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
A buck converter turns 36 V into 18 V for a 25 Ω load. It switches at 40 kHz with L = 220 µH. Find the peak inductor current Imax.
Vin = 36 VVo = 18 Vf = 40 kHzL = 220 µHR = 25 Ω
Imax = A
2.DC-DC convertersHard
A buck converter turns 12 V into 4.8 V at 50 kHz with L = 100 µH. Find the smallest load current Io,min for which it stays in continuous conduction.
Vin = 12 VVo = 4.8 Vf = 50 kHzL = 100 µH
Io,min = A
3.DC motorsMedium
A DC motor (k = 0.6 N·m/A, Ra = 1.5 Ω) runs from 48 V and drives a load torque of 1.2 N·m. Find its speed n in rpm.
V = 48 Vk = 0.6 N·m/ARa = 1.5 ΩT = 1.2 N·m
n = rpm
4.DC motorsHard
A DC motor with k = 0.5 V·s/rad and Ra = 1 Ω must drive a load of 3 N·m at 300 rpm. Find the armature voltage V it needs.
k = 0.5 V·s/radRa = 1 ΩT = 3 N·mn = 300 rpm
V = V
5.AC power & power factorMedium
Two loads are connected in parallel to a 400 V single-phase supply: a 27 kW load at a power factor of 0.8 lagging and a 10 kW load at a power factor of 0.9 leading. Find the total current I drawn from the supply.
V = 400 VP1 = 27 kW, pf1 = 0.8 laggingP2 = 10 kW, pf2 = 0.9 leading
I = A
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 84 kW at a power factor of 0.7 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 200 kVAP1 = 84 kWpf1 = 0.7
ΔP = kW
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.
VL = 480 VIL = 30 Acos φ = 0.82
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.7 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. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 75 kWpf1 = 0.7pf2 = 0.98VL = 415 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 1.231 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 = 18/36 = 0.5
Average and ripple current
Io = 18/25 = 0.72 A
ΔIL = VL·D/(L·f) = 18·0.5/(220·10⁻⁶·40·10³) = 1.023 A
Peak current
Imax = 0.72 + 1.023/2 = 1.231 A
Answer
Imax = 1.231 A
2.DC-DC converters
Answer: Io,min = 0.288 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 = 4.8/12 = 0.4
Ripple current
ΔIL = 4.8·0.6/(100·10⁻⁶·50·10³) = 0.576 A
Boundary load current
Io,min = ΔIL/2 = 0.288 A
Answer
Io,min = 0.288 A
Below this current the converter enters discontinuous conduction and Vo rises above D·Vin.
3.DC motors
Answer: n = 716.2 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 = 1.2/0.6 = 2 A
Back-EMF and speed
E = 48 − 2·1.5 = 45 V
ω = E/k = 75 rad/s → n = 716.2 rpm
Answer
n = 716.2 rpm
4.DC motors
Answer: V = 21.71 V
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
E = 0.5·31.42 = 15.71 V
Armature current
Ia = 3/0.5 = 6 A
Armature voltage
V = 15.71 + 6·1 = 21.71 V
Answer
V = 21.71 V
5.AC power & power factor
Answer: I = 100.2 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 = 27·tan(arccos 0.8) = 20.25 kvar
Q2 = −10·tan(arccos 0.9) = −4.843 kvar
Totals
P = 27 + 10 = 37 kW
Q = 20.25 + (−4.843) = 15.41 kvar
S = √(37² + 15.41²) = 40.08 kVA
Total current
I = S/V = 40080/400 = 100.2 A
Answer
I = 100.2 A
6.AC power & power factor
Answer: ΔP = 96.71 kW
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 the existing load
Q1 = 84·tan(arccos 0.7) = 85.7 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 85.7²) = 180.7 kW
Extra real power
ΔP = 180.7 − 84 = 96.71 kW
Answer
ΔP = 96.71 kW
7.Three-phase power
Answer: P = 20.45 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 = √3·480·30 = 24.94 kVA
Real power
P = S·cos φ = 24.94·0.82 = 20.45 kW
Answer
P = 20.45 kW
8.Three-phase power
Answer: Cph = 377.6 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 = 75·tan(acos 0.7) = 75·1.02 = 76.52 kvar
Q2 = 75·tan(acos 0.98) = 75·0.2031 = 15.23 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 61.29 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.