ELD255 LecTest1 - Set A 2
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Electrical Engineering
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Dec 6, 2023
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ELD255 Lecture Test 1: Set A – Answer Key
Lecture Test 1: Diodes
Last Name:
Answer
First Name:
Key
Student Number: _____-_____-_____
ELD255NAB
Date: June 8, 2023
Score: _____ / 62
+5
Question 1:
Honesty Declaration
Questions 2 and 3:
[6 marks]
Given:
E
= 18 V,
R
1
= 5.6 kΩ and
R
2
= 8.2 kΩ.
2.
Find the source current flowing in this circuit [3]
Since D is reverse-biased, the current will be
I
S
=
E
R
1
+
R
2
=
18
V
5.6
k Ω
+
8.2
k Ω
I
S
=
1.30 mA
3.
Find the voltage across the diode [3]
V
D
=
I
S
R
1
=
1.30
mA
(
5.6
k Ω
)
=
7.28
V
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
1
ELD255 Lecture Test 1: Set A – Answer Key
V
D
=
7.28 V
Questions 4 to 6:
[9 marks]
Determine the following:
4.
Current flowing through
R
1
[3]
I
R
1
=
V
S
−
V
Z
R
1
=
10
V
−
8
V
1.5
k Ω
I
R1
=
1.33 mA
5.
Current flowing through
R
2
[3]
I
R2
=
0 mA
The circuit is open because of the capacitor C.
6.
Current coming from the voltage source [3]
I
S
=
I
R
1
+
I
R
2
=
1.33
mA
+
0
mA
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
2
ELD255 Lecture Test 1: Set A – Answer Key
I
S
=
1.33 mA
Questions 7 to 10:
[14 marks]
For the circuit below, the 1N4728 Zener diode is conducting with
V
Z
= 3.3 V.
Determine the following:
R
S
= 82 Ω,
I
S
= 154.88 mA, and
R
L
= 330 Ω
7.
The load current [3]
I
L
=
V
Z
R
L
=
3.3
V
330
Ω
I
L
=
10 mA
8.
The Zener current [3]
I
Z
=
I
S
−
I
L
=
154.88
mA
−
10
mA
I
Z
=
144.88 mA
9.
The source voltage [4]
E
=
I
S
R
S
+
V
Z
=
154.88
mA
(
82
Ω
)
+
3.3
V
E
=
16 V
10. If the source voltage drops to 15 V, what will be the value of the load current? [4]
When E < V
Z
, Zener diode is OFF
I
L
= I
S
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
3
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ELD255 Lecture Test 1: Set A – Answer Key
I
L
=
I
S
=
E
R
S
+
R
L
=
15
V
82
Ω
+
330
Ω
I
L
=
36.41 mA
Questions 11 to 15:
[15 marks]
The 1N4728 Zener diode used for the circuit below has the following specifications:
V
Z
= 3.3 V,
I
ZK
= 1 mA, and
I
ZM
= 276 mA
Determine the following:
11. The value of load current anytime Zener diode is conducting [3]
I
L
=
V
L
R
L
=
3.3
V
3.3
k Ω
I
L
=
1 mA
12. The minimum source current the Zener diode is conducting [3]
I
Smin
=
I
ZK
+
I
L
=
1
mA
+
1
mA
I
S min
=
2 mA
13. The minimum source voltage the Zener diode is conducting [3]
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
4
ELD255 Lecture Test 1: Set A – Answer Key
V
Smin
=
I
Smin
R
S
+
V
Z
=
2
mA
(
240
Ω
)
+
3.3
V
V
S min
=
3.78 V
14. The maximum source current the Zener diode can safely conduct [3]
I
Smax
=
I
ZM
+
I
L
=
276
mA
+
1
mA
I
S max
=
277 mA
15. The maximum source voltage the Zener diode can safely conduct [3]
V
Smax
=
I
Smax
R
S
+
V
Z
=
277
mA
(
240
Ω
)
+
3.3
V
V
S max
=
69.78 V
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
5
ELD255 Lecture Test 1: Set A – Answer Key
Questions 16 to 21:
[18 marks]
Determine the following for the circuit shown.
16. Primary peak voltage [3]
V
P
(
pk
)
=
V
P
(
rms
)
√
2
=
120
V
√
2
V
P (pk)
=
169.71 V
17. Secondary peak voltage [3]
V
S
(
pk
)
=
V
P
(
pk
)
a
=
169.71
V
6
V
S (pk)
=
28.28 V
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
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ELD255 Lecture Test 1: Set A – Answer Key
18. Load peak voltage [3]
V
L
(
pk
)
=
V
S
(
pk
)
−
V
D
=
28.28
V
−
0.7
V
V
L (pk)
=
27.58 V
19. Average (DC) voltage [3]
V
DC
=
V
L
(
pk
)
π
=
27.58
V
π
V
DC
=
8.78 V
20. Average (DC) current [3]
I
DC
=
V
DC
R
L
=
8.78
V
1
k Ω
I
DC
=
8.78
m
A
=
I
L
21. A 750 µF capacitor is inserted between the rectifier and the load, find the ripple voltage
[3]
V
r
=
I
L
f C
=
8.78
mA
60
Hz
(
750
μF
)
V
r
=
195.11 mVpp
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
7
ELD255 Lecture Test 1: Set A – Answer Key
Questions 22 and 23:
[+5 marks]
If the load needs a regulated dc output voltage of 5 V, what should be the value of
R
1
?
The potentiometer
R
2
is set at 1.675 kΩ.
Using the formula:
V
out
=
1.25
V
[
1
+
R
2
R
1
]
22. Derive the formula for finding the resistor,
R
1
[+3]
V
out
1.25
V
=
1
+
R
2
R
1
V
out
1.25
V
−
1
=
R
2
R
1
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
8
ELD255 Lecture Test 1: Set A – Answer Key
⟹
R
1
=
R
2
V
out
1.25
V
−
1
[derived formula]
23. Find the value of
R
1
for the situation mentioned above [+2]
R
1
=
1.675
k Ω
5
V
1.25
V
−
1
R
1
=
560
Ω
[value]
For the ELD255 Lec Classes of Prof. Ruperto Tumibay
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