Chapter 11: DC Choppers: Principles and Control Strategies — Power Electronics

Where this sits

Part 3 · DC–DC Converters
Chapter 11 of 30 — the first of Part 3

You should already know

Average and RMS of a switched waveform (Chapter 1), transistor switching and gate drives (Chapter 5), and forced commutation (Chapter 4).

By the end you can

Size a chopper's switching frequency and inductance from a ripple specification, choose a control strategy and defend it, and pick the chopper class a four-quadrant load needs.

Time

≈ 55 min reading · ≈ 50 min problems

i What you'll learn

Why a transformer cannot change a DC voltage, and what a chopper does instead.
The one equation that starts Part 3: \( V_o = D\,V_s \), and why \( V_{o(rms)} = \sqrt{D}\,V_s \) is a different number for a reason.
The three control strategies — constant frequency (PWM), variable frequency, and current-limit — and why industry standardised on the first.
What an inductive load does to the picture: ripple current \( \Delta I = \dfrac{V_s D(1-D)}{fL} \), and the continuous / discontinuous boundary.
The step-up chopper, \( V_o = \dfrac{V_s}{1-D} \) — the first circuit in this book that produces more voltage than it is given.
The five chopper classes A to E, and how to read a quadrant diagram to pick one.

Section 11-1

DC Refuses to Be Transformed

Part 2 solved one problem completely: taking AC from the grid and producing an adjustable DC voltage. Part 3 asks a question that sounds easier and turns out to be harder.

You have 400 V DC. You need 150 V DC. What do you do?

On the AC side this would be trivial — a transformer with the right turns ratio, done. On the DC side the transformer is useless. A transformer works because a changing flux induces a voltage in the secondary. DC produces constant flux, constant flux induces nothing, and the primary simply looks like a short circuit made of copper wire.

So the two obvious alternatives are the two bad ones:

Three ways to get 150 V from a 400 V DC source, delivering 10 A
Method	How it works	Power delivered	Power wasted	Efficiency
Series resistor	Drop 250 V across 25 Ω	1.5 kW	2.5 kW as heat	37.5%
Linear regulator	Transistor held partly on, dropping 250 V	1.5 kW	2.5 kW as heat	37.5%
Chopper	Switch fully on 37.5% of the time, fully off the rest	1.5 kW	tens of watts	> 97%

◇Intuition · the tap you cannot half-open

Imagine filling a bucket from a tap that has only two positions: fully open or fully shut. You need half the flow. What do you do?

You open it for one second, shut it for one second, and repeat. Averaged over a minute you have delivered exactly half. And crucially — the tap is never in the half-open position where it would be dissipating energy by throttling the flow.

That is a chopper. The switch is fully on (nearly zero voltage across it, so \( vi \approx 0 \)) or fully off (nearly zero current through it, so \( vi \approx 0 \)). Either way the switch itself does almost no work. The only losses are the small on-state drop and the brief transitions between the two states.

The price is ripple. The output is not a steady 150 V — it is 400 V and 0 V alternating, averaging 150 V. The whole of Part 3 is, in one way or another, about making that average look steady enough to use.

The circuit is disarmingly simple. One controllable switch in series with the load, and one diode across the load.

The step-down chopper. The switch connects the load to the source for a fraction \( D \) of every period; the freewheeling diode gives the load current a path for the rest.

🔑

The duty ratio, and the two averages it produces

\[ D = \frac{t_{on}}{T} = f\,t_{on}, \qquad V_o = D\,V_s, \qquad V_{o(rms)} = \sqrt{D}\;V_s \]

\( D \) runs from 0 to 1, so a step-down chopper can produce any voltage between zero and \( V_s \) — but never more. \( T = 1/f \) is the chopping period and \( f \) the chopping frequency.

▲Common mistake

"Average is \( DV_s \) and RMS is \( DV_s \) too — it's DC, so they're the same."

Wrong \( V_{o(rms)} = D\,V_s \). At \( D = 0.25 \) with \( V_s = 400 \) V this gives 100 V.

Right \( V_{o(rms)} = \sqrt{D}\,V_s = \sqrt{0.25}\,(400) = 200 \) V — twice as large.

Why the two differ. The output is not a steady \( DV_s \); it is \( V_s \) for a fraction \( D \) of the time and zero for the rest. Squaring before averaging weights the tall parts far more heavily:

\[ V_{o(rms)} = \sqrt{\frac{1}{T}\int_0^{DT} V_s^2\,dt} = \sqrt{D V_s^2} = \sqrt{D}\,V_s \]

When each one is the right number. Use the average for anything that responds to volt-seconds — a motor's back-EMF and therefore its speed, a battery being charged, a capacitor's steady voltage. Use the RMS for anything that responds to heating — the \( I^2R \) loss in a purely resistive load, or the rating of a heating element.

The gap between them is exactly the ripple, and it is largest at small \( D \). This is the same average-versus-RMS trap from Chapter 1, and it catches people just as reliably the second time.

For a purely resistive load the rest follows immediately:

Output power \( P_o = \dfrac{V_{o(rms)}^2}{R} = \dfrac{D V_s^2}{R} \)
Average source current \( I_s = \dfrac{P_o}{V_s} = \dfrac{D V_s}{R} \)
Effective input resistance \( R_i = \dfrac{V_s}{I_s} = \dfrac{R}{D} \)

That last one is worth pausing on. As far as the source is concerned, the chopper has turned a fixed \( R \) into an adjustable resistance \( R/D \) — without any adjustable resistor existing anywhere in the circuit, and without dissipating anything to do it.

▣In practice · why your phone charger is warm, not hot

A 20 W phone charger converts 230 V AC to 5 V DC. If it did the last step linearly — dropping the surplus across a transistor — it would waste several times the power it delivers, and would need a heatsink the size of the charger.

Instead it rectifies to about 325 V DC and then chops, at somewhere between 65 kHz and 200 kHz, through a small transformer. At those frequencies the transformer is the size of a sugar cube instead of a house brick, because core size falls roughly with frequency.

Everything in Part 3 scales this way: higher switching frequency buys smaller magnetics, and costs switching loss. That single trade decides most converter designs, and we will meet it in every chapter that follows.

