ELECTRONICS & COMMUNICATION

ELECTRONICS & COMMUNICATION ENGINEERING - Paper I

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Mains Exam Solution

ELECTRONICS & COMMUNICATION


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SECTION-A

Q.1: (a) (i) Calculate the temperature at which silicon (Si) semiconductor tends to behave like ametal. [6 Marks]

(ii) Prove that reverse saturation current approximately doubles for every 10 °C rise intemperature in a semiconductor diode. [6 Marks]

Sol: (i)

EG

C.B.

V.B.

EC

EV

The forbidden energy gap, EG(T) = EGO – T

Let at T = Tc, EG = 0

0 = EGO – BTc

Tc =GO

4E 1.21 eV

3.6 10 ev / k

= 3361.1°k

For T > Tc, V.B. & C.B get concided.

It behaves like a metal.

(ii) The expression for reverse saturation current, I0 is

I0 = p 2ni

p D n A

D DAe nL N L N

Here, A is the cross-sectional area of p-n junction.

Dp and Dn are diffusion constant for holes and electrons respectively.

Lp and Ln are diffusion lengths for holes and electrons respectively.

ND and NA are doping concentration for electrons and holes respectively.

ni is intrinsic concentration and equal to

2in = 3 3G0 G0

0 0T

E VA T exp A T exp

KT V

Hence, I0 = p 3 G0n0

p D n A T

D VDAe A T expL N L N V

The diffusion constants Dp and Dn vary approximately inversely proportional to T for Germaniumdiode. Therefore, temperature dependence of I0 is

I0 =2 G0

T

VKT expV




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where K is temperature independent constant.

For silicon diode, the diffusion current is negligible compared to transition-layer-charge-generation

current. Hence, I0 is found to be proportional to ni instead of 2in . Therefore,

I0 = 1.5 G0

T

VKT exp

2V

In general, the expression for I0 can be written as

I0 = 3/ G0

T

VKT exp

V

where, is idealty factor

VG0 is semiconductor bandgap voltage

T is absolute temperature

VT is thermal voltage equal to TKTVq

Solve for 0dI :dT

0dIdT = 3/ G0qVd KT exp

dT KT

= 3/ 1 3/G0 G0 G0

2qV qV qV3 KT exp KT exp

KT KT hKT

= 3/ 1G0 G0

T

qV V3KT exp TKT TV

= G00

T

V3IT TV

00

1 d II dT

= G0

T

V3T TV

0d ln I

dT = G0

T

V3T TV

or 0ln I = G0

T

V3 TT TV

Consider I01, is saturation current at temperature T and I02 is saturation current at temperatureT T therefore,

02

01

II G0

T

V3exp TT TV

Assume silicon diode at room temperature for T 10 C,

02

01

II G0

T

V3exp 10T TV

Substitude VT = 0.0259 V, VG0 = 1.11 V, T = 300 K and 2.




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02

01

II 2.147

Therefore, depending on , the saturation current approximately doubles for 10°C increase intemperature.

Q.1: (b) (i) Prove that when two resistors are connected in parallel, the equivalent resistance ofthe combination is always smaller than that of smaller resistor. [6 Marks]

(ii) A conductor has resistance 5.4 at 20°C and 7 at 100°C. Determine the resistanceof the conductor at 0 °C. [6 Marks]

Sol: (i) Consider a parallel combination of two resistor r1 and r2 as shown in the figure below.

i1r1

r2

i2i

a bi

Vab

i = i1 + i2

= ab ab

1 2

V Vr r

i = ab1 2

1 1Vr r

abVi

=

1 2

11 1r r

Now, equivalent resistance of the parallel combination

req =

1 2

11 1r r

= 1 2

1 2

r rr r

If, r1 > r2

then, let, r1 = kr2 (k > 1)

Now, req = 2 2

2 2

kr rkr r = 2

k rk 1

k > 1 k 1

k 1

Hence, req < r2 (proved)

(ii) Given, R20°C = 5.4

R100°C = 7R0°C = ?

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Since, RT2 = T1R 1 T

then, R100°C = 20 CR 1 100 20

7 = 5.4 1 80

1 80 =7

5.4

80 =7 1.61

5.4 5.4

=1.6

5.4 80 = 31 3.7 10

54 5

Now, R20°C = 0 CR 1 20 0

5.4 = 30 CR 1 3.7 10 20

R0°C =5.4

1.074 = 5.028

i.e Resistance of conductor at 0°C = 5.028

Q.1: (c) What is line imperfection defect in a crystal? How does it affect the properties of a metal?[12 Marks]

Sol: Line Imperfection DefectLine imperfections (one-dimensional) defects are also called dislocations. They are abrupt changes inthe regular ordering of the atoms along a line called dislocation line in the solid.

They occur in high density and strongly influence the mechanical properties of a material.

They are characterised by Burgers vector (b), whose direction and magnitude can be determined byconstructing a loop around the disrupted region and noticing the extra interatomic spacing needed toclose the loop.

Dislocations occur when an extra incomplete plane is inserted. The dislocation line is at the end of theplane.

(i) Edge dislocations or Taylor Orowan dislocationIt is characterized by a Burger’s vector that is perpendicular to the dislocation line. It may be describedas an edge of an extra plane of atoms within a crystal structure. Thus, regions of compression andtension are associated with an edge dislocation.

Edge dislocation is called ‘positive’ when compressive stresses are present above the dislocation lineand represented as ‘ ’.

If the stress is opposite i.e. compressive stresses exist below the dislocation like, it is considered as‘negative’ edge dislocation and represented as ‘T’.




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Region ofcompression

Region oftension

Positive edgedislocation

Negative edgedislocation

(ii) Screw dislocation or Burger DislocationIt has its dislocation line parallel to the Burger’s vector. A screw dislocation is like a spiral ramp with animperfection line down its axis.

Screw dislocations result when displacing planes relative to each other through shear.

Screw dislocation is positive if Burger’s vector and t-vector are parallel, vice versa. (t-vector a unit vector

representing the direction of dislocation line) and represented by ‘ ’. whereas negative screw dislocation

as “ ”.

Line imperfection defects are accompanied by the distortion in the crystal which varies with the distancefrom the centre of the dislocation.

In this way they affect the mechanical, chemical and electrical properties of the metal.

Q.1: (d) (i) Enumerate te different performance indices based on which an engineer selects aninstrument.

(ii) What is the SI system of units? Mention some (at least four) well-defined unitsmaintained by the international system of units.

Sol: (i) An instrument is selected according to requirement of quality of measurement.

Various specific terms describe the characteristic and quality of measuring instrument definedas:

Accuracy: The degree of agreement of the measured dimensions with its true magnitude.

Resolution: The smallest dimension that can be read on an instrument.

Pricision: Degree to which a measuring instrument gives repeated measurement of thesame standard.

Linearity: The accuracy of the reading of a tool over its full working range.

Amplification: The ratio of measuring instrument output to the input dimension. It is alsocalled magnification.

Calibration: The adjustment or setting of a measuring instrument to give reading that areaccurate with in a reference standard.

The speed of response: How rapidly a measuring instrument indicated a measurement,particularly when some parts are measured in rapid succession.

Stability: An instrument’s capability to maintain its calibration over time.




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(ii) These definitions are adopted all over the world for this system, called the International systemof units and designated by the abbreviation SI. It consists of six base units, two supplementaryunits and 27 derived units.

