chopper can drive a motor but not stop it. A Class B chopper can stop a motor but not drive it. Each is half a drive, and neither is much use on its own.

The obvious fix is to fit both and switch between them with a contactor. That works, and it is horrible: you need two chopper circuits, a changeover contactor big enough for the full motor current, and a control scheme that has to decide when to throw it — during which the motor is briefly connected to nothing at all.

The Class C chopper does something much better. Lay a Class A and a Class B on top of each other and you find they share every node. Merge them and you get two switches and two diodes in a single leg — the familiar half-bridge — that needs no contactor, no changeover, and no decision. The transition from driving to braking happens on its own, mid-cycle, because of which way the current happens to be flowing.

This is the point in the chopper family where the topology stops being a teaching exercise and becomes the circuit that is actually inside real industrial drives.

What is a Class C Chopper?

A Class C chopper — also called a Type C chopper or a two-quadrant chopper — is a DC-DC converter that can move power in either direction between a fixed DC source and an R–L–E load. It is built from four devices:

  • CH1, a switch in series with the source, with D2 connected antiparallel across it;
  • CH2, a switch in shunt across the load, with D1 connected antiparallel across it.

Look at those pairings and the ancestry is obvious. CH1 and D1 are exactly the Class A chopper (series switch, freewheeling diode across the load). CH2 and D2 are exactly the Class B chopper (shunt switch, series diode into the source). The Class C circuit is the two of them superimposed — which is why textbooks describe it as a parallel combination of a Class A and a Class B, and why it is sometimes called a "two-quadrant Class A chopper".

The two switches are driven complementarily: CH1 for a fraction D of each period, CH2 for the remaining (1 − D). And here is the result that makes the whole topology work:

Vo = D · Vs   — always, whichever way the current is flowing Io = (D · Vs − E) / R   — and this may come out either sign

Read those two lines together and the behaviour falls out. The output voltage obeys the same law as a Class A chopper and is always positive. The current, however, is decided by a race between D·Vs and the machine's emf E:

ConditionIoQuadrantWhat the machine doesPower
D·Vs > EpositiveIMotors — the source pushes current inSource → load
D·Vs = Ezeroon the boundaryNeither — it just spinsAlmost none
D·Vs < EnegativeIIBrakes — the machine pushes current outLoad → source
This is the idea worth taking away from the whole page. There is no "motoring mode" and no "braking mode" in a Class C chopper. There is one circuit, one duty ratio, and one equation. Set D so that D·Vs is above the machine's emf and it accelerates; set it below and it brakes. The hardware never notices, and nothing is reconfigured — the current simply changes sign.

Two Quadrants, One Straight Line

On the voltage–current plane, a Class C chopper owns the entire upper half: vo is always positive, io is free to be either sign.

Four-quadrant voltage-current plane with both the first and second quadrants highlighted, showing that a Class C chopper covers the whole upper half plane, with a motoring point at 70 volts and plus 40 amperes, a braking point at 50 volts and minus 40 amperes, and the crossover at 60 volts and zero current.
Figure 1: A Class C chopper owns Quadrants I and II — one operating line through both

The dashed line through both quadrants is the key feature. Rearranging Io = (Vo − E)/R gives Vo = E + IoR — a straight line of slope R that crosses the voltage axis at E. Every operating point the chopper can reach lies on that one line, and changing D slides you along it. To the right of the axis you are motoring; to the left you are braking. There is no boundary to cross, no discontinuity, and no mode change — the line simply passes through zero current and keeps going.

Compare that with what the two single-quadrant pages had to say. A Class A chopper was stuck on the right-hand half of this line and could only ever coast back toward the origin. A Class B chopper was stuck on the left-hand half and could not drive at all. A Class C chopper is the whole line.

What it still cannot do is go below the axis. vo is either 0 or +Vs, so its average can never be negative — the machine can never be made to run backwards. Reversing needs negative voltage, which is Quadrant IV territory, and that is a Class D or Class E chopper.

Block Diagram

Block diagram of a Class C chopper: a DC source feeds a half-bridge leg of switch CH1 with diode D2 antiparallel and switch CH2 with diode D1 antiparallel, driving an output node clamped to either the supply voltage or zero, which feeds an R-L-E machine that can act as motor or generator. Every arrow is double headed because power flows both ways, and a gate controller drives the two switches complementarily with dead time.
Figure 2: Block diagram of a Class C chopper — note every arrow points both ways

Put this beside the Class A and Class B block diagrams and the difference is a single visual fact: every arrow here is double-headed. In Class A they all pointed at the machine; in Class B they all pointed at the battery. Here the chain is the same, the parts count is barely higher, and the energy is free to travel either way along it.