Section recap. A transformer cannot change DC, and dropping the surplus wastes it. A chopper switches fully on and fully off, so the switch dissipates almost nothing: \( V_o = DV_s \), \( V_{o(rms)} = \sqrt{D}V_s \), and the source sees an effective resistance \( R/D \). The cost is ripple.

Section 11-2

Three Ways to Set the Duty Ratio

\( D = t_{on}/T \) is a ratio of two times, so there are two independent ways to change it — vary \( t_{on} \), or vary \( T \). Both are used, plus a third strategy that sets neither directly.

1 Constant-frequency control (PWM / TRC)

What is varied: \( t_{on} \). The period \( T \) is fixed.
Also called: Time Ratio Control, pulse-width modulation.
How it is generated: compare a control voltage against a fixed-frequency sawtooth. Where the control voltage is higher, the switch is on. Raising the control voltage widens the pulse.
Why it dominates: everything downstream is designed for one known frequency — the filter, the magnetics, the EMI mitigation, the current sensor bandwidth.

2 Variable-frequency control (FM / CLC)

What is varied: \( T \). Either \( t_{on} \) or \( t_{off} \) is held fixed.
Also called: frequency modulation, current-limit control in its hysteresis form.
The problem: to move \( D \) over a wide range, \( f \) must move over a wide range too. A filter that works at 5 kHz may be useless at 300 Hz.
The other problem: the frequency sweeps through the audible band on its way, and the magnetics sing.

3 Current-limit (hysteresis) control

What is controlled: not \( D \) at all — the current directly.
Rule: switch on when the load current falls to \( I_{min} \); switch off when it rises to \( I_{max} \). The duty ratio is whatever it has to be.
The gain: the current is bounded by construction. A short circuit cannot produce more than \( I_{max} \), because the switch turns off the moment it is reached.
The cost: the switching frequency is an output of the circuit, not an input — it varies with load, supply voltage and inductance.

Choosing between the three strategies
Criterion	Constant frequency	Variable frequency	Current limit
Filter design	Straightforward — one frequency	Difficult — must work across the range	Difficult
EMI	Concentrated at \(f\) and harmonics; filterable	Spread across a band; harder to filter, but lower peaks	Spread
Audible noise	Avoidable — put \(f\) above 20 kHz	Sweeps through the audible band	Sweeps
Current protection	Needs a separate loop	Needs a separate loop	Inherent
Dynamic response	Limited by loop bandwidth	Limited by loop bandwidth	Cycle-by-cycle — very fast
Typical use	Almost everything	Some resonant and light-load modes	Servo drives, battery chargers, LED drivers

✦Think about it

A designer chooses variable-frequency control with \( t_{on} \) fixed at 100 μs, and needs \( D \) to range from 0.1 to 0.9.

What frequency range does that force? And what does it imply for the output filter?

Work it through

\( D = t_{on}/T = t_{on} f \), so \( f = D/t_{on} \).

At \( D = 0.9 \): \( f = 0.9/100\,\mu s = 9 \) kHz
At \( D = 0.1 \): \( f = 0.1/100\,\mu s = 1 \) kHz

A 9:1 frequency range. A low-pass filter designed to attenuate 9 kHz adequately gives 19 dB less attenuation at 1 kHz — and 1 kHz is exactly where the duty ratio is smallest and the ripple content is worst. The filter must be designed for the lowest frequency, which means it is heavily oversized for most of the operating range.

This is the whole argument for constant-frequency control in one calculation.

Section recap. Constant-frequency PWM varies \( t_{on} \) and is the industry default because the filter and EMI design have one frequency to handle. Variable frequency forces a wide filter and audible noise. Current-limit control regulates the current directly and is inherently protected, at the price of a frequency you no longer choose.

Section 11-3

What an Inductive Load Changes

A resistive load follows the chopped voltage exactly: the current is a rectangular pulse train too. Real loads are almost never resistive. A DC motor, the most common chopper load of all, is a resistance, an inductance and a back-EMF in series.

The inductance changes everything, for the better. It refuses to let the current jump, so instead of a rectangular current we get a ramp up during \( t_{on} \) and a ramp down during \( t_{off} \) — a triangular ripple riding on a DC average.

◇Intuition · the flywheel

Push a heavy flywheel in short bursts and it does not stutter — it spins at a nearly constant speed with a small periodic variation. The mass smooths what your hands could not.

Inductance is electrical inertia. The switch delivers energy in bursts; the inductor smooths it into a nearly steady current. And the two halves of the cycle have distinct jobs:

Switch on. The source is across the load. The voltage available to push current up is \( V_s - E - iR \), so the current rises.
Switch off. The inductor's stored energy keeps the current flowing — it must go somewhere, and it goes round through the freewheeling diode. The load now sees zero volts, so the current decays.

The freewheeling diode is not optional. Without it, turning the switch off would force the inductor current to zero instantaneously, and \( v = L\,di/dt \) with \( di/dt \to \infty \) means an arbitrarily large voltage across the switch. It destroys the switch on the first turn-off. Every inductive-load chopper has one.

In steady state the current returns to the same value every cycle. That single fact gives us the ripple with almost no algebra, provided the period is short compared with the load's time constant \( \tau = L/R \) — which it always is in a well-designed chopper.

Steady state means the volt-seconds balance Over one period the inductor's net change in current is zero, so the average voltage across \( L \) must be zero. The average output voltage \( DV_s \) therefore appears entirely across \( R \) and \( E \): \( DV_s = E + I_o R \).
During the off-time the load terminals sit at zero volts The freewheeling diode is conducting, so the load sees 0 V. The voltage across the inductance is therefore \( -(E + I_oR) = -DV_s \).
So the current falls at a constant rate

Working

\[ \frac{di}{dt} = \frac{-DV_s}{L} \quad \text{for a time } t_{off} = (1-D)T \]
The fall equals the ripple In steady state, what goes up must come down — so the decay during \( t_{off} \) is the whole peak-to-peak ripple.

