Class D Chopper (Two Quadrant Chopper)

Two switches gated together, two diodes on the other diagonal. The current can only flow one way — but the output voltage swings all the way from +Vs to −Vs.

Table of Contents:

Introduction — the other way to get two quadrants

A Class C chopper reaches two quadrants by keeping the voltage positive and letting the current change sign. A Class D chopper reaches two quadrants the opposite way: it keeps the current positive and lets the voltage change sign.

That one swap changes everything downstream. Class C gets Quadrants I and II — driving and braking one way. Class D gets Quadrants I and IV, which is a stranger pair: forward motoring, and braking a machine that is spinning backwards.

It is the least intuitive member of the family and, honestly, the least used on its own. But it is worth understanding properly for two reasons. It is the first chopper here whose output voltage can go negative — the thing every previous page said it could not do — and it is half of a Class E, the full four-quadrant drive that ends this series.

What is a Class D Chopper?

A Class D chopper — Type D, or two-quadrant chopper — uses two switches and two diodes arranged as an H-bridge with the switches on one diagonal and the diodes on the other:

  • CH1 top-left and CH2 bottom-right — and they are gated by one signal, together. There is no complementary pair.
  • D2 bottom-left and D1 top-right, forming the return path.
  • The R–L–E load sits between the two midpoints.

Because both switches move together, the load only ever sees the full supply, one way round or the other:

both switches ON  →  vo = +Vs    both switches OFF  →  vo = −Vs

Averaging those two levels gives the formula that defines the topology:

Vo = (1/T)[Vs·Ton − Vs·Toff] = Vs(Ton−Toff)/T = (2D − 1) · Vs
DutyVoQuadrantWhat it means
D > 0.5positiveIForward motoring — the source drives the machine
D = 0.5zeroon the axisEqual +Vs and −Vs intervals cancel
D < 0.5negativeIVReverse braking — the machine spins backwards and feeds the source

And the current? It has no choice. Trace the circuit and you will find that every path — through the switches or through the diodes — carries current from A through the load to B. io is always positive, whatever D is.

Note what Vo = (2D−1)Vs is really saying: the output can reach ±Vs, a 200 V span from a 100 V battery. Every previous chopper on this site was confined to 0…Vs. That extra reach is not free — it costs double the current ripple, which we come to below.

Quadrants I and IV — the mirror of a Class C

Voltage-current plane with the first and fourth quadrants highlighted, the right-hand half, showing that a Class D chopper keeps the current positive while the voltage takes either sign, with a motoring point at plus 60 volts and 40 amperes and a reverse-braking point at minus 60 volts and 40 amperes.
Figure 1: A Class D chopper owns the right-hand half — io ≥ 0, vo either sign

Hold this next to the Class C quadrant diagram and the symmetry is exact. Class C owns the upper half — voltage fixed positive, current free. Class D owns the right half — current fixed positive, voltage free. Two topologies, two quadrants each, rotated ninety degrees from one another.

Quadrant IV takes a moment to picture, so let us be concrete. vo is negative and io is positive, so the power VoIo is negative — the machine is generating. For current to flow forwards against a negative applied voltage, the machine's own emf must be even more negative, which means it is spinning in reverse. So Quadrant IV is: the machine is running backwards, and the chopper is braking it and returning the energy. Not forward braking (that is Quadrant II, and a Class D cannot reach it) — reverse braking.

Which is why Class D is rarely used alone. A drive that can motor forwards and brake in reverse, but cannot brake forwards or motor in reverse, is an odd combination. Class D's real job is to be understood as the voltage-reversing half of the story — and then combined with Class C's current-reversing half to make a Class E, which does all four.

Block Diagram

Block diagram of a Class D chopper: a DC source feeds an H-bridge whose switch diagonal is gated by one signal and whose diode diagonal provides the return path, applying plus or minus the supply voltage to an R-L-E machine, with double-headed arrows because power flows both ways.
Figure 2: Block diagram of a Class D chopper

One detail here is worth more than it looks: there is a single gate signal. Both switches follow it. That means a Class D chopper has no complementary pair, and therefore — unlike a Class C — no dead time and no shoot-through risk at all. The two switches are never fighting each other, because they are never in opposition. It is a genuine simplification, and one of the few places where Class D is easier than Class C.