Circuit Diagram & Construction

Four power devices in a single leg, feeding the same R–L–E load the previous two pages used.

Circuit diagram of a Class C chopper: a DC source Vs feeds a half-bridge leg consisting of switch CH1 on top with diode D2 antiparallel across it and switch CH2 below with diode D1 antiparallel across it. The midpoint node A feeds an R-L-E branch representing a dc machine armature. Both diodes point upward toward the positive rail.
Figure 3: Class C chopper — power circuit (a half-bridge leg)
CH1 + D1The Class A half. CH1 in series with the source, D1 freewheeling across the load. Together they handle motoring.
CH2 + D2The Class B half. CH2 in shunt across the load, D2 in series into the source. Together they handle braking.
The pairing on the pagePhysically, the parts group differently: CH1 with D2 makes the top pair (both connect node A to Vs), and CH2 with D1 makes the bottom pair (both connect node A to 0). That is why the circuit draws as a half-bridge.
Both diodes point upCathode toward the more positive rail, in each case. This is what lets current of either polarity always find a path — and it is the whole reason there is no discontinuous conduction.
R–L–E — the loadThe same dc machine armature as the Class A and Class B pages: resistance R, inductance L, and emf E proportional to speed.

If this drawing looks familiar, it should: it is one leg of an ordinary half-bridge, the same building block used in inverters, in synchronous buck converters, and in almost every modern power stage. That is not a coincidence — it is why a Class C chopper is so cheap to build. You are not buying an exotic circuit; you are buying a standard half-bridge module and a two-channel gate driver.

Two ways to read the same four parts. Group them by function and you get a Class A (CH1+D1) plus a Class B (CH2+D2). Group them by position and you get a top pair (CH1+D2) and a bottom pair (CH2+D1). The first grouping explains where the topology came from; the second explains how it behaves. Both are worth carrying — the second is the one that makes the waveforms obvious.

Principle of Operation

The controller drives CH1 for DT and CH2 for (1 − D)T, never both together. What happens next is best understood in two steps.

Step 1 — the gate decides the voltage

Whichever pair is gated clamps node A, and it does so regardless of which way the current is flowing:

  • Top pair gated (CH1/D2): node A is tied to Vs. If the current is positive it goes through CH1; if it is negative it goes through D2. Either way, vo = Vs.
  • Bottom pair gated (CH2/D1): node A is tied to 0. If the current is positive it goes through D1; if it is negative it goes through CH2. Either way, vo = 0.

So vo is a clean two-level square wave in every circumstance, and its average is Vo = D·Vs, full stop. The current direction has no vote.

Step 2 — the current decides the device

A switch conducts one way only. When the current reverses, the switch simply cannot carry it — so its antiparallel diode takes over automatically, with no gate signal and no delay. The gate chooses the pair; the current chooses the member.

And that is the entire mechanism. Put the two steps together and you get the behaviour:

Vo = D·Vs  (set by the gates)  →  Io = (Vo − E)/R  (whatever sign that comes out)

Raise D above E/Vs and the applied voltage beats the emf, current flows into the machine and it accelerates. Lower D below E/Vs and the emf wins, current flows out, and it brakes — hard, and back into the battery. The switchover is not commanded; it just happens, in whichever cycle D·Vs crosses E.

Modes of Operation

Four devices, four modes. In the circuits below a highlighted loop shows where current actually flows, green is conducting and greyed dashed is not.

Read them as two pairs. Modes 1 and 2 are the Class A page, verbatim. Modes 3 and 4 are the Class B page, verbatim. That is not an analogy — the loops are literally the same loops.

Mode 1 — CH1 conducting (motoring, vo = Vs)

Class C chopper mode 1: switch CH1 conducts, connecting the source directly to the load. Current flows from the source through CH1, through the R-L-E load and back to the source. Output voltage equals the supply voltage and the current is positive.
Figure 4: Mode 1 — CH1 conducting, vo = Vs, io > 0

The top pair is gated and the current is positive, so CH1 carries it. The battery is bolted to the machine and the surplus voltage (Vs − E − i R) pushes the current up. The inductance stores energy and the source delivers.

vo = Vs    L (dio/dt) = Vs − E − ioR    is = io

This is Mode 1 of the Class A chopper, unchanged.

