mA Volt

Transcription

1 etaile solution of IS 4 (C) Conventional Paer I Sol. (a) (i) Conuctivity is a egree to which a secifie material conucts electricity an gives iea how much smooth flow is of electricity by a carrier. Mobility is egree to which secifie material can move freely an easily an it gives iea about ability of movement of a carrier. (ii) Zener break own occurs in highly oe Zener ioe an it is ue to tunneling henomenon while Avalanche breakown occurs in lightly oe Zener ioe an it is avalanche multilication ue to successive collisions of electrons in eletion region of Zener ioe. Zener occurs at smaller value of break own voltage while avalanche occurs at higher value of break own voltage. (iii) Piezo-electric materials which are insulators become electrically olarize in resence of mechanical stress an rouce voltage which is reversible rocess. Ceramic materials are in organic materials an are combination of metal an non metals which are generally forme by action of heat an subsequent cooling. (iv) irect ban ga: k Minima of C.B coincies with Maxima of VB an here energy is emitte in form of light by hotons Inirect ban ga: k Minima of C.B oes not coincies with Maxima of VB an here energy is emitte in form of heat. (v) Polarisability is the ability of molecules to be olarize an exress as iole moment er unit electric Fiel. Permittivity is the measure of resistance which is encountere when forming an electric fiel in a meium. Unit of olarisability is F-m while ermittivity has no unit. Sol: (b) (i) Given conition that NMOS is in Saturation region I K V V VGS.V &VT V GS T Then I. 3. =.96 ma which is nearly equal to ma I (ii) g KV V m GS T Vgs 3 gm.. =.76mA / V (iii) If V I = mv then new V GS =. Volt so transistor will still remains in saturation region Sol.(c): 3 = I K V V GS T...956mA Outut voltage V V RI 9.956mA 7.Volt W Vj q NA N Since NA N so N N A

2 So here W Vj qn A But qn so W A V j Sol.() : To synthesize a riving oint immittance function z(s) the first ste is to ecomose it into a sum of simler ositive real functions z (s), z (s), z 3 (s), z 4 (s),. z n (s), an then to synthesize these iniviual z(s) as elements of the overall network whose riving oint imeance is z(s). A function is sai to be ositive real function if it satisfies the following conitions:. F(s) is real for real s i.e F(σ) is real. F(s) may have only simle oles on the jw axis with real an ositive resiues 3. ReF(jw) for all w 6 s 3 s Zs 6 s s 6 s (s 6) Sol.(e): t t h t e ut, xt e ut Xs.H s Y s Ys s s t t y t e u t e u t Sol.(f): when clock is high comlete circuit resons similarly to an OPMAP in unity gain feeback configuration when clock is low inut voltage at that time is store on caacitor. By use of OPAMP in feeback loo inut imeance of samle an hol circuit is greatly increase. Figure: CLK + V in C Hol V out Sol. (g): Proagation constant (P) R jlg j C As frequency is not mentione so roblem can t be solve Sol. (h): It is a transucer which uses change in the electrical resistance to measure strain. Here electrical resistance is roortional to instantaneous satial average strain over its surface. Alications :. Vibration measurement. Comression an tension measurement 3. Contractions in muscles in meical science 4. Bloo ressure measurement 5. Use in volumetric ifferential low ressure

3 Temerature comensation in strain gauge: Active gauge Force R ummy or comensating gauge R The active strain gauge is installe on the test secimen while the ummy or comensating gauge is installe on a like iece of material an is not secifie to strain. Sol. 3(a) As n n i n N N but A NA n N n n N n N N 4n n i But n cannot be negative so N n As i i n N N ni 4 ni n N N n i n i can be neglecte. Sol. 3(b) V N N 4ni N neglecting ni N n i V + + 5V 3

4 is R.B an Non conucting is F.B but Non conucting is not in breakown as V B = 5 V but 5 V V Both ioes are in series, in reverse current will flow from N to P while in current will flow from P to N so here I I V VT I I e V VT n = mv=.3466 Volt So here V 5 V V light Sol. 3(c) t n tye S.C x < x > x = cm since it is N-tye SC so major change will be in concentration of holes only an not electrons. Here given So t 3 HP / cm / sec 6 5 / cc n here, x = 34.6 m here; Initially in N tye s.c, holes are minority an only contribution which is ominant is after the following of light. So x = 34.6 m J A/m 3 An I J area I Now, Initially Now Then n ma n N 5 3 n / cm 4.5 /cm 5 3 Change in e is n ' / cm Thus Alying continuity equation. Now, Jn q x Jn am / m I J A n In n 9.376mA 3 4

