Class B Chopper (Second Quadrant Chopper)

Introduction — the problem a Class A chopper leaves behind

A Class A chopper is an excellent way to make a motor go. It is a hopeless way to make one stop.

Turn its duty ratio down and the applied voltage falls, so the motor stops accelerating — but nothing is actively slowing it. It coasts, losing speed to friction and windage on its own schedule. For a fan that is fine. For a 2-tonne forklift on a ramp, a mine hoist with a load hanging off it, or a tram coming into a stop, "it'll slow down eventually" is not an answer.

You could bolt a big resistor across the motor and let it dump its kinetic energy as heat. That is rheostatic braking, it works, and it wastes every joule — the same objection that made us build a chopper instead of a series resistor in the first place. A spinning motor is already a generator; its kinetic energy is genuinely worth something. On a battery vehicle it is worth range.

A Class B chopper catches it. It takes the same two devices a Class A has — one controlled switch, one diode — and rearranges them so the current flows the other way. The machine stops being a motor and starts being a generator, the braking is electrical and controllable, and the energy it gives up is pushed back into the battery it came from.

What is a Class B Chopper?

A Class B chopper — also called a Type B chopper or a second-quadrant chopper — is a DC-DC converter that moves power from a load back into a fixed DC source. It is built from exactly two power devices:

• a controlled switch (CH) connected in shunt, straight across the load, and
• a diode (D) connected in series with the source, oriented so current can only flow into the source, never out of it.

Chopping the shunt switch on and off gives an average terminal voltage of:

Note the (1 − D) where a Class A chopper has a plain D. That is not a typographical curiosity; it is the same swap showing up in the algebra. In a Class A the load sees the supply while the switch is on. In a Class B the switch shorts the load while it is on, so the load only meets the supply while the switch is off. Everything is inverted.

Given that emf, two things are true at all times, and together they define the topology:

• v_o stays positive — it is either 0 (switch shorting the load) or +V_s (diode conducting). Neither state can make it negative.
• i_o stays negative — current only ever flows out of the machine, because the switch and the diode both point that way.

Positive voltage with negative current is the second quadrant, and it means one thing: the power is negative. Energy flows from the load to the source, every single cycle. That is the whole point.

What "Second Quadrant" Means (and the sign convention)

Draw the load's average voltage V_o up the vertical axis and its average current I_o along the horizontal axis, and the plane splits into four quadrants. A Class A chopper occupies the first. A Class B occupies the second, and only the second.

The important observation is what does not change. v_o is still positive — the machine is still spinning forwards, and its emf still has the same polarity it had while motoring. The motor has not reversed. Only the current has. And that alone flips the sign of the power:

A negative number here is not an error to be tidied away. It is the result. It says the machine is delivering power rather than absorbing it, which is precisely what "regenerative braking" means, and it is why a Class B chopper can slow a vehicle down while a Class A cannot.

What it cannot do is equally worth stating. A Class B chopper cannot drive a motor — there is no state in which the source can push current into the load, because the series diode blocks it. It is a braking-only converter. Real drives that need both pair a Class A and a Class B in one box, which is exactly what a Class C chopper is.

Block Diagram

Here is the shape of the whole thing. Read it left to right and you are following the energy: it starts as kinetic energy in a spinning rotor and ends up as charge in a battery.

Hold this next to the Class A block diagram and the relationship is obvious: every arrow has been reversed. There, a fixed source fed a switch which fed a machine which produced torque. Here, a machine gives up its kinetic energy, and a one-way diode hands it to a source that absorbs it. Same five blocks, same parts count, opposite direction of travel.

The two blocks hanging off the main chain do the same jobs as before, with different emphasis. The gate controller still sets D. The chopper switch is now the thing that shorts the load rather than feeding it — which sounds alarming, and is exactly how the circuit builds up the current it later pushes uphill into the battery.

