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 9 V into 2.7 V at a switching frequency of 50 kHz. Find the inductance L that keeps the peak-to-peak inductor ripple current at 0.6 A.
Vin = 9 VVo = 2.7 Vf = 50 kHzΔIL = 0.6 A
L = µH
2.DC-DC convertersHard
An ideal boost converter turns 15 V into 60 V for a 10 Ω load at 40 kHz. The output capacitor is 100 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
Vin = 15 VVo = 60 Vf = 40 kHzC = 100 µFR = 10 Ω
ΔVo = mV
3.DC motorsMedium
A DC motor (k = 0.3 V·s/rad, Ra = 1.2 Ω) runs from 48 V. At full load it draws 2 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 48 Vk = 0.3 V·s/radRa = 1.2 ΩIFL = 2 A
SR = %
4.DC motorsHard
A DC chopper feeds a motor (k = 0.1 V·s/rad, Ra = 0.25 Ω) from a 120 V DC bus. The average armature voltage is D·120 V. Find the duty cycle D that runs the motor at 1000 rpm with a load of 0.6 N·m.
A single-phase motor takes 23 kW at a power factor of 0.75 lagging from a 240 V supply. Find the line current I.
P = 23 kWpf = 0.75V = 240 V
I = A
6.AC power & power factorHard
A 100 V (rms), 60 Hz source has an internal impedance Zs = 2 + j10 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 = 100 V (rms)f = 60 HzZs = 2 + j10 Ω
C = µF
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 12 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
VL = 208 VIL = 12 Acos φ = 0.85
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 200 kW at a power factor of 0.75 lagging from a 690 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 = 200 kWpf1 = 0.75pf2 = 0.92VL = 690 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: L = 63 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 = 2.7/9 = 0.3
Inductance
L = 6.3·0.3/(0.6·50·10³) = 63 µH
Answer
L = 63 µH
2.DC-DC converters
Answer: ΔVo = 1125 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 = 60/10 = 6 A
Output ripple
ΔVo = 6·0.75/(40·10³·100·10⁻⁶) = 1125 mV
Answer
ΔVo = 1125 mV
3.DC motors
Answer: SR = 5.263 %
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 and full-load speed
n₀ = 60·48/(2π·0.3) = 1528 rpm
nFL = 60·(48 − 2·1.2)/(2π·0.3) = 1451 rpm
Speed regulation
SR = (1528 − 1451)/1451·100 = 5.263 %
Answer
SR = 5.263 %
4.DC motors
Answer: D = 9.977 %
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.
Required armature voltage
Va = 0.1·104.7 + 6·0.25 = 11.97 V
Duty cycle
D = 11.97/120 = 9.977 %
Answer
D = 9.977 %
5.AC power & power factor
Answer: I = 127.8 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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 23000/(240·0.75) = 127.8 A
Answer
I = 127.8 A
6.AC power & power factor
Answer: C = 265.3 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* = 2 − j10 Ω
Capacitance
XC = 10 Ω → C = 1/(2π·60·10) = 265.3 µF
Answer
C = 265.3 µF
7.Three-phase power
Answer: P = 3.675 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 = √3·208·12 = 4.323 kVA
Real power
P = S·cos φ = 4.323·0.85 = 3.675 kW
Answer
P = 3.675 kW
8.Three-phase power
Answer: Cph = 169.3 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 = 200·tan(acos 0.75) = 200·0.8819 = 176.4 kvar
Q2 = 200·tan(acos 0.92) = 200·0.426 = 85.2 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 91.18 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.