Solved Problems · Set 33

DC Motor Drives

Part 6 · Electric Drives — the chopper of Set 17 finally meets the machine it was built for. Six problems on the drive that taught everyone how drives work.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 33 — DC Motor Drives

Thirty-two sets have designed converters against assumed requirements. A duty ratio produced a certain voltage, a modulation index reached 415 V because a motor was rated for it — but nothing has asked what the machine does with what it is given, or where the requirement came from in the first place.

The DC machine is where that question is easiest to answer, because its equations are almost trivially simple: torque is proportional to armature current and speed is proportional to armature voltage, with one constant linking both. Everything that vector control will later work hard to achieve in an induction machine, the DC motor does by construction — which is why it defined how drives are built long after better machines existed.

Part 6 · Chapter 25 · 6 solved problems

i Method Recap
  • Two equations describe the separately excited machine, sharing one constant:

    \[ E_b = K_a\Phi\,\omega, \qquad T = K_a\Phi\,I_a, \qquad V_a = E_b+I_aR_a \]
  • Speed follows the armature voltage; torque follows the armature current:

    \[ \omega = \frac{V_a-I_aR_a}{K_a\Phi} \]
  • A chopper's ripple is worst at \(D = 0.5\):

    \[ \Delta I = \frac{D(1-D)V_{dc}}{f_sL_a} \]
  • Regenerative braking needs \(V_a\) below \(E_b\):

    \[ V_a = E_b-I_aR_a, \qquad E_{recovered} = \tfrac12J\omega^2-\int I_a^2R_a\,dt \]
  • A controlled rectifier's ripple grows sharply with firing angle, and the smoothing inductance must hold the current continuous:

    \[ L \ge \frac{\hat V_6}{2\pi f_{ripple}I_{min}} \]
  • Cascade control separates the loops by a decade:

    \[ f_{c,current} \approx 10f_{c,speed}, \qquad \tau_i = \tau_a\ \text{(pole cancellation)} \]
Problem 1CoreThe Machine Itself

A separately excited DC motor is rated 220 V, 50 A, 1500 rpm with \(R_a = 0.25\ \Omega\). Find the back EMF, the machine constant, the rated torque and efficiency, the speed at half torque, and the speed regulation from no load to full load.

1591 1500 Va = 220 V 176 V 132 V 66 N·m T → regulation 6.0%
Parallel, nearly flat curves — voltage sets speed, torque barely disturbs it
Solution

The back EMF at rated conditions:

\[ E_b = V_a-I_aR_a = 220-(50)(0.25) = 207.5\ \text{V} \]

Ninety-four per cent of the terminal voltage is back EMF. The armature drop is a small correction, which is exactly why the speed–torque curve is nearly flat.

The machine constant:

\[ \omega = \frac{2\pi(1500)}{60} = 157.1\ \text{rad/s} \]
\[ K_a\Phi = \frac{E_b}{\omega} = \frac{207.5}{157.1} = 1.321\ \text{V}\!\cdot\!\text{s/rad} \]

One constant, and it does double duty: multiply by speed for back EMF, multiply by current for torque. That is a consequence of energy conservation — the units V·s/rad and N·m/A are the same thing.

Torque and power:

\[ T = K_a\Phi I_a = (1.321)(50) = 66.05\ \text{N}\!\cdot\!\text{m} \]
\[ P_{mech} = E_bI_a = (207.5)(50) = 10.4\ \text{kW} \]
\[ \text{check: } T\omega = (66.05)(157.1) = 10.4\ \text{kW}\ \checkmark \]
\[ \eta = \frac{10{,}375}{11{,}000} = 94.3\% \quad\text{(armature only)} \]

The 625 W of copper loss is the whole armature-circuit story. Field excitation, friction, windage and iron loss are extra — a real machine of this rating runs at about 88%.

Speed at half torque, same terminal voltage:

\[ I_a = 25\ \text{A} \;\Longrightarrow\; E_b = 220-6.25 = 213.75\ \text{V} \]
\[ \omega = \frac{213.75}{1.321} = 161.8\ \text{rad/s} = 1545\ \text{rpm} \]

Halving the load raises the speed by 3% — and it rises because the armature drop halves, not because anything was commanded. The machine self-regulates.

