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 15 V into 10.5 V at a switching frequency of 50 kHz. Find the inductance L that keeps the peak-to-peak inductor ripple current at 0.5 A.
Vin = 15 VVo = 10.5 Vf = 50 kHzΔIL = 0.5 A
L = µH
2.DC-DC convertersHard
An ideal buck converter turns 36 V into 9 V for a 10 Ω load at 50 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 36 VVo = 9 Vf = 50 kHzR = 10 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor (k = 0.4 N·m/A, Ra = 0.4 Ω) runs from 48 V and drives a load torque of 4.8 N·m. Find its speed n in rpm.
V = 48 Vk = 0.4 N·m/ARa = 0.4 ΩT = 4.8 N·m
n = rpm
4.DC motorsHard
A separately excited DC motor (k = 2 V·s/rad at full field, Ra = 2 Ω, V = 240 V) drives a constant load torque of 24 N·m. The field flux is reduced to 70 % of its full value. Find the new speed n.
V = 240 Vk = 2 V·s/radRa = 2 ΩT = 24 N·mΦ = 70 %
n = rpm
5.AC power & power factorMedium
Two loads are connected in parallel to a 400 V single-phase supply: a 22 kW load at a power factor of 0.65 lagging and a 5 kW load at a power factor of 0.7 lagging. Find the overall power factor cos φ seen by the supply.
A single-phase load takes 30 kW at a power factor of 0.75 lagging from a 400 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.
P = 30 kWpf1 = 0.75pf2 = 0.98V = 400 Vf = 60 Hz
C = µF
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 415 V. The power factor is 0.92 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 25 Acos φ = 0.92
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 120 kW at a power factor of 0.8 lagging from a 480 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. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 120 kWpf1 = 0.8pf2 = 0.95VL = 480 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: L = 126 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 = 10.5/15 = 0.7
Inductance
L = 4.5·0.7/(0.5·50·10³) = 126 µH
Answer
L = 126 µH
2.DC-DC converters
Answer: Lcrit = 75 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 = 9/36 = 0.25
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 − 0.25)·10/(250·10³) = 75 µH
Answer
Lcrit = 75 µH
3.DC motors
Answer: n = 1031 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 = 4.8/0.4 = 12 A
Back-EMF and speed
E = 48 − 12·0.4 = 43.2 V
ω = E/k = 108 rad/s → n = 1031 rpm
Answer
n = 1031 rpm
4.DC motors
Answer: n = 1403 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.
Field weakening is how a DC drive runs above its base speed.
5.AC power & power factor
Answer: cos φ = 0.6589
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 of each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 22·tan(arccos 0.65) = 25.72 kvar
Q2 = 5·tan(arccos 0.7) = 5.101 kvar
Totals
P = 22 + 5 = 27 kW
Q = 25.72 + 5.101 = 30.82 kvar
S = √(27² + 30.82²) = 40.98 kVA
Overall power factor
cos φ = P/S = 27/40.98 = 0.6589 lagging
Answer
cos φ = 0.6589
6.AC power & power factor
Answer: C = 337.6 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.
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(41.41°) = 26.46 kvar
Q2 = P·tan φ2 = 30·tan(11.48°) = 6.092 kvar
QC = Q1 − Q2 = 20.37 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 20366/(2π·60·400²) = 337.6 µF
Answer
C = 337.6 µF
7.Three-phase power
Answer: P = 16.53 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 = √3·415·25 = 17.97 kVA
Real power
P = S·cos φ = 17.97·0.92 = 16.53 kW
Answer
P = 16.53 kW
8.Three-phase power
Answer: QC = 50.56 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.