Circuit Diagram & Construction

Four components, and three of them are in exactly the same place as they were on the Class A page. The load branch is drawn identically on purpose — same R, same L, same E, same polarity — so that the only differences you have to absorb are where CH sits and which way D points.

Notice the diode's orientation and what it costs you. Its anode is on the machine side and its cathode on the battery side, so it conducts only when the machine's terminal voltage tries to exceed V_s. While CH is on, the machine's terminal sits at 0 V and the battery holds the other side at +100 V, so the diode is firmly reverse biased and the battery is completely isolated. It is not "off" by control; it is off by physics.

Principle of Operation

The trick a Class B chopper pulls is genuinely clever, and it is worth stating plainly before the mode diagrams, because the circuit looks like it should not work.

Here is the difficulty. The machine is spinning and generating E = 60 V. The battery is at V_s = 100 V. You want current to flow from the 60 V machine into the 100 V battery. Connect them directly and nothing happens — you cannot push charge from a low potential to a high one, any more than water flows uphill. The diode simply blocks and the machine coasts.

So the chopper does it in two steps, using the inductance as a pump:

• CH on — short the machine out through the switch. The battery is disconnected and irrelevant. With its own terminals shorted, the machine's emf drives a rising current through its own winding, and the inductance L soaks up energy. Nothing is delivered anywhere yet; you are cocking the spring.
• CH off — abruptly take the short away. The inductance will not let its current stop, so it generates whatever voltage it takes to keep it going, and that voltage adds to E. Now the terminal voltage is E + L di/dt, which is bigger than the 100 V battery. The diode conducts and the current is forced uphill into the battery.

That is the whole mechanism: store low, release high. The inductance is not a nuisance to be tolerated here — it is the component doing the work. Without it there is no boost, and a Class B chopper simply cannot function.

And the current does not stop when the switch turns back on; it just gets shorted out again and builds a little more. In steady state the rise during T_on exactly cancels the fall during T_off, and a steady average current flows out of the machine, cycle after cycle, all the way into the battery.

Raise D and you spend more of each cycle storing, which as we will see pushes the braking current — and therefore the braking torque — up hard. Lower D and braking eases off. That is the control law, and it is far more aggressive than the Class A one.

Modes of Operation

Two devices, two modes. In the circuits below a highlighted loop shows where current actually flows, a green device is conducting, and a greyed dashed device is not. In both modes the loop runs upward through the load — against the i_o reference arrow, which is why i_o is negative.

Mode 1 — Chopper ON, storing energy (0 ≤ t < T_on)

CH is gated on and short-circuits the machine's terminals. Look at the highlighted loop: the source is not in it. The battery has been cut out of the circuit entirely by the reverse-biased diode.

With the terminals shorted, the only thing left in the loop is the machine itself. Writing KVL around it, with i = |i_o| the magnitude of the current flowing out of E's positive terminal:

E is 60 V and there is nothing opposing it but the winding's own resistance, so di/dt is positive and the current climbs — a rising exponential aimed at E/R = 240 A, the current that would eventually flow if you simply left the machine shorted forever. It never gets close, because the switch reopens after half a millisecond. But it is worth noticing that the target is a very large number: this mode is a deliberate, controlled short circuit of a generator, and the only thing keeping it safe is that it is brief.

During this mode:

• The machine gives up energy — its emf is doing work against R and L, so it feels a retarding torque. Braking is already happening.
• The inductance stores energy, ½Li², building the reserve that will be forced into the battery next.
• The source receives nothing — i_s = 0 for the whole of T_on.
• Some energy is lost as heat in R and never leaves the machine. This is the part of regenerative braking you do not get back.