Speed regulation:

\[ \omega_{NL} = \frac{220}{1.321} = 166.5\ \text{rad/s} = 1590\ \text{rpm} \]
\[ \text{regulation} = \frac{1590-1500}{1500} = 6.0\% \]

Six per cent is entirely due to \(I_aR_a\), so it can be eliminated by a feedback loop that raises \(V_a\) as the current rises — or even by open-loop \(IR\) compensation, adding \(I_aR_a\) to the voltage command. That simple trick is why DC drives held zero-speed-error positions long before encoders were cheap.

The two control handles:

VaryEffectRegion
Armature voltage\(\omega \propto V_a\), \(T_{max}\) constantconstant torque, 0 to base speed
Field current\(\omega \propto 1/\Phi\), \(T \propto \Phi\)constant power, above base speed

Two independent inputs giving two independent regions — and the second is available without any extra power hardware, because the field circuit carries only a few per cent of the machine's current. Set 34 will find that an induction machine achieves the same split with far more effort.

One constant links voltage to speed and current to torque, and the two are independent. That decoupling is free in a DC machine because the commutator holds the armature MMF perpendicular to the field mechanically. Vector control exists to reproduce it in a machine that has no commutator — which is the whole story of Sets 34 and 35 in one sentence.
Answera\(E_b = 207.5\ \text{V},\ K_a\Phi = 1.321\)   b\(T = 66.05\ \text{N}\!\cdot\!\text{m},\ \eta = 94.3\%\)   c 1545 rpm at half torque   d regulation 6.0%
Problem 2CoreFeeding It From a Chopper

The same motor is fed from a buck chopper on a 400 V DC link. Find the duty ratio for rated operation and the maximum speed available, then compute the armature current ripple at 1 kHz and at 5 kHz given \(L_a = 6.5\) mH.

Solution

The duty ratio is set directly by the required armature voltage:

\[ D = \frac{V_a}{V_{dc}} = \frac{220}{400} = 0.55 \]

The chopper of Set 17 has become a speed controller with no change of any kind — the same volt-second balance, the same duty ratio, now interpreted as a speed command.

The maximum speed at rated torque:

\[ D = 1:\ V_a = 400,\ E_b = 400-12.5 = 387.5\ \text{V} \]
\[ \omega = \frac{387.5}{1.321} = 293.3\ \text{rad/s} = 2801\ \text{rpm} \]

Nearly twice base speed, but the machine's commutator and its 220 V insulation were not designed for it. In practice the armature voltage is capped at its rating and the extra speed comes from field weakening — the second control handle from Problem 1.

The current ripple. The chopper output is a square wave of \(V_{dc}\) against a smooth back EMF, so the inductance sees the difference:

\[ \Delta I = \frac{D(1-D)V_{dc}}{f_sL_a} \]
\[ f_s = 1\ \text{kHz}:\ \Delta I = \frac{(0.55)(0.45)(400)}{(1000)\left(6.5\times10^{-3}\right)} = 15.2\ \text{A} \]
\[ \frac{15.2}{50} = 30\% \quad\text{— far too much} \]

Thirty per cent of rated current as ripple, at 1 kHz. Note the \(D(1-D)\) shape, which peaks at \(D = 0.5\) — so mid-speed is the worst case, and this operating point at 0.55 is close to it.

Raising the frequency:

\[ f_s = 5\ \text{kHz}:\ \Delta I = 3.05\ \text{A} = 6.1\% \]
\(f_s\)\(\Delta I\)% of ratedVerdict
500 Hz30.5 A61%discontinuous at part load
1 kHz15.2 A30%excessive
2 kHz7.6 A15%marginal
5 kHz3.05 A6.1%good

Why the ripple matters more than it looks. It costs in four separate ways:

ConsequenceMechanism
Extra copper loss\(I_{rms}^2R_a\) exceeds \(I_{dc}^2R_a\)
Torque ripple\(T \propto I_a\) directly
Commutator sparkingripple current in the commutating coil
Discontinuous conduction at light load\(\Delta I/2 > I_a\) — Problem 4

The third row is specific to DC machines and is what limits their life. A commutator segment carries the ripple as well as the mean, and the resulting sparking erodes both brush and copper — which is why DC drives specify a maximum ripple, typically 5–10%, rather than merely tolerating it.

