# IGNOU MPH-005 Solved Assignment 2024 | MSCPH | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MPH-005 Assignment Question Paper 2024

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.
b) Define the External Quantum Efficiency (EQE) of an LED. Calculate its value for a semiconductor material with refractive index ${n}_{2}=3.6$${n}_{2}=3.6$n_(2)=3.6n_2=3.6${n}_{2}=3.6$; assuming the refractive index of the air $=1$$=1$=1=1$=1$.
c) Draw the symbols of following diodes:
i) Varactor diode, ii) Schottky diode, iii) p-i-n diode, iv) Tunnel diode
Explain the origin of negative resistance in Gunn diode with the help of appropriate E-k diagram. What are the applications of Gunn diode?
2. a) Explain with the help of appropriate diagram, how extremely high current gain can be obtained using the Darlington pair configuration of BJT.
b) Explain with the help of a circuit diagram and appropriate waveforms, how an SRC can be used for motor speed control. Suggest the modification in the circuit to turn it $\mathrm{O}\mathrm{N}$$\mathrm{O}\mathrm{N}$ON\mathrm{ON}$\mathrm{O}\mathrm{N}$ beyond ${90}^{\circ }$${90}^{\circ }$90^(@)90^{\circ}${90}^{\circ }$ phase of the input ac cycle.
3. a) Discuss the effect of various circuit parameters governing the low frequency and high frequency responses of a transistor amplifier. Prove that the high frequency response of an amplifier decreases by $6\text{}\mathrm{d}\mathrm{B}$$6\text{}\mathrm{d}\mathrm{B}$6dB6 \mathrm{~dB}$6\text{}\mathrm{d}\mathrm{B}$ per octave unit.
b) In single ended input – single ended output configuration of a differential amplifier, the input voltage is $5\mathrm{m}\mathrm{V}$$5\mathrm{m}\mathrm{V}$5mV5 \mathrm{mV}$5\mathrm{m}\mathrm{V}$. If the collector resistors ${R}_{C1}={R}_{C2}=51\mathrm{k}\mathrm{\Omega }$${R}_{C1}={R}_{C2}=51\mathrm{k}\mathrm{\Omega }$R_(C1)=R_(C2)=51kOmegaR_{C 1}=R_{C 2}=51 \mathrm{k} \Omega${R}_{C1}={R}_{C2}=51\mathrm{k}\mathrm{\Omega }$, emitter resistor ${R}_{E}=47\mathrm{k}\mathrm{\Omega }$${R}_{E}=47\mathrm{k}\mathrm{\Omega }$R_(E)=47kOmegaR_E=47 \mathrm{k} \Omega${R}_{E}=47\mathrm{k}\mathrm{\Omega }$ and supply voltage is $±10\text{}\mathrm{V}$$±10\text{}\mathrm{V}$+-10V\pm 10 \mathrm{~V}$±10\text{}\mathrm{V}$, calculate the differential gain and output voltage of the circuit.
4 a) Using frequency domain analysis obtain the expressions for the oscillation frequency and the required amplifier gain for a phase shift oscillator circuit.
b) Draw and explain the equivalent $LCR$$LCR$LCRL C R$LCR$ resonant circuit of an oscillating crystal. Write the expression for its series resonant frequency and quality factor. What is the effect of the crystal thickness on its frequency?
c) Describe the five processes involved in the generation of microwaves in a vacuum tube device with the help of an example of a reflex klystron oscillator.
PART B
5. a) Draw the four possible negative feedback configurations of an op amp. Comment on the input and output impedances of these four circuits.
b) Design and draw the circuit using op amps to solve the following simultaneous equations:
$\begin{array}{rl}& 2x+3y=12\\ & -x+3y=3\end{array}$$\begin{array}{r}2x+3y=12\\ -x+3y=3\end{array}${:[2x+3y=12],[-x+3y=3]:}\begin{aligned} & 2 x+3 y=12 \\ & -x+3 y=3 \end{aligned}$\begin{array}{rl}& 2x+3y=12\\ & -x+3y=3\end{array}$
1. a) Describe the factors giving rise to the output offset voltage of an op amp. Explain the method for measuring the output resistance of an op amp.
b) Explain the working of a Schmitt trigger using an op amp with the help of appropriate circuit diagram. Design a Schmitt trigger circuit with hysteresis of $20\mathrm{%}$$20\mathrm{%}$20%20 \%$20\mathrm{%}$ of $±{V}_{SAT}$$±{V}_{SAT}$+-V_(SAT)\pm V_{S A T}$±{V}_{SAT}$.
c) With the help of a circuit diagram explain the working of an active full wave rectifier. What are the advantages of using an op amp in rectifier and filter circuits? What is the limitation of using an op amp in a high pass filter circuit?
2. a) Explain the constant current limiting circuit implemented in a series voltage regulator. How is this circuit modified to achieve a fold-back configuration?
b) What are the advantages of switch mode dc-dc convertor over linear voltage regulator? Explain the working of buck convertor.
3. a) Explain the working of a PLL using a block diagram. Why is the capture range of a PLL always smaller than its lock range?
b) Explain the working of a tracking type ADC. Compare its performance with flash type ADC.
c) List the general purpose registers available in 8085 microprocessor. What is a data pointer? Describe the roles of accumulator and flag register in 8085 .
$$a=b\:cos\:C+c\:cos\:B$$