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

The gate is removed and CH stops conducting, taking the short away. The current in L cannot stop, so L reverses its own voltage to keep it flowing — and that reversed voltage adds to the machine's emf. The terminal voltage momentarily tries to exceed 100 V, the diode becomes forward biased, and the current spills into the battery. Once the diode conducts it clamps the terminal at V_s:

The loop equation now has the battery in it, opposing the machine:

Since V_s (100 V) is larger than E (60 V), the right-hand side is negative: di/dt is negative and the current decays, aiming at (E − V_s)/R = −160 A — a target it is never allowed to reach, because the diode blocks the instant the current tries to pass through zero. In continuous conduction the switch closes again long before that, and the current bottoms out at I_min.

This is the mode that does the useful work. The battery is being charged, at V_s volts, by a machine generating only 60. The inductance is paying the difference.

Why It Is Called a Step-Up Chopper

Class B choppers are routinely described as step-up choppers, and the name confuses people because the obvious reading of the circuit says the opposite: V_o = (1 − D)V_s is smaller than V_s, so surely it steps down?

The answer is that you are looking at the wrong pair of terminals. "Step-up" here describes the direction the energy travels, not the ratio of the terminal voltages. Rearrange the steady-state relation:

Read it from the machine's point of view. The machine sits at E = 60 V. The battery sits at 100 V. The chopper takes energy from the low-voltage side and delivers it to the high-voltage side, and the 1/(1 − D) factor is the boost ratio doing it. That is a step-up conversion, and it is the same 1/(1 − D) you will recognise from a boost converter — because a Class B chopper is a boost converter, with the machine playing the input source and the battery playing the load.

The correspondence is exact, component for component. If you already understand a boost converter you already understand a Class B chopper; the only new idea is that the input source happens to be a spinning machine whose voltage falls as it slows down.

Waveforms Explained in Detail

Everything above comes together here. This is the same machine as the Class A page — a 100 V battery, an armature of 0.25 Ω and 1.5 mH, an emf of 60 V, chopped at 1 kHz — but now braking, at D = 0.5. The traces are computed from the exponential solutions of the two loop equations above, not sketched by hand. Two full periods are shown.

v_o i_o i_CH i_D / i_s

The gate pulse — the only input

A square wave of fixed period T = 1 ms, high for T_on = 0.5 ms and low for T_off = 0.5 ms. The blue shading behind it runs down the whole figure so you can line every other trace up against the ON interval.

But give the two halves their proper names, because unlike a Class A chopper they do completely different jobs: T_on stores (nothing leaves the machine) and T_off returns (everything leaves the machine). The chopper spends half its life apparently doing nothing useful, which is exactly how it gets the current high enough to be useful in the other half.

Output voltage v_o — the inverse of Class A

The terminal voltage is 0 during T_on and V_s during T_off, with nothing in between. Compare that with the Class A trace, which is V_s during T_on and 0 during T_off. It is the same square wave, inverted — the direct visual consequence of moving the switch from series to shunt.

And that inversion is the entire origin of the (1 − D):

Note what this trace never does: it never goes negative. Both of its levels are zero or positive, so the average cannot be negative — Quadrants III and IV are unreachable, exactly as Figure 1 claimed. The chopper's inability to reach them is visible right here in the shape of the waveform.

Load current i_o — negative, and why

This is the trace that defines the page, and the first thing to notice is that the entire waveform sits below the zero line. It never touches it, never crosses it. That is what the second quadrant looks like on an oscilloscope.

Follow one period:

• At t = 0 the switch closes with 31.7 A already flowing out of the machine — left over from the previous cycle. It does not start from zero, and that is the definition of continuous conduction.
• Through T_on the magnitude climbs to 48.3 A — so on the signed plot the trace moves further down, away from zero. The current is growing more negative while the switch is on, which is the visual signature of a Class B chopper.
• Through T_off it decays back toward zero (less negative) as the inductor empties into the battery, arriving at 31.7 A exactly as the next period begins.
• The loop closes. Steady state is the condition that the rise and the fall match.

The curvature is gentle for the same reason as on the Class A page: τ = L/R = 6 ms against a 1 ms period, so you are seeing only the nearly-straight beginning of a slow exponential. Same machine, same time constant, same visual result.

