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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
Central Philippine University
Jaro, Iloilo City
School of Graduate Studies
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 12 π battery supplied the
inductor of 100 ππ» with internal resistance of 10 πΊ. 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 π‘ = 0, determine the equation of π(π‘), ππΏ(π‘), and the
voltage output π0at that time.
Solution:
For Fig. 1.1
Applying KVL:
π β ππ β ππΏ = 0
π β π π(π‘) βπΏππ(π‘)
ππ‘= 0
Applying Laplace Transform:
π
π β π πΌ(π ) β (π πΏπΌ(π ) + πΏπΌ0) = 0
πΌ(π )(π + π πΏ) =π
π β πΏπΌ0
πΌ(π ) = [π
π (π + π πΏ)] β [
πΏπΌ0
(π + π πΏ)] β
1πΏβ
1πΏβ
πΌ(π ) =π
πΏ[
1
π (π + π πΏβ )
] β πΌ0 [1
(π + π πΏβ )
]
Applying Inverse Laplace Transform:
π(π‘) =π
πΏ(1 β π
βπ π‘πΏβ ) β πΌ0π
βπ π‘πΏβ
@ π‘ = 0, ππ πΆπππ ππ, πΌ0 = 0
π(π‘) =π
πΏ(1 + π
βπ π‘πΏβ )
π(π‘) =12 π
100 ππ»(1 + π
βπ π‘πΏβ )
π(π‘) = 120(1 β πβ100(0)) = 0 π΄
ππΏ(π‘) = πΏππ(π‘)
ππ‘= πΏ
π
ππ‘[π
πΏ(1 β π
βπ π‘πΏβ )
β πΌ0πβπ π‘
πΏβ ]
ππΏ(π‘) = ππβπ π‘
πΏβ β π πΌ0πβπ π‘
πΏβ π
@ π‘ = 0, ππ πΆπππ ππ, πΌ0 = 0
ππΏ(π‘) = 12 (πβ10(0)
0.1β ) π
ππΏ(π‘) = 12 π
At Fig. 1.2:
Applying KVL:
ππ 10 + ππ 100πΎ β ππΏ = 0
π 100πΎπ(π‘) + π 10π(π‘) β ππΏ(π‘) = 0
π(π‘) =ππΏ(π‘)
π 100πΎ + π 10
π(π‘) =12 π
100 ππΊ + 10πΊ
π(π‘) = 200ππ΄
π0 = ππΏ(π‘) β π 10π(π‘)
π0 = 12 π β 10(200 ππ΄)
π½π = ππ. πππ π½
2. Given UJT relaxation oscillator shown in figure below prove or derive the formula of a frequency of
oscillation (π) given the following: ππ΅π΅, ππ, ππ, π 1, π 2 and π πΆ .
π‘1 = π 1πΆ ππ ( ππ΅π΅βππ
ππ΅π΅βππ) π‘2 = (π 1 + π 2)πΆ ππ (
ππ΅π΅βππ
ππ΅π΅βππ) π =
1
π‘1+π‘2
Where: ππ΅π΅is the supply voltage ππ is the peak voltage, ππ is the valley voltage
Solution:
Charging and discharging phases for trigger network
The general equation for the charging period is:
ππΆ = ππ + (ππ΅π΅ β ππ) (1 + πβπ‘
π 1πΆβ )
The discharging equation for the voltage ππΆ is:
ππΆ = ππ + (ππ΅π΅ β ππ) (1 + πβπ‘
ππ 1+π 2πΆβ)
The period π‘1 can be determined in the following
manner:
ππΆ(πβππππππ) = ππ + (ππ΅π΅ β ππ) (1 + πβπ‘
π 1πΆ)
ππΆ(πβππππππ) = ππ + (ππ΅π΅ β ππ)
β (ππ΅π΅ β ππ)πβπ‘
π 1πΆ
ππΆ(πβππππππ) = ππ΅π΅ β (ππ΅π΅ β ππ)πβπ‘
π 1πΆ
When ππΆ = ππ, π‘ = π‘1
(ππ΅π΅βππ)
(ππ΅π΅βππ)= π
βπ‘π 1πΆ (Applying In both sides)
βπ‘1
π 1πΆ= ln [
(ππ΅π΅ β ππ)
(ππ΅π΅ β ππ)]
π‘1 = π 1πΆ ln (ππ΅π΅ β ππ
ππ΅π΅ β ππ)
ππΆ(πππ πβππππππ) = ππ + (ππ΅π΅ β ππ) (1 β πβπ‘
π 1πΆ)
ππΆ(πππ πβππππππ) = ππ + ππ΅π΅ β ππ β (ππ΅π΅ β ππ) (1 β πβπ‘
π 1πΆ)
ππΆ(πππ πβππππππ) = ππ΅π΅ β (ππ΅π΅ β ππ) (1 β πβπ‘
π 1πΆ)
When ππΆ = ππ, and π‘ = π‘2
ππ = ππ΅π΅ β (ππ΅π΅ β ππ)πβπ‘
π 1+π 2πΆ
(ππ΅π΅ β ππ)πβπ‘
π 1+π 2πΆ = ππ΅π΅ β ππ
πβπ‘
π 1+π 2πΆ =(ππ΅π΅ β ππ)
(ππ΅π΅ β ππ) (Applying In both sides)
βπ‘
π 1 + π 2πΆ
= ln(ππ΅π΅ β ππ)
(ππ΅π΅ β ππ)
π» = ππ + ππ ; π =π
π»=
π
ππ+ππ
3. Given diode with R-L-C load as shown in the figure below, at time π‘ = 0, determine π(π‘), ππΏ(π‘), ππΆ(π‘),
and the slope of π(π‘). Assumed all initial condition is zero.
