Central Philippine University Jaro, Iloilo City
School of Graduate Studies
Midterm Exam in Power Electronics (EE 721)
In Partial Fulfillment to the Requirements of the Degree of Master in Engineering Major in Electrical Engineering
Submitted by: Rhiza Joi C. Navallasca EE 712 Student
Submitted to: Engr. Ramon A. Alguidano Jr., PEcE EE 721 Professor
Date Submitted: March 15, 2013
1.
The fish in the river of Aganan is now vanishing because of some people doing illegal fishing like poisoning the river and others are by using electricity. The circuit shown below is the schematic diagram of the electrical device, which are used by illegal fisher to catch fish by means of electricity. The battery supplied the inductor of with internal resistance of . The switch has been position as shown in figure 1.1 for a long period of time to allow full charging of an inductor. At an instant, the position of a switch is transfer to another position as shown in figure 1.2. at time , determine the equation of , , and the voltage output at that time.
100
Solution: For Fig. 1.1
10 =0
12
= = ∗ 11⁄⁄ = ( 1⁄) ( 1⁄) = 1−⁄−⁄ @ = 0, −, = 0 = 1 ⁄ = 10012 1−⁄ = 120(1−)= 0
Applying Inverse Laplace Transform:
Applying KVL:
= 0 = 0 = 0
Applying Laplace Transform:
= = −⁄1−⁄ = −⁄ −⁄ @ = 0, −,⁄ = 0 = 12 . = 12 At Fig. 1.2:
Applying KVL:
=0 = 0 = = 100 12 10 = 200 == 1210 200 = .
2.
Where:
− , , , − = − = − = +
Given UJT relaxation oscillator shown in figure below prove or derive the formula of a frequency of oscillation given the following: , and .
is the supply voltage
is the peak voltage,
Solution:
is the valley voltage
−⁄
The discharging equation for the voltage
= 1 ℎ= 1 ℎ= ℎ= = = −− = The period manner:
Charging and discharging phases for trigger network
The general equation for the charging period is:
= 1−⁄ = ln
is:
When
can be determined in the following
,
(Applying In both sides)
= ln
ℎ= 1 ℎ= 1 ℎ= 1 = = = = = Applying In both sides = l n When
,
and
= = = + ;
3.
Given diode with R-L-C load as shown in the figure below, at time and the slope of . Assumed all initial condition is zero.
=0
, determine
, , ,
Solution: Applying KVL
= 0 1 ∫ = 0 = 0 11 = = 1 2 2 = 2 1 2 = 1 = 2
Applying Laplace Transform:
= 2 = = 1∙ = = − sin − cos − sin @ = 0, , = 0, = 0 = − sin = 100 , = 10 , = 0.1 , = 100 1 = 2 = 31,225 = 2 = 5000 = 31,22510010 − sin 31,225 = .− , = = − sin− sin = − cos − sin − sin − cos − cos − sin
Applying Inverse Laplace Transform:
Thus:
Using Derivative of the Product:
@ = 0, , = 0, = 0 = − cos − sin = − cos − sin − = − . = 1 ∫ 1 = ∫ − sin − cos sin − − − − si n cos si n = −sin − cos cos @ = 0, , = 0, = 0 = − sin − cos = 3.20256×10−5000− sin31225 31225− cos3122531225 = .− − = = 1000.303− sin31225 = . − Integration by Parts:
The slope of
is:
4.
Solution:
= 2%
±
Design a power supply circuit using capacitor filter with a maximum ripple factor of at maximum load current of 15 A, and an output voltage of 70 V. Show your solution neatly and clearly showing the standard value of the components, which includes the value of capacitor, the diode rating and the VA rating of the transformer needed.
== →1 = √ 2 →2 = →3 = →4 = 70 = = 15 = 4.67 = √ 3 =√ 3 →5 = = =0.0270= 1.4 =√ 31.4 =2.422.42 = = 4.67 = 0.52 =15 0.15.5522 = 15.52 = √ 2 = √ 2 = 10.97 = .