And look how sensitive that is. The braking current is set by the 10 V difference between E and V_o, across a 0.25 Ω winding. Nudge D from 0.5 to 0.6 and V_o falls to 40 V, so the difference doubles to 20 V and the current doubles to 80 A. The braking torque doubles with a 0.1 change in duty. Hold that thought — it is the subject of the control section, and it is the single most dangerous property of this converter.

Switch and diode currents — the machine's current, shared out

The two device traces are the same waveform cut into two pieces. They are drawn as positive magnitudes, the way a datasheet quotes them, because a switch and a diode each conduct only one way:

i_CH carries the machine's current for the whole ON time, then drops vertically to zero the instant the gate goes low. i_D does the mirror image: flat zero, then a vertical jump to 48.3 A at the same instant, then the decay. The current is handed from switch to diode without itself changing — the inductor forbids it to.

But i_D deserves special attention: it is the only trace on this page that puts charge back into the battery. i_CH does no useful delivering at all; it merely circulates current inside the machine. Everything you recover leaves through the diode.

Note the "≈". As on the Class A page, that relation is not exact — it assumes a ripple-free current. I_s is really (1 − D) times the average of the current over the off-time, and the current is decaying throughout T_off, so its off-time average sits slightly below the whole-period average. Integrating the drawn waveform gives I_s = 19.94 A, not 20.00 — about 0.29 % low. Small here, and not small at all if you shrink the inductor.

These traces are also the datasheet numbers you order parts from, and the split is worth reading:

At high duty — hard braking — the switch carries most of the burden and the diode delivers in short, tall bursts. At low duty it reverses. Neither device gets an easy life across the range, which is why both are sized for the full peak.

Switch voltage v_CH — what sets the part rating

Zero while the switch conducts, and V_s — no more while it blocks.

That bound is the diode's doing. Once D conducts it clamps the terminal to the battery voltage, so the switch sees exactly V_s across it and nothing else. Without that clamp the inductor would drive the node to whatever voltage it liked and destroy the switch on the first turn-off — the same job the freewheeling diode does in a Class A chopper, performed by a diode in a completely different place.

Continuous vs Discontinuous Conduction

Everything so far assumed the current never quite reaches zero. Lower the duty ratio far enough and that assumption breaks — and for a Class B chopper the consequence is not subtle: the braking stops working.

The current bottoms out at I_min. Reduce D and there is less time each cycle to build the current up and more time to dump it into the battery, so I_min falls. At some duty it reaches zero. Below that, the decaying current hits zero part way through the off time; the diode cannot conduct backwards so it blocks, and the current stays at zero until the switch closes again. Three states per period, not two.

Look at the amber gap in the lower pair, and at what v_o does inside it. With no current flowing there is no drop across R and none across L, so the terminals show what is left: the machine's bare emf. v_o falls back to E = 60 V. A third level the two-level analysis never predicted.

The direction of the error is the interesting part, and it is worth comparing against the Class A page. There, discontinuous conduction pulled V_o up toward E. Here it drags V_o down toward E. Same rule in both cases — a zero-current gap exposes the machine's emf and hauls the average toward it — but the ideal line sits below E in one case and above it in the other, so the deviation points opposite ways. One principle, two appearances.

For this machine the boundary sits at D_crit ≈ 0.42, found by solving I_min = 0. Note it is low duty that goes discontinuous here, which is also the opposite end from a Class A chopper — a short T_on gives the machine too little time to build current up, and a long T_off gives the battery plenty of time to take it all back. The general condition is:

Why it matters in practice:

• Gentle braking is unreliable. Light braking means low D, which is exactly where the chopper falls into discontinuous conduction and the braking torque collapses far faster than the duty ratio suggests. Asking for 10 % braking can get you 1 %.
• It gets worse as the vehicle slows. E is proportional to speed. As the machine slows, E falls, the condition above gets harder to satisfy, and regeneration fades out — which is why real electric vehicles blend in friction brakes at low speed. It is not a tuning failure; it is the physics of the topology.
• The control loop's gain shifts underneath a controller that was tuned in continuous conduction.

