Class A Chopper (First Quadrant Chopper)

Introduction — why chop a DC supply at all?

Suppose you have a 100 V battery and a DC motor that you want to run at half speed. The lazy answer is to put a resistor in series and drop the extra 50 V across it. It works — and it throws away half your battery as heat. On a forklift or an electric scooter that is not a rounding error; it is the difference between a shift and half a shift.

A chopper takes a different route. Instead of dropping the unwanted voltage, it removes the unwanted time. A fast electronic switch connects the battery to the motor, disconnects it, connects it again, thousands of times a second. The motor is far too slow and too inductive to follow that flicker, so what it feels is the average. Connect for 70 % of the time and the motor behaves as though it were fed from a 70 V supply — but nothing is burning off 30 V, because an ideal switch dissipates nothing whether it is fully on or fully off.

That is the whole idea of a DC chopper: a variable DC output from a fixed DC input, by switching rather than by resistance. The Class A chopper is the simplest member of the family and the one to learn first, because the other four classes (B, C, D and E) are all built by adding switches and diodes around this same skeleton.

What is a Class A Chopper?

A Class A chopper — also called a Type A chopper or a first-quadrant chopper — is a DC-DC converter made of exactly two power devices:

• a controlled switch in series between the supply and the load (an IGBT, a power MOSFET, or in older equipment a thyristor with forced commutation), and
• a freewheeling diode connected across the load.

The switch is turned on and off repeatedly. The fraction of each cycle that it stays on is the duty ratio, and that single number sets the output:

Because D can only lie between 0 and 1, V_o can never exceed V_s. A Class A chopper is always a step-down converter. And because the switch conducts in only one direction and the diode conducts in only one direction, the load current can never reverse either. Positive voltage, positive current, always — hence "first quadrant".

What "First Quadrant" Actually Means

The chopper classes are named after a picture. Draw the load's average voltage V_o up the vertical axis and its average current I_o along the horizontal axis. Those two axes cut the plane into four quadrants, and each quadrant means something physical for a motor:

A Class A chopper can only ever reach the first of those four. The reason is not a control limitation you could program around — it is physics baked into the components. The switch is a one-way device and the diode is a one-way device, so:

• v_o cannot go negative. When the switch is on, the load sees +V_s. When it is off, the diode clamps the load to 0 V. There is no arrangement of those two states that produces a negative terminal voltage.
• i_o cannot go negative. Current can only enter the load through the switch or through the diode, and both point the same way.

The practical consequence is worth stating bluntly, because it is the honest limitation of this topology: a Class A chopper can speed a motor up, but it cannot actively slow it down. Reduce the duty ratio and the applied voltage drops, but the motor can only coast down on friction and windage. Braking would require the current to reverse so that energy flows back into the battery, and that is Quadrant II — the job of a Class B chopper. Reversing direction needs negative voltage, which is Quadrant IV territory.

Block Diagram

Before the transistors, here is the shape of the whole thing. Five blocks in a line, plus two that hang off the side: the freewheeling diode, which exists purely to give the inductive load somewhere to send its current, and the gate controller, which is where the duty ratio comes from.

Read it left to right and the story is simple: a fixed voltage goes in, a switch chops it into a train of pulses, the load averages those pulses into a smooth-ish current, and the motor turns. Read the arrow directions and you find the constraint: every arrow points one way. Nothing in this chain is capable of carrying energy back toward the battery. That, in one picture, is why it is a one-quadrant converter.

Circuit Diagram & Construction

Now the real circuit. There are only four things in it.

Notice where the diode's polarity comes from. While CH is on, the top node sits at +V_s and the bottom rail at 0, so the diode has its cathode positive with respect to its anode — reverse biased, blocking, out of the circuit. The moment CH opens, the inductor drives the top node below the bottom rail by one diode drop, the diode becomes forward biased, and it conducts. It needs no gate signal and no cleverness; it just responds.

Principle of Operation

The controller runs a fixed clock of period T. Within every period it holds CH on for T_on and off for T_off, where T_on + T_off = T. The ratio D = T_on/T is the one knob the whole converter has.