If the frequency cannot be raised — for a large drive where switching loss binds — the alternative is added inductance:

\[ L_{total} = \frac{D(1-D)V_{dc}}{f_s\Delta I_{target}} = \frac{99}{(1000)(5)} = 19.8\ \text{mH} \]

So 13.3 mH of smoothing inductor added to the armature's 6.5. That inductor is a real cost and it slows the current loop — the time constant rises from 26 ms to 79 — so raising the frequency is preferable whenever the switching loss allows it.

The chopper needed no modification at all to become a drive. Same topology, same volt-second balance, same \(D(1-D)\) ripple peaking at half duty — only the interpretation changed, from output voltage to shaft speed. The one new constraint is the commutator, which cares about ripple in a way a resistor never did.
Answera\(D = 0.55\), up to 2801 rpm   b\(\Delta I = 15.2\ \text{A}\) (30%) at 1 kHz   c\(3.05\ \text{A}\) (6.1%) at 5 kHz
Problem 3Exam levelAll Four Quadrants

The drive must brake regeneratively from 1500 rpm at rated current. Find the required duty ratio and the power returned, then compute the deceleration time and the energy actually recovered for an inertia of \(0.5\ \text{kg}\!\cdot\!\text{m}^2\). Identify the four quadrants and the converter each requires.

Solution

What braking requires. To reverse the current while the speed and back EMF are unchanged, the applied voltage must fall below the back EMF:

\[ I_a = \frac{V_a-E_b}{R_a} < 0 \;\Longleftrightarrow\; V_a < E_b \]
\[ V_a = E_b-I_aR_a = 207.5-12.5 = 195\ \text{V} \;\Longrightarrow\; D = \frac{195}{400} = 0.4875 \]

A change of duty ratio from 0.55 to 0.4875 — six per cent — takes the machine from full motoring to full braking. The transition is that small because the armature drop is that small, which is also why the current loop must be fast and well controlled.

The power returned:

\[ P = V_aI_a = (195)(50) = 9.75\ \text{kW into the link} \]
\[ T_{brake} = K_a\Phi I_a = 66.05\ \text{N}\!\cdot\!\text{m} \]

The same torque as motoring, in the opposite sense. That symmetry is not automatic in other machines and is one reason DC drives dominated hoisting and traction for so long.

The deceleration:

\[ \alpha = \frac{T}{J} = \frac{66.05}{0.5} = 132.1\ \text{rad/s}^2 \]
\[ t = \frac{\omega}{\alpha} = \frac{157.1}{132.1} = 1.19\ \text{s} \]

The energy actually recovered. Start with the stored kinetic energy and subtract what is lost:

\[ E_{kinetic} = \tfrac12J\omega^2 = \tfrac12(0.5)(157.1)^2 = 6169\ \text{J} \]
\[ E_{loss} = I_a^2R_at = (625)(1.19) = 743\ \text{J} \]
\[ E_{recovered} = 6169-743 = 5426\ \text{J} = 88\% \]

Eighty-eight per cent recovered, and the loss is fixed at 625 W for the whole braking period because the current is held constant. Braking harder recovers more, because the same loss runs for less time.

The four quadrants, and what each demands of the converter:

Quadrant\(V_a\)\(I_a\)OperationConverter needed
I++forward motoringbuck chopper
II+forward brakingtwo-quadrant (Set 17)
IIIreverse motoringfull bridge
IV+reverse brakingfull bridge

The H-bridge of Set 17 covers all four with no additions — which is exactly what it was built for, and why that set's four-quadrant chopper was worth working through before any machine appeared.

Where the recovered energy goes, and the complication nobody mentions:

Link supplied byCan absorb regeneration?Solution
Diode bridgeno — diodes blockbraking resistor, or the link overvolts
Batteryyescharges — ideal
Active front endyesreturns it to the supply
Dual converteryesSet 16 — inversion mode

The first row catches most people. A drive fed from an ordinary diode bridge cannot send power back — the 9.75 kW has nowhere to go, so it charges the link capacitor until the drive trips on overvoltage. A braking resistor with its own chopper dumps it as heat, which is standard on general purpose drives and wastes every joule of the 5426.