Result

\[ \Delta I = \frac{D V_s}{L}(1-D)T = \frac{V_s\,D(1-D)}{fL} \]

🔑

Ripple current, and where it is worst

\[ \Delta I = \frac{V_s\,D(1-D)}{f L}, \qquad \Delta I_{max} = \frac{V_s}{4fL} \ \text{ at } D = 0.5 \]

\( D(1-D) \) is a downward parabola peaking at \( D = 0.5 \). Design the inductor for that worst case and every other duty ratio is safe.

Interactive · ripple, frequency and inductance

A 200 V chopper driving a DC motor: \( R = 0.5\,\Omega \), back-EMF \( E = 80 \) V. Watch what happens to the load current as you change the duty ratio, the switching frequency and the inductance — and find the point where conduction goes discontinuous.

Duty ratio D 0.50

Switching frequency 1000 Hz

Load inductance 5.0 mH

Output V_o100 V

Average current40.0 A

Ripple ΔI10.0 A

ConductionContinuous

—

◈Why the ripple vanishes at both ends

Slide the duty ratio to 0.05 or to 0.95 in the figure above and the ripple almost disappears. That surprises people, because \( D = 0.95 \) sounds like "nearly full power", which sounds like "the extreme case".

The reason is that ripple needs both halves of the cycle.

At \( D \to 1 \) the switch is almost never off, so there is almost no time for the current to decay. \( t_{off} \to 0 \), so \( \Delta I \to 0 \). The output is nearly steady DC at \( V_s \).
At \( D \to 0 \) the switch is almost never on, so the current barely gets a chance to rise. \( t_{on} \to 0 \), so \( \Delta I \to 0 \) again.
In between, both intervals are long enough to matter, and their product \( D(1-D) \) is maximised at exactly \( D = 0.5 \).

The design consequence. Size \( L \) using \( \Delta I_{max} = V_s/(4fL) \). If you size it at your nominal duty ratio and the nominal happens to be 0.8, you will be badly undersized the first time the converter passes through 0.5 — which it does on every start-up.

The exact ripple, valid even when the period is not short compared with \( \tau = L/R \), comes from solving the two exponentials and matching them at the boundaries:

\[ \Delta I = \frac{V_s}{R}\cdot\frac{\left(1-e^{-DT/\tau}\right)\left(1-e^{-(1-D)T/\tau}\right)}{1-e^{-T/\tau}} \]

Expand the exponentials for \( T \ll \tau \) and this collapses to \( V_sD(1-D)/(fL) \), which is why the simple form is used for nearly all design work. The exact form matters only for slow, lightly inductive loads.

🔑

The boundary between continuous and discontinuous conduction

\[ \text{Continuous while } I_o > \frac{\Delta I}{2}, \qquad L_{crit} = \frac{V_s D(1-D)}{2 f I_o} \]

Below \( L_{crit} \) — or below the load current that goes with it — the current reaches zero before the next pulse, the diode stops conducting, and the load terminals float up to the back-EMF. \( V_o = DV_s \) is then no longer true.

▲Common mistake

"Discontinuous conduction just means a bit more ripple. The average is still \( DV_s \)."

Wrong Treating \( V_o = DV_s \) as valid at all loads, and being surprised when a lightly-loaded motor runs fast.

Right In discontinuous conduction \( V_o > DV_s \), rising towards \( V_s \) as the load falls towards zero. The relationship becomes load-dependent.

What actually happens. While the current is flowing the diode clamps the load terminals to zero volts. Once the current reaches zero the diode stops conducting, the switch is off, and nothing is connected to the load. The terminals rise to whatever the load itself sits at — for a motor, its back-EMF \( E \).

So the output waveform now has three levels, not two: \( V_s \) during \( t_{on} \), zero while the current decays, and \( E \) for the rest. The average of a three-level waveform is not \( DV_s \).

Why it matters in a real drive. The speed loop was tuned assuming \( V_o = DV_s \), a fixed relationship between the control signal and the output. In discontinuous conduction that gain changes with load — so a loop that was comfortably stable at full load can oscillate at light load. Drives handle this by adding enough inductance to stay continuous down to a defined minimum current, and by detecting the mode and switching gains when they cannot.

Section recap. Inductance converts the chopped voltage into a nearly steady current with triangular ripple \( \Delta I = V_sD(1-D)/(fL) \), worst at \( D = 0.5 \). If the ripple exceeds twice the average current, conduction goes discontinuous, \( V_o \) rises above \( DV_s \), and the control loop's gain changes underneath it.

Section 11-4

The Step-Up Chopper

Everything so far has produced less voltage than it started with. Rearranging three components produces more — the first circuit in this book that does.

Move the inductor into series with the source, put the switch across the line after it, and put the diode in series towards the load.

◇Intuition · the hydraulic ram

A hydraulic ram pump lifts a little water to a great height using the momentum of a lot of water flowing gently downhill. You let the flow build up, then slam a valve shut — and the moving column's inertia produces a pressure surge far above the supply head.

The step-up chopper does exactly this with current:

Switch on. The source is connected straight across the inductor. Current builds up linearly, \( di/dt = V_s/L \). Energy accumulates in the magnetic field. The load is fed from the output capacitor.
Switch off. The current has nowhere to go except through the diode into the load — but it cannot change instantly. The inductor produces whatever voltage it takes to keep it flowing.
That voltage adds to the source voltage, because the inductor is in series with it. The output is therefore always higher than the input.

The inductor's role has completely changed. In the step-down chopper it was a smoothing element — nice to have, in the way. Here it is the energy-transfer mechanism. Remove it and the circuit does nothing at all.

State the principle In steady state the inductor current is the same at the start and the end of every period, so the average voltage across the inductor over one period is zero. If it were not, the current would drift up or down without limit.
Find the inductor voltage while the switch is on The switch connects the inductor straight across the source, so \( v_L = V_s \) for a time \( DT \).
Find it while the switch is off Now the inductor sits between the source and the output, so \( v_L = V_s - V_o \) for a time \( (1-D)T \).
Set the total volt-seconds to zero

Working

\[ V_s(DT) + (V_s - V_o)(1-D)T = 0 \;\Longrightarrow\; V_s = V_o(1-D) \]

\[ V_o = \frac{V_s}{1-D} \]

Volt-second balance is the single most useful tool in Part 3. It works for every converter in the next four chapters, it needs no calculus, and it takes four lines. Learn it here and Chapters 12 and 13 become almost mechanical.

