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 6.3 V at 250 kHz with L = 10 µH. Find the peak-to-peak inductor ripple current ΔIL.
Vin = 9 VVo = 6.3 Vf = 250 kHzL = 10 µH
ΔIL = A
2.DC-DC convertersHard
An ideal buck converter turns 9 V into 3.6 V for a 50 Ω load at 100 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 9 VVo = 3.6 Vf = 100 kHzR = 50 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor (k = 1 N·m/A, Ra = 2 Ω) runs from 240 V and drives a load torque of 10 N·m. Find its speed n in rpm.
V = 240 Vk = 1 N·m/ARa = 2 ΩT = 10 N·m
n = rpm
4.DC motorsHard
A separately excited DC motor on 220 V (Ra = 0.5 Ω) runs at 1200 rpm with an armature current of 15 A. The load changes and the current becomes 22.5 A, with the field unchanged. Find the new speed n2.
V = 220 VRa = 0.5 Ωn1 = 1200 rpmIa1 = 15 AIa2 = 22.5 A
n2 = rpm
5.AC power & power factorMedium
A single-phase motor takes 26 kW at a power factor of 0.6 lagging from a 240 V supply. Find the line current I.
P = 26 kWpf = 0.6V = 240 V
I = A
6.AC power & power factorHard
A 24 V (rms), 1000 Hz source has an internal impedance Zs = 10 + j6 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 24 V (rms)f = 1000 HzZs = 10 + j6 Ω
Pmax = W
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 60 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
VL = 400 VIL = 60 Acos φ = 0.85
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 100 kW at a power factor of 0.8 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 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 100 kWpf1 = 0.8pf2 = 0.9VL = 400 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.756 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 = 6.3/9 = 0.7
Inductor voltage during the on-time
While S is on, the inductor sits between Vin and Vo.
VL = 9 − 6.3 = 2.7 V
Ripple current
ΔIL = VL·D/(L·f) = 2.7·0.7/(10·10⁻⁶·250·10³) = 0.756 A
Answer
ΔIL = 0.756 A
2.DC-DC converters
Answer: Lcrit = 150 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 = 3.6/9 = 0.4
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = (1 − D)·R/(2f)
Critical inductance
Lcrit = (1 − 0.4)·50/(2100·10³) = 150 µH
Answer
Lcrit = 150 µH
3.DC motors
Answer: n = 2101 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 = 10/1 = 10 A
Back-EMF and speed
E = 240 − 10·2 = 220 V
ω = E/k = 220 rad/s → n = 2101 rpm
Answer
n = 2101 rpm
4.DC motors
Answer: n2 = 1179 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 = 220 − 15·0.5 = 212.5 V
E2 = 220 − 22.5·0.5 = 208.8 V
New speed
n2 = 1200·208.8/212.5 = 1179 rpm
Answer
n2 = 1179 rpm
5.AC power & power factor
Answer: I = 180.6 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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 26000/(240·0.6) = 180.6 A
Answer
I = 180.6 A
6.AC power & power factor
Answer: Pmax = 14.4 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.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 10 − j6 Ω
Maximum power
I = Vs/(2Rs) = 24/20 = 1.2 A
Pmax = I²·RL = Vs²/(4Rs) = 24²/(4·10) = 14.4 W
Answer
Pmax = 14.4 W
7.Three-phase power
Answer: P = 35.33 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 = √3·400·60 = 41.57 kVA
Real power
P = S·cos φ = 41.57·0.85 = 35.33 kW
Answer
P = 35.33 kW
8.Three-phase power
Answer: QC = 26.57 kvar
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.