The circuit therefore has exactly two states, and it alternates between them forever:

• CH on — the battery is bolted to the motor. v_o = V_s. Because V_s is bigger than the back emf E, the surplus voltage pushes current up, and the inductance stores energy.
• CH off — the battery is disconnected. The inductance refuses to let the current stop, so it pushes the current around the loop through FD instead. v_o = 0. With no supply to hold it up, the current decays, and the stored energy drains into the motor and the winding resistance.

So the load current is a sawtooth-ish thing that climbs during T_on and falls during T_off. In steady state the two exactly cancel — the rise equals the fall — otherwise the current would drift up or down period after period. Meanwhile the motor's inertia and inductance smooth everything out, and it turns at the speed the average voltage commands.

Raise D and the ON stretch grows, the average voltage rises, more current flows, the motor accelerates. Lower D and the reverse happens. That is the entire control law.

Modes of Operation

With only two devices there are only two modes. In the circuits below, a highlighted loop shows where current is actually flowing, a green device is conducting, and a greyed dashed device is not.

Mode 1 — Chopper ON (0 ≤ t < T_on)

CH is gated on. The battery, the switch and the motor form one closed loop, and the full supply appears across the load:

What happens next follows from the loop equation:

With V_s = 100 V and E = 60 V there is 40 V of surplus, minus the resistive drop, left over to push against L. So di_o/dt is positive and the current rises — not as a straight line but as a rising exponential, heading toward the value it would settle at if you simply left the switch on forever, (V_s − E)/R = 160 A. It never gets remotely close, because the switch opens again after a millisecond. That is why the visible part of the curve looks almost straight: you are seeing the first 2 % of a 6 ms exponential.

Three things happen at once during this mode:

• The source delivers energy — this is the only part of the cycle in which the battery supplies any current at all.
• The motor receives energy, E · i_o of it converted into mechanical work.
• The inductance stores energy, ½Li², building up the reserve that will carry the load through the next mode.

Mode 2 — Chopper OFF, freewheeling (T_on ≤ t < T)

The gate is removed and CH stops conducting. The battery is now out of the picture entirely — look at the highlighted loop and you will see it does not include the source. But the motor still has 47 A flowing in it and an inductor that will not allow that to stop. The inductor reverses the polarity of its own voltage to keep the current going, which forward-biases FD, and the current simply circulates: out of the bottom of the load, along the return rail, up through the diode, back into the top of the load.

The loop equation loses its source term:

Both terms on the right are now negative, so di_o/dt is negative and the current decays — this time aiming at −E/R = −240 A, a target it is never allowed to reach. It would have to pass through zero first, and the diode blocks the moment it tries. In continuous conduction the switch turns back on long before that happens, and the current bottoms out at I_min instead.

During this mode:

• The source delivers nothing — i_s = 0.
• The inductance releases its stored energy, some into the motor as useful work, some into R as heat.
• The switch blocks V_s, because the diode holds the top node at 0 V while the battery holds the other side of CH at V_s. Clean and bounded — that is the second job the diode does.

Waveforms Explained in Detail

Everything above comes together here. This is the same motor throughout the page — a 100 V supply, an armature of 0.25 Ω and 1.5 mH, a back emf of 60 V, chopped at 1 kHz with D = 0.7 — and the traces below are computed from the exponential solutions of the two loop equations, not sketched by hand. Two full periods are shown so you can see the pattern repeat.

v_o i_o i_CH / i_s i_FD

The gate pulse — the only input

The top trace is what the controller produces: a square wave of fixed period T = 1 ms, high for T_on = 0.7 ms and low for T_off = 0.3 ms. The blue shading behind it runs all the way down the figure so you can line every other trace up against the ON interval.

Note what does not change: the period. In the usual constant-frequency scheme, the falling edge slides left and right while the clock stays put. Every other waveform on this page is downstream of that one edge.

Output voltage v_o — a square wave, and that is the point

The load terminal voltage is the crudest trace in the figure and the most important: it is V_s during T_on and 0 during T_off, with nothing in between. It is a two-level square wave that copies the gate exactly, because the switch and the diode between them force the node to one of only two places.