Mode 2 — D1 freewheeling (motoring, vo = 0)

Class C chopper mode 2: diode D1 freewheels, carrying the positive load current in a loop around the load without involving the source. The output voltage is zero and the current decays.
Figure 5: Mode 2 — D1 freewheeling, vo = 0, io > 0

The bottom pair is gated but the current is still positive, so CH2 cannot carry it — D1 does, automatically. The current circulates around the load, the source is cut out, and the current decays.

vo = 0    L (dio/dt) = −E − ioR    is = 0

This is Mode 2 of the Class A chopper, unchanged.

Mode 3 — CH2 conducting (braking, vo = 0)

Class C chopper mode 3: switch CH2 conducts and short-circuits the load, so the machine emf drives a reverse current in a loop through CH2 and the inductance stores energy. The output voltage is zero and the current is negative.
Figure 6: Mode 3 — CH2 conducting, vo = 0, io < 0

The bottom pair is gated and the current is now negative, so CH2 carries it. The machine is short-circuited through the switch; its emf drives the current further negative and the inductance stores energy. Nothing reaches the battery yet.

vo = 0    L (dio/dt) = −E − ioR    is = 0

This is Mode 1 of the Class B chopper, unchanged. Note the loop equation is identical to Mode 2's — of course it is, since vo = 0 in both. Modes 2 and 3 are the same electrical state; only the sign of the current, and therefore which device carries it, differs.

Mode 4 — D2 conducting (braking, vo = Vs)

Class C chopper mode 4: diode D2 conducts and the negative load current is forced up into the DC source, returning energy to the battery. The output voltage equals the supply voltage.
Figure 7: Mode 4 — D2 conducting, vo = Vs, io < 0

The top pair is gated but the current is negative, so CH1 cannot carry it — D2 does. The stored inductive energy plus the machine's emf together beat Vs, and the current is forced uphill into the battery. This is the mode that actually recovers energy.

vo = Vs    L (dio/dt) = Vs − E − ioR    is = io < 0

This is Mode 2 of the Class B chopper, unchanged — and again its loop equation matches Mode 1's, because vo = Vs in both.

There are only two electrical states, not four. vo = Vs (Modes 1 and 4) or vo = 0 (Modes 2 and 3). The four "modes" are just those two states crossed with the two possible current signs, and the sign only decides which device carries it. This is why the maths for a Class C chopper is no harder than for a Class A: there are two loop equations, exactly as before.
Gated pairioConducting devicevoSame as
Top (CH1/D2)> 0CH1VsClass A, switch on
Bottom (CH2/D1)> 0D10Class A, freewheeling
Bottom (CH2/D1)< 0CH20Class B, storing
Top (CH1/D2)< 0D2VsClass B, returning

Waveforms Explained in Detail

Same machine as the previous two pages — 100 V battery, 0.25 Ω and 1.5 mH armature, 60 V emf, 1 kHz — and the traces are computed from the exponential solutions, not sketched. Three duty ratios tell the whole story, and they are chosen deliberately:

DVo = D·Vsvs E = 60 VIoWhat happens
0.770 Vabove+40 AMotoring — and it reproduces the Class A page exactly
0.660 Vequal0.0 AThe crossover — all four devices conduct in one cycle
0.550 Vbelow−40 ABraking — and it reproduces the Class B page exactly
CH1D1CH2D2

Motoring at D = 0.7

Class C chopper waveforms while motoring at duty ratio 0.7: complementary gate pulses for CH1 and CH2, an output voltage square wave between 100 volts and zero averaging 70 volts which is above the 60 volt machine emf, a load current entirely positive between 32.9 and 46.9 amperes averaging plus 40 amperes, a device band showing only CH1 and D1 conducting, and a positive source current averaging 28 amperes meaning the battery is delivering 2804 watts.
Figure 8: Motoring — D = 0.7, Vo = 70 V > E, Io = +40 A

Look at the device band: only CH1 and D1 appear. The current never goes negative, so CH2 and D2 are never called upon — they sit there gated or reverse-biased and carry nothing at all. The circuit is behaving as a pure Class A chopper, and the numbers prove it: Imax = 46.92 A and Imin = 32.93 A are identical to the Class A page's figures, as is the 2,804 W drawn from the battery.

That is worth pausing on. Adding the second switch and second diode cost you nothing while motoring. The Class C chopper is not a compromise between two topologies; it is a Class A chopper that also happens to be able to brake.

