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 inverting buck-boost converter turns 12 V into −12 V at 250 kHz with L = 47 µH. Find the peak-to-peak inductor ripple current ΔIL.
Vin = 12 VVo = −12 Vf = 250 kHzL = 47 µH
ΔIL = A
2.DC-DC convertersHard
A buck converter turns 24 V into 12 V at 50 kHz with L = 68 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
Vin = 24 VVo = 12 Vf = 50 kHzL = 68 µH
Io,min = A
3.DC motorsMedium
A DC motor (k = 0.12 V·s/rad, Ra = 0.4 Ω) runs from 24 V. At full load it draws 5 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 24 Vk = 0.12 V·s/radRa = 0.4 ΩIFL = 5 A
SR = %
4.DC motorsHard
A separately excited DC motor on 240 V (Ra = 0.25 Ω) runs at 1000 rpm with an armature current of 40 A. The load changes and the current becomes 20 A, with the field unchanged. Find the new speed n2.
V = 240 VRa = 0.25 Ωn1 = 1000 rpmIa1 = 40 AIa2 = 20 A
n2 = rpm
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: a 5 kW load at a power factor of 0.7 lagging and a 2 kW heater (unity power factor). Find the total current I drawn from the supply.
A single-phase load takes 25 kW at a power factor of 0.8 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the reactive power QC the capacitor must supply.
P = 25 kWpf1 = 0.8pf2 = 0.9V = 400 Vf = 50 Hz
QC = kvar
7.Three-phase powerMedium
A three-phase induction motor delivers 22 kW of mechanical output power at full load. At this load its efficiency is 88% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.
Pout = 22 kWη = 88%cos φ = 0.82VL = 415 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 100 kW at a power factor of 0.8 lagging from a 415 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 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 = 0.98VL = 415 Vf = 50 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.5106 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 = 12/(12 + 12) = 0.5
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 12 V
Ripple current
ΔIL = VL·D/(L·f) = 12·0.5/(47·10⁻⁶·250·10³) = 0.5106 A
Answer
ΔIL = 0.5106 A
2.DC-DC converters
Answer: Io,min = 0.8824 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 = 12/24 = 0.5
Ripple current
ΔIL = 12·0.5/(68·10⁻⁶·50·10³) = 1.765 A
Boundary load current
Io,min = ΔIL/2 = 0.8824 A
Answer
Io,min = 0.8824 A
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
3.DC motors
Answer: SR = 9.091 %
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 = 60·24/(2π·0.12) = 1910 rpm
nFL = 60·(24 − 5·0.4)/(2π·0.12) = 1751 rpm
Speed regulation
SR = (1910 − 1751)/1751·100 = 9.091 %
Answer
SR = 9.091 %
4.DC motors
Answer: n2 = 1022 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 − 40·0.25 = 230 V
E2 = 240 − 20·0.25 = 235 V
New speed
n2 = 1000·235/230 = 1022 rpm
Answer
n2 = 1022 rpm
5.AC power & power factor
Answer: I = 37.66 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.
Reactive power of each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 5·tan(arccos 0.7) = 5.101 kvar
Q2 = 0 = 0 kvar
Totals
P = 5 + 2 = 7 kW
Q = 5.101 + 0 = 5.101 kvar
S = √(7² + 5.101²) = 8.661 kVA
Total current
I = S/V = 8661/230 = 37.66 A
Answer
I = 37.66 A
6.AC power & power factor
Answer: QC = 6.642 kvar
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 = 25·tan(36.87°) = 18.75 kvar
Q2 = P·tan φ2 = 25·tan(25.84°) = 12.11 kvar
QC = Q1 − Q2 = 6.642 kvar
Answer
QC = 6.642 kvar
7.Three-phase power
Answer: IL = 42.41 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 = 22/0.88 = 25 kW
Line current
IL = Pin/(√3·VL·cos φ) = 25000/(√3·415·0.82) = 42.41 A
Answer
IL = 42.41 A
8.Three-phase power
Answer: QC = 54.69 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.