5 Sol. 3. () VCC V, VC V, V 6V, S,, IC 5mA R R V cc R c R R C V CC V T R e R R Now V Sol. 3.(e) s R R R RI I As C B R.k I ; ' ' is very high Then V V I R R RC.k CC C C C s R R R B TH R.365k RR Now,.365k R R R R R S V V R R I I R T B R Then; R R.349 RR R R.365 R 3.545k R 7.46k B Initially T is switch off then caacitor get charge to V in steay state. So caacitor is charge to Volt. at t= caacitor voltage will remain at volt Now T becomes ON by 4 volt Here V S = Volt an V GS =4 Volt so MOSFT will be relace by resistance then r s on g r g 3 where m m g K V V 5 ma / V m GS T rs 5 Now r s an C will be in arallel an caacitor will ischarge through r s ma / V t/ t/ V(t) V e 5 e where 5 s t ln 5.693s 3466 sec 3.47 n sec 5

6 Sol. 4(a) Sol.4(b) Z transform offers significant avantage relative to time omain roceures. By this we can moel iscrete time hysical systems with linear ifferential equations with constant coefficients one examle is linear time variant igital filter. The Z transform of a ifference equation gives us a goo escrition of the characteristics of the equation an hence of hysical system. In aition transforme ifference equations are algebraic an therefore easier to maniulate. k The Z transform of a samle signal or sequence is efine as : Ys Ys Ys s s s 3s s 3s s s t t y t e u t e u t ( There is misrint in this question) Z f (kt) f (kt)z k Sol.4() B V N only ucate B both V only voters N None Probability of eucate = 4/ Probability of both = Probability of voters = 5 B V N % V B 4% B % B 5% = 3% V = % N= 3% (i) PVoter (ii) / ucate 4 / Not eucate / P Voter 5 / 5 6

7 Sol.4(e) Sol.5(a) (iii) 3 3 P N By uality theorem F f f t F t sgn t j sgn jt s sgn jt s V I I Sol.5.(b) Alying KVL in loo / s I si I I () I,I are in 's' omain Alying KVL in loo si I I I I s I Put () in () I / 3 5 / 3 5 s s 3 s 3t t i t e 5e u t.() It is a arallel arallel combination hence total Y arameter will be ae. [Y] [Y] Y Y Y 7

8 sol.5(c) / 3 / 3 Y / 3 / 3 / 3 / 3 Similarly Y / 3 / 3 4 / 3 / 3 / 3 4 / 3 So Y Y Y So over all Y arameter of above network is : R R R I + + I I + U Y + L Y = Sol. 6 (a) V R 3 V R R I R I ----() 3 3 V R R I R R I ----() 3 3 We gate Z R R, Z R V 3 3 Z R R, Z R R Z As Z 3 3 Z so network is not recirocal. R R3 R3 R R3 R R 3 5 Charge ensity c / m 5 q xx q 4 x 6 x5 x x 4 x 6. N / C F Q. 5. F 9N Sol. 6(b) x 6 6 Q Q V a h h where Q 4 9 V a h h Volt x x

9 where r 5m 9 Q 4nC 4 C Potential V Q 4 r V 9 V 7V 5 Sol.6(c) 5m, Z 3 3m z 5 f 5MHz c 3 6m 6 f 5 6 Length of quarter wave line = =.5m 4 4 Characteristic imeance Z Z.Z 35 Z.3 Z Z Reflection coefficient Z Z VSWR 3 Sol.6() VSWR a b =.3 cm. cm Oerating frequency f =.5 f c Guie wave length g =?, Phase velocity v =? As So a b first five moes will be T, T, T,T an T Cut of frequency f c r r c m n a b For T moe c fc T.3 f.5f c GHz c cm 9 f 9.7 = 6.5 GHz 9

10 g 3.67 fc f c f.5fc 3.67 g 4.cm.7453 v f g 9 v m / s v 4.4 m / s Similarly we calculate guie wavelength ( g ) & hase velocity (v ) for next moes g Moe Guie wavelength Phase velocity v T 4.cm 4. m / s T.6cm 4. m / s T.79cm 4. m / s T.64cm 4. m / s T.9cm 4. m / s Sol.7(a) (i) Absolute error= Actual value-measure value =-= ma (ii) Percentage error = Absoluteerror % %.% Measure value Actual value (iii)relative Accuracy = Absoluteerror Actual value (iv) Percentage accuracy =9% =.9 (v) Precision= Sol. 7(b) Total eflection in beam is given by ev L L ms v where eva v m Here s is searation between eflecting lates an s=5mm Here is istance of screen from centre of lates an =3 cm Here L is length of each eflecting late an L= cm Here =3 cm an V shoul be calculate ev v m 9 9 ev L L ms v 9 a 5 3

11 V L L s V a V Solve V from here an that will be answer : Sol. 7(c) Given R 4k, R4 Sol.7() RT R T ex T T R R ex So er º K R R ex Now 5 R An for R R R ex R So range of resistance 569. to Given, A.5cm.5 m Searation 4 3mm 3 3 m 4 P N / m, eflection =.3 mm = C = 3 PF Caacitance after a ressure of 4 N/m A C A 3PF 3 3 A 3PF3 C' x 3.3 C' PF m