Braking at D = 0.5

Class C chopper waveforms while braking at duty ratio 0.5: the output voltage square wave now averages 50 volts, below the 60 volt machine emf, the load current is entirely negative between minus 48.3 and minus 31.7 amperes averaging minus 40 amperes, the device band shows only D2 and CH2 conducting, and the source current is negative meaning the battery is absorbing about 1994 watts.
Figure 9: Braking — D = 0.5, Vo = 50 V < E, Io = −40 A

Nothing about the circuit has changed. The gates are still complementary, vo is still a clean square wave, and Vo is still exactly D·Vs. The only thing that changed is that 50 V is now less than the machine's 60 V emf — so the emf wins, the current comes out negative, and the device band flips to D2 and CH2.

The source current trace is the payoff: it has gone below zero. The battery is now being charged, at 1,994 W — which, again, is precisely the Class B page's result for the same machine.

Two figures, one circuit, one changed number. Figures 8 and 9 differ only in D — 0.7 against 0.5. Everything else about them, including every component, is identical. One is the Class A page and the other is the Class B page, and the chopper got from one to the other without a contactor, a mode flag, or a single line of reconfiguration logic. That is what "two-quadrant" buys you.

The crossover at D = 0.6 — all four devices in one cycle

Now set D so that D·Vs lands exactly on E: D = E/Vs = 60/100 = 0.6. The average current is zero — but the instantaneous current certainly is not.

Class C chopper waveforms at the crossover duty ratio of 0.6, where the average output voltage of 60 volts exactly equals the machine emf. The load current ripples symmetrically between minus 8 and plus 8 amperes with a zero average, crossing zero twice per period, so all four devices conduct in sequence in a single cycle: D2, then CH1, then D1, then CH2. The current axis is zoomed to plus and minus 12 amperes.
Figure 10: The crossover — D = 0.6, Vo = E, Io = 0. Note the current axis is zoomed to ±12 A. All four devices conduct in one period.

This is the figure that shows what a Class C chopper really is. The average current is zero, but the ripple is still 16.0 A peak-to-peak, so the current spends half of each cycle positive and half negative. Follow the device band round:

  • D2 — the cycle starts with current still negative while the top pair is gated, so the diode pushes it into the battery.
  • CH1 — the current crosses zero and goes positive; CH1 picks it up and the battery starts delivering.
  • D1 — the gates swap; current is still positive so D1 freewheels it.
  • CH2 — the current crosses zero again and goes negative; CH2 takes it.

Four devices, four handovers, one period, and not a single one of them was commanded. The gates only ever chose the pair; the current chose the member every time, automatically, by simply being unable to flow through a switch the wrong way.

A subtlety worth catching: zero average current is not zero loss. The average current here is 0.00 A, so the machine produces no net torque and no net power. But the RMS current is 4.62 A — the ripple is still real current, still heating the winding. The battery quietly supplies 5.3 W to cover it. This is the honest cost of running a chopper at the boundary, and it is invisible if you only ever look at averages.

It also explains why the crossing is so smooth. There is no instant at which the current is "about to reverse" and the controller must act. Around D = 0.6 the current is already reversing twice every millisecond as a matter of routine. Sliding D from 0.7 to 0.5 just shifts how much of each cycle is spent on each side of zero. Nothing discrete ever happens.

Why There Is No Discontinuous Conduction

Both previous pages had a section on discontinuous conduction, and both had to admit it broke their duty-cycle law. This page does not need one, and the reason is worth spelling out because it is one of the topology's biggest practical wins.

Discontinuous conduction happens when the current tries to reverse and nothing can carry it. In a Class A chopper the current decays toward zero during freewheeling; when it arrives, the switch is off and the diode cannot conduct backwards, so the current simply stops and sits at zero until the next pulse. That gap is what exposes E, adds a third voltage level, and wrecks Vo = D·Vs.

In a Class C chopper the current arrives at zero and just keeps going. CH2 is sitting there gated, ready to take it negative; D2 is ready on the other half. There is never a moment without a path, so there is never a gap.

At D = 0.45, same machineClass A chopperClass C chopper
ConductionDiscontinuous — current sits at 0 for part of every cycleContinuous — current flows, negatively
vo levelsVs, 0, E (three)Vs, 0 (two)
D·Vs predicts45 V45 V
Actual Vo61.1 V ✘ law broken45.0 V ✔ exact
Actual Io+4.2 A (feeble motoring)−60 A (firm braking)
ControlGain shifts, loop must copePerfectly linear

Look at the last two rows. At the same duty ratio, on the same machine, the Class A chopper is limping along motoring at 4 A with its control law broken, while the Class C is braking firmly at 60 A with Vo = D·Vs holding exactly. They are not slightly different — they are doing opposite things.

