A buck converter turns 24 V into 7.2 V for a 20 Ω load at 250 kHz, with L = 150 µH and C = 470 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
A DC motor (k = 0.3 V·s/rad, Ra = 1.5 Ω) runs from 24 V. At full load it draws 2 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 24 Vk = 0.3 V·s/radRa = 1.5 ΩIFL = 2 A
SR = %
4.DC motorsHard
A separately excited DC motor on 220 V (Ra = 1.5 Ω) runs at 1200 rpm with an armature current of 20 A. The load changes and the current becomes 10 A, with the field unchanged. Find the new speed n2.
V = 220 VRa = 1.5 Ωn1 = 1200 rpmIa1 = 20 AIa2 = 10 A
n2 = rpm
5.AC power & power factorMedium
A load with impedance Z = 12 + j16 Ω is connected to a 230 V (rms) single-phase supply. Find the apparent power S.
V = 230 VZ = 12 + j16 Ω
S = VA
6.AC power & power factorHard
A single-phase load takes 35 kW at a power factor of 0.75 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.9. Find the reactive power QC the capacitor must supply.
P = 35 kWpf1 = 0.75pf2 = 0.9V = 230 Vf = 60 Hz
QC = kvar
7.Three-phase powerMedium
A three-phase induction motor delivers 15 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 15 kWη = 95%cos φ = 0.85VL = 480 V
IL = A
8.Three-phase powerHard
A balanced three-phase load takes 250 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 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 250 kWpf1 = 0.7pf2 = 0.92VL = 415 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 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 = 7.5/15 = 0.5
Average and ripple current
Io = 7.5/15 = 0.5 A
ΔIL = VL·D/(L·f) = 7.5·0.5/(33·10⁻⁶·250·10³) = 0.4545 A
Peak current
Imax = 0.5 + 0.4545/2 = 0.7273 A
Answer
Imax = 0.7273 A
2.DC-DC converters
Answer: ΔVo = 0.143 mV
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 = 7.2/24 = 0.3
Inductor ripple current
ΔIL = (24 − 7.2)·0.3/(150·10⁻⁶·250·10³) = 0.1344 A
Output ripple
ΔVo = 0.1344/(8·250·10³·470·10⁻⁶) = 0.143 mV
Answer
ΔVo = 0.143 mV
3.DC motors
Answer: SR = 14.29 %
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·0.3) = 763.9 rpm
nFL = 60·(24 − 2·1.5)/(2π·0.3) = 668.5 rpm
Speed regulation
SR = (763.9 − 668.5)/668.5·100 = 14.29 %
Answer
SR = 14.29 %
4.DC motors
Answer: n2 = 1295 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 = 220 − 20·1.5 = 190 V
E2 = 220 − 10·1.5 = 205 V
New speed
n2 = 1200·205/190 = 1295 rpm
Answer
n2 = 1295 rpm
5.AC power & power factor
Answer: S = 2645 VA
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.
Current
|Z| = √(12² + 16²) = 20 Ω
I = V/|Z| = 230/20 = 11.5 A
Apparent power
S = V·I = 230·11.5 = 2645 VA
Answer
S = 2645 VA
6.AC power & power factor
Answer: QC = 13.92 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 = 35·tan(41.41°) = 30.87 kvar
Q2 = P·tan φ2 = 35·tan(25.84°) = 16.95 kvar
QC = Q1 − Q2 = 13.92 kvar
Answer
QC = 13.92 kvar
7.Three-phase power
Answer: IL = 22.34 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 = 15/0.95 = 15.79 kW
Line current
IL = Pin/(√3·VL·cos φ) = 15789/(√3·480·0.85) = 22.34 A
Answer
IL = 22.34 A
8.Three-phase power
Answer: Cph = 915.2 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 = 250·tan(acos 0.7) = 250·1.02 = 255.1 kvar
Q2 = 250·tan(acos 0.92) = 250·0.426 = 106.5 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 148.6 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.