Circuit Diagram & Construction

Circuit diagram of a Class D chopper drawn as an H-bridge: switch CH1 top-left and switch CH2 bottom-right form one diagonal, diode D2 bottom-left and diode D1 top-right form the other, and the R-L-E dc machine load sits horizontally between the two midpoints A and B.
Figure 3: Class D chopper — an H-bridge with switches on one diagonal, diodes on the other
CH1 and CH2The switch diagonal: top-left and bottom-right. Gated together by one signal. When they close, they connect Vs+ to A and B to Vs−.
D1 and D2The diode diagonal: top-right and bottom-left, both pointing up toward Vs+. They are the return path when the switches open — and they are what forces vo negative.
The loadR–L–E between midpoints A and B, drawn horizontally. vo = vA − vB, and the current can only run A → B.
Why an H-bridge?Because reversing a voltage needs both ends of the load to be reachable. Every earlier chopper had one load terminal permanently tied to a rail, which is exactly why none of them could produce a negative output.

Principle of Operation

The controller has one output and the circuit has two states. That is the whole thing.

  • Gate high — CH1 and CH2 both close. A is pulled to Vs, B to 0, so the load sees +Vs. Current flows A→B and rises; the inductance stores energy.
  • Gate low — both switches open. The inductance will not let the current stop, so it keeps pushing A→B. The only path left is out of B, up through D1 into Vs+, back out of Vs− and up through D2 into A. That path reverses the load's connection to the battery: now A sits at 0 and B sits at Vs, so the load sees −Vs.

That second state is the clever bit, and it is worth saying plainly: the load is still connected to the same battery, but backwards. Nobody flipped a switch to do it — the diodes are simply the only path available, and the path they offer happens to enter the battery at the opposite terminal. The current keeps its direction; the voltage flips.

So the load is spending DT of every period connected one way and (1 − D)T connected the other. Its average is the difference between the two, Vs(Ton − Toff)/T, and if the OFF interval is the longer of the two, the average comes out negative.

Notice what the OFF state also does. With the diodes conducting, the current is being forced into the battery against the full Vs, so the machine and the inductance are both being made to work hard against it — that is why the current falls so steeply in this mode, and it is the root of the doubled ripple.

Modes of Operation

Two devices per state, two states. Highlighted loop = where current flows, green = conducting.

Mode 1 — both switches ON (vo = +Vs)

Class D chopper mode 1: both switches CH1 and CH2 conduct, connecting the source across the load. Current flows from the source through CH1, left to right through the R-L-E load, through CH2 and back. The output voltage is plus the supply voltage and both diodes are reverse biased.
Figure 4: Mode 1 — CH1 and CH2 conducting, vo = +Vs
vo = +Vs    L (dio/dt) = Vs − E − ioR    is = +io

The battery is connected straight across the load, the surplus voltage pushes the current up, and the source delivers. Both diodes are firmly reverse biased — each has the full Vs across it the wrong way.

Mode 2 — both switches OFF (vo = −Vs)

Class D chopper mode 2: both switches are off and both diodes conduct. The load current still flows from A to B, then up through diode D1 into the positive rail, through the source and back up through diode D2 to A. This connects the load across the battery the other way round, so the output voltage is minus the supply voltage while the current stays positive and decays.
Figure 5: Mode 2 — D1 and D2 conducting, vo = −Vs, io still positive
vo = −Vs    L (dio/dt) = −Vs − E − ioR    is = −io

Follow the highlighted loop carefully — it is the key to the whole page. The current leaves the load at B, goes up through D1 into the battery's positive terminal, through the battery, and back up through D2 into A. The load's current direction has not changed. But its connection to the battery has been turned around, so vo is now −Vs, and the source current is is negative: the battery is being charged.

With −Vs − E − ioR driving it, the current falls steeply. In a Class A chopper the freewheeling current only had to fight E and R; here it is fighting the full battery voltage as well.

