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 9 V into 3.6 V for a 20 Ω load. It switches at 200 kHz with L = 100 µH. Find the peak inductor current Imax.

Vin​ = 9 VVo​ = 3.6 Vf = 200 kHzL = 100 µHR = 20 Ω
Imax = A
2.DC-DC convertersHard
tiL​Imax​IL​DTT2T

An ideal inverting buck-boost converter turns 48 V into −16 V for a 12 Ω load at 200 kHz. Find the smallest inductance Lcrit for continuous conduction.

Vin​ = 48 VVo​ = −16 Vf = 200 kHzR = 12 Ω
Lcrit = µH
3.DC motorsMedium
T (N·m)n (rpm)Tst​ = ?n0​

A DC motor with k = 0.75 N·m/A and Ra = 1 Ω is connected to 120 V. Find its stall torque Tstall, the torque at zero speed.

V = 120 Vk = 0.75 N·m/ARa = 1 Ω
Tstall = N·m
4.DC motorsHard
+−120 VRa​ = 0.4 ΩIa​: 30 → 45 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 0.4 Ω) runs at 1200 rpm with an armature current of 30 A. The load changes and the current becomes 45 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 0.4 Ωn1 = 1200 rpmIa1 = 30 AIa2 = 45 A
n2 = rpm
5.AC power & power factorMedium
230 VR + jXP = 633 W5.5 A

An inductive load takes 5.5 A and 633 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 = 5.5 AP = 633 W
R = Ω
6.AC power & power factorHard
230 V60 Hzload6 kW, 0.8C47 µFI = ?

A single-phase load takes 6 kW at a power factor of 0.8 lagging from a 230 V, 60 Hz supply. A 47 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 6 kWpf1 = 0.8V = 230 Vf = 60 HzC = 47 µF
I = A
7.Three-phase powerMedium
ABCVL​ = 208 VIL​ = 50 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 50 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 50 Acos φ = 0.85
P = kW
8.Three-phase powerHard
P = 250 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 250 kW at a power factor of 0.65 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 250 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar

3Solutions

1.DC-DC converters

Answer: Imax = 0.234 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 = 3.6/9 = 0.4

Average and ripple current

Io = 3.6/20 = 0.18 A
ΔIL = VL·D/(L·f) = 5.4·0.4/(100·10⁻⁶·200·10³) = 0.108 A

Peak current

Imax = 0.18 + 0.108/2 = 0.234 A

Answer

Imax = 0.234 A
2.DC-DC converters

Answer: Lcrit = 16.88 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 = 16/(48 + 16) = 0.25

Boundary condition

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

Lcrit = (1 − D)²·R/(2f)

Critical inductance

Lcrit = (1 − 0.25)²·12/(2200·10³) = 16.88 µH

Answer

Lcrit = 16.88 µH
3.DC motors

Answer: Tstall = 90 N·m

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.

Stall current

Ia = V/Ra = 120/1 = 120 A

Stall torque

Tstall = 0.75·120 = 90 N·m

Answer

Tstall = 90 N·m
4.DC motors

Answer: n2 = 1133 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 − 30·0.4 = 108 V
E2 = 120 − 45·0.4 = 102 V

New speed

n2 = 1200·102/108 = 1133 rpm

Answer

n2 = 1133 rpm
5.AC power & power factor

Answer: R = 20.93 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 = 633/5.5² = 20.93 Ω

Answer

R = 20.93 Ω
6.AC power & power factor

Answer: I = 30.34 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 = 6·tan(arccos 0.8) = 4.5 kvar
QC = V²·2πf·C = 230²·2π·60·47·10⁻⁶ = 937.3 var = 0.9373 kvar
Q2 = 4.5 − 0.9373 = 3.563 kvar

New apparent power

S2 = √(6² + 3.563²) = 6.978 kVA

New line current

I = S2/V = 6978/230 = 30.34 A

Answer

I = 30.34 A
7.Three-phase power

Answer: P = 15.31 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 = √3·208·50 = 18.01 kVA

Real power

P = S·cos φ = 18.01·0.85 = 15.31 kW

Answer

P = 15.31 kW
8.Three-phase power

Answer: QC = 210.1 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 = 250·tan(acos 0.65) = 250·1.169 = 292.3 kvar
Q2 = 250·tan(acos 0.95) = 250·0.3287 = 82.17 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 210.1 kvar

Answer

QC = 210.1 kvar

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