==10.97→ 6 = √ 2 →7 ==70 ==702. 2.42 42 = 72.42 = √ 2 = 72.√ 422 = 51.21 = 561.77 = 2.4 ; = 2.411.5000 4 = 25714.29 6800= 10,000100 100 = 15=600
Thus, the standard components to be used are; /
/
(2 pieces) and (1 piece) connected in parallel
Complete Design of Power Supply Required in the Problem
D1 18DB05
T1 10TO1CT
+70 v
220 V AC +
+
C1 1uF
C2 1uF
+
+
C4 1uF
C3 1uF
+
+
C6 1uF
C5 1uF
RL
RL
-70 v
155055 = 220 600 == == 6800 = =10010000 100 = 4.7 /
/
5.
Given circuit shown in figure below, determine the average output voltage, the Fourier series expansion of an input current and draw the input current waveform.
Solution:
≤ ≤ =√ 22sin =√ 2 = ∫ ⁄ 2 = ∫⁄ √ 2 sin = 2√ 2 cos⁄⁄ For
;
= 2√ 2 [cos 2⁄3cos ⁄3] = 2√ 2 0.50.5 = 2√ 2 = 2√ √ 22 = 2 = 2√ 2220 = 198.10 = 0 coefficients of Fourier series equal zero,
= 2 ∫ sin = 2 cos = 2 cos0cos = 4 =41, 3, 5, … 1 1 1 1 = sin 3 sin3 5 sin5 7 sin7 9 sin 9 ⋯ 220sin = = √ 210038 = 3.1 sin cos21°sin21°
6.
150°
10°
Design an AC voltage controller using RC triggering circuit with a firing delay angle ranges from -to. Show your solution neatly and clearly, draw your circuit design showing the standard value of the components
∅=0.=4710°
== 0 →1 ∅ = 180°8.33 = 180°10° 8.33 0.4=630.463= 0.47 = 0.0.44637 = 985.1Ω ≈ 1.0 Ω ∅= = 150°
∅ = 8.180°33 = 150°180° 8.33= 6.94 = 0.6.4974 = 14.77 Ω 50 Ω= 14.77Ω1. 0 Ω = 13. 7 7 Ω == 50Ω 1Ω = 0.47 250 /
180
220 220 60 1 = = 2 ∫ ∙ = = 22 ∫+ sin ∙ + = = ∫ sin ∙ = = cos+ = = cos cos cos = cos = = 2 cos ∅=0.=1 52.35° @ = 180 =√ 2 =√ : 2 400 =0= →1 2 = cos 3 3 ∅ = 8.180° 180 = 2√ 2400 cos 52. 3 5° = 8 . 3 3 = 2. 4 2 180° @= 60°= 220 2.42 2.42=0.1 = 0.1 = 24.2 Ω =√ 2 =√ : 2400 : ≈ 2.78 = 2 cos = 0.1 = 24.2 Ω 220 = 2√ 2400 cos ∅ = 60° = 52.35° 7.
Design a DC voltage controller using triggering circuit with an output voltage ranges from to from a supply voltage of , . Show your solution neatly and clearly, draw your circuit design showing the standard value of the components.
Solution:
Design Output:
Step-up Transformer
Therefore:
Standard Value for
==0.1 = 8.180°33 = 180°60° 8.33 = 2.78 = 2.0.718 = 27.8 Ω = 27.8 24.2 = 3.6 Ω
: ≈ Standard Value for
8.
Given a three phase half-wave rectifier shown in figure below. Determine, TUF.
, ,
efficiency, FF, RF, and
Solution:
Input waveform of three-phase half wave rectifier
1 = ∫ ⁄ 1 = 2⁄3 ∫⁄ sin = 32 cos⁄⁄ = 32 cos 5⁄6cos ⁄6 = 32 √ 2 3 √ 23 = 3√ 23 1 = ∫
⁄ 3 = 2 ∫⁄ sin ⁄ 3 = 2 ∫⁄ 12 12 cos2 ⁄ 3 1 = 2 2 sin2⁄ 3 = 2 56 6 12 sin 106 26 3 = 2 23 √ 23
= 2 3 23 √ 23 = = ⁄ 3√ 3 ⁄ 2 = 2 √ 3 23 2 == 0.96.967676% = 2 3 23 √ 23 = 3√ 23 = 1.016 = 1
== 0. 11796.016 1 = 17.96% = 2 = 3 = √ 2 = 41 23 √ 23 = 0.4854 = ⁄ 3 √ 23 = 3 1 2 √ 3 √ 2 4 3 2 == 0.66.664242% ~END~