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 = 25 Ω+−30 V

An ideal boost converter turns 12 V into 30 V for a 25 Ω load. Find the average inductor current IL.

Vin​ = 12 VVo​ = 30 VR = 25 Ω
IL = A
2.DC-DC convertersHard
+−48 VSLC = 220 µFR = 8 Ω−+−112 V

An ideal buck-boost converter turns 48 V into −112 V for an 8 Ω load at 20 kHz. The output capacitor is 220 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.

Vin​ = 48 VVo​ = −112 Vf = 20 kHzC = 220 µFR = 8 Ω
ΔVo = mV
3.DC motorsMedium
T (N·m)n (rpm)3 N·mn0​nFL​

A DC motor (k = 0.3 V·s/rad, Ra = 1.2 Ω) runs from 120 V. At full load it draws 10 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 120 Vk = 0.3 V·s/radRa = 1.2 ΩIFL = 10 A
SR = %
4.DC motorsHard
+−120 VRa​ = 0.4 ΩIa​: 15 → 22.5 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 0.4 Ω) runs at 900 rpm with an armature current of 15 A. The load changes and the current becomes 22.5 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 0.4 Ωn1 = 900 rpmIa1 = 15 AIa2 = 22.5 A
n2 = rpm
5.AC power & power factorMedium
P = 26 kWQ = ?Sφ

A single-phase motor takes 26 kW at a power factor of 0.8 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 26 kWpf = 0.8V = 230 V
Q = kvar
6.AC power & power factorHard
P = 35 kWQ1​S1​φ1​QC​

A single-phase load takes 35 kW at a power factor of 0.75 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the capacitance C needed.

P = 35 kWpf1 = 0.75pf2 = 0.95V = 400 Vf = 50 Hz
C = µF
7.Three-phase powerMedium
ABCVL​ = 690 VIL​ = ?M110 kW, η = 91%cos φ = 0.86

A three-phase induction motor delivers 110 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 110 kWη = 91%cos φ = 0.86VL = 690 V
IL = A
8.Three-phase powerHard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 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 Y. Find the capacitance Cph of each capacitor in µF.

P = 300 kWpf1 = 0.75pf2 = 0.95VL = 480 Vf = 60 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: IL = 3 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 − 12/30 = 0.6

Load current

Io = 30/25 = 1.2 A

Inductor current

IL = Io/(1 − D) = 1.2/0.4 = 3 A

Answer

IL = 3 A
2.DC-DC converters

Answer: ΔVo = 2227 mV

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 = 112/(48 + 112) = 0.7

Load current

Io = 112/8 = 14 A

Output ripple

ΔVo = 14·0.7/(20·10³·220·10⁻⁶) = 2227 mV

Answer

ΔVo = 2227 mV
3.DC motors

Answer: SR = 11.11 %

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 = 60·120/(2π·0.3) = 3820 rpm
nFL = 60·(120 − 10·1.2)/(2π·0.3) = 3438 rpm

Speed regulation

SR = (3820 − 3438)/3438·100 = 11.11 %

Answer

SR = 11.11 %
4.DC motors

Answer: n2 = 876.3 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 − 15·0.4 = 114 V
E2 = 120 − 22.5·0.4 = 111 V

New speed

n2 = 900·111/114 = 876.3 rpm

Answer

n2 = 876.3 rpm
5.AC power & power factor

Answer: Q = 19.5 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.

Angle

φ = arccos(0.8) = 36.87°, tan φ = 0.75

Reactive power

Q = P·tan φ = 26·0.75 = 19.5 kvar

Answer

Q = 19.5 kvar
6.AC power & power factor

Answer: C = 385.2 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 = 35·tan(41.41°) = 30.87 kvar
Q2 = P·tan φ2 = 35·tan(18.19°) = 11.5 kvar
QC = Q1 − Q2 = 19.36 kvar

Capacitance

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

C = QC/(2πf·V²) = 19363/(2π·50·400²) = 385.2 µF

Answer

C = 385.2 µF
7.Three-phase power

Answer: IL = 117.6 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 = 110/0.91 = 120.9 kW

Line current

IL = Pin/(√3·VL·cos φ) = 120879/(√3·690·0.86) = 117.6 A

Answer

IL = 117.6 A
8.Three-phase power

Answer: Cph = 1911 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 = 300·tan(acos 0.75) = 300·0.8819 = 264.6 kvar
Q2 = 300·tan(acos 0.95) = 300·0.3287 = 98.61 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 166 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) = 55323/(377·277.1²) = 1911 µF

Answer

Cph = 1911 µF

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