UMBB/IGEE/EE444
Correction FINAL EXAM POWER ELECTRONICS
Exercise 01(5 pts):
A step-down chopper fed from a 200V source operates a dc motor whose armature EMF is 170 V
and armature resistance is 0.5 Ω. With the magnitude control ratio of 0.4.
- Find the average and RMS value of output voltage and current
Vo=kxVi=0.4x200=80V.
0.75
0.75
Vrms=√kxVi=126.5V
Io=(Vo-E)/R=-180A.
0.75
Io=(Vrms-E)/R=-87A.
0.75
- Motor power
P=VoxIo=-80x180=14,4kw.
0.5
- Determine the quadrant of operation.
V>0 I<0 => 2nd quadrant
0.5
- If the conducting time of IGBT is 200µs, compute the chopping frequency.
Ton=200µs
Ton=Txk=>T=Ton/k=200/0.4=500 µs.
F=1/T=2khz.
0.5
- Determine the steady state motor speed if actual speed is 1000 rpm.
At steady state without a mechanical load EFM will be E=80V.
170V -> 1000rpm
80V -> w
W=80x1000/170=470rpm.
0.5
Exercise 02(7 pts):
A step-up chopper is supplied from three 36V photovoltaic panel.
1- Find the duty ratio of the chopper switch such that the output voltage has a peak value of
350 V.
Vo =Vi/(1-k)=>k=1-Vi/Vo.
K=1-3x36/350=0.69.
1
2- What is the duration of output voltage pulses if the switching frequency is 5 kHz?
Toff=(1-k)xT=(1-k)/f=62 µs.
1
1
This chopper supply a single-phase bridge inverter having square wave output debited on a
220V/400W electric lamp. Determine
1- The output Voltage form (Graph V(t) ).
VA
350V
-350V
2- The RMS value of output voltage
3- Output current
350
1
1
2
1
I=Vrms/R
R=(Vnom)2 /pnom
I=Vrms xpnom /(Vnom)2=350x400/2202=2.89A
1
2
0.5
4- Lamp impedance.
R=(Vnom)2 /pnom=2202/400=121Ω
0.5
5- Real power absorbed by the lamp.
P=Vrms*I=350x2.89=1011.5w
0.5
In order to improve the quality of voltage output, the inverter controlled by PWM techniques.
Determine
2
1- Maximum RMS Value of the output Voltage and current (modulation index m=1).
.m=1 => Vmax=Udc =>Vrms=Udc/√2=248V.
0.5
2- Calculate the modulation index for Vrms=220V.
.m=220x√2/350=0.886.
0.5
3- The current and the power in this case.
I=V/R=220/121=1.81A
0.5
P=VxI=400w.
Exercise 03(6 pts):
:Output voltages of three-phase inverter.
Vector
ABC
VAB
VBC
VCA
VAN/Vdc
VBN /Vdc
VCN /Vdc
0
1
2
3
000
001
011
010
0
0
-Vdc
-Vdc
0
-Vdc
0
Vdc
0
Vdc
Vdc
0
0
-1/3
-2/3
-1/3
0
-1/3
1/3
2/3
0
2/3
1/3
-1/3
4
5
6
110
100
101
0
Vdc
Vdc
Vdc
0
-Vdc
-Vdc
-Vdc
0
1/3
2/3
1/3
1/3
-1/3
-2/3
-2/3
-1/3
1/3
7
111
0
0
0
0
0
0
-0.25 for each mistake
Exercise 04 (2 pts):
1- The voltage VA is the output of analog comparator
This comparator change state at:
3
V
V xR
R
R
2
Comparator output can be VA =0 or VA=Vcc;
2- The voltage VB is the output of analog integrator
1
! "!
V =
(
−
2
)
So
If VA=0:
V =−
V
%&
2
=
! "!
2
+V
(1 −
#$
R
) = 0.095
R
= 0,475
VA=Vcc:
V =
V
2 ! "!
#$
=
2
+V
(1 +
%&
R
) = 0.905
R
= 4,525
0.5
4
1.5
5