The dashed green line is its average, and here is the arithmetic — it really is this simple:

Two facts fall straight out of the shape. First, since the trace never dips below zero, the average cannot be negative — Quadrant III and IV are unreachable, exactly as Figure 1 said. Second, since it never rises above V_s, the average cannot exceed the supply — the chopper only steps down.

It also explains something students often trip over: the output voltage is not smooth and is not supposed to be. Nobody is filtering it. The motor does not respond to v_o; it responds to the current that v_o drives, and the smoothing happens in the winding. Which brings us to the trace that actually matters.

Load current i_o — where the exponentials show up

This is the heart of the figure. Follow one period:

• At t = 0 the switch closes with 32.9 A already flowing — left over from the previous cycle. It does not start from zero, and that is the definition of continuous conduction.
• Through T_on the current climbs along a rising exponential aimed at 160 A, reaching I_max = 46.9 A when the switch opens.
• Through T_off it decays along a falling exponential aimed at −240 A, arriving back at I_min = 32.9 A exactly as the next period begins.
• The loop closes. It has to. If the rise exceeded the fall the current would creep up cycle after cycle; steady state is the condition that they match.

The curvature is real but gentle, and there is a good reason. The load's time constant is τ = L/R = 1.5 mH / 0.25 Ω = 6 ms, while the whole switching period is only 1 ms. You are watching a slow exponential through a very short window, so you only ever see its nearly straight beginning. This is why the textbook straight-line approximation works: when τ >> T the exponentials flatten into ramps.

Three numbers are marked on the trace:

And notice how sensitive that is. The average current is set by a 10 V difference across a 0.25 Ω winding. Nudge D from 0.7 to 0.75 and V_o goes to 75 V — the current does not rise by 7 %, it doubles to 60 A. In a real drive the back emf moves too (the motor speeds up, E rises, and current settles back), which is exactly why real choppers close a current loop around this rather than running open-loop on D.

Switch and diode currents — the load current, shared out

The third and fourth current traces are the same waveform cut into two pieces. Lay them on top of each other and you get i_o back:

i_CH follows the load current for the whole ON time, then drops vertically to zero the instant the gate goes low. i_FD does the mirror image: flat zero, then a vertical jump up to 46.9 A at the same instant, then the decay. Those two vertical edges happen at the same moment and are the same current — the load current being handed from the switch to the diode without ever changing itself. It cannot change; the inductor forbids it.

Because the switch is the only path from the battery, i_CH is also the source current i_s — the battery delivers current for T_on and nothing at all for T_off. That gives the second formula every textbook quotes:

Note the "≈". That relation is not exact, and it is worth seeing why, because it is the same trap that catches people on the loss calculation. I_s is D times the average of i_o over the on-time — and look at the trace: the current is rising throughout T_on, so its on-time average sits a little above its whole-period average I_o. Integrating the drawn waveform gives I_s = 28.04 A, not 28.00. The shortcut assumes a ripple-free current, and is only as good as that assumption — here it is off by 0.15 %, which is negligible; with a smaller inductor it would not be.

Using the honest number, here is the efficiency argument for the whole topology:

They match exactly, and they must — but the reason is prettier than "conservation of energy". Look at the traces: whenever v_o = V_s, i_s = i_o; and whenever v_o = 0, i_s = 0 too. So v_oi_o and V_si_s are the same number at every instant, not merely on average. An ideal switch drops no voltage when closed and passes no current when open, so it dissipates nothing in either state, and nothing is left over to lose.

However you count it, the point stands: the chopper handed the load a lower voltage and a higher current than it drew from the battery, and lost nothing doing it. A series resistor set to give the same 70 V would have burned 1200 W as heat. That is the entire commercial case for choppers.

These two traces are also the datasheet numbers you order parts from. The switch must carry 46.9 A peak with an average of about 28 A; the diode must carry the same peak with an average of about 12 A (I_FD ≈ (1 − D) · I_o, and note that I_CH + I_FD = I_o exactly, since the two pieces add back up to the whole). At low duty those roles swap — the diode ends up doing most of the work, which is why undersizing it is such a common and expensive mistake.

Switch voltage v_CH — what sets the part rating

The bottom trace is the voltage the switch has to stand. Zero while it conducts (an ideal switch is a short), and V_s — no more while it blocks.