There is no freewheeling state in a Class D chopper. Class A had one (vo = 0 through the diode); Class C had one (the bottom pair). Class D does not — the load is always connected to the battery, one way or the other. That is the source of both its unique ability (negative output) and its main drawback (double the ripple).

Waveforms Explained in Detail

Same supply and armature as the other three chopper pages — 100 V, 0.25 Ω, 1.5 mH, 1 kHz — computed from the exponential solutions. Class D needs two machine emfs, because Quadrant IV means the machine is turning backwards:

EDVoIoQuadrant
Motoring+50 V (forward)0.8+60 V+40 AI
Reverse braking−70 V (reverse)0.2−60 V+40 AIV
CH1 + CH2D1 + D2

Quadrant I at D = 0.8

Class D chopper waveforms in the first quadrant at duty 0.8: one gate signal, an output voltage that jumps between plus 100 and minus 100 volts averaging plus 60 volts, a load current entirely positive between 29.2 and 50.5 amperes averaging plus 40 amperes with 21.3 amperes of ripple, a device band alternating between the switch pair and the diode pair, and a source current that is positive during the on time and negative during the off time, averaging plus 24 amperes.
Figure 6: Quadrant I — E = +50 V, D = 0.8, Vo = +60 V, Io = +40 A

Three things to read off this figure. First, vo never touches zero — it slams between +100 and −100 V, spending 80 % of the time high, which averages to +60 V.

Second, the current stays positive but ripples hard: 21.3 A peak-to-peak on a 40 A average. Compare that with the 14 A a Class A chopper produced on the same armature. Look at the steepness of the falling edge — that is the load being driven backwards into the battery.

Third, look at is. It is positive while the switches conduct and negative while the diodes do — the battery is delivering for part of every cycle and being charged back for the rest, even though the drive is motoring. The net is +24.1 A, so it delivers 2,409 W overall, but a Class D chopper churns energy back and forth through its source every single cycle. That is a real cost: it means more RMS current in the battery and its wiring than the net power suggests.

Quadrant IV at D = 0.2

Class D chopper waveforms in the fourth quadrant at duty 0.2 with the machine spinning in reverse at minus 70 volts of emf: the output voltage now averages minus 60 volts, the load current is still entirely positive averaging plus 40 amperes with the same 21.3 amperes of ripple, and the source current now averages minus 24 amperes, meaning the battery is absorbing about 2391 watts.
Figure 7: Quadrant IV — E = −70 V, D = 0.2, Vo = −60 V, Io = +40 A

Put Figures 6 and 7 side by side and the comparison is striking. The current traces are almost identical — same positive band, same 21.3 A ripple, same +40 A average. The voltage trace has simply inverted: now it is low for 80 % of the period instead of high, averaging −60 V.

And the source current has flipped with it — now averaging -23.9 A, so the battery absorbs 2,391 W. The machine, spinning backwards, is being braked and its energy recovered.

The ripple is identical in both quadrants — 21.33 A against 21.33 A. That is not a coincidence: as on every other page in this series, the ripple depends on the rails the inductor is switched between, not on the machine's emf. E cancels out of the subtraction entirely.

Why the Ripple Is Double

This is Class D's defining weakness and it deserves its own section, because it follows directly from the thing that makes the topology special.

Every other chopper in this series swings its switching node between Vs and 0 — a 100 V span. A Class D chopper swings the load between +Vs and −Vs — a 200 V span. Ripple is volt-seconds divided by inductance, so doubling the voltage swing doubles the ripple:

Class A / B / C:  Δio ≈ Vs · D(1−D) · T / L Class D:        Δio ≈ 2 · Vs · D(1−D) · T / L

For our armature at D = 0.8 the exact expression gives 21.33 A and the approximation gives 21.33 A — agreeing closely, and both about 1.5× what a Class A managed at its own worst point. At D = 0.5 the Class D ripple peaks at VsT/2L = 33.3 A.

Which costs you, in three ways:

  • More heating. Ripple raises Irms above Io, and the copper loss follows Irms². In the worked example the ripple adds about 9 W of pure waste.
  • Torque ripple, since torque follows current.
  • A bigger inductor or a faster clock to get it back — and both cost something.
The honest summary: a Class D chopper buys the ability to reverse its output voltage, and pays for it with double the ripple and a source current that reverses every cycle. If you do not actually need a negative output voltage, this is a bad trade — which is exactly why a Class C is the more common two-quadrant choice.

