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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 48 V into 33.6 V for a 2.5 Ω load. It switches at 40 kHz with L = 68 µH. Find the peak inductor current Imax.

Vin​ = 48 VVo​ = 33.6 Vf = 40 kHzL = 68 µHR = 2.5 Ω
Imax = A
2.DC-DC convertersHard
+−15 VLSC = 22 µFR = 2 Ω+−60 V

An ideal boost converter turns 15 V into 60 V for a 2 Ω load at 40 kHz. The output capacitor is 22 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.

Vin​ = 15 VVo​ = 60 Vf = 40 kHzC = 22 µFR = 2 Ω
ΔVo = mV
3.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 0.3 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 120 Vk = 0.3 V·s/rad
n0 = rpm
4.DC motorsHard
+−24 VRa​ = 0.5 ΩM+−EωT

A small DC motor has an armature resistance of 0.5 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 24 VRa = 0.5 Ω
Pmax = W
5.AC power & power factorMedium
120 VZ7 + j24 ΩI

A load with impedance Z = 7 + j24 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.

V = 120 VZ = 7 + j24 Ω
Q = var
6.AC power & power factorHard
240 V50 Hzload11 kW, 0.65C150 µFI = ?

A single-phase load takes 11 kW at a power factor of 0.65 lagging from a 240 V, 50 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 11 kWpf1 = 0.65V = 240 Vf = 50 HzC = 150 µF
I = A
7.Three-phase powerMedium
ABCVL​ = 690 VIL​ = 15 Acos φ = 0.82lagging

A balanced three-phase load draws a line current of 15 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 15 Acos φ = 0.82
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 0.75 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 = 250 kWpf1 = 0.75pf2 = 0.9VL = 400 Vf = 50 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: Imax = 15.29 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 = 33.6/48 = 0.7

Average and ripple current

Io = 33.6/2.5 = 13.44 A
ΔIL = VL·D/(L·f) = 14.4·0.7/(68·10⁻⁶·40·10³) = 3.706 A

Peak current

Imax = 13.44 + 3.706/2 = 15.29 A

Answer

Imax = 15.29 A
2.DC-DC converters

Answer: ΔVo = 25568 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 = 1 − 15/60 = 0.75

Load current

Io = 60/2 = 30 A

Output ripple

ΔVo = 30·0.75/(40·10³·22·10⁻⁶) = 25568 mV

Answer

ΔVo = 25568 mV
3.DC motors

Answer: n0 = 3820 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.

No-load speed

ω₀ = V/k = 120/0.3 = 400 rad/s
n0 = 60·400/(2π) = 3820 rpm

Answer

n0 = 3820 rpm
4.DC motors

Answer: Pmax = 288 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 = 12 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·0.5) = 288 W

Answer

Pmax = 288 W

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

5.AC power & power factor

Answer: Q = 553 var

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.

Current

|Z| = √(7² + 24²) = 25 Ω
I = V/|Z| = 120/25 = 4.8 A

Reactive power

Q = I²·X = 4.8²·24 = 553 var

Answer

Q = 553 var
6.AC power & power factor

Answer: I = 62.35 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 = 11·tan(arccos 0.65) = 12.86 kvar
QC = V²·2πf·C = 240²·2π·50·150·10⁻⁶ = 2714 var = 2.714 kvar
Q2 = 12.86 − 2.714 = 10.15 kvar

New apparent power

S2 = √(11² + 10.15²) = 14.96 kVA

New line current

I = S2/V = 14965/240 = 62.35 A

Answer

I = 62.35 A
7.Three-phase power

Answer: P = 14.7 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 = √3·690·15 = 17.93 kVA

Real power

P = S·cos φ = 17.93·0.82 = 14.7 kW

Answer

P = 14.7 kW
8.Three-phase power

Answer: Cph = 659.2 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 = 250·tan(acos 0.75) = 250·0.8819 = 220.5 kvar
Q2 = 250·tan(acos 0.9) = 250·0.4843 = 121.1 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 99.4 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) = 33133/(314.2·400²) = 659.2 µF

Answer

Cph = 659.2 µF

ExamProblems · examproblems.com/courses/power-electronics