Solved: USQ ELE3805 Power Electronics Assignment 1 Solution, 2018

This document provides detailed solutions to a Power Electronics assignment, specifically addressing questions related to resistor-capacitor circuits, inductor behavior, delta-star winding configurations, and IGBT switching losses. The solutions involve calculating mean and RMS values, peak-to-peak currents, power losses, duty ratios, and total harmonic distortion, applying fundamental electrical engineering principles and formulas. The assignment covers various aspects of power electronics, including circuit analysis, waveform sketching, and component characteristics, offering a comprehensive understanding of the subject matter. Desklib is a valuable resource for students seeking similar solved assignments and study materials.

Question 1:
The resistor voltage Vr is given below.
The resistor and the capacitor are assumed to be ideal. r = 10Ω, C = 1800μF.
When, the resistor is consuming power during Ton the capacitor is getting charged but during
Toff the capacitor voltage will appear in inverse polarity.
Now, the mean value of |Vc| = (Vc * Toff)/(Ton + Toff)
Now, Vc = Vr as the same voltage drop occurs at Toff across Capacitor.
So, mean value of |Vc| = 5.3 * (1-0.12) = 4.664 Volts.
Vc
Ton Toff Ton Toff

Question 2:
Given that the peak value of current i = 3.44 Amps.
Now, at Ton the inductor gets charged and discharges in the Toff period.
Hence, the rms value of i = 3.44*(1-0.19) = 2.7864 Amps.
Question 3:

Given that the line current I = 8.3 Amps and duty cycle α = π/24.
Now, delta to star ratio = 6:1.
So, phase current in star = (8.3*1)/(6+1) = 1.19 Amps.
So, mean value of |ib| = 1.19*π/24 = 0.156 Amps = 156 mA.
Question 4:
Given, I = 16.3 Amps and the duty cycle α = π/7.
Delta to star ratio = 8:1.
So, phase current in star ib = 16.3*(1/9) = 1.81 Amps.
phase current in star ia = 1.81*(π/7) = 0.812 Amps.
Hence, peak to peak value of iba = 2(1.81-0.812) = 2*0.998 = 1.996 Amps.

Question 5:
a= 14 V, L = 0.015H and R = 15 Ω.
The equation of iL = (V/R)*(1-e^(-R/L)t) = (14/15)*(1-e^(-15/0.015)t)
Now, the current at t = 0.0077 sec is iL = (14/15)*(1-e^(-15/0.015)0.0077) = 0.9329 Amps.
Question 6:

a = 6V,
L=0.006H and R=8Ω
The equation of iL = (V/R)*(1-e^(-R/L)t) = (6/8)*(1-e^(-8/0.006)t)
Now, at t = 0.0147 sec iL = (6/8)*(1-e^(-8/0.006)0.0147) = 0.75 Amp.
Question 7:
Given: E = 300V; RL = 18Ω; Gate to source voltage = 11V.
Now, using KVL the voltage drop at RL = VRL = 18 -11 = 7 volts.
Hence, power loss at load = VRL^2/ RL = 7^2/18 = 49/18 = 2.72 watts.
Question 8:
Expression of switching loss in IGBT is given by,
P(switching loss) = (Eon + Eoff)*fs
Here, Eon = Vds = 300 volts, Eoff = VGs = 15 Volts, fs = 17.2 kHz
P(switching loss) = (300+ 15)*17.2*10^3 = 5418 kW.

Question 9:
Given that,
Vd = 200V, Vo =400V, I = 3.15A ,
L=0.2H, f = 20kHz , D=0.25
The expression of peak value of inductor current iL at discontinuous conduction mode is
given by,
iL = ((Vd- Vo)/Lf)*D = ((200-400)*0.25)/(0.2*20*10^3) = -0.0125 Amps.
Question 10:
Given that,
Vd =61V, Vo =23V, Io = 9.2A, L=2mH
The condition for operating in the boundary of continuous and discontinuous conduction is
given by,
L = Lcdm = (D(Vd- Vo))/(2fVo) => f = 2VoL/(D(Vd- Vo)) = (2*23*2*10^(-3))/(0.25(61-
23)) = 9.68 kHz.
Question 11:
Given,
V =168V, E=55V, I = 9.9A
The expression of duty ratio D for a full converter operating in bipolar mode is given by,
D = (E/Vd) = 55/168 = 0.327.

Question 12:
In this case also as the DC to DC converter is step down converter, hence duty ratio
D = E/Vd = 33/108 = 0.306.
Question 13:
The percentage total harmonic distortion is given by,
THD = Vrms without fundamental components/Vrms with fundamental components.
Now, by method of Fourier transform the equation reduces to
THD = sqrt(π^2/8 -1) = 0.483 = 48.3%
Question 14:
Given, VAB = 587 Volts, Id = 9.7 Amps(rms)
Now, as the 25th harmonic is the odd harmonic and the peak value of all the harmonics are
equal to the load current Id hence Ipeak(25th harmonic) = 9.7*sqrt(2) = 13.72 Amps.
Question 15:

Inductor voltage = VAB*(1/1+7) = 400/8 = 50 Volts.
Now, the secondary side is 12 pulse while the primary side is 3 pulse.
Hence, voltage amplified = 50*(12/3) = 200 volts.
Hence, the peak to peak voltage across inductor = 200*2*sqrt(2) = 568.69 volts.