The Narrow Duty Window

There is a constraint here that the tidy formula Vo = (2D−1)Vs hides, and it catches people out.

The current can only be positive. But Io = (Vo − E)/R, so positive current requires Vo > E, which means:

(2D − 1)Vs > E  →  D > ½(1 + E/Vs)

For our motoring case, E = +50 V and Vs = 100 V, so D must exceed ½(1 + 0.5) = 0.75. The entire usable range is D = 0.75 to 1.0 — a window just 0.25 wide, and at D = 0.8 we are already sitting near the bottom of it drawing 40 A.

DVoIo = (Vo−50)/0.25Valid?
0.90+80 V+120 AYes — but a lot of current
0.80+60 V+40 AYes — the worked example
0.75+50 V0 AThe boundary — Vo = E
0.500 V−200 ANo — impossible; conduction breaks down

That last row is not a rounding error, it is a contradiction. The formula asks for −200 A and the circuit has no path for negative current, so what actually happens is that conduction becomes discontinuous: the current hits zero part way through the off time, all four devices block, and vo collapses to E. The neat straight line breaks, exactly as it did on the Class A page.

So Vo = (2D−1)Vs is exactly true — as long as current is flowing. Whether it flows is decided by the machine's emf, not by the formula. This is why the control curve below is drawn as a property of the chopper alone, and why real drives close a current loop rather than trusting D.

Key Formulas (with derivation)

Two loop equations, as always. Continuous conduction and ideal devices assumed; τ = L/R.

Average output voltage

Vo = (1/T)[VsTon − VsToff] = (2D − 1)Vs    range: −Vs … +Vs

Average current and the quadrant

Io = (Vo − E)/R = ((2D−1)Vs − E)/R    must be > 0, so D > ½(1 + E/Vs)

Peak and valley current

With K1 = (Vs−E)/R, K2 = (−Vs−E)/R, a = e−DT/τ, b = e−(1−D)T/τ:

Imax = [K1(1−a) + K2 a(1−b)] / (1 − ab)    Imin = K2(1−b) + b Imax

Ripple — the doubled one

Δio ≈ 2 Vs D(1−D) T / L    →  Δio,max = VsT / 2L  at D = 0.5

Independent of E, like every other chopper here — but twice the size, because the swing is 2Vs.

Source side and power

Is ≈ (2D − 1) Io    (exact only if io were ripple-free)

The source carries +io for DT and −io for (1−D)T, which is why the (2D−1) appears again. Power balance is exact for ideal devices:

Pin = VsIs = 〈voio〉 = E·Io + Io,rms²R

Device ratings

all four devices block Vs    all four carry Imax    — but Imax is higher here, because the ripple is doubled

Control & the Vo–D Characteristic

Average output voltage against duty ratio for a Class D chopper: a single straight line from minus 100 volts at duty zero, through zero at duty 0.5, to plus 100 volts at duty one. Below 0.5 the chopper is in the fourth quadrant and above it in the first, with operating points marked at duty 0.2 giving minus 60 volts and duty 0.8 giving plus 60 volts.
Figure 8: Vo = (2D−1)Vs — the only chopper here whose line crosses zero

One straight line, running the full −Vs to +Vs. Every other chopper in this series produced a line confined to 0…Vs; this one passes clean through zero at D = 0.5. That single fact is the entire reason Class D exists.

Note what the chart does not show: the current. Vo depends only on D, so this curve is the same for any machine — but which parts of it you can actually use depends entirely on E, as the duty window showed. Combine the two and the practical picture is a chopper whose voltage law is beautifully simple and whose usable range is narrow and emf-dependent. Hence, again: close a current loop and let the controller find D.

Worked Example

Vs = 100 V, R = 0.25 Ω, L = 1.5 mH (τ = 6 ms), f = 1 kHz — the same armature as the rest of the series, at two operating points.