Base units of SI the seven base units which form the basis of the now universally acceptedinternational systems of units, in their present status are defined below:

1. Metre(m): It is unit of length. The metre is the length equal to 1,65,763.73 wave lengths invacuum of radiation corresponding to the transition between the level 2p10 and 5d5 (orange redline) of the krypton 86 atom (excited at the triple point of nitrogen, 63.15° kelvin).

2. Kilogramme (kg): It is the unit of mass. A kilogramme is equal to the mass of the internationalprototype of mass. This prototype is a cylinder of platinum irridium alloy.

3. Second(s): It is the unit of time. A second is defined as the duration of 9,192,631,770 periodsof radiation corresponding to the transition between the two hyperfine levels of the ground stateof the cesium 133 atom.

4. Ampere (A): It is the unit of electric current. The ampere is that constant current which, ifmaintained in two straight parallel conductors of infinite length, of negligible cross-section, andplaced one metre apart in vacuum, would produce between them a force equal to 2×10–7 newtonper metre length.

5. Kelvin (K): It is the unit of temperature. The Kelvin unit of themodynamic temperature, is thefraction 1/273.16 of the thermodynamic temperature of triple point of water.

6. Candela (cd): It is the unit of luminous intensity. It is the luminous intensity, in a perpendiculardirection, of a surface of 1/600,000 square metre of a blackbody at the temperature of freezingplatinum under a pressure of 101,325 newton per square metre.

7. Mole (mol): The mole is the amount of substance of a system which contains as manyelementary entities as there are atoms in 0.012kg of carbon 12. It should be noted that whenevermole is used, the elementary entities must be specified, which may be atoms, molecules, ions,electrons, other particles, or specified groups of such particles.

Supplementary units: Two supplementary units have been added to the basic six SI units. Theyare (i) radian for the plane angles, and (ii) steradian for the solid angles.

8. Radian (rad): Plane angle subtended by anr arc of a circle equal in length to the radius of thecircle.

9. Steradian (sr): Solid angle subtended at the centre of a sphere by the surface whose area isequal to the square of the radius of the sphere.

Q.1: (e) In the circuit shown in the figure below, I = 1mA is a DC current and vin (t) is a sinusoidalvoltage with small amplitude.

I=1 mA

V (t)o

C = 10 nFv (t)in




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Sol:

v (t)in

I = 1 mA

C = 10 nF

v (t)0

Using small signal AC model

v (t) = asin tin

v (t)0

rdI 1

j Cwhere T

dVrI

Using, KCL:

0 in

d

v t v tr

= 1

v0(t) – v in(t) = VT

v0(t) = v in(t) + 0.025

For cut off frequency,

|rd| = |XC|

TVI =

1C

Taking I = 1 mA

251

=1C

= 91 1

25C 25 10

fH =7100 10

25 2

= 6.36 MHz

Q.2: (a) (i) Find the equilibrium hole concentration P0 at 300K of Si sample doped with phosphorusimpurity if Fermi level energy (EF) of doped Si is 0.407 eV more than intrinsic levelenergy (Ei). Given ni = 1.5 × 1010 atoms/cm3 and kT = 0.0259eV. [10 Marks]

(ii) A filter capacitor C is used to smooth out the pulses from the full wave rectifier asshown in the figure below: [10 Marks]

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Vout

RL= 10 k

220 V50 HzAC

C

Find the value of C so that the steady current supply to load RL can be maintained.

Sol: (i) Equilibrium hole concentration p0 = ? at T = 300°K

Donor (P) impurity

Efn – Efi = 0.467eV

ni = 1.5 × 1010 atoms/cm3

KT = 0.0259eV

Efn – Efi = kT D

i

Nnn

0.407 = 0.0259 D

i

Nnn

ND = 1.002×1017 /cm3

nN ND

holes po = 2102

i17

N

1.5 10nn 1 10

= 2.25 × 103 /cm3

(ii) the ripple factor of full-wave bridge rectifier with capacitive filter is given as

r ='RMS

DC 0 L

V 1V 4 3f C.R

...(i)

where, f0 is operating frequency

RL is load resistor

The ripple factor for full-wave bridge rectifier is 0.483.

Substitude r = 0.483 in equation (i),

0.483 =0 L

14 3f CR

C = 31

4 3 0.483 50 10 10

C = 0.59 × 10–6

C = 0.59 F




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Q.2: (b) (i) What is the principle of nanomagnetism? Based on the specific properties ofnanomagnetism, write its applications in engineering field. [12 Marks]

(ii) Classify insulating materials according to their temperature stability limit and givefew examples of each grade. [8 Marks]

Sol: (i) Principle of Nanomagnetism: Nanomagnetism deals with the magnetic properties ofobjects that have at least one dimension in the nanoscopic range (1nm to 100nm).

It is s phenomehon of a submicrometric system that represents spontaneous magnetizationat zero applied magnetic field (Remanence).

The small size of nanomagnetic prevents the formation of magnetic domains.

At low temperatures, they represent quantum phenomenon such as microscopic spintunnelling.

At larger temperatures, the magnetization undergoes random thermal fluctuations (superparamagnetism) which presents a limitation on the use of nanomagnets for permanentinformation storage. e.g grains of ferromagnetic materials (iron, cobalt, nickel, tramitionmetals, titanium, vanadium, chromium, manganese) or rare earth (Gadolinium, Europium,Erbium)

The phenomenon of zero field magnetization reqires three conditions.(i) A ground state with finite spin.(ii) A magnetic anisotrophy energy barrier.(iii) Long spin relaxation time

Applications in engineering field

Active component ferrofluids

Recording tape

Flexible disk recording media

Biomedical materials

Catalysts

Hard disk recording media

Permanent magnets

Magnetic refrigerators

Research tools in materials phyiscs, geology, biology and medicine.

(ii) Classification of insulating materials according to their temperature stability limit.

(a) Class A: Maximum allowable temperature is 105°C or 221F eg. Cotton, Silk, Impregnatedpaper

(b) Class B: Maximum allowable temperature 130°C or 266F e.g mica, glassfibre, asbestos etc.

(c) Class C: Maximum allowable temperature 180°C or 356F. e.g Mica, porcelain, glass, quartz.

(d) Class E: Maximum allowable temperature 120°C or 248F. It consists of materials possessingdegree of thermal stability at temperature 15°C higher than class-A materials.

(e) Class F: Maximum allowable temperature 155°C or 311F. It consists of materials possessingdegree of thermal stability at temperature 25°C higher than class-B materials. e.gmica, glass fibre, asfestoe etc.

(f) Class H: Maximum allowable temperature 180°C or 356F. eg. Silicone elastomer, combinationof mica, glass-fibre, asbestos etc.




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Q.2: (c) (i) What are the signal conditioning requirements for measurements with strain gauges?How can you compensate erros due to temperature in strain gauge measurements?

[10 Marks]

(ii) A piezoelectric transducer is subjected to a force of 6N. The dimension of thetransducer is given as 6mm × 6mm × 1.3mm. The charge sensitivity and the dielectricconstant of the transducer are given as 160pC/N and 1250 × 10–11 F/m respectively.Calculate the voltage generated and the deflection caused to the surface. The young’smodulus of elasticity of the material is given as 12 × 106 N/m2. [10 Marks]

Sol: (i) Strain gauge measurements are complex and several factors can affect measurementperformance. To ensure accurate strain measurements, proper selection and use of thebridge, signal conditioning, wiring and data acquisition components are required.

1. Bridge completion to complete the required circuitry for quarter and half bridge straingauges. Strain gauge signal conditioners typically provide half bridge completion networksconsisting of two high-precision reference resistors. Figure below shows the wiring of ahalf-bridge strain gauge circuit to a conditioner with completion resistors R1 and R2.