Key Formulas (with derivation)

All of this follows from the two loop equations already written down, and all of it assumes continuous conduction and ideal devices. Throughout, i is the magnitude of the current flowing out of the machine (so i_o = −i), and τ = L/R.

Average output voltage

Average braking current

In steady state the inductor's average voltage over one period is zero, so the machine's emf is shared between R and the average terminal voltage alone:

Regeneration only happens while this is positive, which needs E > (1 − D)V_s. That is the condition for braking to exist at all — and it is why you can always brake at some duty, no matter how slow the machine, as long as you are willing to push D high enough.

Peak and valley current

Solve the ON exponential and the OFF exponential, then force i(0) = i(T). With a = e^−DT/τ and b = e^{−(1−D)T/τ}:

Peak-to-peak current ripple

Subtract the two. The E/R terms cancel, and what is left is startlingly familiar:

That is the exact same expression as the Class A chopper's ripple — identical, symbol for symbol. Two different circuits, two different duty-ratio laws, two different quadrants, and the same ripple. It is not a coincidence: in both converters the inductor is switched between the same two rails, 0 and V_s, and it is that swing which sets the ripple. Which end the energy is flowing towards makes no difference.

Two things fall out, exactly as they did for Class A:

• The ripple does not depend on E at all — the machine's speed shifts the whole current waveform up or down but does not change the ripple riding on it.
• When τ >> T, expand e^−x ≈ 1 − x and it collapses to the textbook form:

For our example the exact formula gives 16.66 A and the approximation gives 16.67 A — close, because τ/T = 6. Note the worked example sits at D = 0.5, which is the worst-case ripple point.

Source side and power

The energy balance, on the other hand, is exact for ideal devices:

E · I is exact because E is a constant, so the average of E i really is E times the average of i. The copper loss, though, is I_rms²R and never I²R — I_rms exceeds I whenever there is ripple, and using the wrong one quietly under-reports the loss.

Boost relation

Conduction boundary

Device ratings

Control — and Why Braking Current Runs Away

This section is the one that matters if you ever build the thing. A Class B chopper is not merely a Class A with the sign changed; its control characteristic is far more vicious, and the reason is visible the moment you plot it.

The blue curve is well behaved. Above D_crit it sits exactly on the ideal (1 − D)V_s line; below it, discontinuous conduction flattens it toward E. Nothing alarming.

Now read the red curve, which is the one that produces torque. It is not a gentle ramp. It is barely awake below D_crit, reaches 40 A at D = 0.5, and then climbs at 400 A per unit of duty — because I = (E − (1 − D)V_s)/R and R is only a quarter of an ohm. The whole useful braking range of this drive is compressed into roughly D = 0.42 to D = 0.6:

This is why a Class B chopper is essentially never run open-loop on duty ratio. Real regenerative-braking drives close a current loop around it: the operator's brake demand becomes a current reference, a controller measures the actual armature current, and D is whatever the loop needs it to be — moment to moment, as E falls with speed. The duty ratio stops being the control input and becomes an internal variable. The chopper is still doing exactly what this page describes; it is just no longer you choosing D.

The same time-ratio-control options as Class A exist in principle (constant frequency with variable T_on, or variable frequency), and constant-frequency PWM is what is actually used, for the same reasons: predictable ripple, predictable losses, a filter you can design once. But the outer loop is the real story here.

Worked Example

The example used everywhere on this page, worked start to finish — and deliberately the same machine as the Class A worked example, so the two can be read against each other. A DC machine is coupled to a 100 V battery through a Class B chopper switching at 1 kHz. The armature is 0.25 Ω and 1.5 mH, and at the speed of interest it generates 60 V. The duty ratio is 0.5.

