An ideal inverting buck-boost converter turns 48 V into −16 V for a 12 Ω load at 200 kHz. Find the smallest inductance Lcrit for continuous conduction.
Vin = 48 VVo = −16 Vf = 200 kHzR = 12 Ω
Lcrit = µH
3.DC motorsMedium
A DC motor with k = 0.75 N·m/A and Ra = 1 Ω is connected to 120 V. Find its stall torque Tstall, the torque at zero speed.
V = 120 Vk = 0.75 N·m/ARa = 1 Ω
Tstall = N·m
4.DC motorsHard
A separately excited DC motor on 120 V (Ra = 0.4 Ω) runs at 1200 rpm with an armature current of 30 A. The load changes and the current becomes 45 A, with the field unchanged. Find the new speed n2.
V = 120 VRa = 0.4 Ωn1 = 1200 rpmIa1 = 30 AIa2 = 45 A
n2 = rpm
5.AC power & power factorMedium
An inductive load takes 5.5 A and 633 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 230 VI = 5.5 AP = 633 W
R = Ω
6.AC power & power factorHard
A single-phase load takes 6 kW at a power factor of 0.8 lagging from a 230 V, 60 Hz supply. A 47 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 6 kWpf1 = 0.8V = 230 Vf = 60 HzC = 47 µF
I = A
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 50 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.
VL = 208 VIL = 50 Acos φ = 0.85
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 250 kW at a power factor of 0.65 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 250 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: Imax = 0.234 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 = 3.6/9 = 0.4
Average and ripple current
Io = 3.6/20 = 0.18 A
ΔIL = VL·D/(L·f) = 5.4·0.4/(100·10⁻⁶·200·10³) = 0.108 A
Peak current
Imax = 0.18 + 0.108/2 = 0.234 A
Answer
Imax = 0.234 A
2.DC-DC converters
Answer: Lcrit = 16.88 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 = 16/(48 + 16) = 0.25
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 − 0.25)²·12/(2200·10³) = 16.88 µH
Answer
Lcrit = 16.88 µH
3.DC motors
Answer: Tstall = 90 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 = 120 A
Stall torque
Tstall = 0.75·120 = 90 N·m
Answer
Tstall = 90 N·m
4.DC motors
Answer: n2 = 1133 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 = 120 − 30·0.4 = 108 V
E2 = 120 − 45·0.4 = 102 V
New speed
n2 = 1200·102/108 = 1133 rpm
Answer
n2 = 1133 rpm
5.AC power & power factor
Answer: R = 20.93 Ω
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.
Resistance from the real power
R = P/I² = 633/5.5² = 20.93 Ω
Answer
R = 20.93 Ω
6.AC power & power factor
Answer: I = 30.34 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 = 6·tan(arccos 0.8) = 4.5 kvar
QC = V²·2πf·C = 230²·2π·60·47·10⁻⁶ = 937.3 var = 0.9373 kvar
Q2 = 4.5 − 0.9373 = 3.563 kvar
New apparent power
S2 = √(6² + 3.563²) = 6.978 kVA
New line current
I = S2/V = 6978/230 = 30.34 A
Answer
I = 30.34 A
7.Three-phase power
Answer: P = 15.31 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 = √3·208·50 = 18.01 kVA
Real power
P = S·cos φ = 18.01·0.85 = 15.31 kW
Answer
P = 15.31 kW
8.Three-phase power
Answer: QC = 210.1 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.