🔑

Step-up chopper

\[ V_o = \frac{V_s}{1-D} = V_s\,\frac{T}{t_{off}}, \qquad I_s = \frac{I_o}{1-D} \]

Note the second relation. Power is conserved, so if the voltage is multiplied by \( 1/(1-D) \), the input current must be multiplied by the same factor. A boost converter always draws more current than it delivers.

Voltage gain of a step-up chopper, and why the top end is fiction
Duty ratio D	Gain \(1/(1-D)\)	Output from 100 V	Input current for 10 A out	Realistic?
0.00	1.0	100 V	10 A	Yes
0.50	2.0	200 V	20 A	Yes
0.80	5.0	500 V	50 A	Getting hard
0.90	10.0	1000 V	100 A	Rarely
0.99	100.0	10 kV	1000 A	No

▲Common mistake

"Set \( D = 0.99 \) and you get a hundredfold voltage gain — free high voltage."

Wrong Reading \( V_o = V_s/(1-D) \) as a promise that holds all the way to \( D = 1 \).

Right The gain peaks somewhere around \( D \approx 0.8 \) to \( 0.9 \) and then collapses. Practical boost ratios stop at about 4 or 5.

Why the formula lies. It was derived assuming an ideal, lossless circuit. Put the winding resistance \( r_L \) back in and the gain becomes

\[ \frac{V_o}{V_s} = \frac{1}{1-D}\cdot\frac{1}{1 + \dfrac{r_L}{R(1-D)^2}} \]

The second factor is near 1 at low \( D \) and crushes the first as \( D \to 1 \), because \( (1-D)^2 \) in the denominator blows up. Physically: the input current is \( I_o/(1-D) \), so at \( D = 0.99 \) you would need a hundred times the output current flowing through the inductor and the switch — and \( I^2r_L \) rises with the square of that. The copper loss overwhelms the output long before the theoretical gain is reached.

What to do instead when you need a large ratio: use a transformer. That is precisely what Chapter 14's isolated converters are for — a turns ratio gives you the gain without pushing the duty ratio into the region where it destroys itself.

▣In practice · regenerative braking

An electric vehicle slowing down has a motor acting as a generator, producing perhaps 200 V. The battery it must charge sits at 400 V. Current will not flow from a lower voltage into a higher one.

A step-up chopper solves it: the motor is the source, the battery is the load, and \( D \approx 0.5 \) lifts 200 V to 400 V so the energy flows back into the pack instead of into the brake discs.

Notice what the duty ratio now controls. Not the output voltage — the battery fixes that. It controls the current, and therefore the braking torque. Push \( D \) up and the vehicle brakes harder. This is the same converter running the same equation, with the roles of the fixed and free variables exchanged — and it is worth getting comfortable with, because it recurs throughout Parts 6 and 7.

Section recap. Volt-second balance across the inductor gives \( V_o = V_s/(1-D) \) in four lines, and the same tool will handle every converter in Part 3. The gain is real up to about 4 or 5; beyond that, winding resistance and the \( I_o/(1-D) \) input current defeat it, and the answer is a transformer, not a larger duty ratio.

Section 11-5

Four Quadrants: The Chopper Classes

The step-down chopper of Section 11-1 can only produce positive voltage and positive current. That is fine for a fan and useless for a lift, which must also lower a load and brake a descent.

The convention is the same as for the dual converter in Chapter 10: plot output voltage against output current, and each quadrant is a mode of operation.

The four quadrants of chopper operation. Which quadrants a load needs determines which chopper class you must build.

The five chopper classes
Class	Quadrants	Devices	What it can do	Typical load
A	I	1 switch, 1 diode	Step-down only. Power flows source → load.	Fan, pump, unidirectional traction
B	II	1 switch, 1 diode	Step-up only. Power flows load → source. Regeneration.	A generator or braking motor feeding a battery
C	I + II	2 switches, 2 diodes	Motor and brake in one direction. Current may reverse; voltage may not.	Hoists, EV traction, lifts
D	I + IV	2 switches, 2 diodes	Voltage may reverse; current may not. Reverses field, not armature.	Field-reversal DC drives
E	I–IV	4 switches, 4 diodes (H-bridge)	Everything: both directions, motoring and braking in each.	Servo drives, rolling mills, robotics

◈Why Class C is one leg, and how to read any of them

A Class C chopper is just two Class A choppers sharing a load — the same half-bridge leg you already met in Chapter 5, and the same one you will meet again as an inverter leg in Chapter 16. Being able to read a leg is worth more than memorising five circuits.

The upper switch connects the load to \( +V_s \). It can only push current into the load.
The lower switch connects the load to 0 V. It can only pull current out of the load — that is, allow it to build up in the reverse direction.
Each switch has an antiparallel diode, which carries current in the direction its switch cannot.
So the leg as a whole handles either direction of current, but the output voltage is always between 0 and \( +V_s \) — never negative. Hence quadrants I and II only.

To get negative voltage you need a second leg, and the load between the two midpoints. That is Class E, the H-bridge, and it is why four quadrants costs four switches.

The one rule you cannot break: never turn both switches in a leg on at the same time. That connects \( +V_s \) directly to 0 V through two devices — a shoot-through fault, destructive in microseconds. Every practical gate driver enforces a dead time of a few hundred nanoseconds to a few microseconds between one switch turning off and the other turning on. The same rule, and the same dead time, appear in every inverter in Part 4.

✦Think about it

A goods lift must raise a load, lower it under control, and hold it stationary. Its motor turns one way going up and the other way coming down.

Which quadrants does it visit, and what is the minimum chopper class?

Reason it through

Going up, accelerating. Motor drives the load upward: voltage positive, current positive — Quadrant I.

Going up, but decelerating near the top. The motor must resist the motion it is already producing: current reverses while the rotation direction (and so the voltage polarity) does not — Quadrant II.

Coming down. The rotation reverses, so the back-EMF reverses: voltage negative — Quadrant III.

