Conventional Paper-I Solutions:(ECE)

Transcription

1 Conentional Paper-- 13 olutions:(ece) ol. 1(a)(i) n certain type of polar dielectric materials permanent electric diploes are oriented in specific direction een in absence of external electric field. o these materials hae polarization in absence of electric field. Adantage of Ferro electric oer dielectric is that Ferro electric has ery high alue ( ) of dielectric constant in compare to normal dielectric. ol. 1(a)(ii) water is example of a dielectric and has ery high relatie permittiity alue equal to 8 while example of Ferro electric is BaTiO 3 which has alue of relatie permittiity of the order of. ol. 1(b) ilicon is preferred oer Germanium for a fabrication of P-N junction transistors because: (i) Higher Energy gap in i than Ge. (ii) Temperature limit of i is 15 C while in Ge it is only 1 C (iii)eakage current is of the order of na in case of i while it is order of micro amp in Ge (i) i has higher break down strength than Ge () Oxide of i is io is insulator which is used in fabrication for isolation. (i) i is aailable in abundance and is cheaper than Ge (ii)elatie cost of electronic grade is 1 in i while for Ge it is high. ol. 1(c) Fourier transforms of Autocorrelation function is power spectral density. XX XX (w) PD e d Power spectral density of j j e e d j j e e d e e d 1 1 j j XX ol.1 (d) Apply to KC at (1) & () XX XX Power spectral density V1 V1 1 V V 4 (1) V1V 36 Because of super node concept () V1 3V1 36 V 4V V 5V 6 (3) 1 V V 36 (4) 1 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

2 V V 1 4V 1V Q1 Q ol. 1(e) 1 & ole charge density for both. et r is the distance at which potential is 5 Volt. (n spherical it is assumed that whole charge is placed at centre of cell). et r is the distance at which potential is calculated then Q1 Q 5 4 r 4 r Q & Q arealready known 1 c ol.1 (f) Cut-off frequency for parallel plate air filled waeguide fc a.m c 1 a 1cm Gien & c 3 1 cm f fc GHz.5GHz For Dominant mode: 1 Any signal those frequency is more than.5 GHz is passed through parallel plate wae guide only 1GHz is passed through to parallel plate wae guide.1ghz & 1GHz are not passed through the parallel plate waeguide. ol. 1(g) t will be simple circle. t is ery easy question ol. 1(h) (a)(i) et V 1 is the oltage at inerting terminal Apply to KC at node (1) 1 i 1 o ol. 1(h) (a)(ii) ol. 1(h) (b)(i) that s a circuit inerted to input signal. i 1 switch is OFF. KC at node (1) 1 o 1 i i i when 1 is OFF & ON i 5sin1 t ; That s a circuit act as oltage follower. 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

3 ol. 1(h) (b)(ii) ol. (a) when 1 is OFF & OFF 1 i 1 o i 1 5sin1 t when 1 is ON & is ON o When 1 ON & OFF 1 i i 5sin1 t Electrical equialent of quartz crystal C M C s = eries inductance which represents Mass of quartz crystal s = eries esistance which represents the damping ratio of Quartz crystal. C(C )= eries capacitance which represents the spring const. C M (C P ) = Parallel capacitance which represents to electrostatic capacitance between two plate of quartz crystal s series frequency fs sc scm p Parallel frequency fp CsC p Cs Cp s s Cs Cp Cs Cp 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

4 ol. (b) Gien data C s = eries capacitance = 1 F C p = Parallel capacitance = 1 F s = 1 mh = equialent inductances 1 1 f s = 53.9Hz 6 3 C s s f p = 1 CC s p s C s C p =711.76Hz CC where 51 F C C s p 6 6 s p ol. (c) Thyristor (also called silicon controlled rectifier or C) consists of alternate p and n layers (i.e., p- n-p-n) forming three p-n junctions. The anode terminal is outside the p layer. A contact welded to inner p layer (i.e., p) forms the gate. ts symbol is as shown in Figure. 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

