Identify the converter (buck, boost or buck-boost) and the duty cycle D from the voltage ratio.
Use volt-second balance on the inductor: the average inductor voltage over a period is zero.
Find the inductor voltage while the switch is on, then ΔIL = VL·D/(L·f).
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
Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
Use V = E + Ia·Ra to link voltage, current and back-EMF.
Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
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
A buck converter turns 9 V into 3.6 V for a 4 Ω load. It switches at 500 kHz with L = 33 µH. Find the peak inductor current Imax.
Vin = 9 VVo = 3.6 Vf = 500 kHzL = 33 µHR = 4 Ω
Imax = A
2.DC-DC convertersHard
A buck converter turns 36 V into 27 V at 500 kHz with L = 10 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
Vin = 36 VVo = 27 Vf = 500 kHzL = 10 µH
Io,min = A
3.DC motorsMedium
A DC motor draws 8 A from a 240 V supply. Its armature resistance is 1 Ω and its rotational (friction, windage and iron) losses are 111 W. Find its efficiency η.
V = 240 VIa = 8 ARa = 1 ΩProt = 111 W
η = %
4.DC motorsHard
A small DC motor has an armature resistance of 2.5 Ω and runs from 120 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 120 VRa = 2.5 Ω
Pmax = W
5.AC power & power factorMedium
An inductive load takes 8 A and 816 W from a 120 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find X.
V = 120 VI = 8 AP = 816 W
X = Ω
6.AC power & power factorHard
A single-phase load takes 30 kW at a power factor of 0.6 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.
P = 30 kWpf1 = 0.6pf2 = 0.98V = 400 Vf = 60 Hz
Inew = A
7.Three-phase powerMedium
A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 37 kWη = 92%cos φ = 0.86VL = 480 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.7 lagging from a 415 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 = 60 kWpf1 = 0.7pf2 = 0.95VL = 415 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 0.9655 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 = 3.6/4 = 0.9 A
ΔIL = VL·D/(L·f) = 5.4·0.4/(33·10⁻⁶·500·10³) = 0.1309 A
Peak current
Imax = 0.9 + 0.1309/2 = 0.9655 A
Answer
Imax = 0.9655 A
2.DC-DC converters
Answer: Io,min = 0.675 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 = 27/36 = 0.75
Ripple current
ΔIL = 27·0.25/(10·10⁻⁶·500·10³) = 1.35 A
Boundary load current
Io,min = ΔIL/2 = 0.675 A
Answer
Io,min = 0.675 A
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
3.DC motors
Answer: η = 90.89 %
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.
Powers
Pin = 240·8 = 1920 W
Pdev = 232·8 = 1856 W
Pout = 1856 − 111 = 1745 W
Efficiency
η = 1745/1920·100 = 90.89 %
Answer
η = 90.89 %
4.DC motors
Answer: Pmax = 1440 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 = 60 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 120²/(4·2.5) = 1440 W
Answer
Pmax = 1440 W
At this point the efficiency is only 50 %, so real motors run well above half speed.
5.AC power & power factor
Answer: X = 7.902 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 = 816/8² = 12.75 Ω
Reactance from the impedance
|Z| = V/I = 120/8 = 15 Ω
X = √(|Z|² − R²) = √(15² − 12.75²) = 7.902 Ω
Answer
X = 7.902 Ω
6.AC power & power factor
Answer: Inew = 76.53 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 = 30·tan(53.13°) = 40 kvar
Q2 = P·tan φ2 = 30·tan(11.48°) = 6.092 kvar
QC = Q1 − Q2 = 33.91 kvar
New line current
Inew = P/(V·cos φ2) = 30000/(400·0.98) = 76.53 A
Answer
Inew = 76.53 A
7.Three-phase power
Answer: IL = 56.25 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 = 37/0.92 = 40.22 kW
Line current
IL = Pin/(√3·VL·cos φ) = 40217/(√3·480·0.86) = 56.25 A
Answer
IL = 56.25 A
8.Three-phase power
Answer: Cph = 639 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 = 60·tan(acos 0.7) = 60·1.02 = 61.21 kvar
Q2 = 60·tan(acos 0.95) = 60·0.3287 = 19.72 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 41.49 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.