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​ΔI = ?DTT2T

An ideal boost converter turns 48 V into 64 V at 50 kHz with L = 150 µH. Find the peak-to-peak inductor ripple current ΔIL.

Vin​ = 48 VVo​ = 64 Vf = 50 kHzL = 150 µH
ΔIL = A
2.DC-DC convertersHard
+−12 VSL = 47 µHC = ?R+−7.2 V

A buck converter turns 12 V into 7.2 V at 500 kHz with L = 47 µH. Find the output capacitance C that limits the peak-to-peak output ripple to 10 mV.

Vin​ = 12 VVo​ = 7.2 Vf = 500 kHzL = 47 µHΔVo = 10 mV
C = µF
3.DC motorsMedium
T (N·m)n (rpm)30 N·mn0​nFL​

A DC motor (k = 1.2 V·s/rad, Ra = 1.2 Ω) runs from 240 V. At full load it draws 25 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 240 Vk = 1.2 V·s/radRa = 1.2 ΩIFL = 25 A
SR = %
4.DC motorsHard
+−110 VRa​ = 1.5 ΩM+−En = ?T = 6 N·m

A separately excited DC motor (k = 1 V·s/rad at full field, Ra = 1.5 Ω, V = 110 V) drives a constant load torque of 6 N·m. The field flux is reduced to 75 % of its full value. Find the new speed n.

V = 110 Vk = 1 V·s/radRa = 1.5 ΩT = 6 N·mΦ = 75 %
n = rpm
5.AC power & power factorMedium
230 V121 kW, 0.6524 kW, 0.95I

Two loads are connected in parallel to a 230 V single-phase supply: a 21 kW load at a power factor of 0.65 lagging and a 4 kW load at a power factor of 0.95 leading. Find the overall power factor cos φ seen by the supply.

V = 230 VP1 = 21 kW, pf1 = 0.65 laggingP2 = 4 kW, pf2 = 0.95 leading
cos φ =
6.AC power & power factorHard
120 V60 HzRs​ = 8 ΩXs​ = j15 ΩZL​ = ?

A 120 V (rms), 60 Hz source has an internal impedance Zs = 8 + j15 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.

Vs = 120 V (rms)f = 60 HzZs = 8 + j15 Ω
C = µF
7.Three-phase powerMedium
ABCVL​ = 415 VIL​ = 75 Acos φ = 0.9lagging

A balanced three-phase load draws a line current of 75 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 75 Acos φ = 0.9
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 0.75 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 0.95 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 = 0.75pf2 = 0.95VL = 480 Vf = 60 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: ΔIL = 1.6 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 − 48/64 = 0.25

Inductor voltage during the on-time

While S is on, the inductor sits directly across Vin.

VL = 48 V

Ripple current

ΔIL = VL·D/(L·f) = 48·0.25/(150·10⁻⁶·50·10³) = 1.6 A

Answer

ΔIL = 1.6 A
2.DC-DC converters

Answer: C = 3.064 µF

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 = 7.2/12 = 0.6

Inductor ripple current

ΔIL = (12 − 7.2)·0.6/(47·10⁻⁶·500·10³) = 0.1226 A

Capacitance

C = 0.1226/(8·500·10³·10·10⁻³) = 3.064 µF

Answer

C = 3.064 µF
3.DC motors

Answer: SR = 14.29 %

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 and full-load speed

n₀ = 60·240/(2π·1.2) = 1910 rpm
nFL = 60·(240 − 25·1.2)/(2π·1.2) = 1671 rpm

Speed regulation

SR = (1910 − 1671)/1671·100 = 14.29 %

Answer

SR = 14.29 %
4.DC motors

Answer: n = 1248 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.

Weakened motor constant

k₂ = 0.75·1 = 0.75 V·s/rad

New armature current

Ia = 6/0.75 = 8 A

New speed

ω = (110 − 8·1.5)/0.75 = 130.7 rad/s → n = 1248 rpm

Answer

n = 1248 rpm

Field weakening is how a DC drive runs above its base speed.

5.AC power & power factor

Answer: cos φ = 0.7325

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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 21·tan(arccos 0.65) = 24.55 kvar
Q2 = −4·tan(arccos 0.95) = −1.315 kvar

Totals

P = 21 + 4 = 25 kW
Q = 24.55 + (−1.315) = 23.24 kvar
S = √(25² + 23.24²) = 34.13 kVA

Overall power factor

cos φ = P/S = 25/34.13 = 0.7325 lagging

Answer

cos φ = 0.7325
6.AC power & power factor

Answer: C = 176.8 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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 8 − j15 Ω

Capacitance

XC = 15 Ω → C = 1/(2π·60·15) = 176.8 µF

Answer

C = 176.8 µF
7.Three-phase power

Answer: P = 48.52 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 = √3·415·75 = 53.91 kVA

Real power

P = S·cos φ = 53.91·0.9 = 48.52 kW

Answer

P = 48.52 kW
8.Three-phase power

Answer: Cph = 424.6 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 0.75) = 200·0.8819 = 176.4 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 110.6 kvar

Capacitance per phase

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

V = 480 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 36882/(377·480²) = 424.6 µF

Answer

Cph = 424.6 µF

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