## MPH-005 Sample Solution 2024

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.
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$${n}_{2}=3.6$n_(2)=3.6n_2=3.6${n}_{2}=3.6$; assuming the refractive index of the air $=1$$=1$=1=1$=1$.
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:
$EQE=\frac{\text{Number of photons emitted}}{\text{Number of electrons injected}}×100\mathrm{%}$$EQE=\frac{\text{Number of photons emitted}}{\text{Number of electrons injected}}×100\mathrm{%}$EQE=(“Number of photons emitted”)/(“Number of electrons injected”)xx100%EQE = \frac{\text{Number of photons emitted}}{\text{Number of electrons injected}} \times 100\%$EQE=\frac{\text{Number of photons emitted}}{\text{Number of electrons injected}}×100\mathrm{%}$
EQE is a critical parameter in evaluating the performance of an LED, as it takes into account both the internal quantum efficiency (the efficiency with which electrons and holes recombine to generate photons) and the extraction efficiency (the efficiency with which the generated photons are extracted from the LED).
To calculate the EQE for a semiconductor material with a refractive index ${n}_{2}=3.6$${n}_{2}=3.6$n_(2)=3.6n_2 = 3.6${n}_{2}=3.6$ and assuming the refractive index of air ${n}_{1}=1$${n}_{1}=1$n_(1)=1n_1 = 1${n}_{1}=1$, we need to consider the extraction efficiency. The extraction efficiency is influenced by the critical angle ${\theta }_{c}$${\theta }_{c}$theta _(c)\theta_c${\theta }_{c}$ for total internal reflection at the interface between the semiconductor and air. The critical angle is given by Snell’s law:
$\mathrm{sin}\left({\theta }_{c}\right)=\frac{{n}_{1}}{{n}_{2}}$$\mathrm{sin}\left({\theta }_{c}\right)=\frac{{n}_{1}}{{n}_{2}}$sin(theta _(c))=(n_(1))/(n_(2))\sin(\theta_c) = \frac{n_1}{n_2}$\mathrm{sin}\left({\theta }_{c}\right)=\frac{{n}_{1}}{{n}_{2}}$
Substituting the given values:
$\mathrm{sin}\left({\theta }_{c}\right)=\frac{1}{3.6}$$\mathrm{sin}\left({\theta }_{c}\right)=\frac{1}{3.6}$sin(theta _(c))=(1)/(3.6)\sin(\theta_c) = \frac{1}{3.6}$\mathrm{sin}\left({\theta }_{c}\right)=\frac{1}{3.6}$
${\theta }_{c}=\mathrm{arcsin}\left(\frac{1}{3.6}\right)$${\theta }_{c}=\mathrm{arcsin}\left(\frac{1}{3.6}\right)$theta _(c)=arcsin((1)/(3.6))\theta_c = \arcsin\left(\frac{1}{3.6}\right)${\theta }_{c}=\mathrm{arcsin}\left(\frac{1}{3.6}\right)$
The fraction of light that can escape the semiconductor material (extraction efficiency) is approximately given by:
$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left({\theta }_{c}\right)\right)$$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left({\theta }_{c}\right)\right)$EE=1-(1)/(2)(1+cos^(2)(theta _(c)))EE = 1 – \frac{1}{2} \left(1 + \cos^2(\theta_c)\right)$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left({\theta }_{c}\right)\right)$
Substituting the value of ${\theta }_{c}$${\theta }_{c}$theta _(c)\theta_c${\theta }_{c}$:
$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)$$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)$EE=1-(1)/(2)(1+cos^(2)(arcsin((1)/(3.6))))EE = 1 – \frac{1}{2} \left(1 + \cos^2\left(\arcsin\left(\frac{1}{3.6}\right)\right)\right)$EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)$
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,
$EQE\approx EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)=\frac{25}{648}$$EQE\approx EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)=\frac{25}{648}$EQE~~EE=1-(1)/(2)(1+cos^(2)(arcsin((1)/(3.6))))=(25)/(648)EQE \approx EE = 1 – \frac{1}{2} \left(1 + \cos^2\left(\arcsin\left(\frac{1}{3.6}\right)\right)\right)=\frac{25}{648}$EQE\approx EE=1-\frac{1}{2}\left(1+{\mathrm{cos}}^{2}\left(\mathrm{arcsin}\left(\frac{1}{3.6}\right)\right)\right)=\frac{25}{648}$
$EE=0.038$$EE=0.038$EE=0.038EE=0.038$EE=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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