QuantityQuadrant I (motoring)Quadrant IV (reverse braking)
Machine emf E+50 V (forward)−70 V (reverse)
Duty D0.80.2
Vo = (2D−1)Vs+60 V−60 V
Io = (Vo−E)/R+40.00 A+40.00 A
Imax / Imin50.49 / 29.16 A50.84 / 29.51 A
Ripple Δio21.33 A21.33 A  (identical)
Io,rms40.47 A40.47 A
Source current Is+24.09 A-23.91 A
Battery power VsIs+2,409 W delivering-2,391 W absorbing
Shaft power E·Io+2,000 W-2,800 W
Copper loss Io,rms²R409.5 W409.5 W
BalanceVsIs − E·Io − Io,rms²R = 0 in both columns ✔

Read the Quadrant IV column as a story: the machine is spinning backwards and generating 2,800 W. Of that, 409 W is burnt heating its own winding, and 2,391 W reaches the battery. The chopper itself loses nothing ideally.

And the ripple penalty, made concrete. The copper loss here is 409 W. If the current were perfectly smooth at 40 A it would be 400 W. So the ripple alone is costing about 9 W — roughly 2 % more winding loss than the useful current requires, on a Class A chopper's 1  … roughly four times the excess a Class A produces on the same machine. It is small in absolute terms here, but it scales with the square of the ripple.

Advantages & Disadvantages

Advantages

  • The output voltage can go negative — the first chopper in this series that can, and the whole reason it exists.
  • Full ±Vs range from one supply: a 200 V span from a 100 V battery.
  • No dead time, no shoot-through. One gate signal for both switches, never a complementary pair — a genuine simplification over a Class C.
  • Simple control law: Vo = (2D−1)Vs, a single straight line through the full range, independent of the machine.
  • Regenerates in Quadrant IV, recovering energy from a reverse-running machine.
  • Half of a Class E — understanding it is the step to the full four-quadrant drive.

Disadvantages

  • Double the current ripple of every other chopper here, because the load swings ±Vs instead of Vs…0. This is the big one.
  • No freewheeling state. The load is always tied to the battery, so there is no gentle zero-volt interval to coast through.
  • The source current reverses every cycle, so the battery and its wiring see far more RMS current than the net power implies — and the DC link needs real capacitance to cope.
  • An awkward quadrant pair. Forward motoring plus reverse braking is rarely what a real machine actually needs.
  • A narrow, emf-dependent duty window — D must exceed ½(1 + E/Vs) or conduction breaks down.
  • Four devices for two quadrants, the same count as a Class C, but with worse ripple.

Applications

  • As one half of a Class E four-quadrant drive — by far its most common real role.
  • Loads that must be actively pushed both ways, where a reversible voltage is needed but the current naturally stays one-directional.
  • Reverse-braking applications — lowering an overhauling load, where the machine turns backwards and its energy is recovered.
  • Teaching and analysis, as the cleanest demonstration that an average output voltage can be made negative from a single positive supply.
  • Certain excitation and magnet supplies that need a fast, reversible voltage across an inductive winding whose current is unidirectional.

Be honest about the selection rule: if you only need Quadrants I and II — driving and braking one way — use a Class C. It uses the same four devices with half the ripple. Reach for a Class D only when you specifically need the output voltage to reverse.

Class D vs C (and the rest of the family)

Class CClass D
QuadrantsI & III & IV
What is fixedvo ≥ 0io ≥ 0
What swingsthe currentthe voltage
TopologyHalf-bridge (one leg)H-bridge (switch/diode diagonals)
VoD·Vs  (0 … Vs)(2D−1)·Vs  (−Vs … +Vs)
vo levelsVs, 0+Vs, −Vs (no zero!)
GatingComplementary pairOne signal, both together
Dead time needed?Yes — shoot-through riskNo
RippleVsD(1−D)T/L2× that
Freewheeling state?YesNo
Discontinuous conductionNeverYes, outside the duty window
Source currentZero for part of the cycleReverses every cycle
Typical useThe standard industrial driveHalf of a Class E

And the family:

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 diodesOutput voltage reverses; brakes a reverse-running machine.
EI, II, III & IV4 switches + 4 diodesFull four-quadrant: motor and brake in both directions.