So the second switch buys two things, not one. The obvious one is Quadrant II. The less-obvious one, and arguably the more useful day to day, is that Vo = D·Vs becomes exactly true over the entire range 0 ≤ D ≤ 1, with no discontinuous region to bend the curve, shift the loop gain or surprise the controller at light load. For a drive that has to be well-behaved at every operating point, that alone can justify the extra switch.

The one caveat: this holds because both switches are gated complementarily. If you stop gating CH2 — some simple controllers do, to save a driver, or during start-up — the circuit collapses back into a plain Class A chopper, discontinuous conduction and all. The hardware alone does not give you continuous conduction; the gating does.

Dead Time — the one thing that can destroy it

CH1 and CH2 must never conduct at the same time. If they do, they form a direct short circuit from Vs straight to ground through two switches, with nothing but stray resistance to limit it. This is called shoot-through, and it destroys both devices in microseconds. It is the single most important practical fact about a Class C chopper, and it is a hazard that neither a Class A nor a Class B has, because neither has two switches in a leg.

Ideally the gates are exact complements: CH1 off at the same instant CH2 goes on. Real switches do not oblige. A power device takes finite time to turn off — often longer than it takes to turn on — so a genuinely complementary command produces a brief window where the outgoing switch has not finished turning off while the incoming one has already started conducting. That window is a short circuit.

The fix is dead time: a deliberate gap, typically a few hundred nanoseconds to a couple of microseconds, inserted between one switch turning off and the other turning on, during which both gates are low.

And here the topology rescues itself rather elegantly. During dead time the load current still has to go somewhere — and the diodes are still there. Whichever diode matches the current's direction picks it up instantly, with no gate needed. If the current is positive, D1 conducts; if negative, D2 does. The node still gets clamped to a sensible rail, and the current never notices the gap.

What dead time costs. During the gap the node is set by the current's direction rather than by your gate signal, so vo is briefly not what you asked for. That introduces a small error into Vo = D·Vs, and its sign depends on the direction of the current — which means it flips as the drive crosses from motoring into braking. This is the well-known dead-time distortion, and precision drives compensate for it explicitly. The idealised waveforms on this page assume zero dead time, which is the standard teaching simplification and is entirely reasonable when the dead time is well under 1 % of the period — 1 µs of dead time in a 1 ms period is a 0.1 % effect.

Key Formulas (with derivation)

There are only two electrical states, so there are only two loop equations — and they are exactly the ones from the Class A page. Everything follows. Throughout, io is signed (positive into the load) and τ = L/R.

Average output voltage

Vo = (1/T)[Vs·Ton + 0·Toff] = D · Vs    valid for ALL 0 ≤ D ≤ 1

No caveat, no discontinuous exception. This is the only page in the chopper series where that sentence can be written without qualification.

Average load current — and the quadrant

Vo = IoR + E  →  Io = (D·Vs − E) / R Io > 0 ⇔ D > E/Vs  (motoring)    Io < 0 ⇔ D < E/Vs  (braking)

The crossover duty is therefore just the ratio of the emf to the supply:

D0 = E / Vs = 60/100 = 0.6    — and it moves as the machine's speed changes

Peak and valley current

Solve both exponentials and force i(0) = i(T). With a = e−DT/τ and ab = e−T/τ:

Imax = (Vs/R) · (1 − e−DT/τ) / (1 − e−T/τ) − E/R Imin = (Vs/R) · (eDT/τ − 1) / (eT/τ − 1) − E/R

These are the same expressions as the Class A chopper's — but with a crucial difference in how they are read. On the Class A page, an Imin that came out negative meant the formula had broken and you had to switch to the discontinuous branch. Here, a negative Imin is simply a negative current, and the formula stays valid. Removing that restriction is exactly what the second switch bought.

Peak-to-peak ripple

Δio = (Vs/R) · (1 − a)(1 − b) / (1 − ab)  ≈  Vs · D(1 − D) · T / L

The same expression for the third time in this series — Class A, Class B and Class C all share it, because in every case the inductor is swung between the same two rails, 0 and Vs. It is independent of E, and it peaks at D = 0.5 with Δio,max = VsT/4L.

Source side and power

Is = (1/T) ∫0Ton io dt    Is ≈ D · Io  (exact only if io were ripple-free)

Is is signed, and its sign is the whole story: positive means the battery is delivering, negative means it is being charged. The power balance is exact for ideal devices, because voio and Vsis are equal at every instant (when the top pair conducts both equal Vsio; when the bottom pair conducts both are zero):

Pin = VsIs = 〈voio〉 = E·Io + Io,rms²R

Read the signs. When motoring, all three terms are positive: the battery delivers, the shaft takes, the winding heats. When braking, Pin and E·Io both go negative but Io,rms²R stays stubbornly positive — the copper loss never reverses, which is exactly why you never recover everything you put in.

