EL-304: Electronic Devices-1
Dr. Faiz Ahmad
EL-304: Electronic Devices-1
Assignment 1
Instructor: Dr. Faiz Ahmad
Due Date: October 2, 2025
Instructions
• Solve all problems below. Show all steps clearly for full credit.
• Submit your solutions in hand-written papers.
• Total points: 50 (10 points per problem).
Problems
1. Explain the role of valence electrons in determining the electrical properties of a
semiconductor like silicon. Compare the atomic structure of silicon (atomic number
14) with that of carbon (atomic number 6), focusing on the number of valence
electrons and their impact on conductivity. (10 points)
2. A silicon diode has a reverse saturation current Is = 2×10−12 A at 300 K. Calculate
the current I when a forward bias
] V is applied, using kT /q ≈ 0.0259 V.
[ of
( qVV )= 0.7
Use the diode equation I = Is exp kT − 1 . (10 points)
3. Analyze a circuit with a 15 V source, a 2 kΩ resistor, and a diode modeled with a
constant voltage drop of 0.7 V in series. Determine the current through the diode
and the voltage across the resistor. (10 points)
4. Light Emitting Diodes (LEDs) emit light of specific wavelengths which correspond
to the bandgap energy of the semiconductor material used. The bandgap energy
Eg is related to the emission wavelength λ by the equation:
Eg =
hc
λ
where:
• Eg is the bandgap energy in electron volts (eV),
• h = 6.626 × 10−34 J·s (Planck’s constant),
• c = 3 × 108 m/s (speed of light),
• λ is the wavelength in meters,
• 1 eV = 1.602 × 10−19 J.
Given the following LED colors and their central wavelengths:
• Red: 650 nm
• Green: 530 nm
• Blue: 470 nm
Fall 2025
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EL-304: Electronic Devices-1
Dr. Faiz Ahmad
• Violet: 405 nm
(a) Calculate the bandgap energy Eg (in eV) corresponding to each color wavelength.
(b) Explain how the bandgap energy affects the color of light emitted by the LED.
5. Schottky Diode (Lecture 8)
A Schottky diode has a saturation current IS = 5 × 10−9 A and an ideality factor
η = 1 at 300 K (kT /q ≈ 0.0259 V). Calculate the[current
( I)when] a forward bias of
qV
V = 0.25 V is applied, using the equation I = IS exp ηkT
− 1 . (10 points)
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