Name :_____________________________
Registration Number
BL.EN.U4
Amrita School of Engineering
Amrita Vishwa Vidyapeetham, Bengaluru Campus
Department of ECE
Event: Q1
Branch/ Semester/ Section: ECE/II/H, I & J
Course code/ Course Title: 23ECE111/Electronic Devices and Circuits
Duration: 15 minutes
Q.NO
1[2]
Date:14/02/2025
2[2]
3[1]
4[1]
5[2]
6[2]
Set: B
Max. Marks:10
Total [10]
Sign of
Faculty
Sign of
student
Marks
Obtained
CO
Course Outcomes
CO1
CO2
Characterise semiconductor diodes
Analyse diode-based circuits
CO3
CO4
Design diode-based circuits for specific applications
Understand the operation of transistors
1
Find the values of I and V in the circuit shown in Fig. 1.
[2]
[CO2,
BTL3]
Fig. 1
V = 3-0.7 = 2.3 V (1 mark)
2
I = [2.3-(-8)]/2k = 5.15mA (1 mark)
For the circuit shown in Fig. 2 using ideal diodes, find the labeled values of voltage and [2]
current.
[CO1,
BTL3]
1
Name :_____________________________
Registration Number
BL.EN.U4
Fig. 2
3
4
5
If the reverse saturation current at 30oC is 6nA, then the reverse saturation current at
60oC is ______________.
48nA
Mention the advantage and disadvantage of graphical analysis using exponential
method.
Graphical analysis aids in the visualization of circuit operation. However, the effort
involved in performing such an analysis, particularly for complex circuits, is too great to
be justified in practice.
Assuming the availability of diodes for which vD = 0.71 V at iD = 1 mA and n = 1,
design a circuit that utilizes four diodes connected in series, in series with a resistor R
connected to a 8-V power supply. The voltage across the string of diodes is to be 3.1V.
[1]
[CO1,
BTL1]
[1]
[CO1,
BTL2]
[2]
[CO2,
BTL3]
2
Name :_____________________________
Registration Number
6
BL.EN.U4
The zener diode shown in Fig. 3 has VZ = 15 V and RL = 15 Ω. The voltage across the [2]
load stays at 15 V as long as IZ is maintained between 180 mA and 2.5 A. Find the value [CO3,
of R so that Eo remains 15 V while input Ei is free to vary between 22 V to 28V.
BTL3]
Fig. 3
The zener current will be minimum (i.e. 180mA) when the input voltage is minimum
(i.e. 22 V). The load current stays at constant value,
IL = VZ/RL = 15/15 = 1000 mA. 1 mark
R = (Ei – Eo)/[(Iz)min + (IL)max)] = (22-15)/(180+1000) = 7/1180 = 5.93 Ω 1 mark
Course Outcome / Bloom’s Taxonomy Level (BTL) Mark Distribution Table
CO
Marks
BTL
Marks
CO1
4
BTL 1
01
CO2
4
BTL 2
01
CO3
2
BTL 3
08
CO4
--
BTL 4
--
3