Device ratings

all four devices block Vs    all four must carry |I|max    — identical to a Class A or Class B driving the same machine

Which is why a Class C chopper is built from an off-the-shelf half-bridge module: the ratings are ordinary, and the parts already come packaged in exactly this configuration.

Control & the Vo–Io Characteristic

Average output voltage and average load current plotted against duty ratio for a Class C chopper on twin axes. Both are perfectly straight lines over the whole range from zero to one, with no discontinuous region. The current crosses zero at duty 0.6 where the output voltage equals the 60 volt machine emf, separating a braking region below from a motoring region above.
Figure 11: Vo and Io against D — two straight lines, no exceptions

This is the tidiest control characteristic in the whole chopper family. Both lines are dead straight across the entire range of D, and the current line passes through zero at D0 = 0.6 without so much as a kink. Above it you are motoring, below it you are braking, and the transition is a point on a line rather than an event.

Compare that with the Class A characteristic, which bends away below D = 0.62 as it enters discontinuous conduction, and the Class B characteristic, which does the same at the other end. A Class C chopper has neither pathology.

That said, the current line is steep — slope Vs/R = 400 A per unit of duty. Moving D by 0.05 moves the current by 20 A. So, exactly as with a Class B, you do not run this open-loop on duty ratio. Real drives close a current loop: the operator's torque demand becomes a signed current reference, a PI controller measures the actual armature current and works out D for itself. Command +40 A and it motors; command −40 A and it brakes; sweep the reference smoothly through zero and the drive follows it through without noticing.

This is why Class C is the workhorse. A single signed current reference commands the whole upper half-plane, through one linear plant, with no mode logic anywhere in the controller. That property — not the extra quadrant by itself — is what makes it the default choice for real DC drives.

Worked Example

The same machine as the Class A and Class B pages: Vs = 100 V, R = 0.25 Ω, L = 1.5 mH (τ = 6 ms), E = 60 V, f = 1 kHz. One circuit, three duty ratios.

QuantityD = 0.7 (motoring)D = 0.6 (crossover)D = 0.5 (braking)
Vo = D·Vs70 V60 V50 V
vs E = 60 Vaboveequalbelow
Io = (Vo−E)/R+40.00 A0.00 A-40.00 A
Imax46.92 A+7.95 A-31.67 A
Imin32.93 A-8.04 A-48.33 A
Ripple Δio13.99 A15.99 A16.66 A
Io,rms40.20 A4.62 A40.29 A
Current crosses zero?NoYes, twice/cycleNo
Devices usedCH1, D1all fourCH2, D2
Source current Is+28.04 A+0.053 A-19.94 A
Battery power VsIs+2,804 W delivering+5.3 W-1,994 W absorbing
Shaft power E·Io+2,400 W0.0 W-2,400 W
Copper loss Io,rms²R404.1 W5.33 W405.8 W
Balance checkVsIs − E·Io − Io,rms²R = 0 in all three columns ✔
QuadrantIboundaryII

Three lines deserve attention. First, the D = 0.7 column is the Class A worked example, digit for digit — same peak, same valley, same 2,804 W. Second, the D = 0.5 column is the Class B worked example, likewise — same 1,994 W returned to the battery. One circuit reproduced both pages exactly, and the only thing that changed between the columns was a single number in the controller.

Third, look at the crossover column. Shaft power is 0.0 W — the machine is doing nothing at all — yet the battery still hands over 5.3 W, and the copper loss is 5.33 W. That is the ripple current heating the winding for no return. Small here, but it is the reason a drive parked at zero torque is not a drive drawing zero power.

The whole series, in one comparison. Same machine, same battery, throughout:
  • Class A at D = 0.7: battery delivers 2,804 W. Cannot brake.
  • Class B at D = 0.5: battery absorbs 1,994 W. Cannot motor.
  • Class C: does both of the above, from one circuit, by changing D from 0.7 to 0.5.

