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
An ideal buck converter turns 48 V into 12 V at a switching frequency of 250 kHz. Find the inductance L that keeps the peak-to-peak inductor ripple current at 0.6 A.
Vin = 48 VVo = 12 Vf = 250 kHzΔIL = 0.6 A
L = µH
2.DC-DC convertersHard
An ideal boost converter turns 48 V into 64 V for a 4 Ω load at 200 kHz. The output capacitor is 10 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
Vin = 48 VVo = 64 Vf = 200 kHzC = 10 µFR = 4 Ω
ΔVo = mV
3.DC motorsMedium
A DC motor (k = 0.15 N·m/A, Ra = 1.2 Ω) runs from 24 V and drives a load torque of 0.3 N·m. Find its speed n in rpm.
V = 24 Vk = 0.15 N·m/ARa = 1.2 ΩT = 0.3 N·m
n = rpm
4.DC motorsHard
A separately excited DC motor on 240 V (Ra = 1 Ω) runs at 1800 rpm with an armature current of 12 A. The load changes and the current becomes 4.8 A, with the field unchanged. Find the new speed n2.
V = 240 VRa = 1 Ωn1 = 1800 rpmIa1 = 12 AIa2 = 4.8 A
n2 = rpm
5.AC power & power factorMedium
A load with impedance Z = 20 + j15 Ω is connected to a 230 V (rms) single-phase supply. Find the real power P it absorbs.
V = 230 VZ = 20 + j15 Ω
P = W
6.AC power & power factorHard
A 10 V (rms), 1000 Hz source has an internal impedance Zs = 25 + j20 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.
Vs = 10 V (rms)f = 1000 HzZs = 25 + j20 Ω
C = µF
7.Three-phase powerMedium
A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 88% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 45 kWη = 88%cos φ = 0.84VL = 480 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 50 kW at a power factor of 0.75 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 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 50 kWpf1 = 0.75pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: L = 60 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 = 12/48 = 0.25
Inductance
L = 36·0.25/(0.6·250·10³) = 60 µH
Answer
L = 60 µH
2.DC-DC converters
Answer: ΔVo = 2000 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 − 48/64 = 0.25
Load current
Io = 64/4 = 16 A
Output ripple
ΔVo = 16·0.25/(200·10³·10·10⁻⁶) = 2000 mV
Answer
ΔVo = 2000 mV
3.DC motors
Answer: n = 1375 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.
Armature current
Ia = T/k = 0.3/0.15 = 2 A
Back-EMF and speed
E = 24 − 2·1.2 = 21.6 V
ω = E/k = 144 rad/s → n = 1375 rpm
Answer
n = 1375 rpm
4.DC motors
Answer: n2 = 1857 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 = 240 − 12·1 = 228 V
E2 = 240 − 4.8·1 = 235.2 V
New speed
n2 = 1800·235.2/228 = 1857 rpm
Answer
n2 = 1857 rpm
5.AC power & power factor
Answer: P = 1693 W
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| = √(20² + 15²) = 25 Ω
I = V/|Z| = 230/25 = 9.2 A
Real power
P = I²·R = 9.2²·20 = 1693 W
Answer
P = 1693 W
6.AC power & power factor
Answer: C = 7.958 µF
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.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 25 − j20 Ω
Capacitance
XC = 20 Ω → C = 1/(2π·1000·20) = 7.958 µF
Answer
C = 7.958 µF
7.Three-phase power
Answer: IL = 73.22 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 = 45/0.88 = 51.14 kW
Line current
IL = Pin/(√3·VL·cos φ) = 51136/(√3·480·0.84) = 73.22 A
Answer
IL = 73.22 A
8.Three-phase power
Answer: Cph = 306.2 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 = 50·tan(acos 0.75) = 50·0.8819 = 44.1 kvar
Q2 = 50·tan(acos 0.9) = 50·0.4843 = 24.22 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 19.88 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.