That "no more" is the diode's doing. With FD conducting, the top node is clamped to 0 V, so the switch sees exactly the battery voltage across it and nothing else. Without the diode this trace would show a destructive inductive spike instead of a clean 100 V step. So you specify the switch for V_s plus margin — typically a 200 V part for a 100 V bus, to cover stray-inductance ringing at the switching edges that this idealised drawing does not show.

Continuous vs Discontinuous Conduction

Everything so far assumed the current never quite reaches zero. Lower the duty ratio far enough and that assumption breaks — with real consequences.

The current bottoms out at I_min. Reduce D and I_min falls, and at some duty ratio it hits zero. Below that, the decaying current reaches zero part way through the off time. The diode cannot carry current backwards, so it blocks, and the current stays at zero until the switch turns on again. The circuit now has three states per period, not two.

Look at the amber gap in the lower pair, and then look at what v_o does inside it. With no current flowing there is no drop across R and no drop across L, so the load terminals show you what is left: the back emf itself. v_o jumps up to E = 60 V. The output voltage has a third level that the two-level analysis never predicted.

And that wrecks the formula everyone memorises:

For this motor the boundary sits at D_crit ≈ 0.62, found by solving I_min = 0. It is not a universal constant — it moves with L, with R, with the switching frequency, and with speed, because E moves with speed. The general condition is:

Why it matters in practice:

• The control loop's gain changes. Open-loop, a controller calibrated on V_o = D·V_s under-delivers duty and the speed sits high. Closed-loop drives cross this boundary and see their plant gain shift underneath them.
• Torque gets rough. Torque follows current, and current that collapses to zero every cycle produces pulsating torque — audible, and hard on gearboxes.
• Peak-to-average ratio gets ugly. The same average current delivered in short bursts means more RMS heating for the same useful work.

Key Formulas (with derivation)

Everything here follows from the two loop equations already written down. All of it assumes continuous conduction and ideal devices.

Average output voltage

RMS output voltage

Worth knowing because it is not the same as the average — the ratio V_o,rms/V_o = 1/√D is the form factor, and it is what determines the extra heating the chopping itself causes.

Average load current

In steady state the inductor's average voltage over one period is zero, so the average of v_o is shared between R and E alone:

Peak and valley current

Solve the ON exponential and the OFF exponential, then force i(0) = i(T). Writing τ = L/R:

Peak-to-peak current ripple

Subtract the two. The E terms cancel, and what is left is unexpectedly tidy:

Two things fall out of that. First, the ripple does not depend on E at all — the motor's speed changes the average current but not the ripple around it. Second, when τ >> T you can expand e^−x ≈ 1 − x and it collapses to the form every textbook quotes:

So the ripple is worst at half duty and vanishes at both extremes — which makes sense, since at D = 0 and D = 1 nothing is being chopped at all. For our example the exact formula gives 13.99 A and the approximation gives 14.00 A — close, because τ/T = 6.

Source side

The source only sees current while the switch conducts, so I_s is the on-time slice of i_o. The exact statement, and the familiar approximation to it:

Power balance, on the other hand, is exact for ideal devices — because v_oi_o and V_si_s are equal at every instant, not just on average:

That last one is a nice way to see what a chopper "is" from the battery's point of view: an electronically adjustable resistance, tuned by D — except that unlike a real resistor, it delivers the energy instead of burning it.

Conduction boundary

Device ratings

Control Methods & the V_o–D Curve

D = T_on/T is a ratio of two numbers, so there is more than one way to vary it. Classical texts group the options as time-ratio control (TRC) and current-limit control (CLC).

Whichever you use, the promise is V_o = D·V_s — a straight line. Here is how well that promise is kept for our actual motor:

Above D_crit the two curves are indistinguishable: in continuous conduction V_o = D·V_s is not an approximation, it is exact. Below D_crit the real curve peels away upward, because those zero-current gaps let the back emf show through and lift the average. As D approaches zero the output flattens toward E rather than falling to zero.

This is the honest shape of the control characteristic, and it explains a classic symptom: a drive that tracks its speed reference beautifully at high demand and then refuses to slow down properly at low demand, sitting stubbornly faster than commanded. Nothing is broken. The chopper has simply crossed into discontinuous conduction, where the law it was tuned on no longer holds.