+–

R1

R2

V +–

Exc+

Exc–

IN–

IN+

R4

R4

Straingauges

Signal Conditioner

2. Bridge excitation to power the wheatstone bridge circuitry. Strain gauge signal conditionerstypically provide a constant voltage source to power the bridge.

3. Remote sensing or excitation sensing to compensate for errors in excitation voltage fromlong lead wires. There are two common methods of remote sensing. With feedback remotesensing, we connect extra sense wires to the point where the excitation voltage wiresconnect to the bridge circuit. The extra sense wires sense to regulate the excitation supplyto compensate for lead losses and deliver the needed voltage at the bridge. An alternativeremote sensing scheme uses a seperate measurement channel to measure directly theexcitation voltage delivered across the bridge.

4. Signal amplification to increase measurement resolution and improve signal to noise ratio.Strain gauges conditioners usually include amplifiers to boost the signal level to increasemeasurement resolution and improve signal-to-noise ratios.

5. Offset nulling to balance the bridge to output 0V when no strain is applied. There are a fewdifferent ways that a system can handle this initial offset voltage. These are softwarecompensation, offset-nulling circuit and buffered offset nulling.

6. Shunt calibration to verify the output of the bridge to a known, excepted value. Shuntcalibration involves simulating the input of strain by changing the resistance of an arm inthe bridge by some known amount. The is accomplished by shunting, or connecting, a largeresistor of known value across one arm of the bridge, creating a known R. The outputof the bridge can then be measured and compared to the expected voltage value.

Self compensated strain gauges are specially developed to compensate for the temperature




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behaviour of certain materials by their own temperature behaviour. This means that theycounteract the apparent strain (and thus the temperature induced expansion of the measuringbody). Therefore a strain gauge with a temperature response suitable for the material of themeasuring body is selected.

(ii) Given, F = 6 N

Dimension = 6 mm × 6 mm × 1.3 mm

Charge sensitivity = d = 160 pC/N

Dielectric constant of transducer, = 1250 × 10–11 F/m

Young’s modulus of elasticity, Y = 12 × 106 N/m2

We have to calculate voltage generated E and deflection caused to the surface t

Voltage generated:

Voltage generated, E =p

Q Charge=C Capacitance between electodes

E =dF

At

= 12 3

11 6160 10 6 1.3 10

1250 10 6 6 10

=0.0277 × 102 = 2.77 volt

Now, young’s modulus of elasticity,

Y = F / AStress F.t.

Strain t / t t.A

t = 3

6 6F.t 6 1.3 10Y.A 12 10 6 6 10

t = 0.018 × 10–3 m = 0.018 m

i.e., deflection, t = 0.018 m

Q.3: (a) (i) Calculate the range of base resistance (RB) so that transistor T1 never operates insaturation region: [10 Marks]

+

–

+–

20 V

10 K

DC=50

T1RB

10V

(ii) A amplifier has a bandwidth of 500kHz and voltage gain of 100. What should be theamount of negative feedback if the amplifier bandwidth is extended to 5MHz? Whatwill be the new gain after negative feedback is introduced? [10 Marks]

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Sol: (i)

IC

+VBE

RB IB+10V

+20V

10 K

–

+

–

VCE

IC

2

1

Loop (1)–10 + IB RB + VBE = 0

IB RB = 10 – 0.7 = 9.3 ... (1)Loop (2) – 20 + 10 IC +VCE = 0

VCE = 20 – 10ICFor transistor always in the active region

VCE > VCE,sat = 0.2

20 – 10 IC 0.2

IC 20 0.2 1.98 mA

10

IB = CI 1.9850

IB 0.0396 mA

From (1) RB =B

9.3I

RB B9.3 R 234.8 K

0.0396

(ii) Amplifier has bandwidth of 500kHz and voltage gaing of 100

BW = 500kHz

A = 100

New bandwidth is, Bwnew = 5MHz

We know, Bwnew = BW 1 A

1 A =6

35 10

500 10

= 10




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Hence, amount of feedback, 1 A is 10

We know, Anew = A

1 A for negative feedback

Anew =10010

= 10

Therefore, new gain is 10.

Q.3: (b) (i) What are HTSC? Write the important applications of superconducting materials andHTSC. [10 Marks]

(ii) “A ceramist can alter the properties of ceramic” Justify the statement. [10 Marks]

Sol: (i) The high temperature superconductors (HTSC) represent a new class of materials which bearextra ordinary superconducting and magnetic properties and great potential for wide rangingtechnological applications. However “ordinary” or metallic superconductors usually have transitiontemperatures of about 30K (–243.2 °C) whereas HTSC have been observed with transitiontemperatures as high as 138K(–135°C) called high temperature superconductors on basis ofcritical temperature greater than boiling temperature of liquid Nitrogen (77K or –196°C)

Applications of HTSC1. Magnetic levitation

2. Power transmission

3. Superconducting magnets in generators

4. Energy storage devices

5. Particle accelerations

6. Rotating machinery

7. Magnetic seperations

Applications of superconducting materials1. Particle accelerometers2. Generators3. Transportation4. Power transmission5. Electric motors6. Military7. Computing8. Medical

(ii) A ceramic is a solid material comprising an inorganic compound of metal, non-metal or metalloidatoms primarily held in ionic and covalent bonds. e.g earthenware, porcelain, brick. Whereas acerenist is a person who makes ceramic objects and/or artworks.

The crystallinity of ceramic materials range from highly oriented to semicrystalline, vitrified andoften completely amorphos (e.g glasses). Most often, fired ceramics are either vitrified or semi-vitrified as is the case with earthenware, stone ware and porcelain. Varying crystallinity andelectron composition in the ionic or covalent bonds cause most ceramic materials to be goodthermal and electrical insulators.




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The ceramic family covers vast properties like high melting temperatures, high hardness, poorconductivity, high moduli of elasticity, chemical resistance and low ductility etc.

Therefore, from the selection of reed materials their combination and further processing a ceramistcan very vastly alter the properties of ceramic products.

Q.3: (c) (i) Explain, with necessary diagrams, how you can detect the proximity of an object.[12 Marks]

(ii) The spring constant and seismic mass of an accelerometer are 3300 N/m and 5×10–2

kg respectively. The maximum displacement is 0.25 m (before the mass hits thestops). Calculate (1) the maximum measurable acceleration in g and (2) the naturalfrequency. [8 Marks]

Sol: (i) Proximity sensor: A sensor that can be used for detecting the presence of objectssurrounding it without having any physical contact is termed as a proximity sensor. This canbe done using the electromagnetic field or electromagnetic radiation beam in which the fieldor return signal changes in the event of the presence of any object in its sorrounding.

This object sensed by the proximity sensor is termed as target. There are different proximitysensors are used for different targets. Such as capacitive proximity sensor or photoelectricproximity sensor, magnetic proximity sensor and so on.

Target

Electric coilOscillator

Voltage regulator

Trigger

Output

To loadProximity sensor circuit diagram

when target is detectedThe inductive proximity sensor circuit is used for detecting the metal objects and the circuitdetect any objects other than metal. Whenever, this field is distributed by detecting anymetal objects then an eddy current will be generated that circulated with in the target.

Due to this load will be caused on the sensor that decreases the electromagnetic fieldamplitude. If the object is moved towards the proximity sensor, then the eddy current willincrease accordingly. Thus, the load on the oscillator will increase, which decreases the fieldamplitude.

