MPH-005 Solved Assignment 2024 | IGNOU

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MPH-005 Solved Assignment 2024 | IGNOU Sample Solution
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mph-005-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mph-005-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

  1. a) What are the assumption made while deriving the ideal diode equation? Explain the effect of temperature on the breakdown voltage of the diode.
Answer:
The ideal diode equation, also known as the Shockley diode equation, describes the current-voltage relationship of a diode. The equation is derived under several assumptions:
  1. Low-Level Injection: The injected minority carrier concentration is much smaller than the majority carrier concentration in the respective regions of the diode.
  2. Complete Depletion Approximation: The space-charge region (depletion region) is fully depleted of free carriers, and the electric field is only present in this region.
  3. Quasi-Neutral Regions: Outside the depletion region, the semiconductor is neutral, and the electric field is negligible.
  4. Thermal Equilibrium: The diode is in thermal equilibrium, meaning that the Fermi level is constant throughout the device.
  5. No Recombination in the Depletion Region: It is assumed that there is no significant recombination of carriers in the depletion region.
  6. Ideal Diode: The diode is considered to be ideal, with no series resistance, no shunt resistance, and no leakage current when reverse biased.
Effect of Temperature on Breakdown Voltage:
The breakdown voltage of a diode is the voltage at which the diode begins to conduct significantly in the reverse direction, leading to a breakdown of the junction. The breakdown voltage is affected by temperature in the following ways:
  1. Zener Breakdown: In Zener breakdown, which occurs in diodes with a low breakdown voltage, the breakdown voltage decreases with increasing temperature. This is due to the increased thermal energy that helps carriers to tunnel through the potential barrier.
  2. Avalanche Breakdown: In avalanche breakdown, which occurs in diodes with a higher breakdown voltage, the breakdown voltage increases with increasing temperature. This is because the increased thermal energy leads to a decrease in the mean free path of the carriers, requiring a higher electric field to initiate the avalanche multiplication process.
In summary, the temperature has a significant effect on the breakdown voltage of a diode, with the direction of the effect depending on the breakdown mechanism (Zener or avalanche).
b) Define the External Quantum Efficiency (EQE) of an LED. Calculate its value for a semiconductor material with refractive index n 2 = 3.6 n 2 = 3.6 n_(2)=3.6n_2=3.6n2=3.6; assuming the refractive index of the air = 1 = 1 =1=1=1.
Answer:
External Quantum Efficiency (EQE) of an LED is defined as the ratio of the number of photons emitted by the LED to the number of electrons injected into the LED. It is a measure of how efficiently an LED converts electrical power into light and is expressed as a percentage. Mathematically, it can be represented as:
E Q E = Number of photons emitted Number of electrons injected × 100 % E Q E = Number of photons emitted Number of electrons injected × 100 % EQE=(“Number of photons emitted”)/(“Number of electrons injected”)xx100%EQE = \frac{\text{Number of photons emitted}}{\text{Number of electrons injected}} \times 100\%EQE=Number of photons emittedNumber of electrons injected×100%
EQE is a critical parameter in evaluating the performance of an LED, as it takes into account both the internal quantum efficiency (the efficiency with which electrons and holes recombine to generate photons) and the extraction efficiency (the efficiency with which the generated photons are extracted from the LED).
To calculate the EQE for a semiconductor material with a refractive index n 2 = 3.6 n 2 = 3.6 n_(2)=3.6n_2 = 3.6n2=3.6 and assuming the refractive index of air n 1 = 1 n 1 = 1 n_(1)=1n_1 = 1n1=1, we need to consider the extraction efficiency. The extraction efficiency is influenced by the critical angle θ c θ c theta _(c)\theta_cθc for total internal reflection at the interface between the semiconductor and air. The critical angle is given by Snell’s law:
sin ⁡ ( θ c ) = n 1 n 2 sin ⁡ ( θ c ) = n 1 n 2 sin(theta _(c))=(n_(1))/(n_(2))\sin(\theta_c) = \frac{n_1}{n_2}sin⁡(θc)=n1n2
Substituting the given values:
sin ⁡ ( θ c ) = 1 3.6 sin ⁡ ( θ c ) = 1 3.6 sin(theta _(c))=(1)/(3.6)\sin(\theta_c) = \frac{1}{3.6}sin⁡(θc)=13.6
θ c = arcsin ⁡ ( 1 3.6 ) θ c = arcsin ⁡ 1 3.6 theta _(c)=arcsin((1)/(3.6))\theta_c = \arcsin\left(\frac{1}{3.6}\right)θc=arcsin⁡(13.6)
The fraction of light that can escape the semiconductor material (extraction efficiency) is approximately given by:
E E = 1 − 1 2 ( 1 + cos 2 ⁡ ( θ c ) ) E E = 1 − 1 2 1 + cos 2 ⁡ ( θ c ) EE=1-(1)/(2)(1+cos^(2)(theta _(c)))EE = 1 – \frac{1}{2} \left(1 + \cos^2(\theta_c)\right)EE=1−12(1+cos2⁡(θc))
Substituting the value of θ c θ c theta _(c)\theta_cθc:
E E = 1 − 1 2 ( 1 + cos 2 ⁡ ( arcsin ⁡ ( 1 3.6 ) ) ) E E = 1 − 1 2 1 + cos 2 ⁡ arcsin ⁡ 1 3.6 EE=1-(1)/(2)(1+cos^(2)(arcsin((1)/(3.6))))EE = 1 – \frac{1}{2} \left(1 + \cos^2\left(\arcsin\left(\frac{1}{3.6}\right)\right)\right)EE=1−12(1+cos2⁡(arcsin⁡(13.6)))
Assuming the internal quantum efficiency is 100% (all electron-hole pairs recombine to produce photons), the EQE can be approximated as equal to the extraction efficiency (EE). Thus,
E Q E ≈ E E = 1 − 1 2 ( 1 + cos 2 ⁡ ( arcsin ⁡ ( 1 3.6 ) ) ) = 25 648 E Q E ≈ E E = 1 − 1 2 1 + cos 2 ⁡ arcsin ⁡ 1 3.6 = 25 648 EQE~~EE=1-(1)/(2)(1+cos^(2)(arcsin((1)/(3.6))))=(25)/(648)EQE \approx EE = 1 – \frac{1}{2} \left(1 + \cos^2\left(\arcsin\left(\frac{1}{3.6}\right)\right)\right)=\frac{25}{648}EQE≈EE=1−12(1+cos2⁡(arcsin⁡(13.6)))=25648
E E = 0.038 E E = 0.038 EE=0.038EE=0.038EE=0.038
This calculation provides an estimate of the EQE based on the refractive indices of the semiconductor material and air.
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