Advantages & Disadvantages

Advantages

  • Motoring and regenerative braking from one circuit, with no contactor, no changeover and no mode logic.
  • Seamless, bumpless transition. The current slides through zero; nothing discrete happens. Torque can be commanded continuously from full drive to full brake.
  • No discontinuous conduction. Vo = D·Vs holds exactly for every duty ratio, so the plant stays linear and the control loop stays well behaved at all loads.
  • Energy is recovered rather than burned in a braking resistor.
  • Cheap for what it does. One extra switch and one extra diode over a Class A — and it is a standard half-bridge, so it comes as an off-the-shelf module with a two-channel driver.
  • Same device ratings as a Class A or Class B for the same machine: all four block Vs and carry Imax.
  • Costs nothing while motoring — it performs identically to a Class A chopper, as Figure 8 shows.
  • Ideal for a current loop: one signed reference commands the entire upper half-plane.

Disadvantages

  • Shoot-through is now possible. Two switches in a leg means dead time is mandatory, and getting it wrong destroys both devices. Neither Class A nor Class B can fail this way.
  • Dead-time distortion puts a small error into Vo = D·Vs whose sign flips with the current's direction — precision drives must compensate for it.
  • Still one direction only. vo cannot go negative, so the machine can never run in reverse. That needs a Class D or Class E.
  • More parts and a more complex gate drive — two channels, and the high-side driver needs an isolated or bootstrapped supply.
  • The supply must absorb the returned energy, exactly as for a Class B. A battery can; a plain rectifier cannot.
  • Ripple current flows even at zero torque, so the winding is heated at every operating point.
  • Braking still fades at low speed, because E falls with speed and there is eventually nothing left to recover.

Applications

  • General industrial DC drives — the default choice wherever a motor runs one way but must accelerate and brake under control. This is the topology's home ground.
  • Traction: trams, metro stock, trolleybuses and mine locomotives, which spend their lives alternating between accelerating and braking in one direction.
  • Battery vehicles — forklifts, warehouse tugs, airport tractors: motor to move, regenerate to stop, all from one leg off the traction battery.
  • Hoists, cranes and lifts raising a load (motoring) and lowering it (regenerating), where gravity does the work on the way down.
  • Conveyors and inclined belts, which motor uphill and are driven by their own load downhill.
  • Test rigs and dynamometers needing precise bidirectional torque control through zero.
  • One leg of a Class E (four-quadrant) drive — two Class C legs make an H-bridge, so this is also the building block for full reversing drives.

The rule: if the load runs in one direction but must be actively slowed, a Class C chopper is almost certainly the right answer. It is the point at which the extra switch stops being a luxury and starts being cheaper than the alternatives.

Class C vs A, B, D and E

Class AClass BClass C
QuadrantsIIII & II
DevicesCH1 + D1CH2 + D2all four
VoD·Vs(1−D)·VsD·Vs
io+ only− onlyeither sign
Can motor?YesNoYes
Can brake?NoYesYes
Discontinuous conductionYes, at low DYes, at low DNever
Vo law exact?Only in CCMOnly in CCMAlways
Load needs its own emf?NoYesOnly to brake
Shoot-through possible?NoNoYes — needs dead time
Ripple ΔioIdentical in all three — (Vs/R)(1−a)(1−b)/(1−ab)
Device ratingsIdentical in all three — block Vs, carry Imax

And the family as a whole:

ClassQuadrantsDevicesWhat it can do
AI1 switch + 1 diodeMotoring one way. Step-down.
BII1 switch + 1 diodeRegenerative braking one way. Step-up into the source.
CI & II2 switches + 2 diodesMotor and brake in the same direction. One half-bridge leg.
DI & IV2 switches + 2 diodesAverage output voltage can go negative, so it can brake in the reverse direction.
EI, II, III & IV4 switches + 4 diodesFull four-quadrant: motor and brake in both directions. Two Class C legs — an H-bridge.

The progression is now clear. Class C = Class A + Class B in one leg. Take two Class C legs, put the machine between their midpoints, and you have an H-bridge — which is a Class E, and can do all four quadrants. Class C is not just a step in the family; it is the unit the four-quadrant drive is made of.

Frequently Asked Questions – FAQs

Because it covers the whole upper half of the voltage-current plane. The output voltage is always positive — it is either 0 or +Vs, so its average cannot be negative — but the current is free to take either sign. Positive current with positive voltage is Quadrant I (motoring); negative current with positive voltage is Quadrant II (braking). It cannot reach Quadrants III or IV, because that would need a negative output voltage, which this circuit cannot produce.

It doesn't — and that is the point. There is no switching, no mode and no reconfiguration. The gates always do the same thing: CH1 for DT, CH2 for the rest. That fixes Vo = D·Vs. The current is then whatever (Vo − E)/R happens to be. Set D above E/Vs and the current comes out positive and the machine motors; set it below and the current comes out negative and it brakes. The devices sort themselves out automatically, because a switch cannot conduct backwards and its antiparallel diode takes over when the current reverses.