The trigger block in the proximity sensor circuit is used to monitor the amplitude of theoscilator and at paricular levels (predetermine levels) the trigger circuit switchs on or off thesensor. If the object is moved away from the proximity sensor, then the amplitude of theoxillator will increase.




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targetabsent

target presnet Target absent

(ii) Given, for accelerometer,

Seismic mass, M = 5 × 10–2 kg

Spring constant, K = 3300 N/m

Maximum displacement, xm = ±0.25 m

(1) Maximum measurable acceleration,

am = 2n mx

where, n = natural frequency in rad/s

= K / M

= 23300

5 10

= 66000= 256.90 rad/s

So, am = 2mx

= (256.90)2 × (0.25)

= 16499.4 m/s2

16500 m/s2

(2) Natural frequency,

f = n 256.9 40.90 Hz2 2

Q.4: (a) (i) Consider a differential amplifier as shown in the figure

DifferentialAmplifier

V1

V2

1

2

Voutput

Where the first set of signals is v1 = 50 V , v2 = 50 V and the second set of signals

is v1 = 1050 V , v2 = 950 V . If the common mode rejection ratio is 100, calculate thepercentage difference in output voltage obtained for the two sets of input signals.

[10 Marks](ii) 1. Repeat part (a) (i), if the common mode rejection ratio is 105. [3 Marks]

2. Draw the conclusion by comparing part a(i) and part (a) (ii)(1). [2 Marks]

(iii) Explain photovoltaic potential in short. [5 Marks]

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Sol: (i) The output voltage is given as

V0 = ADMVd + ACMVCM

Where ADM = Difference mode gain

ACM = Common mode gain

Vd = Difference mode input

VCM = Common mode input

CMRR = DM

CM

A 100A

Therefore, ADM = 100 ACM

Vd = v1 – v2 and VCM = 1 2v v2

For the first set of signals,

Vd = 50 V 50 V = 100 V

VCM =50 V 50 V

2

= 0

V0 = CM CM100A 100 V A 0

V0 = 0.01ACM

For the second set of signals

dV = 1050 V 950 V

= 100 V

CMV =1050 V 950 V 1000 V

2

V0 = CM CM100A 100 V A 100 V

= 0.01 ACM + 0.001ACM

= 0.011 ACM

%Difference = 0 0

0

V V 100V

= CM CM

CM

0.011A 0.01A 1000.01 A

= 0%

(ii) (1) The common mode rejection ratio is 105

CMRR = DM

CM

AA = 105

Or ADM = 105 ACM

For the first set of signals




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vd = 100 V and VCM = 0

V0 = 105 ACM × 100 V = 10ACM

For the second set of signals

dV = 100 V and CMV = 1000 V

0V = 5CM CM10 A 100 V A 1000 V

= 10ACM + ACM = 11ACM

% difference = 0 0

0

V VV

= CM CM

CM

11A 10A 10010A

= 10%

(2) Percentage change in output is independent of CMRR.

(iii) Photovoltaic potential: When a PV module is exposed to the sunlight the incoming photoncreates a hole (positive) and (negative) electron pair. This leads to availability of a potential across aPV cell. As a single PV cell output is very low, they are generally connected in an array of series andparallel combination to achieve the desired output voltage.

LoadN-type silicon

electron

Hole

Anti-Reflecting

coating

Lightenergy

V

P-type silicon

Q.4: (b) (i) What is polarization mechanism in dielectric material? Explain active and passivedielectrics with suitable examples. [10 Marks]

(ii) Explain cermets. How are they different from fibre reinforced composites? Write fourapplications of each. [10 Marks]

Sol: (i) Polarization: The application of an electric field to an dielectric material causes a displacementof electric charges, giving rise to the creation or reorientation of the dipoles in the material.The phenomenon of creation and reorientation of the dipoles is called polarization.

Mathematically, the total dipole moment per unit volume is called polarization

(P)

P = Tp

volumewhere, Tp = Total dipole moment, coulomb-metre

The centres of gravity of positive charges and negative charges coincide in neutral atomsand symmetric molecules. Application of an electric field causes relative displacement ofthese charges, leading to the creation of dipoles and hence polarization.

There are four basic polarization mechanisms :

1. Electronic polarization : Electronic polarization is found in materials in which there isno interaction among individual molecules. In inert gases the interaction among the atomsis negligible. This is because the distance between the molecules is large enough (monoatomicgases).




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The dipole moment induced by the field, indp

= 30(4 R )E

This dipole is induced by the field and never existed in the absence of the field.

The induced dipole moment is proportional to the field strength and proportionality factor e

is called electronic polarizability. If a gas has N such atoms per m3, then electronic polarization,

eP =

eN E

2. Ionic polarization: This type of polarization occurs in materials possessing ionic bondbetween two dissimilar atoms. Ionic materials such as NaCl and HCl exhibit ionic polarization.These materials have permanent dipole moment even in absence of external field.These materials exhibit both electronic and ionic polarization

3. Orientational polarization : This type of polarization occurs in materials having partlyionic bonds or covalent bond of ionic character. The existence of a permanent moment ispurely a matter of molecular geometry. Ex – H2O, N2O, CO2

4. Space charge polarization : Space charge polarization is the result of lattice vacanciesin the dielectrics. Effect of these lattice vacancies is localized accumulation of chargeswhich induce image charges on electrodes and gives rise to a dipole moment called spacecharge polarization. The space charge polarization is also known as interfacial polarization.

Total polarization, PT = Pi + Pe + P0 + Ps

where, Ps = interfacial polarization

(1) Active dielectrics: When a dielectric material is kept in an external electric field if it acceptsthe electricity then it is known as active dielectric material. Thus, active dielectrics can easilyadopt themselves to stone the electrical energy in it.

e.g. dry air

(2) Passive dielectrics: They restrict the flow of electrical energy so that they act as insulatorse.g glass, mica, rubber, etc.

(ii) Cermets: A cermet is a composite materials composed of ceramic and metallic materials. Ideally,a cermet is designed to have the optimal properties of both a ceramic, such as high temperatureresistance and hardness, and those of a metal such as plastic deformation. The metal is used asa binder for an oxide, boride or carbide. Cerments are used in the manufacture of resistors(especially potentiometer), capacitors and other electronic components which may experiencehigh temperature.

Difference with fibre reinforced composites: Fibre reinforced composites are compsed offibres embedded in matrix material. Such a material is called a discontinous fibre, if its propertiesvary with fibre length. On the other hand when length of the fibre is such that any further increasethe elastic modulus of composite, the composite is considered to be continuous fibre reinforced.

Fibres are small in diameter and when pushed axially, they bend easily although they have verygood tensile properties.

Applications of cermets1. Ceramic to metal joints and seals eg. construction of vacuum tubes.

2. Bioceramics eg. biomedical materials.

3. Transportation eg. friction materials for brakes and clutches

4. Machining on cutting tools




Mains Exam Solution



IES M

ASTER

Applications of fibre reinforced composites1. Aerospace and defence

2. Automative

3. Construction

4. Marine

5. Oil and gas industries

Q.4: (c) (i) What are the elements of a generalized data acquisition system? Draw a multi-channeldata acquisition system using single A/D converter and briefly explain its working.

[14 Marks]

Sol: (c) (i) Data Acquisition System : Data-acquisition systems are used to measure and record analogsignals is basically two different ways.

(i) Signal which originate from direct measurement of electrical quantities. These signals maybe D.C. or A.C. voltages, frequency or resistance etc.

(ii) Signals which originate from use of transducers.A generalized diagram of a digital data acquisition system is shown below.