Coming down under load. Gravity is doing the work, so the motor must brake to stop the descent running away: current is positive while voltage is negative — Quadrant IV. This is the interesting one, and it is where a lift spends much of its time.

Answer: all four, so Class E.

And notice what quadrants II and IV mean physically: power is flowing backwards, out of the load. It has to go somewhere. Either back into the supply (regenerative braking, which needs a supply that can accept it) or into a resistor (dynamic braking, which just makes heat). Choosing between those two is a genuine design decision in every lift installation.

Section recap. Class A motors, Class B regenerates, Class C does both in one direction using a single half-bridge leg, Class D reverses voltage but not current, and Class E — the H-bridge — does all four quadrants with four switches. Read the load's quadrant requirement first; the class follows from it.

Section 11-6

A Note on Thyristor Choppers

Everything above assumed a switch that turns off when you tell it to. A thyristor does not — as Chapter 4 established, an SCR conducting DC will never turn off on its own, because there is no natural current zero to wait for.

So a thyristor chopper needs a forced commutation circuit: a pre-charged capacitor, sometimes an inductor, and often an auxiliary thyristor, arranged to drive the main device's current momentarily to zero and hold it in reverse for longer than \( t_q \).

Forced commutation methods used in thyristor choppers
Method	Mechanism	Named after	Weakness
Voltage commutation	A charged capacitor is thrown across the main thyristor, reverse-biasing it	Jones, Morgan chopper	Output voltage overshoots to \(2V_s\); commutation time depends on load current
Current commutation	An \(LC\) circuit rings a reverse current through the main thyristor until the net current is zero	—	More components; \(LC\) must be sized for the worst-case load
Load commutation	The load itself is resonant enough to reverse the current	—	Only works for a narrow load range

▣In practice · why this section is short

Thyristor choppers dominated DC traction for decades — the London Underground 1973 stock, Indian Railways DC locomotives, countless trolleybuses. They were remarkable machines. They are also, for new designs, obsolete.

What the commutation circuit costs:

Components. A capacitor, an inductor and an auxiliary thyristor per main device, all rated for the full circuit — often larger than the main thyristor itself.
Frequency. The commutation interval sets a floor on the period. Thyristor choppers rarely exceed 400 Hz to 1 kHz, so the filter inductors are physically enormous.
Loss. Energy is dumped from the commutation capacitor every single cycle, whether the load needs it or not.
Reliability. If commutation fails once, the main thyristor stays on, and the chopper becomes a permanent short across the supply.

The IGBT made all of this unnecessary. Remove the gate signal and the device turns off — no capacitor, no auxiliary device, no minimum period. Switching frequencies moved from hundreds of hertz to tens of kilohertz, and the magnetics shrank by an order of magnitude.

Why learn it at all, then? Two reasons. First, a great deal of installed traction and industrial plant still runs this way and still needs engineers who understand it. Second, it explains why the rest of Part 3 looks the way it does: every converter from here on assumes a self-commutating switch, and the freedom that assumption grants is easiest to appreciate once you have seen its absence.

Section recap. A thyristor chopper needs a forced-commutation circuit, which limits it to a few hundred hertz and adds components larger than the thyristor itself. Self-commutating devices removed the constraint, and everything from Chapter 12 onwards assumes one.

Section 11-7

Worked Examples

1 Duty ratio, average and RMS

Problem. A step-down chopper operates from 220 V DC into a 10 Ω resistive load at 1 kHz with \( t_{on} = 300\ \mu\text{s} \). Find \( D \), the average and RMS output voltages, the output power, and the effective input resistance.

Solution. \( T = 1/1000 = 1000\ \mu\text{s} \).

Working

\[ D = \frac{300}{1000} = 0.3 \]

\[ V_o = 0.3(220) = 66\ \text{V} \qquad V_{o(rms)} = \sqrt{0.3}\,(220) = 120.5\ \text{V} \]

\[ P_o = \frac{V_{o(rms)}^2}{R} = \frac{120.5^2}{10} = 1452\ \text{W} \]

\[ I_s = \frac{P_o}{V_s} = \frac{1452}{220} = 6.6\ \text{A}, \qquad R_i = \frac{220}{6.6} = 33.3\ \Omega = \frac{R}{D} \]

Note the RMS is nearly twice the average — and it is the RMS that determines the heating of a resistive load. Using 66 V would have underestimated the power by a factor of \( 1/0.3 \).

2 Sizing the inductance from a ripple specification

Problem. A 200 V chopper drives a DC motor with \( R = 0.5\ \Omega \) and back-EMF 80 V, switching at 1 kHz. The ripple current must not exceed 10% of the 40 A rated current at any duty ratio. Find the minimum inductance.

Solution. "At any duty ratio" means design for the worst case, \( D = 0.5 \).

Working

\[ \Delta I_{max} = 0.10 \times 40 = 4\ \text{A} \]

\[ \Delta I_{max} = \frac{V_s}{4fL} \;\Longrightarrow\; L = \frac{V_s}{4f\,\Delta I_{max}} = \frac{200}{4(1000)(4)} = 12.5\ \text{mH} \]

Check the operating point. At \( D = 0.5 \), \( V_o = 100 \) V, so \( I_o = (100-80)/0.5 = 40 \) A — rated current, as intended. With \( \Delta I = 4 \) A the minimum current is \( 40 - 2 = 38 \) A, comfortably continuous.

Now the design lever. Raise the switching frequency to 4 kHz and the required inductance falls to 3.1 mH — a quarter of the iron and copper. This is the trade named in Section 11-1, and it is why converter frequencies rose steadily as switching devices improved.

Confirm with the explorer above: set D = 0.50, f = 1000 Hz, L = 12.5 mH and read ΔI ≈ 4 A.

3 The discontinuous boundary

Problem. For the drive of Example 2 with \( L = 12.5 \) mH, find the load current below which conduction becomes discontinuous at \( D = 0.5 \). What speed does the motor reach at no load?

Solution. Conduction stays continuous while the average exceeds half the ripple.