Class C and Class D are the two halves of the same idea. C reverses current, D reverses voltage. Give the H-bridge four switches instead of two — so both diagonals can be driven — and you can reverse both at once. That is a Class E, and it completes the family.

Frequently Asked Questions – FAQs

They get two quadrants in opposite ways. A Class C keeps the voltage positive and lets the current reverse, so it covers Quadrants I and II — driving and braking in one direction. A Class D keeps the current positive and lets the voltage reverse, so it covers Quadrants I and IV — forward motoring and reverse braking. Same device count, perpendicular capabilities.

Because when the switches open, the diodes connect the load across the same battery the other way round. The inductance keeps the current flowing A to B; the only path available takes it out of B, up through D1 into the battery's positive terminal, and back through D2 into A. So terminal A now sits at 0 and B sits at Vs, making vo = vA − vB = −Vs. Nothing switched the battery around — the diodes are simply the only path, and it happens to be a reversed one. This needs an H-bridge; every earlier chopper had one load terminal bolted to a rail, which is why none of them could do it.

Because the load swings between +Vs and −Vs — a 200 V span — where every other chopper swings between Vs and 0, a 100 V span. Ripple is volt-seconds over inductance, so twice the swing gives twice the ripple: Δio ≈ 2Vs·D(1−D)·T/L, peaking at VsT/2L at D = 0.5. It is the direct price of having no zero-voltage freewheeling state, and it is Class D's main practical drawback.

Because they are on the same diagonal of the H-bridge, not in the same leg. CH1 is top-left and CH2 is bottom-right, so closing both connects the battery across the load; opening both hands the current to the diode diagonal. They are never in series across the supply, so they cannot short it — which means a Class D needs no dead time and has no shoot-through risk, unlike a Class C where the two switches share a leg.

Negative voltage with positive current, so the power is negative and the machine is generating. For forward current to flow against a negative applied voltage, the machine's own emf must be even more negative — which means it is spinning in reverse. So Quadrant IV is reverse braking: the machine is running backwards and the chopper is slowing it and returning the energy to the supply. Note it is not forward braking — that is Quadrant II, and a Class D cannot reach it.

The +Vs and −Vs intervals are equal, so they cancel and Vo = 0. But note this is not a quiet state: vo is still slamming between +100 and −100 V every cycle, and the ripple is at its worst — VsT/2L. Whether any current actually flows depends on the machine: with E = +50 V the formula asks for −200 A, which is impossible, so conduction goes discontinuous instead. D = 0.5 only gives a genuine zero-current standstill if the machine's emf is also zero.

Because the source is in the loop in both states. While the switches conduct it delivers current (is = +io); while the diodes conduct the current is pushed back into it (is = −io). The net is Is ≈ (2D−1)Io. Contrast that with a Class A or C, where the source simply sees zero current during freewheeling. The practical consequence is that a Class D's battery and DC link carry a lot more RMS current than its net power suggests, so the link capacitance has to be sized for it.

Because there is no path for it. Trace every route through the circuit: the switch diagonal carries current A→B, and so does the diode diagonal. Nothing in the bridge can carry B→A, because the diodes point the wrong way and the other diagonal has no devices on it at all. That is precisely what a Class E fixes — it puts switches on both diagonals, so the current can reverse too, giving all four quadrants.

Because the current must stay positive, which needs Vo > E, i.e. D > ½(1 + E/Vs). With E = +50 V on a 100 V supply, that means D must exceed 0.75 — the whole usable range is 0.75 to 1.0. Push D lower and the formula asks for negative current, which the circuit cannot supply, so conduction goes discontinuous and Vo = (2D−1)Vs stops being true. And since E moves with speed, the window moves too — which is why real drives close a current loop rather than commanding D directly.

Almost always a Class C. Both use two switches and two diodes, but a Class C gives you Quadrants I and II — motoring and braking in the same direction — which is what nearly every real drive actually wants, and it does it with half the ripple, a freewheeling state, no discontinuous conduction and a source current that never reverses. A Class D's advantage is one specific thing: its output voltage can go negative. If you do not need that, it is the worse trade. If you need all four quadrants, go straight to a Class E.

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