Auxiliary equipment andsystem programer

Signal conditioningequipment

Multiplexer Signalconverter

A/Dconverter

DigitalrecorderTransducer

The varioius components and their functions are described below:1. Transducer : They converts a physical quantity into an electrical signal which is acceptable by

the data acquisition system (DAS).2. Signal conditioning equipment : It is an excitation and amplification system for passive

transducers. It may be an amplification system for active transducers.3. Multiplexer : Multiplexing is the process of sharing a single channel with more than one input.

Thus a multiplexer accepts multiple analog inputs and connects them sequantialy to onemeasuring instruement.

4. Singnal converter : It translates the analog signal to form acceptable by the analog to digitalconverter.

5. A/D converter : It converts analog voltage to its equivalent digital from.6. Digital recorders : Records of information in digital form.7. Auxiliary equipment : This contains devices for system programing functions and digital data

processing, some of the typical funcitons are done by auxillary equipment are linearizationand limit comparision of signals.

(ii) Explain, with a diagram, the operation of a force balance current telemetreing system.

[6 Marks]

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Mains Exam Solution

IES M

ASTER

Sol.: (ii) Force balance current telemetry system is shown below:

Power supply

F.Bforcecoil

N

P

Bourden tube

Milliammetercalibrated

in N/m2

A

S

Channels

S

P

In this system, a part of the current output is “Fed back” to oppose the motion of the input variable.The system is operated by a Bourdon tube which rotates the “Feedback force coil” that in turnchanges the “Flux linkage” between primary (P) and secondary (S) coils. The change in fluxlinkage varies the amplitude of the amplifier. The ouput signal is connected to the “feedback (FB)force coil “which produces force that opposes the bourden tube input.

SECTION-B

Q.5: (a) (i) Find the value of resistance R in the figure below if the voltmeter reads 12V andammeter reads 0.100A [6 Marks]

R

V

A

R =2A

R =10kV

(ii) Repeat part (a)(i) for the circuit arrangement given in the figure below: [6 Marks]

A

V

R

R =10 V

R = 2A

Sol: (i) Given data,

Vv = 12V, IA = 0.1A, Rv = 10k , RA = 2

A

V Iv

IA

R IR




Mains Exam Solution



IES M

ASTER

IA = IR + IV

0.1 = v v

v

V VR R

0.1 =12 12R 10 1000

12R = 0.0988

R = 121.46

(ii) Given data,

Vv = 12V, IA = 0.1A, Rv = 1010k , RA = 2

V

AR = 2A

RVR VA

R = 10kV

Let, VR and VA be the voltage drop across resistor R and the ammeter.

then, Vv = VR + VA

12 = IAR + IARA

12 = (0.1)R + (0.1)(2)

R = 118

Q.5: (b) Copper has an atomic radius of 0.1278mm. Calculate the atomic density (number of atomsper unit) in (1 0 0) plane of copper (FCC). [12 Marks]

Sol: Sol:In the plane (1 0 0) for an FCC unit cell

For this (1 0 0) plane there is one atom at each of the four cube corners, each of which is share withfour adjacent unit cells, while the centre atom lies entirely within the unit cell. Thus, there is theequivalence of 2 atoms associated with this FCC (1 0 0) plane. The planner section represented in theabove figure is square, where in the side length are equal to the unit cell edge length 2R 2 and thus,

the area of this square is just 2 22R 2 8R .

Hence, the planner density for this (1 0 0) plane is just

=Number of atoms centred on (1 0 0) plane

Area of (1 0 0) plane




Mains Exam Solution

IES M

ASTER

= 2 22 atoms 1

8 R 4R

Given: R = 0.1278 nm

=14

9 21 653.3 10

4 (0.1278 10 )

Q.5: (c) How can you convert a galvanometer into an ammeter and a voltmeter? A PMMCgalvanometer of 6 resistance reads up to 60mA. Determine the value of the resistance(i) when connected in parallel to enable the instrument to read up to 1.20A and (ii) whenconnected in series to enable it to read 12V. [12 Marks]

Sol: A galvanometer is converted into ammeter by connecting a very small resistance across it.

RP

I ,Rm m

IP

Where, Im = Galvanometer rated current

Rm = Galvanometer resistance

Rp = Resistance connected across galvanometer

Also, a galvanometer is converted into voltmeter by connecting a high resistance in series.

I , Rm m RS

Where, Rs = Resistance connected in series with galvanometer.

(i) For ammeter

RP

6 60 mA1.2A

IP

Here, 1.2 = 0.060 + Ip Ip = 1.2 – 0.06 = 1.14A

Now, applying Kirchoff’s voltage law in above loop,

We get, 6 × 60 × 10–3 = 1.14 × Rp

Rp =36 60 10

1.14

= 0.3158

pR 0.3158




Mains Exam Solution



IES M

ASTER

(ii) For voltmeter:

6 , 60mARS

12 V

12 = (6 × 60 × 10–3) + (60 × 10–3 × Rs)

12 = 60 × 10–3 (6 + Rs)

6 + Rs = 312 200

60 10

Rs = 200 – 6 = 194

sR 194

Q.5: (d) In the circuit given below, vc(0–) = v0 , while the inductor is not charged. The switch thatis initially open is closed at t = 0. Also L = C in terms of numerical value. Find R (positivevalue) so that the circuit is critically damped. [12 Marks]

R

L

+

–

0 C

Sol: Given data, vc(0–) = v0

i(0–) = 0 (Initially inductor is not charged)

L = C (Numerical value)R

+

–

v0L

Transforming the given network into s-domain, (for t > 0)

sL+–

1sC

vs

0

R

i(s)

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Mains Exam Solution

IES M

ASTER

Now,

i(S) =

0vs

1R SLsC

= 02

v1s L sRC

=0

2

vR 1L s sL LC

i(S) = 02

v / LR 1s sL LC

Comparing the characteristic equation

2

C

R 1s SL L

= 0

with the standard characteristic equation for a second order system i.e.,2 2

n ns 2 s = 0

one get,

n =1LC

and, n2 =RL

=R 1 LCL 2

=R C2 L

For critical damping, the damping ratio should be,

= 1

R C2 L

= 1

LR 2C

L = C

R = 2

Q.5: (e) In the feedback circuit shown in the figure below, hfe is very large. Indentify the type offeedback, and (i) Find the feedback factor and overall transresistance with feedback and

(ii) overall voltage gain, 0vs

s

vAv

[12 Marks]




Mains Exam Solution



IES M

ASTER

V = 7.5CC

R = 1 kC

R = 10 kF

C 2 8C 1 8R = 1 K3

US

Sol: hfe = very large

Coupling capacitance 1 2C , C

i.e. short circuit

Is

R = 1 ks

+Vs

R = 10 kf

R = 1 kC

+7.5 V

V0

If

Ii

Using Rf is the output voltage V0 is being simple i.e., voltage feedback.