Working

\[ I_{o(crit)} = \frac{\Delta I}{2} = \frac{4}{2} = 2\ \text{A} \]

\[ \text{At } I_o = 2\ \text{A}: \quad E = V_o - I_oR = 100 - 2(0.5) = 99\ \text{V} \]

Below 2 A the picture changes. The current reaches zero within each period, the diode stops conducting, and the load terminals float up to \( E \). With no load at all the motor accelerates until its back-EMF nearly equals \( V_s \) — not \( DV_s \).

So the no-load speed corresponds to \( E \to 200 \) V rather than 100 V: roughly double the speed the duty ratio was asking for. This is not a subtle effect, and it is why drives specify a minimum load or add inductance to guarantee continuous conduction across the working range.

4 Step-up chopper for regenerative braking

Problem. An EV motor generates 180 V while braking. The battery is at 360 V. The step-up chopper has \( L = 250\ \mu\text{H} \) and switches at 20 kHz. Find the duty ratio, and the ripple in the inductor current if 50 A is delivered to the battery.

Working

\[ V_o = \frac{V_s}{1-D} \;\Longrightarrow\; 360 = \frac{180}{1-D} \;\Longrightarrow\; 1-D = 0.5 \;\Longrightarrow\; D = 0.5 \]

\[ I_L = \frac{I_o}{1-D} = \frac{50}{0.5} = 100\ \text{A} \quad \text{(inductor and motor current)} \]

\[ \Delta I_L = \frac{V_s D}{fL} = \frac{180(0.5)}{20000(250\times10^{-6})} = \frac{90}{5} = 18\ \text{A} \]

Reading the answer. The inductor carries 100 A with 18 A of ripple — 18%, high but normal for a boost inductor, which is usually the physically largest passive component in the converter.

The engineering point. The motor delivers \( 180 \times 100 = 18 \) kW and the battery receives \( 360 \times 50 = 18 \) kW. Same power, different voltage and current — which is exactly what the transformer could not do on DC, and what the chopper does instead.

5 Choosing a class

Problem. A conveyor runs in one direction only but must stop quickly when the emergency stop is pressed, dumping the belt's kinetic energy. It is fed from a rectifier that cannot accept reverse power. Which chopper class, and what else is needed?

Solution — in three steps.

Which quadrants? Running forward is Quadrant I. Braking while still moving forward reverses the current but not the voltage — Quadrant II. Reverse rotation is never required, so quadrants III and IV are not needed.
Which class? Quadrants I and II is exactly Class C — a single half-bridge leg with two switches and two diodes.
Where does the braking energy go? This is the part that is easy to miss. Class C lets the current reverse, so the power flows back out of the motor into the DC link. The rectifier cannot pass it on to the grid, so the link capacitor absorbs it — and its voltage rises until something fails.

What else is needed: a braking chopper. A resistor and a switch across the DC link, turned on when the link voltage exceeds a threshold (typically about 1.15 times nominal). The braking energy becomes heat in the resistor instead of overvoltage on the capacitor.

Size the resistor from the kinetic energy: \( E_k = \tfrac12 J\omega^2 \) must be absorbed within the stopping time, so \( P_{avg} = E_k / t_{stop} \), with the resistance set by \( V_{link}^2/R \le P_{peak} \). This arrangement is standard in industrial drives and is worth recognising on a schematic.

Section 11-8

Summary & Formula Sheet

Chapter 11 in five sentences:

DC cannot be transformed, and dropping the surplus wastes it — so we switch fully on and fully off instead, and the average is \( DV_s \).
Constant-frequency PWM is the default because one known frequency makes the filter, magnetics and EMI design tractable.
Inductance turns the chopped voltage into a nearly steady current with ripple \( V_sD(1-D)/fL \), worst at \( D = 0.5 \).
If the ripple exceeds twice the average, conduction goes discontinuous and \( V_o = DV_s \) stops being true.
Volt-second balance gives the step-up gain \( 1/(1-D) \) in four lines, and will give every converter in Part 3 just as quickly.

Formula sheet · Chapter 11

Duty ratiodefinition

\( D = \dfrac{t_{on}}{T} = f\,t_{on} \)

Step-down outputcontinuous conduction only

\( V_o = D V_s \)

Step-down RMSfor heating of a resistive load

\( V_{o(rms)} = \sqrt{D}\,V_s \)

Effective input resistanceresistive load

\( R_i = R/D \)

Ripple current\(T \ll L/R\); use for design

\( \Delta I = \dfrac{V_s D(1-D)}{fL} \)

Worst-case rippleat \(D = 0.5\) — size \(L\) here

\( \Delta I_{max} = \dfrac{V_s}{4fL} \)

Exact rippleany \(T/\tau\), \(\tau = L/R\)

\( \Delta I = \dfrac{V_s}{R}\dfrac{(1-e^{-DT/\tau})(1-e^{-(1-D)T/\tau})}{1-e^{-T/\tau}} \)

Critical inductanceboundary of continuous conduction

\( L_{crit} = \dfrac{V_s D(1-D)}{2 f I_o} \)

Steady-state currentmotor load, continuous

\( I_o = \dfrac{D V_s - E}{R} \)

Step-up outputideal; real gain caps near 4–5

\( V_o = \dfrac{V_s}{1-D} \)

Step-up input currentpower balance

\( I_s = \dfrac{I_o}{1-D} \)

Step-up with winding lossshows why large \(D\) fails

\( \dfrac{V_o}{V_s} = \dfrac{1}{1-D}\cdot\dfrac{1}{1 + r_L/[R(1-D)^2]} \)

Key terms

Chopper: A DC–DC converter that switches a source on and off rapidly so the average output is a controlled fraction (or multiple) of the input.
Duty ratio, \(D\): On-time divided by period. The single control variable of every converter in Part 3.
Chopping frequency, \(f\): Switching repetitions per second. Higher \(f\) means smaller magnetics and larger switching loss.
Freewheeling diode: The diode across an inductive load that carries the current while the switch is off. Omitting it destroys the switch.
Ripple current, \(\Delta I\): Peak-to-peak variation of the load current within one switching period. Maximum at \(D = 0.5\).
Continuous conduction: The load current never reaches zero. The condition under which \(V_o = DV_s\) is valid.
Discontinuous conduction: The current reaches zero within each period. \(V_o\) rises above \(DV_s\) and becomes load-dependent.
Volt-second balance: In steady state, the average voltage across an inductor over one period is zero. The main analysis tool of Part 3.
Shoot-through: Both switches of a leg conducting simultaneously, shorting the supply. Prevented by dead time.
Dead time: A deliberate gap between one switch turning off and its complement turning on, sized to cover the slowest device turn-off.
Braking chopper: A resistor and switch across the DC link that dissipate regenerated energy when the supply cannot accept it.

