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
tiL​Imax​ = ?Imin​Io​DTT2T

A buck converter turns 15 V into 6 V for a 4 Ω load. It switches at 100 kHz with L = 330 µH. Find the peak inductor current Imax.

Vin​ = 15 VVo​ = 6 Vf = 100 kHzL = 330 µHR = 4 Ω
Imax = A
2.DC-DC convertersHard
tiL​Imax​IL​DTT2T

An ideal boost converter turns 12 V into 30 V for a 2 Ω load at 25 kHz. Find the smallest inductance Lcrit for continuous conduction.

Vin​ = 12 VVo​ = 30 Vf = 25 kHzR = 2 Ω
Lcrit = µH
3.DC motorsMedium
T (N·m)n (rpm)4.5 N·mn0​n = ?

A DC motor (k = 0.75 N·m/A, Ra = 1.5 Ω) runs from 220 V and drives a load torque of 4.5 N·m. Find its speed n in rpm.

V = 220 Vk = 0.75 N·m/ARa = 1.5 ΩT = 4.5 N·m
n = rpm
4.DC motorsHard
+−110 VRs​ = ?Ra​ = 1 ΩM+−En = 1000 rpmT = 9.6 N·m

A DC motor (k = 0.8 V·s/rad, Ra = 1 Ω) on a fixed 110 V supply drives a load of 9.6 N·m. Find the series resistance Rs that slows it down to 1000 rpm.

V = 110 Vk = 0.8 V·s/radRa = 1 ΩT = 9.6 N·mn = 1000 rpm
Rs = Ω
5.AC power & power factorMedium
120 VR + jXP = 330 W5.5 A

An inductive load takes 5.5 A and 330 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 = 5.5 AP = 330 W
R = Ω
6.AC power & power factorHard
400 V50 Hzload5 kW, 0.65C33 µFI = ?

A single-phase load takes 5 kW at a power factor of 0.65 lagging from a 400 V, 50 Hz supply. A 33 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 5 kWpf1 = 0.65V = 400 Vf = 50 HzC = 33 µF
I = A
7.Three-phase powerMedium
ABCVL​ = 690 VIL​ = 50 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 50 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 50 Acos φ = 0.88
P = kW
8.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 200 kW at a power factor of 0.8 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 200 kWpf1 = 0.8pf2 = 0.9VL = 400 Vf = 50 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: Imax = 1.555 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 = 6/15 = 0.4

Average and ripple current

Io = 6/4 = 1.5 A
ΔIL = VL·D/(L·f) = 9·0.4/(330·10⁻⁶·100·10³) = 0.1091 A

Peak current

Imax = 1.5 + 0.1091/2 = 1.555 A

Answer

Imax = 1.555 A
2.DC-DC converters

Answer: Lcrit = 3.84 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 = 1 − 12/30 = 0.6

Boundary condition

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

Lcrit = D(1 − D)²·R/(2f)

Critical inductance

Lcrit = 0.6(1 − 0.6)²·2/(225·10³) = 3.84 µH

Answer

Lcrit = 3.84 µH
3.DC motors

Answer: n = 2687 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.

Armature current

Ia = T/k = 4.5/0.75 = 6 A

Back-EMF and speed

E = 220 − 6·1.5 = 211 V
ω = E/k = 281.3 rad/s → n = 2687 rpm

Answer

n = 2687 rpm
4.DC motors

Answer: Rs = 1.185 Ω

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.

Current and back-EMF

Ia = 9.6/0.8 = 12 A
E = 0.8·104.7 = 83.78 V

Series resistance

Ra + Rs = (110 − 83.78)/12 = 2.185 Ω
Rs = 1.185 Ω

Answer

Rs = 1.185 Ω

The resistor wastes 170.7 W; a chopper controls the speed with far less loss.

5.AC power & power factor

Answer: R = 10.91 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 = 330/5.5² = 10.91 Ω

Answer

R = 10.91 Ω
6.AC power & power factor

Answer: I = 16.3 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.

Reactive power before and after

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

Q1 = 5·tan(arccos 0.65) = 5.846 kvar
QC = V²·2πf·C = 400²·2π·50·33·10⁻⁶ = 1659 var = 1.659 kvar
Q2 = 5.846 − 1.659 = 4.187 kvar

New apparent power

S2 = √(5² + 4.187²) = 6.522 kVA

New line current

I = S2/V = 6522/400 = 16.3 A

Answer

I = 16.3 A
7.Three-phase power

Answer: P = 52.59 kW

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.

Apparent power

S = √3·VL·IL = √3·690·50 = 59.76 kVA

Real power

P = S·cos φ = 59.76·0.88 = 52.59 kW

Answer

P = 52.59 kW
8.Three-phase power

Answer: Cph = 352.4 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 = 200·tan(acos 0.8) = 200·0.75 = 150 kvar
Q2 = 200·tan(acos 0.9) = 200·0.4843 = 96.86 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 53.14 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 400 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 17712/(314.2·400²) = 352.4 µF

Answer

Cph = 352.4 µF

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