Regeneration is easy in the machine and hard in the supply. Reversing 10 kW of power flow took a 6% change in duty ratio — but unless the link can absorb it, the energy charges the capacitor until the drive trips. Whether a drive regenerates is decided by its front end, not by its motor bridge.
Answera\(D = 0.4875\), 9.75 kW returned   b\(t = 1.19\ \text{s}\)   c 5426 J of 6169 J recovered — 88%   d full bridge for all four quadrants
Problem 4Exam levelWhen the Current Goes Discontinuous

The same motor is instead fed from a three-phase full converter on a 415 V supply. Find the firing angle for rated speed, compute the sixth-harmonic ripple voltage at that angle, and determine the smoothing inductance needed to hold conduction continuous down to 10% of rated current. Explain what happens without it.

Solution

The firing angle:

\[ V_{dc0} = \frac{3\sqrt2\,V_{LL}}{\pi} = 1.35(415) = 560.4\ \text{V} \]
\[ \cos\alpha = \frac{220}{560.4} = 0.3925 \;\Longrightarrow\; \alpha = 66.9^\circ \]

A large firing angle, and Set 14 established what that implies for the supply: a displacement factor of only 0.393, and a substantial reactive burden. Chopper-fed drives avoid this entirely — their input rectifier runs uncontrolled.

The ripple voltage, and its dependence on \(\alpha\). The dominant component is the sixth harmonic of the supply at 300 Hz:

\[ \alpha = 0:\ \hat V_6 = 32.0\ \text{V} \]
\[ \alpha = 66.9^\circ:\ \hat V_6 = 177.2\ \text{V} \]

A factor of 5.5. This is the fact that makes controlled-rectifier drives awkward: the ripple is smallest where the output is largest, and grows dramatically as the drive is turned down — so the worst ripple occurs at low speed, where the machine is least able to filter it and where the mean current is smallest.

The inductance required. Conduction stays continuous while the mean current exceeds half the peak-to-peak ripple:

\[ L \ge \frac{\hat V_6}{2\pi f_6I_{min}} = \frac{177.2}{2\pi(300)(5)} = 18.8\ \text{mH} \]
\[ L_{added} = 18.8-6.5 = 12.3\ \text{mH} \]

Three times the armature's own inductance, added purely to keep the current continuous at light load. That inductor is physically large — it carries 50 A — and it is a real fraction of the drive's cost.

What happens without it. When the current becomes discontinuous the whole analysis changes:

\[ \text{DCM: } V_a > V_{dc0}\cos\alpha \;\Longrightarrow\; \omega\ \text{rises} \]
ConsequenceDetail
Speed rises sharply at light loadthe machine's EMF holds the terminal up between pulses
Regulation collapsesfrom 6% to tens of per cent
Loop gain changesthe transfer function is entirely different in DCM
Torque pulsationcurrent reaches zero every 60°
Commutator wearlarge peak-to-mean current ratio

The same nonlinearity Set 13 found in a semiconverter and Set 20 in a buck-boost — a converter in DCM is a different plant, with a different gain and a different pole. A speed loop tuned in CCM becomes underdamped or unstable when the drive drops into DCM, which happens exactly when the load is removed.

Chopper against controlled rectifier:

PropertyControlled rectifierChopper from a diode bridge
Ripple frequency300 Hz5 kHz — whatever is chosen
Smoothing inductance12 mH addedoften none
Input displacement factor\(\cos\alpha\) — 0.39 herefixed, near unity
Current loop bandwidth~30 Hz (limited by 300 Hz sampling)~500 Hz
Regenerationneeds a dual converterneeds a bridge and a receptive link
Device count6 thyristors6 diodes + 1–4 switches
Power ceilingvery highmoderate

The fourth row is decisive for performance. A phase-controlled converter can only correct its output once per 60° of the supply — six times per cycle — so its current loop bandwidth is capped near 30 Hz however good the controller. A 5 kHz chopper reaches 500 Hz, which is why servo-quality DC drives are always chopper-fed.

