Power Electronics and Drives: practice problems and exam questions

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 V_o = D·V_in, boost V_o = V_in/(1 − D), inverting buck-boost |V_o| = V_in·D/(1 − D). An ideal converter keeps the power: V_in·I_in = V_o·I_o. While S is on the inductor current rises by ΔI_L = V_L·D·T/L, with V_L = V_in − V_o for the buck and V_in for the boost and buck-boost. The output ripple is ΔV_o = ΔI_L/(8fC) for the buck and I_o·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔI_L/2.

DC motor: the armature is a resistance R_a in series with the back-EMF E, so V = E + I_a·R_a. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·I_a, 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·I_a = T·ω; the input power is V·I_a, and the copper loss I_a²R_a and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − R_a·T/k², from the no-load speed V/k to zero at the stall torque k·V/R_a. At standstill E = 0, so the starting current V/R_a is large unless a resistor limits it.