An ideal inverting buck-boost converter turns 9 V into −13.5 V for a 2 Ω load at 50 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 9 VVo = −13.5 Vf = 50 kHzR = 2 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor with k = 0.8 N·m/A and Ra = 1.5 Ω is connected to 120 V. Find its stall torque Tstall, the torque at zero speed.
V = 120 Vk = 0.8 N·m/ARa = 1.5 Ω
Tstall = N·m
4.DC motorsHard
A DC chopper feeds a motor (k = 0.75 V·s/rad, Ra = 0.25 Ω) from a 240 V DC bus. The average armature voltage is D·240 V. Find the duty cycle D that runs the motor at 400 rpm with a load of 30 N·m.
A single-phase motor takes 4 kW at a power factor of 0.7 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 4 kWpf = 0.7V = 230 V
Q = kvar
6.AC power & power factorHard
A 12 V (rms), 1000 Hz source has an internal impedance Zs = 2 + j10 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 12 V (rms)f = 1000 HzZs = 2 + j10 Ω
Pmax = W
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 75 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 75 Acos φ = 0.82
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 75 kW at a power factor of 0.8 lagging from a 400 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 Y. Find the capacitance Cph of each capacitor in µF.
P = 75 kWpf1 = 0.8pf2 = 0.98VL = 400 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 0.3205 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 = 4.5/15 = 0.3
Average and ripple current
Io = 4.5/20 = 0.225 A
ΔIL = VL·D/(L·f) = 10.5·0.3/(33·10⁻⁶·500·10³) = 0.1909 A
Peak current
Imax = 0.225 + 0.1909/2 = 0.3205 A
Answer
Imax = 0.3205 A
2.DC-DC converters
Answer: Lcrit = 3.2 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 = 13.5/(9 + 13.5) = 0.6
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 − 0.6)²·2/(250·10³) = 3.2 µH
Answer
Lcrit = 3.2 µH
3.DC motors
Answer: Tstall = 64 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 = 120/1.5 = 80 A
Stall torque
Tstall = 0.8·80 = 64 N·m
Answer
Tstall = 64 N·m
4.DC motors
Answer: D = 17.26 %
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.
Required armature voltage
Va = 0.75·41.89 + 40·0.25 = 41.42 V
Duty cycle
D = 41.42/240 = 17.26 %
Answer
D = 17.26 %
5.AC power & power factor
Answer: Q = 4.081 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(0.7) = 45.57°, tan φ = 1.02
Reactive power
Q = P·tan φ = 4·1.02 = 4.081 kvar
Answer
Q = 4.081 kvar
6.AC power & power factor
Answer: Pmax = 18 W
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.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 2 − j10 Ω
Maximum power
I = Vs/(2Rs) = 12/4 = 3 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·2) = 18 W
Answer
Pmax = 18 W
7.Three-phase power
Answer: P = 44.21 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 = √3·415·75 = 53.91 kVA
Real power
P = S·cos φ = 53.91·0.82 = 44.21 kW
Answer
P = 44.21 kW
8.Three-phase power
Answer: Cph = 816.1 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 0.8) = 75·0.75 = 56.25 kvar
Q2 = 75·tan(acos 0.98) = 75·0.2031 = 15.23 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 41.02 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.