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 VLSCR = 10 Ω+−60 V

An ideal boost converter turns 36 V into 60 V for a 10 Ω load. Find the average inductor current IL.

Vin​ = 36 VVo​ = 60 VR = 10 Ω
IL = A
2.DC-DC convertersHard
tiL​Imax​Io​DTT2T

A buck converter turns 9 V into 3.6 V at 200 kHz with L = 47 µH. Find the smallest load current Io,min for which it stays in continuous conduction.

Vin​ = 9 VVo​ = 3.6 Vf = 200 kHzL = 47 µH
Io,min = A
3.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 0.5 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 120 Vk = 0.5 V·s/rad
n0 = rpm
4.DC motorsHard
+−220 VRa​ = 0.8 ΩM+−En = ?T = 10 N·m

A separately excited DC motor (k = 1 V·s/rad at full field, Ra = 0.8 Ω, V = 220 V) drives a constant load torque of 10 N·m. The field flux is reduced to 80 % of its full value. Find the new speed n.

V = 220 Vk = 1 V·s/radRa = 0.8 ΩT = 10 N·mΦ = 80 %
n = rpm
5.AC power & power factorMedium
240 VR + jXP = 1824 W9.5 A

An inductive load takes 9.5 A and 1824 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 240 VI = 9.5 AP = 1824 W
R = Ω
6.AC power & power factorHard
P = 50 kWQ1​S1​φ1​QC​

A single-phase load takes 50 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 0.98. Find the new line current Inew.

P = 50 kWpf1 = 0.75pf2 = 0.98V = 400 Vf = 50 Hz
Inew = A
7.Three-phase powerMedium
ABCVL​ = 400 VIL​ = 10 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 10 Acos φ = 0.88
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 0.7 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 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 = 0.7pf2 = 0.92VL = 400 Vf = 60 Hz
Cph = µF

3Solutions

1.DC-DC converters

Answer: IL = 10 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 − 36/60 = 0.4

Load current

Io = 60/10 = 6 A

Inductor current

IL = Io/(1 − D) = 6/0.6 = 10 A

Answer

IL = 10 A
2.DC-DC converters

Answer: Io,min = 0.1149 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

Ripple current

ΔIL = 3.6·0.6/(47·10⁻⁶·200·10³) = 0.2298 A

Boundary load current

Io,min = ΔIL/2 = 0.1149 A

Answer

Io,min = 0.1149 A

Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.

3.DC motors

Answer: n0 = 2292 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 = 120/0.5 = 240 rad/s
n0 = 60·240/(2π) = 2292 rpm

Answer

n0 = 2292 rpm
4.DC motors

Answer: n = 2507 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 = 0.8·1 = 0.8 V·s/rad

New armature current

Ia = 10/0.8 = 12.5 A

New speed

ω = (220 − 12.5·0.8)/0.8 = 262.5 rad/s → n = 2507 rpm

Answer

n = 2507 rpm

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

5.AC power & power factor

Answer: R = 20.21 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 = 1824/9.5² = 20.21 Ω

Answer

R = 20.21 Ω
6.AC power & power factor

Answer: Inew = 127.6 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 = P·tan φ1 = 50·tan(41.41°) = 44.1 kvar
Q2 = P·tan φ2 = 50·tan(11.48°) = 10.15 kvar
QC = Q1 − Q2 = 33.94 kvar

New line current

Inew = P/(V·cos φ2) = 50000/(400·0.98) = 127.6 A

Answer

Inew = 127.6 A
7.Three-phase power

Answer: P = 6.097 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 = √3·400·10 = 6.928 kVA

Real power

P = S·cos φ = 6.928·0.88 = 6.097 kW

Answer

P = 6.097 kW
8.Three-phase power

Answer: Cph = 820.9 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 0.7) = 250·1.02 = 255.1 kvar
Q2 = 250·tan(acos 0.92) = 250·0.426 = 106.5 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 148.6 kvar

Capacitance per phase

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

V = 400 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 49517/(377·400²) = 820.9 µF

Answer

Cph = 820.9 µF

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