A controlled rectifier's ripple is worst exactly where it can least be tolerated. The sixth-harmonic amplitude grows 5.5-fold from \(\alpha = 0\) to 67°, so turning the drive down increases the ripple while reducing the mean current — a double movement towards discontinuous conduction that a large smoothing inductor exists solely to prevent.
Answera\(\alpha = 66.9^\circ\)   b\(\hat V_6 = 177\ \text{V}\), against 32 V at \(\alpha = 0\)   c\(L = 18.8\ \text{mH}\) — add 12.3 mH   d without it, speed rises and the loop gain changes
Problem 5ChallengeTwo Loops, One Inside the Other

Design cascade control for the chopper-fed drive: an inner current loop at 500 Hz and an outer speed loop at 50 Hz. Find the PI parameters for the current loop by pole cancellation, then show why the speed loop's proportional gain implies that the drive spends most transients in current limit.

Solution

Why cascade rather than one loop. The armature current is what must be limited, and it is also what produces torque:

\[ \text{speed error} \to \textbf{current reference} \to \text{voltage} \to \text{current} \to \text{torque} \]

Putting the current in an inner loop makes limiting it trivial — simply clamp the reference — and it linearises the plant seen by the speed loop, which then controls torque directly rather than voltage. Every modern drive of every type uses this structure.

The current-loop plant:

\[ \frac{I_a(s)}{V_a(s)} = \frac{1}{R_a+sL_a} = \frac{1/R_a}{1+s\tau_a}, \qquad \tau_a = \frac{L_a}{R_a} = 26\ \text{ms} \]

A single pole at 6.1 Hz. The back EMF also feeds in, but it varies at mechanical speed — hundreds of times slower than the current loop — so it is treated as a slowly varying disturbance that the integrator rejects.

Pole cancellation. Place the PI's zero on the plant's pole:

\[ \tau_i = \tau_a = 26\ \text{ms} \;\Longrightarrow\; L(s) = \frac{K_pV_{dc}}{R_a\tau_as} \]
\[ \omega_c = 2\pi(500) = 3142\ \text{rad/s} \;\Longrightarrow\; K_p = \frac{\omega_cR_a\tau_a}{V_{dc}} = 0.0511 \]
\[ K_i = \frac{K_p}{\tau_a} = 1.96 \]

Cancelling the pole leaves a pure integrator, whose phase margin is 90° before other delays are counted — the PWM transport delay and the sampling take perhaps 25° of that, leaving a comfortable 65°. Set 23 built the same argument for a voltage regulator.

The speed-loop plant, with the current loop treated as unity:

\[ \frac{\omega(s)}{I_a(s)} = \frac{K_a\Phi}{Js} = \frac{1.321}{0.5s} = \frac{2.642}{s} \]
\[ \omega_{c,speed} = 2\pi(50) = 314\ \text{rad/s} \;\Longrightarrow\; K_{p,\omega} = \frac{314}{2.642} = 119\ \text{A per rad/s} \]

Read that gain again. It demands 119 A for every rad/s of error, against a machine rated at 50:

\[ \text{error of } 0.42\ \text{rad/s (4 rpm)} \;\Longrightarrow\; \text{full rated current} \]

So any speed change larger than four revolutions per minute saturates the current reference immediately. The linear design describes only the last few rpm of any transient; everything before that is the drive accelerating at its current limit.

Which is exactly the intended behaviour:

Phase of a speed stepWhat governs
Large errorcurrent limit — constant acceleration
Approaching the targetspeed loop comes out of saturation
Final settlingthe linear design above
\[ \text{acceleration in limit} = \frac{T_{max}}{J} = \frac{66.05}{0.5} = 132\ \text{rad/s}^2 \]

Time from standstill to 1500 rpm is therefore 1.19 s, set entirely by the current limit and the inertia — not by any controller gain. The speed loop's job is to place the limit and then land accurately, not to shape the acceleration.

The one thing this structure must handle:

\[ \text{integrator wind-up during saturation} \;\Longrightarrow\; \text{overshoot on arrival} \]

With the current reference clamped for over a second, the speed integrator accumulates error the whole time and then must unwind, producing a large overshoot. Anti-windup — freezing or back-calculating the integrator while the output is limited — is not an optional refinement in a cascade drive; it is required for the loop to behave at all.

