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
+−12 VLSCR = 10 Ω+−20 V

An ideal boost converter turns 12 V into 20 V for a 10 Ω load. Find the average inductor current IL.

Vin​ = 12 VVo​ = 20 VR = 10 Ω
IL = A
2.DC-DC convertersHard
tiL​Imax​IL​DTT2T

An ideal buck converter turns 24 V into 12 V for a 20 Ω load at 50 kHz. Find the smallest inductance Lcrit for continuous conduction.

Vin​ = 24 VVo​ = 12 Vf = 50 kHzR = 20 Ω
Lcrit = µH
3.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 2 V·s/rad runs from 240 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 240 Vk = 2 V·s/rad
n0 = rpm
4.DC motorsHard
+−240 VD = ?Ra​ = 1.2 ΩM+−En = 900 rpmT = 10 N·m

A DC chopper feeds a motor (k = 1 V·s/rad, Ra = 1.2 Ω) from a 240 V DC bus. The average armature voltage is D·240 V. Find the duty cycle D that runs the motor at 900 rpm with a load of 10 N·m.

Vdc = 240 Vk = 1 V·s/radRa = 1.2 ΩT = 10 N·mn = 900 rpm
D = %
5.AC power & power factorMedium
120 VR + jXP = 270 W4.5 A

An inductive load takes 4.5 A and 270 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 120 VI = 4.5 AP = 270 W
R = Ω
6.AC power & power factorHard
P = 10 kWQ1​S1​φ1​QC​

A single-phase load takes 10 kW at a power factor of 0.65 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the capacitance C needed.

P = 10 kWpf1 = 0.65pf2 = 0.95V = 230 Vf = 50 Hz
C = µF
7.Three-phase powerMedium
ABCVL​ = 400 VIL​ = ?M37 kW, η = 94%cos φ = 0.82

A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.

Pout = 37 kWη = 94%cos φ = 0.82VL = 400 V
IL = A
8.Three-phase powerHard
P = 150 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 150 kW at a power factor of 0.65 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.65pf2 = 0.95VL = 415 Vf = 60 Hz
QC = kvar

3Solutions

1.DC-DC converters

Answer: IL = 3.333 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 − 12/20 = 0.4

Load current

Io = 20/10 = 2 A

Inductor current

IL = Io/(1 − D) = 2/0.6 = 3.333 A

Answer

IL = 3.333 A
2.DC-DC converters

Answer: Lcrit = 100 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 = 12/24 = 0.5

Boundary condition

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

Lcrit = (1 − D)·R/(2f)

Critical inductance

Lcrit = (1 − 0.5)·20/(250·10³) = 100 µH

Answer

Lcrit = 100 µH
3.DC motors

Answer: n0 = 1146 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.

No-load speed

ω₀ = V/k = 240/2 = 120 rad/s
n0 = 60·120/(2π) = 1146 rpm

Answer

n0 = 1146 rpm
4.DC motors

Answer: D = 44.27 %

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.

Required armature voltage

Va = 1·94.25 + 10·1.2 = 106.2 V

Duty cycle

D = 106.2/240 = 44.27 %

Answer

D = 44.27 %
5.AC power & power factor

Answer: R = 13.33 Ω

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.

Resistance from the real power

R = P/I² = 270/4.5² = 13.33 Ω

Answer

R = 13.33 Ω
6.AC power & power factor

Answer: C = 505.7 µF

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 = 10·tan(49.46°) = 11.69 kvar
Q2 = P·tan φ2 = 10·tan(18.19°) = 3.287 kvar
QC = Q1 − Q2 = 8.404 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 8404/(2π·50·230²) = 505.7 µF

Answer

C = 505.7 µF
7.Three-phase power

Answer: IL = 69.29 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 = 37/0.94 = 39.36 kW

Line current

IL = Pin/(√3·VL·cos φ) = 39362/(√3·400·0.82) = 69.29 A

Answer

IL = 69.29 A
8.Three-phase power

Answer: QC = 126.1 kvar

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 = 150·tan(acos 0.65) = 150·1.169 = 175.4 kvar
Q2 = 150·tan(acos 0.95) = 150·0.3287 = 49.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 126.1 kvar

Answer

QC = 126.1 kvar

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