Check yourself

Test Yourself

Chapter 11 · six questions answers hidden until you ask

A chopper feeds a 20 Ω heater from 400 V at \( D = 0.25 \). A colleague computes the power as \( 100^2/20 = 500 \) W. What did they get wrong, and what is the right answer?

They used the average output voltage where the RMS was required.

\( V_o = 0.25(400) = 100 \) V is correct as an average, but a heater responds to \( I^2R \) heating, and that depends on the RMS:

\[ V_{o(rms)} = \sqrt{0.25}\,(400) = 200\ \text{V}, \qquad P = \frac{200^2}{20} = 2000\ \text{W} \]

Four times the answer they got — a heater sized on 500 W would fail immediately.

The general rule. \( P = V_{o(rms)}^2/R = DV_s^2/R \) for a resistive load. Note it is linear in \( D \), not quadratic, which is a useful sanity check: at \( D = 1 \) it must give \( V_s^2/R \), and it does.

When would 100 V have been the right number? If the load had been a battery being charged, or a motor whose speed follows the average. The load's physics decides which average you need — always.

Two designers both need \( \Delta I \le 5 \) A from a 300 V chopper. One doubles the inductance; the other doubles the switching frequency. Both succeed. What differs?

\( \Delta I = V_sD(1-D)/(fL) \) depends on the product \( fL \), so doubling either halves the ripple. Electrically the two solutions are identical. Physically they are not.

Doubling the inductance:

Roughly doubles the inductor's core volume and mass, and its cost.
Increases winding resistance, so more conduction loss.
Slows the current loop — a larger \( L \) means the current changes more slowly, so the drive responds more sluggishly to a torque demand.

Doubling the frequency:

Costs nothing in materials — the same inductor, run faster.
Roughly doubles the switching loss in the transistor, so a larger heatsink or a faster device.
Worsens EMI, and moves the spectrum where the filter must work.
Keeps the loop fast.

Which is right depends on where you have margin. If the heatsink is already full, add inductance. If the enclosure is full, raise the frequency. In practice the choice is usually made by the switching device: modern silicon-carbide MOSFETs made high frequency cheap, and inductors shrank accordingly.

The one thing you cannot do is raise both without limit — switching loss rises linearly with \( f \) and eventually consumes the efficiency the chopper existed to provide.

Why does removing the freewheeling diode from an inductive-load chopper destroy the switch, when removing it from a resistive-load chopper does nothing?

Because a resistor's current can stop instantly and an inductor's cannot.

Resistive load. \( i = v/R \). When the switch opens, \( v \) goes to zero and \( i \) goes to zero at the same moment. Nothing objects. The diode has nothing to do and is often omitted.

Inductive load. The current \( I_o \) is flowing and the inductor insists it continues. When the switch opens, that current must find a path. Without a diode the only path is through the switch itself, which is trying to become an open circuit.

\[ v_L = L\frac{di}{dt} \]

As the switch's resistance rises, \( di/dt \) becomes large and negative, so \( v_L \) becomes large and positive — and it adds to the supply voltage across the switch. The voltage rises until something breaks down, which in practice is the switch's avalanche rating, in well under a microsecond.

The energy involved. \( \tfrac12 L I_o^2 \) has to go somewhere. With a diode it circulates and decays gently in the load resistance. Without one it is dissipated in the switch's tiny die area in a fraction of a microsecond.

Worth knowing: this is the same physics as an ignition coil, which uses it deliberately — open the contact breaker and the collapsing field produces tens of kilovolts. Excellent for a spark plug, fatal for a transistor.

A boost converter is specified for \( V_s = 24 \) V and \( V_o = 400 \) V. Is this a reasonable single-stage design?

No. The required gain is \( 400/24 = 16.7 \), which needs \( D = 1 - 1/16.7 = 0.94 \).

Three things go wrong at once:

Input current. \( I_s = I_o/(1-D) = 16.7\,I_o \). For 1 A out, 16.7 A must flow through the inductor and switch. Conduction loss scales with \( I^2 \), so it is 280 times what the output current alone would suggest.
Off-time. At \( D = 0.94 \) and, say, 100 kHz, \( t_{off} = 0.6\ \mu\text{s} \). The diode must recover and the whole energy transfer must complete inside that window. Switching and reverse-recovery losses become the dominant loss term.
Sensitivity. \( dV_o/dD = V_s/(1-D)^2 \). At \( D = 0.94 \) that is \( 24/0.0036 \approx 6700 \) V per unit duty — so a 1% error in duty ratio moves the output by 67 V. The loop is operating on a knife edge.

What to do instead:

A flyback or forward converter with a 1:8 turns ratio, running at a comfortable \( D \approx 0.5 \). The transformer provides the gain; the duty ratio only trims it. This is Chapter 14.
Two cascaded boost stages, each with a gain of about 4 at \( D = 0.75 \). Two stages of 95% each still beat one stage of 70%.

The rule of thumb worth carrying: keep a single boost stage below a gain of about 4, and reach for a transformer beyond it.

A Class C chopper's two switches are driven by complementary signals from a microcontroller. The designer sees no need for dead time — the signals are exact complements. What happens?

The converter destroys itself, usually within the first few switching cycles.

Why "exact complements" is not enough. The gate signals may change at the same instant, but the devices do not. A power MOSFET or IGBT turns on faster than it turns off, typically by a factor of two or more, because turn-off requires removing stored charge:

Turn-on delay plus rise: perhaps 50–150 ns.
Turn-off delay plus fall, including the IGBT current tail: perhaps 200–800 ns.

So when the signals swap, the incoming device is fully on while the outgoing one is still conducting. For that overlap the two devices form a direct short from \( +V_s \) to 0 V, limited only by stray inductance.