Worked Example

The example used everywhere on this page, worked start to finish. A DC motor is fed from a 100 V battery through a Class A chopper switching at 1 kHz. The armature is 0.25 Ω and 1.5 mH, and at the speed of interest it generates a back emf of 60 V. The duty ratio is 0.7.

Three lines are worth pausing on. 2,804 W in, 2,804 W out — the chopper itself consumed nothing, which is the whole reason it exists. 2,400 W mechanical + 404 W copper = 2,804 W, so the load's own books balance too — but only once the copper loss is computed from I_o,rms, not I_o. Everything the battery gave up either turned the shaft or warmed the winding; none of it was thrown away to set the speed.

Advantages & Disadvantages

Advantages

• Very high efficiency. The switch is either fully on or fully off, so ideally it dissipates nothing. Real implementations sit in the high 90s.
• The simplest chopper possible. One controlled switch and one diode — the smallest part count, the cheapest gate drive, the fewest ways to fail.
• Smooth, stepless control. D is continuously adjustable, so the output is too — no taps, no contactors, no steps.
• Fast response. The output follows a change in D within a switching period; the current follows within a few.
• No commutation problems with modern devices. IGBTs and MOSFETs turn off on command, unlike the thyristors that made early choppers so awkward.
• The foundation of the family. Understand this and Classes B through E are just this circuit with more of the same parts.

Disadvantages

• One quadrant only. No regenerative braking, no reversing. The motor coasts to a stop on friction — which for a loaded hoist or a heavy vehicle is not an acceptable answer.
• Step-down only. V_o can never exceed V_s, because D cannot exceed 1.
• The output voltage is chopped, not smooth. Fine for a motor whose inductance filters it; useless for a load that needs clean DC without a filter added.
• Current ripple means torque ripple and extra heating. Managing it costs either switching frequency (loss) or inductance (weight, cost).
• Discontinuous conduction at low duty breaks V_o = D·V_s and shifts the loop gain, exactly where fine control is often wanted most.
• It generates EMI. Fast edges chopping tens of amps radiate and conduct interference; filtering and layout are real engineering work, not an afterthought.

Applications

• DC motor speed control in one direction — the classic use, and the reason the topology exists.
• Battery-powered vehicles — forklifts, golf carts, warehouse tugs, mobility scooters: the battery is fixed at 24, 48 or 72 V and the chopper sets the speed.
• Traction drives for trolleybuses, trams and mine locomotives, where the supply is a fixed DC catenary or third rail. (Modern traction adds more quadrants for braking; Class A alone is the motoring half.)
• Conveyors, pumps, fans and blowers on DC supplies — loads that only ever run one way and never need to be braked electrically.
• DC power supplies and battery chargers — the same power stage with an LC filter is a buck converter.
• Resistive control — heaters, DC lighting dimming, electroplating and welding supplies, wherever a fixed DC rail must be turned down efficiently.
• Teaching and lab work, because it is the cleanest possible demonstration of averaging by switching.

The selection rule is short: if the load only ever runs one way and never needs electrical braking, a Class A chopper is enough. The moment you want to slow it down under control, or recover its energy, you have outgrown this topology and need Class B, C or E.

Class A vs the Other Chopper Classes

Each class buys another quadrant with more silicon. Class A is the floor; the rest are built on it.

A useful way to hold it in your head: Class B is Class A turned around — same two components, rearranged so the current flows the other way and the machine feeds the battery instead of drawing from it. Put one of each in the same box and you have Class C, which can drive and brake without ever reversing. Two Class C legs make an H-bridge, and that is Class E, which can do everything.

Frequently Asked Questions – FAQs

Related Topics

• Class B Chopper (Second Quadrant) — regenerative braking, the mirror of this page
• Class C Chopper (Two Quadrant) — this page and Class B in one half-bridge leg
• DC-DC Converters (Choppers) — Overview
• Buck (Step-Down) Converter — the same power stage with an LC filter
• DC-DC Duty Cycle Calculator
• Buck Converter Calculator
• Buck Converter Simulation
• IGBT — the usual chopper switch
• All Power Electronic Converters