Solution:
Applying KVL
π β ππ β ππΏ β ππΆ = 0
π β π π(π‘) βπΏππ(π‘)
ππ‘β [
1
πΆβ« π(π‘)
π‘
0
ππ‘ + π0] = 0
Applying Laplace Transform:
π
π β π πΌ(π ) β (π πΏπΌ(π ) + πΏπΌ0) β (
πΌ(π )
π πΆ+
π0
π ) = 0
[πΌ(π ) (π + π πΏ +1
π πΆ) =
π
π β πΏπΌ0 β
π0
π ]
π
πΏ
πΌ(π ) [π 2 +π
πΏπ +
1
πΏπΆ] =
π
π β π πΏ0 β
π0
πΏ
πΌ(π ) [π 2 +π
πΏπ + (
π
2πΏ)
2
β (π
2πΏ)
2
+1
πΏπΆ] =
π β π0
πΏβ π πΌ0
πΌ(π ) [(π +π
2πΏ)
2
+ (β1
πΏπΆβ (
π
2πΏ)
2
)
2
] =π β π0
πΏβ π πΌ0
ππ = β1
πΏπΆβ (
π
2πΏ)
2
(πππππππ πππππ’ππππ¦)
πΌ =π
2πΏ (πππππππ πππππ’ππππ¦)
πΌ(π )[(π + πΌ)2 + (ππ)2] =π β π0
πΏβ π πΌ0
πΌ(π ) =π β π0
πΏ[
1
(π + πΌ)2 + (ππ)2] βππ
ππβ πΌ0 [
π + πΌ β πΌ
(π + πΌ)2 + (ππ)2]
πΌ(π ) =π β π0
πΏ[
ππ
(π + πΌ)2 + (ππ)2] β πΌ0 [π + πΌ
(π + πΌ)2 + (ππ)2] +πΌπΌ0
ππ[
ππ
(π + πΌ)2 + (ππ)2]
Applying Inverse Laplace Transform:
π(π‘) =π β π0
πππΏπβπΌπ‘ sin ππ π‘ β πΌ0πβπΌπ‘ cos πππ‘ β
πΌπΌ0
πππβπΌπ‘ sin ππ π‘
@ π‘ = 0, ππ πΆπππ ππ, π0 = 0, πΌ0 = 0
π(π‘) =π
πππΏπβπΌπ‘ sin ππ π‘
π = 100 π, πΏ = 10 ππ», πΆ = 0.1 ππΉ, π = 100πΊ
ππ = β1
πΏπΆβ (
π
2πΏ)
2
= 31,225 π»π§
πΌ =π
2πΏ= 5000
Thus:
π(π‘) =100 π
(31,225 π»π§)(10 ππ»)πβ5000π‘ sin 31,225π‘ π΄
π(π) = π. πππππβπππππ π¬π’π§ ππ, ππππ π¨
ππΏ(π‘) = πΏππ(π‘)
ππ‘= πΏ
π
ππ‘(
π β π0
πππΏπβπΌπ‘ sin πππ‘ β πΌ0πβπΌπ‘ sin πππ‘)
Using Derivative of the Product:
ππΏ(π‘) =π β π0
ππ
[πππβπΌπ‘ cos ππ π‘ β πΌπβπΌπ‘ sin ππ π‘] β πΏπΌ0[πππβπΌπ‘ sin ππ π‘ β πΌπβπΌπ‘ cos ππ π‘]
+πΌπΏπΌ0
ππ
[πππβπΌπ‘ cos ππ π‘ β πΌπβπΌπ‘ sin ππ π‘]
@ π‘ = 0, ππ πΆπππ ππ, π0 = 0, πΌ0 = 0
ππΏ(π‘) =π
ππ
[πππβπΌπ‘ cos ππ π‘ β πΌπβπΌπ‘ sin ππ π‘]
ππΏ(π‘) = π [πβπΌπ‘ cos ππ π‘ βπΌπ
πππβπΌπ‘ sin ππ π‘]
π½π³(π) = ππππβπππππ ππ¨π¬ ππππππ β ππ. πππππβπππππ π¬π’π§ ππππππ π½
ππΆ(π‘) =1
πΆβ« π(π‘)ππ‘ + π0
π‘
0
ππΆ(π‘) =1
πΆβ« [
π β π0
πππΏπβπΌπ‘ sin ππ π‘ β πΌ0πβπΌπ‘ cos ππ π‘ β
πΌπΌ0
ππsin ππ π‘] ππ‘
π‘
0
+ π0
