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 boost converter turns 48 V into 80 V for a 20 Ω load. Find the average inductor current IL.
Vin = 48 VVo = 80 VR = 20 Ω
IL = A
2.DC-DC convertersHard
A buck converter turns 36 V into 25.2 V at 250 kHz with L = 100 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
Vin = 36 VVo = 25.2 Vf = 250 kHzL = 100 µH
Io,min = A
3.DC motorsMedium
A DC motor draws 40 A from a 110 V supply. Its armature resistance is 0.2 Ω and its rotational (friction, windage and iron) losses are 163 W. Find its efficiency η.
V = 110 VIa = 40 ARa = 0.2 ΩProt = 163 W
η = %
4.DC motorsHard
A DC motor (k = 0.1 V·s/rad, Ra = 0.5 Ω) on a fixed 24 V supply drives a load of 0.2 N·m. Find the series resistance Rs that slows it down to 300 rpm.
An inductive load takes 10 A and 900 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 = 10 AP = 900 W
X = Ω
6.AC power & power factorHard
A 24 V (rms), 1000 Hz source has an internal impedance Zs = 20 + j10 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 = 20 + j10 Ω
Pmax = W
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 25 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.92 lagging. Find the total real power P drawn by the load in kW.
VL = 690 VIL = 25 Acos φ = 0.92
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 50 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 = 50 kWpf1 = 0.75pf2 = 0.92VL = 690 Vf = 60 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: IL = 6.667 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 − 48/80 = 0.4
Load current
Io = 80/20 = 4 A
Inductor current
IL = Io/(1 − D) = 4/0.6 = 6.667 A
Answer
IL = 6.667 A
2.DC-DC converters
Answer: Io,min = 0.1512 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 = 25.2/36 = 0.7
Ripple current
ΔIL = 25.2·0.3/(100·10⁻⁶·250·10³) = 0.3024 A
Boundary load current
Io,min = ΔIL/2 = 0.1512 A
Answer
Io,min = 0.1512 A
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
3.DC motors
Answer: η = 89.02 %
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 = 110·40 = 4400 W
Pdev = 102·40 = 4080 W
Pout = 4080 − 163 = 3917 W
Efficiency
η = 3917/4400·100 = 89.02 %
Answer
η = 89.02 %
4.DC motors
Answer: Rs = 9.929 Ω
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.
Current and back-EMF
Ia = 0.2/0.1 = 2 A
E = 0.1·31.42 = 3.142 V
Series resistance
Ra + Rs = (24 − 3.142)/2 = 10.43 Ω
Rs = 9.929 Ω
Answer
Rs = 9.929 Ω
The resistor wastes 39.72 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: X = 7.937 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 = 900/10² = 9 Ω
Reactance from the impedance
|Z| = V/I = 120/10 = 12 Ω
X = √(|Z|² − R²) = √(12² − 9²) = 7.937 Ω
Answer
X = 7.937 Ω
6.AC power & power factor
Answer: Pmax = 7.2 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* = 20 − j10 Ω
Maximum power
I = Vs/(2Rs) = 24/40 = 0.6 A
Pmax = I²·RL = Vs²/(4Rs) = 24²/(4·20) = 7.2 W
Answer
Pmax = 7.2 W
7.Three-phase power
Answer: P = 27.49 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 = √3·690·25 = 29.88 kVA
Real power
P = S·cos φ = 29.88·0.92 = 27.49 kW
Answer
P = 27.49 kW
8.Three-phase power
Answer: Cph = 42.34 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 0.92) = 50·0.426 = 21.3 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 22.8 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.