5 V- Characteristics of Thyristor: The -i characteristic of a thyristor is shown in Figure. The behaior of a Thyristor can be explained by two transistor model as shown in figure. Thyristor can be turned on by applying a positie gate signal. Other triggering methods are d/dt triggering, high temperature triggering and light triggering. when a thyristor is triggered the gate loses control. t can be turned off by decreasing the current to less than holding current. The turn on time of thyristor is less than about 3 µ-s and turn off time is between 1-3 µ-s. ol. (d) Direct band gap semiconductor: n this type of semiconductor there is no change in alue of k and hence no change in momentum and kinetic energy during excitation from CB to VB. ince no change in momentum and kinetic energy and hence these particles will be photons and energy will be emitted in form of light. Examples are GaAs(Emits light in ),GaAsP(Emits light in isible region). ndirect band gap semiconductor: n this type of semiconductor there is change in alue of k and hence there is change in momentum and kinetic energy during excitation from CB to VB. ince there is change in momentum and kinetic energy and hence these particles will emit energy in form of heat. Examples are Ge&i. ol.(e) ilicon diode cannot be used in ED because ilicon is indirect band gap emiconductors which can t generate light or photon and it releases heat during transition from conduction band to alance Band.ED will generate light only if material is direct band gap semiconductor. i can be used in photo-diode because photo diode is light detector and when light is falling on photo diodes then new EHPs are created and alue of reerse current is increased in proportion to falling light. 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

6 ol. (f) max for silicon photo diode 1.4 m E e Eg e Energy gap = 1.11 ev(at room temperature) for max g. 1.4 max 1.1 m 1.1 Q (in meter) Now responsiity () 1.4 For Maximum alue of Quantum efficiency should be 1% so max =1.1% for i. ol. (g) The basic idea behind magnetically leitated train is there is no physical contact between train and rail. ince electron moes efficiently through a super conductor because there is no resistnace.n similar manner MAGEV (Magnetically eitated) trains moe more efficiently because there is no friction between wheel and track. Trains can be made to "float" on strong superconducting magnets, irtually eliminating friction between the train and its tracks. f Electromagnets are used then size will be high and they will waste much of electrical energy as heat. uper conductors are used for generation of strong magnetic fields by use of Meissner effect. uperconductors used in MAGEV systems hae been of low temperature ariety for this they must operate below iquid Helium temperature (4. K). ol.3 (a) s Z 5 15 But min Zmax Z Zmax min so 3.75K 4max Z ma 5mA 35mA Zmin max 3 max max max 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

7 ol.3 (b) (a) 11V (b) rms D is diode resistance = 1K V rms D ma Peak load current ( m ) m (c) DC load current 11 A 15.49mA mA (d) AC load current rms alueof load current 1 m (e) Vm m( D ) DC diode oltage =59.54Volt ol.3(c) +6 3 =76.45mA When B 1.4K 1 3 C E 6 Apply KV from 6V to 6V in left most T x ince C E C 1mA C m amp 11.4 E E C For other transistor: (For calculation of 1 ): 1= (Base current is neglected) o 1 = =4.3mA For other transistor: (For calculation of ): 1= (Base current is neglected) 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

8 o = =3.5mA For other transistor: (For calculation of 3 ): 1= (Base current is neglected) o 3 = =.6mA ol.3 (d) Value of oltage gain for Current source load inertor is: Gain K n 1 ref 1/ K W 1 n. D 1 1A / V W 1m 1m 1 1A ol.4 (a) Energy of signal x t dt at x t e u t Gien Energy of signal e at 1 dt a 1 x t X d Energy of signal x = Fourier transform of x(t) 1 x t is e e dt a j at j Fourier transform of 1 x( ) a According condition gien in question: x d Gien a d a a tan 1 a a a 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

9 ol.4 (b)(i) ol.4 (b)(ii) tan tan a a a a 1.95 tan tan(85.5) 1.7 x1 t x1 t x1 t j x t 1 x e by using time shifting property j x t 1 x e j x t 1 x e x t 1 x e j j j x t 1 x t 1 X e X e X cos ol.4 (b) (iii) ol.4(c)(i) x t x 4t 5? 5j x t 5 x e By using time shifting property 1 x 4t 5 X e 4 4 x 5j 4 t d x t 1 3 x t d x t dt dt X j X By using time scaling property By using time differentiation property d x t 1 j e j X dt 4t y t 3e u t & x t u t h t? Taking aplace transforms on both sides. y(t) x(t) h(t) Y s X s H s Ys or Hs Xs st Y s e y t dt Xs 1 s 4t st 3e e dt 3 s 4 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