Integration by Parts:
ππΆ(π‘) =π β π0
πππΏπΆ[βπΌπβπΌπ‘ sin πππ‘ β πππβπΌπ‘ cos πππ‘
(πΌ)2 + (ππ)2] β
πΌ0
πΆ[πππβπΌπ‘ sin πππ‘ β πΌπβπΌπ‘ cos πππ‘
(πΌ)2 + (ππ)2]
+πΌπΌ0
ππ[βπΌπβπΌπ‘ sin πππ‘ β πππβπΌπ‘ cos πππ‘
(πΌ)2 + (ππ)2] + π0
@ π‘ = 0, ππ πΆπππ ππ, π0 = 0, πΌ0 = 0
ππΆ(π‘) =π
πππΏπΆ[(πΌ)2 + (ππ)2][βπΌπβπΌπ‘ sin πππ‘ β πππβπΌπ‘ cos ππ π‘ + ππ]
ππΆ(π‘) = 3.20256 Γ 10β3[β5000πβ5000π‘ sin 31225π‘ β 31225πβ5000π‘ cos 31225π‘ + 31225]
π½πͺ(π) = βππ. ππππβπππππ π¬π’π§ ππππππ β ππππβπππππ ππ¨π¬ πππππ + πππ π½
ππ (π‘) = π π(π‘)
ππ (π‘) = 100(0.303πβ5000π‘ sin 31225π‘)
π½πΉ(π) = ππ. ππ πβπππππ π¬π’π§ ππππππ π½
The slope of π(π‘) is:
4. Design a power supply circuit using capacitor filter with a maximum ripple factor of (π = 2%) at maximum
load current of 15 A, and an output voltage of Β±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.
Solution:
πΌπ· = πππππ ππ’πππππ‘ πππ‘πππ
πΌπ· = πΌπ β (1)
πΌπ =πΌπ
β2β (2)
πΌπ = πΌπ·πΆ + πΌππ β (3)
πΌππ =ππ(π)
π πΏβ (4)
π πΏ = ππΏ
π πΏ =ππ·πΆ
πΌπ·πΆ=
70 π
15 π΄= 4.67 πΊ
πππππ =
πππ
β3
πππ = β3πππππ β (5)
π =πππππ
ππ·πΆ
πππππ = πππ·πΆ = (0.02)(70) = 1.4 π
πππ = β3(1.4 π) = 2.42 π
πΌππ =ππ(π)
π πΏ=
2.42 π
4.67 πΊ= 0.52 π΄
πΌπ = 15 π΄ + 0.52 π΄ = 15.52 π΄
πΌπ =πΌπ
β2=
15.52 π΄
β2= 10.97 π΄
ππ = ππ. ππ π
ππ΄ = ππ πΌπ β (6)
πΌπ = 10.97 π΄
ππ =ππ
β2β (7)
ππ = ππ·πΆ + πππ
ππ·πΆ = 70 π
πππ = 2.42 π
ππ = 70 + 2.42 = 72.42 π
ππ =ππ
β2=
72.42 π
β2= 51.21 π
ππ΄ = 561.77 ππ΄
πΆ =2.4 πΌπ·πΆ
πππππ
;
πΆ =2.4(15000)
1.4
πΆ = 25714.29 ππΉ
Thus, the standard components to be used are;
πΆ = 10,000 ππΉ/100 π (2 pieces) and
6800 ππΉ/100 π (1 piece) connected in parallel
π·ππππ π ππ‘πππ = 15 π΄ π΅πππππ
πππππ ππππππ π ππ‘πππ = 600 ππ΄
Complete Design of Power Supply Required in the Problem
π1 = 220 π πππππππ¦
55 β 0 β 55 π π ππππππππ¦
600 ππ΄ π π‘ππ πππ€π
