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 64 V at 500 kHz with L = 33 µH. Find the peak-to-peak inductor ripple current ΔIL.
Vin = 48 VVo = 64 Vf = 500 kHzL = 33 µH
ΔIL = A
2.DC-DC convertersHard
An ideal inverting buck-boost converter turns 24 V into −24 V for a 10 Ω load at 40 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 24 VVo = −24 Vf = 40 kHzR = 10 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor draws 5 A from a 240 V supply. Its armature resistance is 1.5 Ω and its rotational (friction, windage and iron) losses are 47 W. Find its efficiency η.
V = 240 VIa = 5 ARa = 1.5 ΩProt = 47 W
η = %
4.DC motorsHard
A DC motor (k = 0.8 V·s/rad, Ra = 1 Ω) on a fixed 240 V supply drives a load of 20 N·m. Find the series resistance Rs that slows it down to 300 rpm.
A single-phase motor takes 16 kW at a power factor of 0.6 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 16 kWpf = 0.6V = 230 V
Q = kvar
6.AC power & power factorHard
A 100 kVA single-phase transformer supplies a load of 32 kW at a power factor of 0.65 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 = 100 kVAP1 = 32 kWpf1 = 0.65
ΔP = kW
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 10 Acos φ = 0.9
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 120 kW at a power factor of 0.75 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 120 kWpf1 = 0.75pf2 = 0.92VL = 480 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.7273 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 − 48/64 = 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/(33·10⁻⁶·500·10³) = 0.7273 A
Answer
ΔIL = 0.7273 A
2.DC-DC converters
Answer: Lcrit = 31.25 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 = 24/(24 + 24) = 0.5
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = (1 − D)²·R/(2f)
Critical inductance
Lcrit = (1 − 0.5)²·10/(240·10³) = 31.25 µH
Answer
Lcrit = 31.25 µH
3.DC motors
Answer: η = 92.96 %
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 = 240·5 = 1200 W
Pdev = 232.5·5 = 1163 W
Pout = 1163 − 47 = 1116 W
Efficiency
η = 1116/1200·100 = 92.96 %
Answer
η = 92.96 %
4.DC motors
Answer: Rs = 7.595 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 = 20/0.8 = 25 A
E = 0.8·31.42 = 25.13 V
Series resistance
Ra + Rs = (240 − 25.13)/25 = 8.595 Ω
Rs = 7.595 Ω
Answer
Rs = 7.595 Ω
The resistor wastes 4747 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: Q = 21.33 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.
Angle
φ = arccos(0.6) = 53.13°, tan φ = 1.333
Reactive power
Q = P·tan φ = 16·1.333 = 21.33 kvar
Answer
Q = 21.33 kvar
6.AC power & power factor
Answer: ΔP = 60.74 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 = 32·tan(arccos 0.65) = 37.41 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(100² − 37.41²) = 92.74 kW
Extra real power
ΔP = 92.74 − 32 = 60.74 kW
Answer
ΔP = 60.74 kW
7.Three-phase power
Answer: P = 6.469 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 = √3·415·10 = 7.188 kVA
Real power
P = S·cos φ = 7.188·0.9 = 6.469 kW
Answer
P = 6.469 kW
8.Three-phase power
Answer: QC = 54.71 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.