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 12 V into 3 V for a 5 Ω load. It switches at 40 kHz with L = 330 µH. Find the peak inductor current Imax.

Vin​ = 12 VVo​ = 3 Vf = 40 kHzL = 330 µHR = 5 Ω
Imax = A
2.DC-DC convertersHard
tiL​Imax​Io​DTT2T

A buck converter turns 15 V into 7.5 V at 200 kHz with L = 220 µH. Find the smallest load current Io,min for which it stays in continuous conduction.

Vin​ = 15 VVo​ = 7.5 Vf = 200 kHzL = 220 µH
Io,min = A
3.DC motorsMedium
+−240 VRa​ = 0.8 ΩIa​ = 30 AM+−EωT

A DC motor draws 30 A from a 240 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 194 W. Find its efficiency η.

V = 240 VIa = 30 ARa = 0.8 ΩProt = 194 W
η = %
4.DC motorsHard
+−48 VRa​ = 0.8 ΩM+−EωT

A small DC motor has an armature resistance of 0.8 Ω and runs from 48 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 48 VRa = 0.8 Ω
Pmax = W
5.AC power & power factorMedium
230 VR + jXP = 4037 W19.5 A

An inductive load takes 19.5 A and 4037 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 230 VI = 19.5 AP = 4037 W
R = Ω
6.AC power & power factorHard
P = 40 kWQ1​S1​φ1​QC​

A single-phase load takes 40 kW at a power factor of 0.6 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.

P = 40 kWpf1 = 0.6pf2 = 0.98V = 240 Vf = 60 Hz
C = µF
7.Three-phase powerMedium
ABCVL​ = 480 VZ = 15 + j20 Ωper phase

A balanced Δ-connected load with an impedance of 15 + j20 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 15 + j20 Ω per phase
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 0.65 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 200 kWpf1 = 0.65pf2 = 0.92VL = 415 Vf = 60 Hz
QC = kvar

3Solutions

1.DC-DC converters

Answer: Imax = 0.6852 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 = 3/12 = 0.25

Average and ripple current

Io = 3/5 = 0.6 A
ΔIL = VL·D/(L·f) = 9·0.25/(330·10⁻⁶·40·10³) = 0.1705 A

Peak current

Imax = 0.6 + 0.1705/2 = 0.6852 A

Answer

Imax = 0.6852 A
2.DC-DC converters

Answer: Io,min = 0.04261 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 = 7.5/15 = 0.5

Ripple current

ΔIL = 7.5·0.5/(220·10⁻⁶·200·10³) = 0.08523 A

Boundary load current

Io,min = ΔIL/2 = 0.04261 A

Answer

Io,min = 0.04261 A

Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.

3.DC motors

Answer: η = 87.31 %

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 = 240·30 = 7200 W
Pdev = 216·30 = 6480 W
Pout = 6480 − 194 = 6286 W

Efficiency

η = 6286/7200·100 = 87.31 %

Answer

η = 87.31 %
4.DC motors

Answer: Pmax = 720 W

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.

Power as a function of the back-EMF

P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 24 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 48²/(4·0.8) = 720 W

Answer

Pmax = 720 W

At this point the efficiency is only 50 %, so real motors run well above half speed.

5.AC power & power factor

Answer: R = 10.62 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 = 4037/19.5² = 10.62 Ω

Answer

R = 10.62 Ω
6.AC power & power factor

Answer: C = 2082 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 = 40·tan(53.13°) = 53.33 kvar
Q2 = P·tan φ2 = 40·tan(11.48°) = 8.122 kvar
QC = Q1 − Q2 = 45.21 kvar

Capacitance

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

C = QC/(2πf·V²) = 45211/(2π·60·240²) = 2082 µF

Answer

C = 2082 µF
7.Three-phase power

Answer: P = 16.59 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.

Phase voltage

Vph = VL = 480 V

Phase current

|Z| = √(15² + 20²) = 25 Ω
Iph = 480/25 = 19.2 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·19.2²·15 = 16.59 kW

Answer

P = 16.59 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

8.Three-phase power

Answer: QC = 148.6 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 = 200·tan(acos 0.65) = 200·1.169 = 233.8 kvar
Q2 = 200·tan(acos 0.92) = 200·0.426 = 85.2 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 148.6 kvar

Answer

QC = 148.6 kvar

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