An ideal inverting buck-boost converter turns 36 V into −108 V for a 10 Ω load at 25 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 36 VVo = −108 Vf = 25 kHzR = 10 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor (k = 1.2 V·s/rad, Ra = 1.2 Ω) runs from 110 V. At full load it draws 10 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 110 Vk = 1.2 V·s/radRa = 1.2 ΩIFL = 10 A
SR = %
4.DC motorsHard
A separately excited DC motor on 24 V (Ra = 0.2 Ω) runs at 1440 rpm with an armature current of 4 A. The load changes and the current becomes 8 A, with the field unchanged. Find the new speed n2.
V = 24 VRa = 0.2 Ωn1 = 1440 rpmIa1 = 4 AIa2 = 8 A
n2 = rpm
5.AC power & power factorMedium
A single-phase motor takes 4 kW at a power factor of 0.85 lagging from a 230 V supply. Find the line current I.
P = 4 kWpf = 0.85V = 230 V
I = A
6.AC power & power factorHard
A single-phase load takes 5 kW at a power factor of 0.8 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.
P = 5 kWpf1 = 0.8pf2 = 0.98V = 230 Vf = 50 Hz
Inew = A
7.Three-phase powerMedium
A balanced Y-connected load with an impedance of 16 + j12 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.
VL = 415 VZ = 16 + j12 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.75 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 60 kWpf1 = 0.75pf2 = 0.9VL = 480 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 1.396 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 = 19.2/48 = 0.4
Average and ripple current
Io = 19.2/20 = 0.96 A
ΔIL = VL·D/(L·f) = 28.8·0.4/(330·10⁻⁶·40·10³) = 0.8727 A
Peak current
Imax = 0.96 + 0.8727/2 = 1.396 A
Answer
Imax = 1.396 A
2.DC-DC converters
Answer: Lcrit = 12.5 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 = 108/(36 + 108) = 0.75
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 − 0.75)²·10/(225·10³) = 12.5 µH
Answer
Lcrit = 12.5 µH
3.DC motors
Answer: SR = 12.24 %
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 = 60·110/(2π·1.2) = 875.4 rpm
nFL = 60·(110 − 10·1.2)/(2π·1.2) = 779.9 rpm
Speed regulation
SR = (875.4 − 779.9)/779.9·100 = 12.24 %
Answer
SR = 12.24 %
4.DC motors
Answer: n2 = 1390 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 = 24 − 4·0.2 = 23.2 V
E2 = 24 − 8·0.2 = 22.4 V
New speed
n2 = 1440·22.4/23.2 = 1390 rpm
Answer
n2 = 1390 rpm
5.AC power & power factor
Answer: I = 20.46 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) = 4000/(230·0.85) = 20.46 A
Answer
I = 20.46 A
6.AC power & power factor
Answer: Inew = 22.18 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 = 5·tan(36.87°) = 3.75 kvar
Q2 = P·tan φ2 = 5·tan(11.48°) = 1.015 kvar
QC = Q1 − Q2 = 2.735 kvar
New line current
Inew = P/(V·cos φ2) = 5000/(230·0.98) = 22.18 A
Answer
Inew = 22.18 A
7.Three-phase power
Answer: P = 6.889 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 = 415/√3 = 239.6 V
Phase current
|Z| = √(16² + 12²) = 20 Ω
Iph = 239.6/20 = 11.98 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·11.98²·16 = 6.889 kW
Answer
P = 6.889 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 109.9 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 = 60·tan(acos 0.75) = 60·0.8819 = 52.92 kvar
Q2 = 60·tan(acos 0.9) = 60·0.4843 = 29.06 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 23.86 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.