It is connected in input as shunt

Ii = Is – If i.e. shunt feedback

So, it is voltage-shunt feedback




Mains Exam Solution

IES M

ASTER

DC analysis:

R = 10 kf

R = 1 kC

+7.5 V

IL

IC+

– 1 k

V = 0.7BE

+–

IB

I IE C

C1 & C2 are opened

As

I = CI 0

Loop (1)

–7.5 + 1.IC + 0.7 = 0

IC = 6.8 mA

E CI I = 6.8 mA

rC =T

E

V 26 mV 3.821I 6.8 mA

'il e eh r r = 3.824

hil = 0.0038

hfe =

AC analysis: DC voltage same = 0 V

i.e. S/C

Is

R = 1 ks

+Vs

R = 10 kfV0

If

IiViR = 1 kC




Mains Exam Solution



IES M

ASTER

Av = 0

i

VV

using miller’s theorem

on Rf

IsR = 1 ks

+Vs

V0

Ii

R = 1 kC

V

101 A

If

CE Amplifier |AV| >> 1

V

11 1A

So,V

10 1011

A

R = 1 ks

+Vs

V0

Vi

–0.91k

V

101 A

'iR '

iR

Voltage gain AV = 0 fe L

i ie

V h RV h

=0.91 239.2

0.0038

Input ressitance

Ri = hie = 0.0038 as

V

101 A =

10 0.042 K1 239.2

'iR = i

V

10 R 0.042 0.042 K1 A

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Mains Exam Solution

IES M

ASTER

Rs

+Vs

'iR

Is Vi

i

s

VV =

'i

i s

RR R

=0.042

1 0.042

AVs =0 0 i i

Vs i s s

V V V VAV V V V

= –239.2 × 0.04 = –9.64

Is = i'i

VR

Rm = '0 0V i'

s i i

V VA R

I V / R

= –239.2 × 0.042 = –10.046 K

Q.6: (a) (i) A voltage source delivers 4A when the load connected is 5 and 2A when the loadis 20 . What is the maximum power it can deliver? Also calculate power transferefficieny with RL = 5 and power transfer efficiency when it delivers 50W..

[10 Marks]

(ii) Find the Thevenin equivalent of the circuit at a-b:

a

b

+

–

Vab2

6V +–

8 4

V, I

Sol: (i) Given data, IL = 4A for RL = 5

and IL = 2A for RL = 20

Consider the internal resistance of the voltage source of ‘r’

+– RL

IL

V

r




Mains Exam Solution



IES M

ASTER

Case (i) V = IL (RL + r)

V = 4 (5 + r) = 20 + 4r

V = 20 + 4r ...(i)

Case (ii) V = IL (RL + r)

V = 2 (20 + r) = 40 + 2r

V = 40 + 2r ...(ii)

From, equation (i) and (ii)

40 + 4r = 40 + 2r

2r = 20 r = 10

and, V = 20 + 4 × 10 = 60V

or V = 40 + 2 × 10 = 60V

Hence, voltage source is, V = 60V and r = 10

For maximum power, (for constant source resistance)

RL = r = 10

then, i =V2r =

602 10

= 3A

then, Pmax = i2 RL = (3)2 × 10 = 90W

Power transfer efficiency

=Load power 100%

Source power

Case (i) For RL = 5

i =L

Vr R =

60 4A10 5

then, PL(load) = i2FL = 16 × 5 = 80W

and Ps(source) = Vi = 60 × 4 = 240W

T = L

s

P 100%P

= 80 100% 33.33%

240

Case(ii) PL = 50W

i2RL = 50

i =L

VR r =

L

60R 10

2

L2L

60R

F 10= 50

3600RL = 2L L50 R 20R 100

2L L50R 2600R 5000 = 0




Mains Exam Solution

IES M

ASTER

RL = 50 ,2

For, RL = 50 , i =60 1A60

then, T = L

s

PP =

50 83.33%60 1

For RL = 2 i =60 5A12

then, T = L

s

PP =

50 10060 5

= 16.67%

(ii)

+–

8 V /2ab 4 O a

+

V =Vab th

–

Vab

2

bKVL in loop (1)

–6 – 8 × abV2

+ 0 + Vab = 0

–6 – 3Vab = 0 Vab = –2V

Vth = Vab = –2V

For Rth: 6V source is S/L

Vab

2

8 4V /2ab

+–

+

–

Vab

Pth

IV

b

I+

a I

KVL in loop (2)

abV1 4I 8 I2

= 0 & Vab = 1




Mains Exam Solution



IES M

ASTER

–1 + 4I + 8I + 4×1 = 0

12I = –3 I = 14

Rth = 1 4I

Thevenin equation Rth = 4

+– V = – 2Vth==

+–2V

– 4 a

b

a

b

Q.6: (b) (i) For the circuit shown in the figure below, it is known that the voltage across the

capacitor is C t 20 sin 2t V6

for t 0. Compute and plot the instantaneous

power absorbed by the capacitor and energy stored by the capacitor over [0,t]i (t)c

N 5 mF+–

V (t)C

[8 Marks]

(ii) The op-amp in the figure assumed to be ideal 1 2R 20 k m R 40 k and

C 10 F :

C

R2

R1

+–v (t)in +

–+

–v (t)out

(1) Use nodal to construct a first-order differential equation describing the input-outputrelationship of the voltage.

(2) Laplace transform your equation of (1) and solve out s in terms of in s and

C 0 .

(3) If 2tin t 2e u t V and C 0 0, and find out t . [12 Marks]




Mains Exam Solution

IES M

ASTER

Sol: (i) vC = 2Dsin 2t V t 06

C = 5 mF = 5 × 10–3 F

iC = 3cdvC 5 10 20 2cos 2t

dt 6

iC = 0.2cos 2t A6

iC+

–VC

Instantenous power absorbed by capacitor

pC(t) = vC(t) iC(t)

= 4sin 2t cos 2t6 6

= 2sin2 2t6

pC(t) = 2sin 4t3

Energy during time t = 0 to t = t

WC(t) = t

ct 0

p t dt

=t

02 sin 4t dt

3

=t

0

2 cos 4t4 3

=1 cos 4t cos2 3 3

= 1 1 cos 4t2 2 3

=1 1 2cos 4t Jouls4 3

(ii)

–

+

V = 0p

i2 R2

I 0n

V 0n

i + iC 2

– +VC

C iC

i1

R1

+V (t)i

–

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Mains Exam Solution



IES M

ASTER

1. Nodal on nV 0

1 C 2i i i = 0

i1 + i2 + iC = 0

0 Ci

1 2

V 0 dVV 0 CR R dt

= 0

VC = V0 – Vn = V0

0 0i

1 2

dV VVCdt R R

= 0 ...(i)

2.

0 0

0

dV VC

dt R = i

1

VR

Using Laplace transform

00 0

2

V sC sV s V 0

R =

i

1

V sR

02

1sC V sR

= i

C1

V sCV 0

R ..(ii)

3. If ti CV t 2e u t V, V 0 0

Vi(s) =2

s 2R1 = 20 × 103, R2 = 40 × 103, C = 10–5 F

So, (1 +sCR2)V0(s) = 2i

1

R V sR

5 301 s 10 40 10 V s =

22s 2

01 04s V s = 4s 2

V0(s) = 4

1 0.4s s 2

= 4 1

0.4 s 2.5 s 2

V0(s) = 110

s 2 s 2.5

=10 1 10.5 s 2 s 2.5

=1 120

s 2.5 s 2

V0(t) = 2.5t 2t20 e e u t V




Mains Exam Solution

IES M

ASTER

Q.6: (c) A certain person with poor television reciption and no access to cable TV intends to usethe amplifier in the figure shown below as booster ampifier between his antenna and histelevision:

Antenna

R =300 ANT

C1 810 K 10 K

–12V

C3 8

C2 8

12 V

6.8 K

R =10 KL

i/P resistanceof TV

The transistor has high frequency capacitances with Cbc = 4 pF, Cbe = 2 pF, Cce = 1 pF and100. Calculate low-pass dominant pole frequency and determine whether this amplifier

performs adequately. (Assume VT = 26 mV at room temperature)[20 Marks]

Sol: For dc analysis capacitius are opened.