Because D is defined as the duty of the TOP switch, CH1, and the top pair (CH1 or D2) clamps the node to Vs. So the node sits at Vs for a fraction D of every period and at 0 for the rest, giving D·Vs — the same convention as the Class A page. A Class B chopper's D refers to its shunt switch, which clamps the node to 0, so its duty counts the other interval and its law comes out as (1−D)·Vs. The circuits are consistent; only the naming of D differs. A Class C at D = 0.5 behaves exactly like a Class B at its own D = 0.5, because the two definitions happen to coincide at one half.

Because there is always a device that can carry current of either polarity. Discontinuous conduction happens when the current reaches zero and nothing can take it further — in a Class A chopper the switch is off and the freewheeling diode cannot conduct backwards, so the current stops and sits at zero. In a Class C, CH2 is gated and ready to take the current negative, and D2 handles the other half of the cycle, so the current simply passes through zero and keeps going. That is why Vo = D·Vs is exactly true for every duty ratio here. The catch: this depends on both switches actually being gated. Stop driving CH2 and the circuit reverts to a plain Class A, discontinuous conduction included.

Dead time is a deliberate gap — typically a few hundred nanoseconds to a couple of microseconds — inserted between one switch turning off and the other turning on, during which both gates are low. It matters because CH1 and CH2 are in series across the supply: if they ever conduct together they short the battery through two switches and both are destroyed in microseconds. That is shoot-through, and real devices turn off more slowly than they turn on, so a perfectly complementary gate command would cause it. During the dead time the load current is picked up automatically by whichever diode matches its direction, so nothing is interrupted — but the node voltage is briefly set by the current rather than by your command, which introduces a small dead-time distortion into Vo.

Yes, and it happens whenever the average current is small enough that the ripple carries it across zero — around D = E/Vs. In the worked example at D = 0.6, the average current is zero but the ripple is still about 16 A peak-to-peak, so the current is positive for part of the cycle and negative for the rest. The sequence is D2 → CH1 → D1 → CH2: the diode returns the negative current, CH1 picks it up once it goes positive, D1 freewheels it when the gates swap, and CH2 takes it once it goes negative again. Four handovers per period, none of them commanded.

Functionally yes — CH1 with D1 is the Class A and CH2 with D2 is the Class B, and the page's mode diagrams are literally the two pages' loops. But it is better than the sum of its parts in two ways. First, they share every node, so merging them costs no extra components beyond the second switch and diode, and it needs no contactor to change over. Second, the merge removes discontinuous conduction entirely, which neither parent could do alone. Physically the parts also regroup: CH1 pairs with D2 as the top pair and CH2 with D1 as the bottom pair, which is why it draws as an ordinary half-bridge.

No. The output voltage is either 0 or +Vs, so its average can never be negative, and a DC machine needs negative armature voltage to run backwards. A Class C covers Quadrants I and II — driving and braking in one direction only. Reversing needs Quadrant III or IV, which means a Class D (two quadrants including negative voltage) or a Class E (all four). A Class E is simply two Class C legs forming an H-bridge, with the machine between the midpoints.

Essentially nothing. While the current stays positive, only CH1 and D1 ever conduct — CH2 and D2 carry no current at all. The peak, valley, ripple and input power come out identical to a Class A chopper driving the same machine, as the worked example on this page shows digit for digit. What you do pay for is real but modest: a second gate driver channel with a floating high-side supply, mandatory dead time, and the risk of shoot-through if you get it wrong.

Because Io = (D·Vs − E)/R has a slope of Vs/R = 400 A per unit of duty in the worked example, so a duty change of 0.05 moves the current by 20 A. Worse, E moves with speed, so a fixed D does not even give a fixed torque. Closing a current loop fixes both: the torque demand becomes a signed current reference and a PI controller finds D for itself. Because the plant is perfectly linear across the whole range — no discontinuous region anywhere — one loop covers motoring, braking and everything in between, with no gain scheduling and no mode logic.

No. At D = E/Vs the average current is zero, so the shaft power E·Io is zero and the machine produces no net torque — but the ripple current is still there, about 16 A peak-to-peak in the worked example. That gives an RMS current of roughly 4.6 A, which still heats the winding, and the battery quietly supplies about 5 W to cover it. Averages hide it; the RMS does not. It is the reason a drive parked at zero torque is not a drive consuming zero power.

Related Topics