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 48 V into −16 V at 100 kHz with L = 220 µH. Find the peak-to-peak inductor ripple current ΔIL.
Vin = 48 VVo = −16 Vf = 100 kHzL = 220 µH
ΔIL = A
2.DC-DC convertersHard
An ideal buck converter turns 15 V into 10.5 V for a 4 Ω load at 100 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 15 VVo = 10.5 Vf = 100 kHzR = 4 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor with k = 0.6 N·m/A and Ra = 1.2 Ω is connected to 240 V. Find its stall torque Tstall, the torque at zero speed.
V = 240 Vk = 0.6 N·m/ARa = 1.2 Ω
Tstall = N·m
4.DC motorsHard
A small DC motor has an armature resistance of 1.2 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 24 VRa = 1.2 Ω
Pmax = W
5.AC power & power factorMedium
A single-phase motor takes 15 kW at a power factor of 0.75 lagging from a 400 V supply. Find the line current I.
P = 15 kWpf = 0.75V = 400 V
I = A
6.AC power & power factorHard
A single-phase load takes 45 kW at a power factor of 0.8 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.
P = 45 kWpf1 = 0.8pf2 = 0.98V = 240 Vf = 60 Hz
Inew = A
7.Three-phase powerMedium
A balanced Δ-connected load with an impedance of 12 + j16 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.
VL = 480 VZ = 12 + j16 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 75 kWpf1 = 0.7pf2 = 0.95VL = 690 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.5455 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 = 16/(48 + 16) = 0.25
Inductor voltage during the on-time
While S is on, the inductor sits directly across Vin.
VL = 48 V
Ripple current
ΔIL = VL·D/(L·f) = 48·0.25/(220·10⁻⁶·100·10³) = 0.5455 A
Answer
ΔIL = 0.5455 A
2.DC-DC converters
Answer: Lcrit = 6 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
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 − 0.7)·4/(2100·10³) = 6 µH
Answer
Lcrit = 6 µH
3.DC motors
Answer: Tstall = 120 N·m
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.
Stall current
Ia = V/Ra = 240/1.2 = 200 A
Stall torque
Tstall = 0.6·200 = 120 N·m
Answer
Tstall = 120 N·m
4.DC motors
Answer: Pmax = 120 W
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.
Power as a function of the back-EMF
P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 12 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·1.2) = 120 W
Answer
Pmax = 120 W
At this point the efficiency is only 50 %, so real motors run well above half speed.
5.AC power & power factor
Answer: I = 50 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) = 15000/(400·0.75) = 50 A
Answer
I = 50 A
6.AC power & power factor
Answer: Inew = 191.3 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 before and after
The real power P does not change; only the reactive power drawn from the supply does.
Q1 = P·tan φ1 = 45·tan(36.87°) = 33.75 kvar
Q2 = P·tan φ2 = 45·tan(11.48°) = 9.138 kvar
QC = Q1 − Q2 = 24.61 kvar
New line current
Inew = P/(V·cos φ2) = 45000/(240·0.98) = 191.3 A
Answer
Inew = 191.3 A
7.Three-phase power
Answer: P = 20.74 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.
Phase voltage
Vph = VL = 480 V
Phase current
|Z| = √(12² + 16²) = 20 Ω
Iph = 480/20 = 24 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·24²·12 = 20.74 kW
Answer
P = 20.74 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: QC = 51.86 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.