10 Hs 3s 3 s 4 4 s 4 s 4 4 Hs 31 s 4 4t h t 3 t 4e u t ol.4(c)(ii) y t? x t Hs t e u t 3s s 4 xt *ht y t XsHs Y s Gien data st t 1 x s x te dt e u tdt (s 1) x s Ys 1 & Hs 3s s 1 s 4 3s s 4s 1 4t t y t 4e u t e u t 4 1 s 4 s 1 ol.4(d) qn x n k qn 1 (i) 4 yn qn qn 1 k ii 5 Taking to z-transforms of equation (i) k Q z x z Q z z 4 1 Q z x z kz iii Take to z-transforms of equation (ii) k Y z Q z Q z z 5 1 k Y z Q z 1 z i 5 1 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

11 k 1 X z Yz 1 z k 5 1 z 4 1 For BBO stability: k 1 4 ol. 5(a) Maximum Power transfer theorems: According to maximum power transfer theorems power transferred to load resistance is maximum when load resistance is equal to power source resistance. s Power transferred to where V Power transfer to V Differentiate with respect to dp d V 4 For maximum or minima dp d o At this cond n either maxima or minima will take place so to find out dp V d dp V d V V Put 3 dp d 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

12 V V dp d is negatie o at P is maximum alue hence one can proe aboe condition. ol.5 (b) f is fixed and is aried that s a formula is not a maximum power transfer condition to a load resistance. s Power to load resistance = where V / V Power to load resistance = For aried differentiation of P with respect to P. dp V d 4 For maxima or minima o Either or oad can t be zero not dp d and is also not possible because source resistance cannot be negatie. o that = is not the condition for ariable for maximum power transfer to load resistance. ol.5 (c) j1 A B j j1 For determination of alue of Z for maximum power transfer to Z open Z & find out Z TH across it. Here j1 and 1 ohm will be in parallel and resultant will be in series with j1 and then finally in parallel with j ohm. Zth j 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

13 For maximum power transfer to Z Z Z* TH Z j ol.5 (d) 3V x i i x Apply the KV in oop (1) & oop () KV in loop (1) 1i 5 i i i 5i (i) i i x 1 3 x x x 5 i i 1 1 1i 1i ii 1i 1i 1 1 i i1 iii 5i 5i 1 (A) tandard equation of h parameter models are: h i h h i h i1 5i1 1 1i 1 (B) By taking equation (A)& equation (B) 1 h11 1 ; h1 1 h1 1; h 1 1 h11 1 ; h1 1 h1 1; h 1 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

14 ol.6 (a) Just calculate the field due to power lines on telephone lines and then flux and then oltage difference. ol.6 (b) n = 5 Turns B=?? 6cm 4cm n = 1 Turns =?? A Magnetic field intensity generated by coil A at centre is: N 11 =1.5A/m H1.r 4 Magnetic field intensity generated by coil B at centre of coil A is: H N a 56.(r a ) (6 1 ) 3/ 3/ A/m f total field at the centre of coil A is zero then H 1 +H = Now sole alue of and it should be 33.5 Amp. ol.6(c) F B Ba N Value of force is Now alue of work done is: W b a F.dl W B a.rda W B.rd B r J Time taken for N reolution is 6 sec then for one reolution it would be 6 sec. N W Br NBr o Power = P N t 6 3 NBr By putting the alues P.34 Watt f P=1.5 W then NBr 53.14B B 3.89 T ol.6 (d) Equation of E is E Ecos(wt z)ax Equation for H H cos(wt z)a y 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

15 Where E H j j 6 7 j j j j Further you can sole it calculate magnitude of H ol.7 (a) This is standard and basic question: 4 ig ig V ig ig gain V V 1 V By concept of wheat stone bridge (easy deriation) V V 4 1 Now by use of OPAMP V V1 1 V A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,

16 ol.7 (b) 5 P 1K + V x 1K o 1K + o 5mV Value of x x 5m 1k A V V 3mV mV Now p.5 A 1k f optical power is P in and esponsiity of photo diode is then Then photo current P related to incident optical power is P.P where in Qmeter Now A/W.69 A/W But o Pin A.5A.69 P W P in 3.6 Watt ol.7(c).5a W.69A P ext Q VG.71.43V 1V o can besoled Now Apply KV equation: in = V CE +1 Here V CE alue will be low so output will be at logic low leel 8A/11, Ground Floor- Jia arai New Delhi 16 Ph.: ,