The speed loop's linear design describes only the last few rpm. A gain of 119 A per rad/s saturates on a four-rpm error, so every real transient is constant-current acceleration set by \(T_{max}/J\) — and the controller's most important feature is not its bandwidth but its anti-windup.
Answera\(\tau_i = 26\ \text{ms},\ K_p = 0.051,\ K_i = 1.96\)   b\(K_{p,\omega} = 119\) — saturates on a 4 rpm error   c acceleration \(= 132\ \text{rad/s}^2\) in limit
Problem 6ChallengeAn Elevator

Design the drive for a passenger elevator: 1000 kg worst-case imbalance between car and counterweight, 1.5 m/s, a 0.5 m sheave through a 20:1 gearbox at 95% efficiency. Find the motor rating, identify which quadrant each operating case falls in, and specify the converter and its front end.

Solution

The mechanical requirement:

\[ F = mg = (1000)(9.81) = 9810\ \text{N} \]
\[ T_{sheave} = Fr = (9810)(0.25) = 2453\ \text{N}\!\cdot\!\text{m} \]
\[ \omega_{sheave} = \frac{v}{r} = \frac{1.5}{0.25} = 6.0\ \text{rad/s} \]

Referred to the motor:

\[ T_{motor} = \frac{2453}{(20)(0.95)} = 129\ \text{N}\!\cdot\!\text{m} \]
\[ \omega_{motor} = (6.0)(20) = 120\ \text{rad/s} = 1146\ \text{rpm} \]
\[ P = (129)(120) = 15.5\ \text{kW} \]

A 15 kW machine at about 1150 rpm — and note that this is the worst-case rating. A properly counterbalanced elevator, with the counterweight at car plus half payload, spends most of its life at a fraction of it.

Now the quadrants, which is where an elevator becomes interesting:

CaseHeavier sideDirectionQuadrantPower flow
Full car, upcarupImotoring
Full car, downcardownIVgenerating
Empty car, upcounterweightupIIgenerating
Empty car, downcounterweightdownIIImotoring

All four quadrants, in ordinary passenger service — not as an occasional braking event but as half of every journey. An elevator is the textbook four-quadrant load, and it is why hoisting applications drove four-quadrant converter development.

Which fixes the converter:

\[ \text{full H-bridge on the armature: } \pm V_a,\ \pm I_a \]
ElementSpecificationReason
Armature converterfull H-bridge, 4-quadrantall four cases above
Switching frequency5 kHzripple 6% — Problem 2
Current loop500 HzProblem 5
Speed loop50 HzProblem 5
Front endactive, or braking resistorhalf of all journeys regenerate
Position controlouter loop, encoderfloor levelling to ±3 mm
Brake controlsequenced with torquehold torque before releasing

The last row is a safety requirement rather than a control one. The mechanical brake must not release until the drive is already producing holding torque, or the car drops before the loop catches it — the characteristic lurch of a badly commissioned lift.

Why the front end matters so much here. Unlike a fan or a conveyor, an elevator regenerates constantly:

\[ \text{typical recovery} \approx 25\text{–}35\%\ \text{of consumed energy} \]
Front endRegenerated energyConsequence
Diode bridge alonenowhere to goovervoltage trip
Diode bridge + braking resistordissipated as heatmachine room heats up; energy wasted
Active front endreturned to the supply~30% energy saving
Common DC bus, multiple liftsone car feeds anotherbest in a bank of lifts

The last row is worth noting. In a building with several lifts on a shared DC bus, a descending full car directly supplies an ascending one — the energy never touches the supply at all, and no active front end is needed to capture most of it.

Why a modern lift is not DC. Everything above works, and almost nobody builds it:

\[ \text{brushes and commutator} \;\Longrightarrow\; \text{scheduled maintenance in a shaft head} \]

A DC machine needs brush inspection and commutator servicing, in a machine room at the top of a building, for the life of the installation. A gearless permanent-magnet synchronous machine under vector control delivers the same four-quadrant torque with no brushes, no gearbox and higher efficiency — which is Set 36. The DC drive taught the control structure; the structure outlived the machine.