π‘ππππ ππππππ
πΆ1 = πΆ2 = πΆ3 = πΆ4 = 10 000 ππΉ/100 π
πΆ5 = πΆ6 = 6800 ππΉ/100 π
π πΏ = 4.7 πΊ
220 V AC
+70 v
-70 v
RL
RL
+
C61uF
+
C51uF
+
C41uF
+
C31uF
+
C21uF
+
C11uF
D118DB05T1
10TO1CT
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:
For π
3β€ ππ‘ β€
2π
3
π0 = β2ππΏ sin ππ‘; ππ = β2ππΏ
ππ·πΆ =2
πβ« π0(π‘)ππ‘
π‘
0
ππ·πΆ =2
πβ« β2ππΏ sin ππ‘ π(ππ‘)
2π3β
π3β
ππ·πΆ =2β2ππΏ
π[β cos ππ‘]π
3β
2π3β
ππ·πΆ =2β2ππΏ
π[β cos 2π
3β + cos π3β ]
ππ·πΆ =2β2ππΏ
π[β(β0.5) + 0.5]
ππ·πΆ =2β2ππΏ
π=
2β2ππ
β2π=
2ππ
π
ππ·πΆ =2β2(220π)
π= 198.10 π
ππ coefficients of Fourier series equal zero, ππ = 0
ππ =2
πβ« πΌπ sin πππ‘ π(ππ‘)
π
0
ππ =2πΌ0
ππ[β cos πππ‘]0
π
ππ =2πΌ0
ππ[cos 0 β cos ππ]
ππ =4πΌ0
ππ
πππ π = 1, 3, 5, β¦
π(π ) =4πΌ0
π(sin ππ‘ +
1
3sin 3ππ‘ +
1
5sin 5ππ‘ +
1
7sin 7ππ‘ +
1
9sin 9ππ‘ + β―)
πΌ0 =π0
π + πππΏ=
β2(220) sin ππ‘
100 + π38
πΌ0 = 3.1 sin ππ‘(cos β21Β° + sin β21Β°) π΄
6. Design an AC voltage controller using RC triggering circuit with a firing delay angle ranges from 10Β°-to-
150Β°. Show your solution neatly and clearly, draw your circuit design showing the standard value of the
components
β πππ = 10Β°
πΆ = 0.47 ππΉ
πΏππ‘ π 2 = 0
π(πππ) = π 1πΆ β (1)
π‘1
β πππ=
8.33
180Β°
π‘1 =10Β°
180Β°(8.33ππ )
π‘1 = 0.463ππ
0.463ππ = π 1(0.47ππΉ)
π 1 =0.463ππ
0.47ππΉ= 985.1Ξ© β 1.0 πΞ©
β πππ₯ = 150Β°
π 2 = πππ₯
π‘2
β πππ₯=
8.33 ππ
180Β°
π‘2 =150Β°
180Β°(8.33) = 6.94ππ
(π 1 + π 2) =6.94ππ
0.47 ππΊ= 14.77 πΞ©
π 2 = 14.77 πΞ© β 1.0 πΞ© = 13.77 πΞ©
50 πΞ© π€πππ ππ π’π ππ
π 1 = 1πΞ©
π 1 = 50πΞ©
πΆ = 0.47 ππΉ/250 π
7. Design a DC voltage controller using triggering circuit with an output voltage ranges from 180 ππ·πΆ to
220 ππ·πΆ from a supply voltage of 220 ππ΄πΆ, 60 π»π§. Show your solution neatly and clearly, draw your
circuit design showing the standard value of the components.
