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
+−36 VSL = ?CR+−14.4 V

An ideal buck converter turns 36 V into 14.4 V at a switching frequency of 250 kHz. Find the inductance L that keeps the peak-to-peak inductor ripple current at 0.8 A.

Vin​ = 36 VVo​ = 14.4 Vf = 250 kHzΔIL = 0.8 A
L = µH
2.DC-DC convertersHard
+−12 VLSC = 100 µFR = 2 Ω+−20 V

An ideal boost converter turns 12 V into 20 V for a 2 Ω load at 40 kHz. The output capacitor is 100 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.

Vin​ = 12 VVo​ = 20 Vf = 40 kHzC = 100 µFR = 2 Ω
ΔVo = mV
3.DC motorsMedium
+−VRa​M+−En = 960 rpmT = ?

A DC motor delivers 2.2 kW of mechanical power at 960 rpm. Find the shaft torque T.

P = 2.2 kWn = 960 rpm
T = N·m
4.DC motorsHard
+−120 VRa​ = 0.5 ΩIa​: 30 → 12 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 0.5 Ω) runs at 1440 rpm with an armature current of 30 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 0.5 Ωn1 = 1440 rpmIa1 = 30 AIa2 = 12 A
n2 = rpm
5.AC power & power factorMedium
P = 12 kWQ = ?Sφ

A single-phase motor takes 12 kW at a power factor of 0.9 lagging from a 240 V supply. Find the reactive power Q it draws.

P = 12 kWpf = 0.9V = 240 V
Q = kvar
6.AC power & power factorHard
P1​ = 66 kWQ1​S1​φ

A 150 kVA single-phase transformer supplies a load of 66 kW at a power factor of 0.75 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 150 kVAP1 = 66 kWpf1 = 0.75
ΔP = kW
7.Three-phase powerMedium
ABCVL​ = 415 VNZ = 3 + j4 Ωper phase

A balanced Y-connected load with an impedance of 3 + j4 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 3 + j4 Ω per phase
P = kW
8.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 120 kW at a power factor of 0.65 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.65pf2 = 0.98VL = 415 Vf = 50 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: L = 43.2 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 = 14.4/36 = 0.4

Inductance

L = 21.6·0.4/(0.8·250·10³) = 43.2 µH

Answer

L = 43.2 µH
2.DC-DC converters

Answer: ΔVo = 1000 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 − 12/20 = 0.4

Load current

Io = 20/2 = 10 A

Output ripple

ΔVo = 10·0.4/(40·10³·100·10⁻⁶) = 1000 mV

Answer

ΔVo = 1000 mV
3.DC motors

Answer: T = 21.88 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.

Speed in rad/s

ω = 2π·960/60 = 100.5 rad/s

Torque

T = P/ω = 2200/100.5 = 21.88 N·m

Answer

T = 21.88 N·m
4.DC motors

Answer: n2 = 1563 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 − 30·0.5 = 105 V
E2 = 120 − 12·0.5 = 114 V

New speed

n2 = 1440·114/105 = 1563 rpm

Answer

n2 = 1563 rpm
5.AC power & power factor

Answer: Q = 5.812 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(0.9) = 25.84°, tan φ = 0.4843

Reactive power

Q = P·tan φ = 12·0.4843 = 5.812 kvar

Answer

Q = 5.812 kvar
6.AC power & power factor

Answer: ΔP = 72.25 kW

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 the existing load

Q1 = 66·tan(arccos 0.75) = 58.21 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(150² − 58.21²) = 138.2 kW

Extra real power

ΔP = 138.2 − 66 = 72.25 kW

Answer

ΔP = 72.25 kW
7.Three-phase power

Answer: P = 20.67 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 = 415/√3 = 239.6 V

Phase current

|Z| = √(3² + 4²) = 5 Ω
Iph = 239.6/5 = 47.92 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·47.92²·3 = 20.67 kW

Answer

P = 20.67 kW

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

8.Three-phase power

Answer: Cph = 714.2 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 = 120·tan(acos 0.65) = 120·1.169 = 140.3 kvar
Q2 = 120·tan(acos 0.98) = 120·0.2031 = 24.37 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 115.9 kvar

Capacitance per phase

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

V = 415 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 38643/(314.2·415²) = 714.2 µF

Answer

Cph = 714.2 µF

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