An elevator uses all four quadrants in normal service, not as an exception. A full car descending and an empty car ascending both generate — half of every round trip — so the front end is not an efficiency refinement but a functional requirement. Half the journeys simply do not work without somewhere for the power to go.
Answera\(T = 129\ \text{N}\!\cdot\!\text{m}\) at 1146 rpm, 15.5 kW   b all four quadrants in normal service   c H-bridge plus an active front end or braking resistor
Formulas

Key Formulas

QuantityRelationNotes
Back EMF\(E_b = K_a\Phi\omega\)
Torque\(T = K_a\Phi I_a\)Same constant
Armature equation\(V_a = E_b+I_aR_a\)Steady state
Speed\(\omega = \dfrac{V_a-I_aR_a}{K_a\Phi}\)
Speed regulation\(\dfrac{I_aR_a}{E_b}\)6.0% here
Chopper drive\(V_a = DV_{dc}\)
Armature ripple\(\Delta I = \dfrac{D(1-D)V_{dc}}{f_sL_a}\)Worst at \(D = 0.5\)
Braking voltage\(V_a = E_b-I_aR_a\)\(V_a < E_b\)
Braking time\(t = J\omega/T\)Constant current
Energy recovered\(\tfrac12J\omega^2-I_a^2R_at\)88% here
Rectifier drive\(V_a = 1.35V_{LL}\cos\alpha\)Six-pulse
Ripple growth\(\hat V_6\): 32 V at \(\alpha=0\), 177 V at 67°5.5×
Smoothing inductance\(L \ge \dfrac{\hat V_6}{2\pi f_6I_{min}}\)18.8 mH here
Armature time constant\(\tau_a = L_a/R_a\)26 ms
Current PI\(\tau_i = \tau_a,\ K_p = \dfrac{\omega_cR_a\tau_a}{V_{dc}}\)Pole cancellation
Speed plant\(K_a\Phi/(Js)\)Pure integrator
Acceleration in limit\(T_{max}/J\)Not set by gains
Pitfalls

Common Mistakes

  1. Using terminal voltage instead of back EMF for speed. The \(I_aR_a\) drop is the entire speed regulation — Problem 1.

  2. Ignoring the commutator's ripple limit. Ripple current causes sparking and brush wear, independent of the loss — Problem 2.

  3. Evaluating chopper ripple at the operating duty rather than the worst case. It peaks at \(D = 0.5\) — Problem 2.

  4. Assuming a drive can regenerate because its bridge can. A diode front end blocks the return path entirely — Problems 3 and 6.

  5. Computing recovered energy from kinetic energy alone. The armature loss runs for the whole braking period — Problem 3.

  6. Using the \(\alpha = 0\) ripple to size a smoothing inductor. At 67° it is 5.5 times larger — Problem 4.

  7. Keeping a CCM-tuned speed loop when the drive enters DCM. The plant transfer function changes completely — Problem 4.

  8. Expecting a controlled rectifier to give servo bandwidth. Six corrections per supply cycle caps the current loop near 30 Hz — Problem 4.

  9. Tuning a speed loop as though it stays linear. It saturates on a four-rpm error — Problem 5.

  10. Omitting anti-windup. With the current clamped for over a second, the integrator guarantees a large overshoot — Problem 5.

Looking Ahead

Six problems and the drive structure that every later set will reuse is complete. One constant linked voltage to speed and current to torque; the chopper of Set 17 became a speed controller with no modification; four-quadrant operation turned out to need a 6% change in duty ratio and a front end that can accept the power; and cascade control put a fast current loop inside a slow speed loop, with the current limit — not the gains — setting how fast the machine actually accelerates.

All of it depended on the commutator. The DC machine decouples flux from torque mechanically: brushes hold the armature MMF perpendicular to the field whatever the rotor does, so \(I_a\) commands torque directly. That mechanism is also the machine's fatal flaw — brushes wear, commutators need servicing, and the peripheral speed of the commutator caps the rating.

An induction machine has none of those problems and none of that convenience. Its rotor current is not supplied but induced, its flux and torque are produced by the same stator current, and there is no way to command one without disturbing the other — at least, not without doing considerable work.

Next: Set 34 — Induction Motor Drives and V/f Control, where the equivalent circuit gives a breakdown torque 2.8 times rated, constant V/f holds the flux but loses 25% of that torque at half frequency because the stator resistance does not scale, slip frequency emerges as the real control variable, and a 22 kW pump drive pays for itself in under a year.