Solution:
π0(π·πΆ) = ππ·πΆ =1
2π[β« π0
2π
0
β π(ππ‘)]
π0(π·πΆ) = ππ·πΆ =2
2π[β« ππ
π+πΌ
πΌ
sin ππ‘ β π(ππ‘)]
π0(π·πΆ) = ππ·πΆ =ππ
π[β« sin ππ‘ β π(ππ‘)
π+πΌ
πΌ
]
π0(π·πΆ) = ππ·πΆ =ππ
π[β cos ππ‘]πΌ
π+πΌ
π0(π·πΆ) = ππ·πΆ =ππ
π[β cos(π + πΌ) + cos πΌ]
cos(π + πΌ) = β cos πΌ
Therefore:
π0(π·πΆ) = ππ·πΆ =2ππ
πcos πΌ
@ππ·πΆ = 180 π
ππ = β2ππ = β2(400 π)
πΌ ππ πππ’ππ π‘π:
ππ·πΆ =2ππ
πcos πΌ
180 =2β2(400 π)
πcos πΌ
πΌ = 60Β° (πππ₯πππ’π ππππππ πππππ)
@ππ·πΆ = 220 π
ππ = β2ππ = β2(400 π)
πΌ ππ πππ’ππ π‘π:
ππ·πΆ =2ππ
πcos πΌ
220 π =2β2(400 π)
πcos πΌ
πΌ = 52.35Β° (ππππππ’π ππππππ πππππ)
β πππ = 52.35Β°
πΆ = 0.1 ππΉ
π 2 = 0
π(πππ) = π 1πΆ β (1)
π‘1
β πππ=
8.33 ππ
180Β°
π‘1 =52.35Β°
180Β°(8.33 ππ ) = 2.42 ππ
2.42 ππ = π 1(0.1ππΉ)
π 1 =2.42 ππ
0.1ππΉ= 24.2 πΞ©
Standard Value for π 1:
π 1 β ππ ππ΄
(π 1 + π 2) =2.78 ππ
0.1 ππΉ= 24.2 πΞ©
β πππ = 60Β°
Design Output:
Step-up Transformer
πΆ = 0.1 ππΉ
π 2 = πππ₯
π‘2 =8.33 ππ
180Β°
π‘1 =60Β°
180Β°(8.33 ππ ) = 2.78 ππ
(π 1 + π 2) =2.78 ππ
0.1ππΉ= 27.8 πΞ©
π 2 = 27.8 β 24.2 = 3.6 πΞ©
Standard Value for π 2:
π 1 β ππ ππ΄
8. Given a three phase half-wave rectifier shown in figure below. Determine, ππ·πΆ , ππππ , efficiency, FF, RF, and
TUF.
Solution:
Input waveform of three-phase half wave rectifier
ππ·πΆ =1
πβ« π(π‘)
π‘
0
ππ‘
ππ·πΆ =1
2π3β
β« ππ
5π6β
π6β
sin ππ‘ π(ππ‘)
ππ·πΆ =3ππ
2π[β cos ππ‘]π
6β
5π6β
ππ·πΆ =3ππ
2π[β cos 5π
6β + cos π6β ]
ππ·πΆ =3ππ
2π[β (
ββ3
2) +
β3
2]
ππ·πΆ =3β3 ππ
2π
ππππ = β1
πβ« π(π‘)2ππ‘
π‘
0
ππππ = β3
2πβ« ππ
2 sin2 ππ‘ π(ππ‘)
5π6β
π6β
ππππ = β3ππ
2πβ« [
1
2β
1
2cos 2ππ‘] π(ππ‘)
5π6β
π6β
ππππ = β3ππ
2
2π[ππ‘ β
1
2sin 2ππ‘]
π6β
5π6β
ππππ = β3ππ
2
2π[(
5π
6β
π
6) β
1
2(sin
10π
6β
2π
6)]
ππππ = β3ππ
2
2π[2π
3+
β3
2]
ππππ =ππ
2β
3
π[2π
3+
β3
2]
π· =ππ·πΆ
ππ΄πΆ
π· =
π2π·πΆ
π πΏβ
π2πππ
π πΏβ
π· =(
3β3ππ2π
β )2
(ππ2
β2π3 +
β32 )
2
π· = 0.9676
π· = 96.76%
πΉπΉ =ππππ
ππ·πΆ
πΉπΉ =
ππ2
β3π [
2π3 +
β32 ]
3β3 ππ2π
πΉπΉ = 1.016
π πΉ = βπΉπΉ2 β 1
π πΉ = β1.0162 β 1
π πΉ = 0.1796
π πΉ = 17.96%
πππΉ =ππ·πΆ
2ππ΄
πππΉ =
π2π·πΆ
π πΏβ
3ππ πΌπ
ππ =ππ
β2
πΌπ = πΌπβ1
4π[2π
3+
β3
2]
πΌπ = 0.4854 πΌπ
πΌπ =ππ
π πΏβ
πππΉ =
(3β3ππ
2π )
2
π πΏ
β
3 (ππ
β2β ) (β
14π [
2π3 +
β32 ]
πππ πΏ
)
πππΉ = 0.6642
πππΉ = 66.42%
~END~
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