B = 100, VT = 26mV

IB

10k

D

V = 0.7BEI + IB C

10k

–12V

6.8k

12V

Loop (1) B B L10I 0.7 10 I I 12 = 0

IC = BI = 100IBHence 10IB + 10*101IB = 11.3

IB =11.3 mA1020 = 11.1 A

IC = 100IB = 11.1*100 1.11mA

1020

hil = Lr = er = T

E

VI

= T3

B

V 26mVI 11.1*10 mA = 2.34 k




Mains Exam Solution



IES M

ASTER

hft = B = 100

AC analysis coupling capacitances and s/c

10k Ri

RL

6.8kB

RL

C VE

ViVin

R = 0.3kant

1R = 6.8/10 = 4.1K

Ri = hil = 2.34K

Gain at mid frequencies it junction capacitances are opened

Ri

RL=4.1k

V i

V0

Au0 = Lfe

i

RhR

=4.1100 *

2.34 = –173

At high frequencies Xc =c

1w

So junction capacitances can’t be opened.

Cbe

B

CV0

Cbc

Vi

E

R =4.1kLCce

Using Miller theorem on Cbc

RLCce

V0

Vi

C (1–A )bc v Cbc

bcv

1C 1A




Mains Exam Solution

IES M

ASTER

As uA 1

bc bcv

1C 1 CA

Vi

RLC0

V0

RiCi

D bc ceC C C = 4 +1 = 5pFCi = Cbe + (1 – Av)Cbc = 2 + (1 – Av)4

As uA 173Ci = 2 + 174*4 = 698 PF

As Ci >> C0 so dominantLow pass frequency is due to Ci

10k

0.3kVin Vi

CiR=2.34ki

V i0.3k

10||2.34=1.9k C = 698pFi

Vin 0.3k

1.94k Vth R = 0.3||1.94 = 0.26k

th

C = 698pFi

R = 0.26kth

Vi

Cut-off frequency

fc =th i

12 R C = 12

3 121 0.87 *10

2z * 0.26 *10 * 698 *10

= 870GHz




Mains Exam Solution



IES M

ASTER

Q.7: (a) In the circuit shown in the figure below, |V1| = 200 V, V2 = 2000 0 V and |I| = 12 A. Thetotal power absorbed by the circuit is 1.8 kW. Find R1, X1 and X2:

V2V1

++

––

I

R1X1

– j 20

I R = 10 2 X2

[20 Marks]Sol:

I2zIz

+

–

1V

+

–

2V

2z = 2 2 2 2 2R jX 10 jX z

z2 =2 1 22 2

X100 X tan10

I = 2

2 2 2

V 200 0z z

= 22

200z

| I | = 22

200 12 z 16.67z

z2 = 22100 X 16.67

X2 = 13.33

2 = 1 13.33tan 53.1310

I = 12 –53.13 A

Power absorbed by R2

P2 = I2R2 = 122 × 10 = 1440W = 1.44 kW

So power absorbed by z

P1 = P – P2 = 1.8 – 1.44 = 0.36 kW




Mains Exam Solution

IES M

ASTER

R1–jX1

–j20

=

–j20

I I

1z

1z = R1 – jX1

1z =

1

1 1z j20

z = z –

V = I z 12 –53.13 z

V = 12 V 53.13 I 12 53.13

Power absorbed by z i.e., P1

P1 = VIcos

Angle between V & I

P1 = 12z 12cos 360 W

zcos = 2.5

As 1V = 2V I z

= 200 0 12z – 53.13

= 200 12zcos 53.13 j12z sin 53.13

So, 21V = 2 2200 12zcos 53.13 12zsin 53.13

2002 = 2 2200 12z 2 12z 200cos 53.13

12z 12z 400cos 53.13 = 0

12 z cos 53.13400

= 0

As z = 2.5

cos from (1)

0.03 2.5 cos 53.13 sin sub53.13cos

= 0

20.75 0.6cos 0.68sin cos = 0




Mains Exam Solution



IES M

ASTER

1 cos20.4 2sin cos 0.6 0.0752

0.4sin2 0.3 0.3cos2 = 0.075

sin2 0.75cos2 = 0.9375

Let i = Rcos , 0.75 = Rsin

R = 21 0.75 1.25

= 1 0.75tan 36.8701

So, R sin2 cos cos2 sin = 0.9375

1.25sin 2 = 0.9375

sin 2 36.870 = 0.75

2 36.87 = 48.6°

= 42.73°

z =2.5 2.5 3.4

cos cos42.73

As1z

=1

1 1z j20

1

3.4 – 42.73 =1

1 1z j20

1

1z = 0.3 42.73 j0.05

= 0.22 + j0.254

1z =1

0.22 j0.254

= 1.95 – j2.25

1z = 1 1R jX 1.95 j2.25

R1 = 11.95 , X 2.25




Mains Exam Solution

IES M

ASTER

Q.7: (b) (i) Determine the voltage across the resistor in the circuit of the figure shown below

using phasor concept for 1 t 20 cos t 53.13 V , 2 t 19.68 sin

t 152.8 V and 3 t 4.215 cos t 71.61 V :

+ +– –

+–V (t)1

V (t)2 V (t)3

+

–U (t)R

[6 Marks](ii) Find the Laplace transform of the sawtooth sketched in the figure below:

f (t)

1

1

t

[4 Marks](iii) The figure below shows a simplified model of an autio amplifier containing and ideal

transformer. The input voltage is at 2 kHz with a magnitude 1 V r.m.s. The load is aloudspeaker, represented by 4 resistance:

+– +–

+

–

200 Vin Vin

900 a :1

4

Model of an amplifier loudspeaker

[10 Marks]Sol: (i) VR = V1 + V2 + V3

Phasor RV = 1 2 3V V V

Phasor rms & phase angle

cos x = sin 90

V1(t) = 20cos t 53.13 V

= 20sin t 143.13 V




Mains Exam Solution



IES M

ASTER

1V =20 143.13 V

2 let sin t as reference

V2(t) = 19.68sin t 152.8 V

2V =19.68 152.8 V

2

V3(t) = 4.215cos t 71.61 V

= 4.215sin t 161.61 V

3V =4.215 161.61 V

2

TVI

=25 mV 251mA

RV = 1 20 143.13 19.68 152.8 4.215 161.612

= 1 43.6537.5 j22.33 149.222 2

VR(t) = 43.64sin t 149.22 V

(ii) f(t) = t for 0 < t < 1

= t u t u t 1

= tu t tu t 1

= tu t t 1 1 u t 1

= tu t t 1 u t 1 u t 1

1 tu t = 21s

1 t 1 u t 1 =s

2es

1 u t 1 =se

s

F(s) =s s

2 21 e e

ss s

(iii) Thevenins equivalent across primary of transformer

+–200Vin+–Vin

900

V = Vth oc

+

–

a

bThevenins voltage

Vth = –200Vin




Mains Exam Solution

IES M

ASTER

For Rth

200V =0inV =0inR =900th

900

IL

R = 4L V =–200Vth in

+–

R = 900th

Vin = 1Vrms, Vth = –200V

IL = th

th L

V 200 AP R 904

Power PL = 2L LI R =

2200 * 4904

= 0.196W = 196mWa! 1

R'L

RL

With x-mor connected

1

2

NN =

1a

Load resistance referrred to primary

LR =

L2

2 1

RN N

= a2RL = 52 * 4 = 100

~+

–Vth

I'2

R = 900 th

R = 100'L

LI =

th

th L

VR R

=200

900 100

= –0.2A

LP = 2L LI R = 20.2 *100 = 4W

Iv1 = s150I *

50 150 = 0.25Is1




Mains Exam Solution



IES M

ASTER

Only Is2 , IS1 = 0 (O/C) Vs3 = 0VI01Is1

50 100Is1

–Io2

150 Is250 50

Io2

100

Is2

Io2

–I02 = s2 02 s150I * I 0.75I

5 250

Only, Vs3, Is1 = 0, Is2 = 0KVL in loop (1)