The line to pause on is the balance. The machine gives up 2,400 W of mechanical power; 406 W is wasted heating its own winding; and 1,994 W arrives in the battery. Nothing is lost in the chopper itself — an ideal switch is either a short or an open, and neither dissipates. Every watt that does not reach the battery was lost in the machine's own resistance, which no amount of clever switching can avoid.

Advantages & Disadvantages

Advantages

• The energy is recovered, not burned. Roughly 83 % of the machine's mechanical output goes back into the battery instead of into a resistor bank. On an electric vehicle that is directly more range.
• Genuine, controllable braking torque — smooth and stepless, set by the duty ratio, with no friction material to wear out.
• Very simple. One controlled switch and one diode — the same part count and the same voltage and current ratings as a Class A chopper.
• It boosts, so a machine whose emf has dropped well below the battery voltage can still return energy. Braking does not stop the moment E falls under V_s.
• No extra hardware for the braking effort itself — no resistor bank, no contactor to switch it in, and no heat to get rid of.
• Combines trivially with a Class A to make a Class C — the natural upgrade path, at the cost of one more switch and diode.

Disadvantages

• It cannot motor. Braking only. On its own it is half a drive.
• The supply must be able to absorb the returned energy. A battery can; a plain diode rectifier cannot, and the DC-link voltage will climb until something breaks.
• The load must contain its own emf. There is nothing to regenerate from a passive R–L load.
• The braking current runs away with duty ratio — 40 A at D = 0.5, 240 A at D = 1, where the switch simply shorts the machine. It practically demands a closed current loop.
• Regeneration fades at low speed, because E falls with speed and conduction goes discontinuous. Friction brakes are still needed to actually stop.
• Discontinuous conduction at low duty breaks V_o = (1−D)V_s exactly where gentle braking is wanted.
• Nothing is recovered from the winding loss — I_rms²R is gone regardless.

Applications

• Regenerative braking on battery vehicles — forklifts, warehouse tugs, golf carts, electric scooters and milk floats: the battery is right there and delighted to be charged.
• Traction braking on trams, trolleybuses, metro stock and mine locomotives, where a train slowing into every station is a large and very repetitive source of recoverable energy.
• Hoists, cranes and lifts lowering a load — gravity is doing work on the machine, and a Class B chopper turns that into charge instead of brake-shoe heat.
• Overhauling loads generally: conveyor belts running downhill, centrifuges and flywheels being run down under control.
• Battery-to-battery energy transfer — the same circuit boosts from a lower-voltage battery to a higher-voltage one, with no machine involved at all.
• The braking half of a Class C or Class E drive, which is where most real Class B choppers actually live — rarely alone, almost always as one half of a two- or four-quadrant converter.

The selection rule: use a Class B wherever a load only ever needs to be slowed, never driven, and the supply can take the energy back. If the same machine must also motor, you have outgrown it and want a Class C.

Class B vs Class A (and the rest of the family)

The two are mirror images, and laying them side by side is the fastest way to learn either:

And the family as a whole. Each class buys another quadrant with more silicon:

Now the family tree makes sense. Class C is literally a Class A and a Class B sharing one leg — the Class A switch on top, the Class B switch underneath, each with the other's diode across it. That is just an ordinary half-bridge, which is why a Class C costs so little more than a Class A. Two Class C legs make an H-bridge, and that is Class E, which does everything.

Frequently Asked Questions – FAQs

Related Topics

• Class A Chopper (First Quadrant) — the motoring mirror of this page
• Class C Chopper (Two Quadrant) — this page and Class A in one half-bridge leg
• DC-DC Converters (Choppers) — Overview
• Boost (Step-Up) Converter — the same power stage
• Buck (Step-Down) Converter
• DC-DC Duty Cycle Calculator
• PWM Active Front-End Rectifier — for regenerating into the mains
• IGBT — the usual chopper switch
• All Power Electronic Converters