The current involved. With a 400 V link and perhaps 100 nH of loop inductance, \( di/dt = V/L = 4000 \) A/μs. Even a 300 ns overlap admits well over a kiloamp — far beyond any device's rating.

The fix is a deliberate dead time: both signals low for a few hundred nanoseconds to a few microseconds between transitions. Every gate-driver IC provides it, most microcontroller PWM peripherals generate it in hardware, and the value is set from the worst-case turn-off time over temperature.

The cost of dead time is a small distortion — during the gap, the output voltage is set by the load current direction rather than by the control signal. Harmless in a chopper; a well-known nuisance in an inverter, where it distorts the output waveform. We will meet it again in Chapter 18.

Why do choppers not use phase control, when Part 2 used it for everything?

Because phase control needs something to be in phase with, and DC has no phase.

What phase control actually depended on. In Part 2 the supply crossed zero 100 times a second. That gave us two gifts:

A reference — a repeating instant from which to measure a delay angle \( \alpha \).
A turn-off mechanism — the current naturally fell to zero every half cycle, so a thyristor switched itself off without help.

A DC supply provides neither. There is no zero crossing to measure from and none to turn a device off with.

So the control variable changes. Instead of when within a cycle to switch (angle \( \alpha \)), we control what fraction of a cycle to be on (duty ratio \( D \)), with the cycle now defined by the converter itself rather than by the grid.

What that independence is worth. The chopper picks its own frequency. Free of the 50 Hz grid, it can run at 20 kHz or 200 kHz — and since magnetic component size falls roughly with frequency, the filter shrinks from a floor-standing choke to something the size of a coin. That single freedom is why a modern laptop adapter fits in a pocket.

The parallel worth noticing: \( \alpha \) and \( D \) play the same role. Both are a single scalar between limits that sets the output voltage. If you were comfortable with \( V_{dc} = 1.35V_L\cos\alpha \), you already have the right mental model for \( V_o = DV_s \).

The explorer shows ripple falling as duty ratio approaches 0.95. Does that mean a converter running at \( D = 0.95 \) needs no inductor?

No — and the reasoning behind the question is a common and expensive error.

What is true: at \( D = 0.95 \), \( \Delta I = V_s(0.95)(0.05)/(fL) \), which is only 19% of the ripple at \( D = 0.5 \). In that steady state the inductor is barely working.

Why it does not follow that you can shrink it:

Start-up passes through \( D = 0.5 \). Every time the converter powers up, ramps, or recovers from a transient, the duty ratio sweeps the whole range. An inductor sized for \( D = 0.95 \) sees five times its design ripple on the way, and may saturate.
Saturation is not graceful. A saturating inductor's inductance collapses, so \( di/dt = v/L \) increases sharply, which saturates it further. The current runs away within a single switching period, faster than most protection responds.
The input voltage moves. \( D \) is not a free choice — it is whatever the loop needs to hold the output. If \( V_s \) rises 30%, the loop reduces \( D \) accordingly, and you may land near 0.5 whether you planned to or not.

The design rule: size the inductor for the worst case across the whole operating envelope — which for a step-down chopper is \( D = 0.5 \) — and then verify the saturation current with margin above the peak, not the average.

The one exception is a converter whose duty ratio is genuinely constrained by construction, such as a fixed-ratio bus converter. Those exist, and they do use smaller inductors. But it must be a designed constraint, not an observation about the nominal operating point.

Practice

Problems

Three habits to carry through Part 3:

Check the conduction mode before using \( V_o = DV_s \). It is a continuous-conduction result, and light load is exactly when it fails.
Size magnetics at \( D = 0.5 \), not at the nominal operating point.
Use volt-second balance rather than memorised gain formulas. It is faster and it generalises.

Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design questions worth discussing in a tutorial.

A step-down chopper on 110 V DC feeds a 5 Ω resistive load at 2 kHz with \( t_{on} = 150\ \mu\text{s} \). Find \( D \), \( V_o \), \( V_{o(rms)} \), the output power and the effective input resistance.
A 240 V chopper drives a load of \( R = 1\ \Omega \), \( L = 8 \) mH, \( E = 100 \) V at 500 Hz with \( D = 0.6 \). Find the average current, the ripple current, and the maximum and minimum instantaneous currents.
For the drive of Problem 2, find the switching frequency at which the ripple would be halved, and the inductance that would achieve the same result at the original frequency.
A 400 V chopper must limit ripple to 8% of a 25 A rating at all duty ratios, switching at 5 kHz. Find the minimum inductance and state the duty ratio at which the requirement binds.
A step-up chopper raises 48 V to 150 V. Find \( D \), and the inductor current if 8 A is delivered to the load.
A 300 V chopper drives a motor with \( R = 0.4\ \Omega \), \( L = 6 \) mH, \( E = 150 \) V at \( D = 0.6 \) and 1 kHz. Find the load current below which conduction goes discontinuous, and explain what happens to the motor speed below it.
Show that for a step-up chopper the input current ripple and the inductor current ripple are the same quantity, and explain why this makes the input filter of a boost converter easier to design than that of a buck converter.
A designer proposes a variable-frequency chopper with \( t_{off} \) fixed at 200 μs, requiring \( D \) from 0.2 to 0.8. Find the frequency range, and comment on the filter and audible-noise consequences.
A Class C chopper drives a hoist. Sketch the current path in all four combinations of switch state and current direction, and identify which device conducts in each. Hence explain why the antiparallel diodes are not optional.
A 415 V DC link feeds a conveyor drive that must stop a 40 kg·m² inertia from 1500 rpm in 3 seconds. Estimate the average braking power, size a braking resistor, and state its peak power rating if the chopper triggers at 480 V.
A boost converter is required to produce 380 V from a solar panel whose voltage falls from 320 V at low irradiance to 180 V at high current. Find the duty-ratio range required, comment on whether a single stage is sensible, and propose an alternative if it is not.
An engineer must choose between a 1 kHz thyristor chopper and a 15 kHz IGBT chopper for a 100 kW traction application. Compare them on filter size, efficiency, audible noise, component count and failure mode, and make a recommendation with reasons. Under what circumstances would the thyristor version still be the right answer?