Vs3

50

– +

I03 = s3s3

V 0.005V200

So, I0 = I01 + I02 + I03

= 0.25Is1 + 0.75Is3 + 0.005Vs3

Q.7: (c) (i) Implement the logic function shown below with a static CMOS gate:

Out = ABC ABC ABC [10 Marks]

(ii) A certain counting type 12 bit ADC operates with FSR = 0 to 10 V and clock frequencyfclk = 1 MHz. Determine the dynamic range of the converter, conversion time, conversionrate and Nyquist frequency of the converter. [10 Marks]

Sol: (i) Given: Out = ABC ABC ABC

Simpligying the expression, we get

Out = ABC ABC ABC

= AB C C ABC

= AB ABC

= B A AC




Mains Exam Solution

IES M

ASTER

= B A A A C

= B A C

The circuit for exspression out with static (MOS gate) is shown below:

A C

B

Vad

B C

A

Out

(ii) Given, x = 12, fCLK = 1 MHz

TCLK = 1 sec

Resolution = FS

nV

2 1 = 12

10 2.44 mV2 1

Conversion time, TC = (2n – 1)TCLK

= 122 1 1 sec

= 4.095 m sec

Conversion rate =C

1T

= 244 conversion/sec

Nyquist rate = n1

2 1 T = 244 MHz

Q.8: (a) A shunt generator delivers 50 kW at 250 V when running at 400 r.p.m. The armature andfield resistance are 0.02 and 50 respectively. Calculate the speed of the machinewhen running as a shunt motor and taking 50 kW at 250 V. Given, total voltage drop in thebrushes is 2V. [20 Marks]

Sol:

R = 0.02aR = 50f

If

Ia

I

+

–

E

+

–

V

As a Generator, V = 250 V




Mains Exam Solution



IES M

ASTER

P = VI = 50 kW

I =350 10

250 = 200 AA

E = V + IaRa + VB

If =f

VR =

25050 = 5A

Ia = I + If = 200 + 5 = 205 A

E = V + VB + IaRa = 250 + 2 + 205 × 0.02

= 256.1 V

E = nK N , Speed N = 400 rpm

Constant nk =EN =

256.1400 = 0.64 V/rpm

As a motor :

Rf

If

Ia

I

+

–

E

+

–

V=250V

Electrical input, Pin = VI

50 × 103 = 250 × I I = 200A

If =f

V 250 5AR 50

Ia = fI I = 200 – 5 = 195A

E = V – VB – IaRa

= 250 – 2 – 195 × 0.02

= 244.1 V

Speed, N =n

Ek =

244.10.64 = 381.4 rpm




Mains Exam Solution

IES M

ASTER

Q.8: (b) (i) The linear resistive circuit shown in the figure below has four independent sources.Three of them have fixed value, only one is3 is adjustable:

+–

+–

is–1

vs1

vs2

vOut is3

+

–

Linear resistivenetwork withdependent

sources

The table shows the four sets of measurements taken in a laboratory:

s3 outi mA V V1 062 105 ?? 0

Complete the last two rows of the table. For the data in row 3, find the power deliveredby the current source is3. [12 Marks]

(ii) Consider the three-source circuit of the figure below:

+–

iout

is1 50 150 is2

vs3

Compute Iout using superposition theorem.Sol: (i)

Vs2

Vs1

Vs3

Vs4

N V0

For lienar network

V0 = 1 s 2 s 3 s 4 s1 2 3 4C V C V C I C I

Where, C1, C2, C3, C4 are constants

s s s1 2 4V ,V ,I fixed so

1 s 2 s 4 s1 2 4C V C V C I = C constant




Mains Exam Solution



IES M

ASTER

V0 = 3 s3C C I

For s3I = 1 mA, V0 = 6 V

6 = C + C3 ...(i)

For s3I = 2 mA, V0 = 10 V

10 = C + 2C3 ...(ii)

Using (1) and (2) (C + 2C3) – (C + C3) = 10 – 6

C3 = 4 C = 6 – C3 = 2

So, V0 = s32 4I

For s3I = 5 mA

V0 = 2 + 4 × 5 = 22 V

For V0 = 0

0 = s s3 32 4I I 0.5

For row-3 s3I = 02 mA, V 22 V

5 mA = Is3V = 22 VD

+

–

By s3I s3P = – 22 × 5 = – 110 mW

= – 0.11 W

Power delivered = 0.11 W

(ii)

50 150Is1 Is2

– +

Vs3Io

Using super position.

Only, Is1, Vs3 = 0V(s/c), Is2 = OA (0/C)




Mains Exam Solution

IES M

ASTER

Q.8: (c) (i) For the circuit shown in the figure below, determine the output voltage if the input

3in t 100 sin 2 10 t mV. Assume that the op-amp is an ideal op-amp and

MOSFET parameters are 2n incos 100 A / V , 1 V , W 10 m and L = 2.5 m :

20 k

+– 4V

–

+V (t)0

2V

vin (t)

[15 Marks](ii) Draw a block diagram of a 4 to 2 encoder. Label all inputs and outputs. How is the

4 to 2 encoder different from 4 to 1 multiplexer ? [5 Marks]Sol: (i) NMOS : µnCox = 100 µA/V2 = 0.1 mA/V2

Vt = 1V, W 10 4L 2.5

2n n ox

WK C 0.1 4 0.4 mA / VL

–

+

ID – +

20K

V0

+– 4V

V = 4VP

ID

S+V(t)i

G

D

V = n V = 4P

+– 2V

VD = Vn = 4V ; Vs = –2V

VDS = VD – VS = 4 – (–2) = 6V

VG = VG = 30.1sin t V 2 10

VGs = Va – Vs = 2 0.1sin t

Vas – Vt = 2 0.1sin t 1 1 0.1sin t V

As VDS > VGs – Vt, so NMOS in saturation




Mains Exam Solution



IES M

ASTER

2 2D n Gs t

1 1I K V V 0.4 1 0.1sin t2 2

2DI 0.2 1 0.1sin t mA

Vo – Vn = 20 × ID 2oV 4 4 1 0.1sin t V

2oV 4 4 1 0.1sin t

= 24 1 1 0.2sin t 0.01sin t

= 24 2 0.2sin t 0.01sin t V

3 2oV (t) 8 1 0.1sin2 10 t 0.005sin tV

(ii) The block diagram of a 4 to 2 encoder is shown below :

Y0

Y1

X3X2

X1X0

Inputs Outputs

Block Diagram of 4:2 Encoder

Encoder4:2

0

001

0

010

0

100

0

011

0

101

1

000

X2X3 X1

Truth TableThe block diagram of 4 to 1 multiplexer is shown below :

S0

Y

D 3D 2

D1D0

InputOutputs

S1

4:1MUX.

Enable E

0

011

0

101 D3

D2

D1

D0

Select Table Output

Truth TableDifference : The 4 to 1 multiplexer decodes the input through select lines and allows only one of themto get through to the output whereas in 4 to 2 bit binary encoder only one input is the value of 1 (formtruth table of 4 to 2 encoder) then the corresponding binary code associated with that enabled inputis displayed at the outputs.

ELECTRONICS & COMMUNICATION

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