Power Electronics and Drives: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

DC-DC converters

Buck
Vo = D·Vin
Boost
Vo = Vin/(1 − D)
Inverting buck-boost
|Vo| = Vin·D/(1 − D)
Ideal power balance
VinIin = VoIo
Inductor ripple
ΔIL = VL,on·D/(L·f); buck Vin − Vo, boost and buck-boost Vin
Average inductor current
buck Io; boost and buck-boost Io/(1 − D)
Output ripple
buck ΔIL/(8fC); boost and buck-boost Io·D/(fC)
CCM boundary
buck (1 − D)R/(2f); boost D(1 − D)²R/(2f); buck-boost (1 − D)²R/(2f)

Method

  1. Identify the converter (buck, boost or buck-boost) and the duty cycle D from the voltage ratio.
  2. Use volt-second balance on the inductor: the average inductor voltage over a period is zero.
  3. Find the inductor voltage while the switch is on, then ΔIL = VL·D/(L·f).
  4. For the output ripple, find the charge that flows in and out of the capacitor; for the CCM boundary, set the average inductor current equal to ΔIL/2.

DC motors

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Method

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

2Practice problems

1.DC-DC convertersMedium
+−24 VLSCR = 10 Ω+−48 V

An ideal boost converter turns 24 V into 48 V for a 10 Ω load. Find the average inductor current IL.

Vin​ = 24 VVo​ = 48 VR = 10 Ω
IL = A
2.DC-DC convertersHard
tiL​Imax​IL​DTT2T

An ideal inverting buck-boost converter turns 9 V into −9 V for a 12 Ω load at 25 kHz. Find the smallest inductance Lcrit for continuous conduction.

Vin​ = 9 VVo​ = −9 Vf = 25 kHzR = 12 Ω
Lcrit = µH
3.DC motorsMedium
+−220 VRa​ = 1 ΩIa​ = 30 AM+−EωT

A DC motor draws 30 A from a 220 V supply. Its armature resistance is 1 Ω and its rotational (friction, windage and iron) losses are 456 W. Find its efficiency η.

V = 220 VIa = 30 ARa = 1 ΩProt = 456 W
η = %
4.DC motorsHard
+−120 VRa​ = 0.25 ΩIa​: 25 → 12.5 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 0.25 Ω) runs at 1200 rpm with an armature current of 25 A. The load changes and the current becomes 12.5 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 0.25 Ωn1 = 1200 rpmIa1 = 25 AIa2 = 12.5 A
n2 = rpm
5.AC power & power factorMedium
P = 40 kWQSφ

A single-phase motor takes 40 kW at a power factor of 0.65 lagging from a 400 V supply. Find the line current I.

P = 40 kWpf = 0.65V = 400 V
I = A
6.AC power & power factorHard
P = 40 kWQ1​S1​φ1​QC​

A single-phase load takes 40 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the reactive power QC the capacitor must supply.

P = 40 kWpf1 = 0.65pf2 = 0.9V = 240 Vf = 60 Hz
QC = kvar
7.Three-phase powerMedium
ABCVL​ = 415 VIL​ = ?M7.5 kW, η = 93%cos φ = 0.8

A three-phase induction motor delivers 7.5 kW of mechanical output power at full load. At this load its efficiency is 93% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.

Pout = 7.5 kWη = 93%cos φ = 0.8VL = 415 V
IL = A
8.Three-phase powerHard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.75 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.75pf2 = 0.98VL = 480 Vf = 60 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: IL = 9.6 A

DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.

Duty cycle

D = 1 − 24/48 = 0.5

Load current

Io = 48/10 = 4.8 A

Inductor current

IL = Io/(1 − D) = 4.8/0.5 = 9.6 A

Answer

IL = 9.6 A
2.DC-DC converters

Answer: Lcrit = 60 µH

DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.

Duty cycle

D = 9/(9 + 9) = 0.5

Boundary condition

Set the average inductor current equal to half the ripple current and solve for L.

Lcrit = (1 − D)²·R/(2f)

Critical inductance

Lcrit = (1 − 0.5)²·12/(225·10³) = 60 µH

Answer

Lcrit = 60 µH
3.DC motors

Answer: η = 79.45 %

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Powers

Pin = 220·30 = 6600 W
Pdev = 190·30 = 5700 W
Pout = 5700 − 456 = 5244 W

Efficiency

η = 5244/6600·100 = 79.45 %

Answer

η = 79.45 %
4.DC motors

Answer: n2 = 1233 rpm

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Back-EMF before and after

E1 = 120 − 25·0.25 = 113.8 V
E2 = 120 − 12.5·0.25 = 116.9 V

New speed

n2 = 1200·116.9/113.8 = 1233 rpm

Answer

n2 = 1233 rpm
5.AC power & power factor

Answer: I = 153.8 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 40000/(400·0.65) = 153.8 A

Answer

I = 153.8 A
6.AC power & power factor

Answer: QC = 27.39 kvar

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = P·tan φ1 = 40·tan(49.46°) = 46.77 kvar
Q2 = P·tan φ2 = 40·tan(25.84°) = 19.37 kvar
QC = Q1 − Q2 = 27.39 kvar

Answer

QC = 27.39 kvar
7.Three-phase power

Answer: IL = 14.02 A

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Electrical input power

Pin = Pout/η = 7.5/0.93 = 8.065 kW

Line current

IL = Pin/(√3·VL·cos φ) = 8065/(√3·415·0.8) = 14.02 A

Answer

IL = 14.02 A
8.Three-phase power

Answer: Cph = 468.9 µF

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 60·tan(acos 0.75) = 60·0.8819 = 52.92 kvar
Q2 = 60·tan(acos 0.98) = 60·0.2031 = 12.18 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 40.73 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 480/√3 = 277.1 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 13577/(377·277.1²) = 468.9 µF

Answer

Cph = 468.9 µF

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