HW2solns

# HW2solns - Prob1 k Vo V s For simplicity we will first find...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Prob1) k Vo / V s For simplicity we will first find V1/Vs. 1 1 1 10000 j10000 V1 10000 j10000 1 j j100 100 j100 100 1 10000 j10000 Vs 20000 j10000 2 j 100 100 1 1 100 j100 100 j100 Then we can find Vo/V1: Vo 100 1 V1 100 j100 1 j And we have: V0 Vo V1 1 1 j 1 j 0 .2 0 .4 j Vs V1 Vs 1 j 2 j 3 j So we have found the phasor gain(K) k 1 j 11 1 1 Arc tan Arc tan .44 630 3 j 9 1 1 3 Z in And for Zin we have: 1 1 1 j100 100 j100 100 200 100 j The input voltage would be divided due to the elements impedances: Vo ( Z (C ) || Z ( R )) Vin ( Z (C ) || Z ( R)) Z ( L) Z ( L ) j L 1 Z (C ) jC Z ( R) R Z ( R) || Z (C ) 1 1 j C R R 1 jRC R 1 j RC Vin Vo R j L 1 j RC Vo R Vin R RLC j L 2 R L Vo Arc tan Vin 2 ( R 2 RLC) 2 (L) 2 R RLC Vo L V Cos t Arc tan 2 2 2 2 0 R RLC ( R RLC ) ( L ) R A ) I D1 I D 2 We know that: V D1on V D 2 on .7v so we have: V1 VD1on .7v , V2 V1 VD 2 on 0 then we have: IR V2 (2) 2 R R 2 .8 K 2.5mA So for making I D1 I D 2 we should have I D 2 2.5mA and R B ) I D1 0.2 I D 2 Since I D1 I D 2 5mA then we should have: I D1 .83mA and I D 2 4.16mA Again V1 and V2 would be the same as part A and we have: R 2v .48 K 4.16mA C ) I D1 5I D 2 The same as part B but now we have I D 2 .83mA and I D1 4.16mA R 2v 2 .4 K .83mA Prob4) For the circuit at right, show that the output voltage out V is proportional to the natural logarithm of the input voltage. Assume that the input voltage is always positive such that the diode operates well into forward conduction where: I d I seVd / nV T Soloution: IR ID Vd Vout ( Vin , I D I s e nVT ID R Vout ) Vin I se R ( Vout ) nVT Vout nVT ln( Vin ) R.I s Prob 5) A ) RL=4K Vz=3.9v IL V z 3 .9 v 0.975mA R L 4k And by KCL we know that I z I R1 I RL I R1 20v 3.9v 1.34mA so I z 1.34mA .97mA 0.37mA 12k And the power dissipation of the zener diode: Pz I z Vz 0.37mA 3.9v 1.44mW B ) RL=10K Similar to part A we have: IL Vz 3 .9 v 0.39mA R L 10k I z I R1 I RL I R1 20v 3.9v 1.34mA and I z 1.34mA 0.39mA 0.95mA 12k Pz I z V z 0.95mA 3.9v 3.7mW Prob 6) Vz1=2.3 Vz2=5.6 Vd=.7 This problem has a point, the right branch of the circuit would have current only if we have Vo (0.7v 5.6v) 6.3 which is the sum of Vdon and Vz and similarly the left branch would have current only if we have Vo (0.7v 2.3) 3 . otherwise D1 or D2 would be off and that branch would be open circuit. For 10 Vin 6.3 We have: right branch has current and left branch is open circuit IR 0 Vin 6.3 0 Vin 6.3 R2 R 2.5k V 6.3 V 6.3 4Vin 6.3 Vo Vin R I R Vin 0.5k in Vin in 2.5k 5 5 Which would result to 9.26 Vo ‐6.3 For 6.3 Vin 3 Both of branches are open circuit due to the reason mentioned and so Vo Vin For 3 Vin 10 The right branch is open circuit and the left branch has current IR Vin 3 Vin 3 R1 R 1.5k V 3 V 3 2Vin 3 Vo Vin R I R Vin 0.5k in Vin in 1.5k 3 3 Which would result to : 3 Vo 7.66 Prob 7) Here we know that I c I R I L for all the times. And from KVL we have: Vc R I R L dI L dt Also we know that: I c C dVc notice that – sign is because during time the voltage of C is dt decreasing. Then we have: dVc d 2Vc Vc R C ( ) L C ( 2 ) dt dt S= RC R 2 C 2 4 LC 2 LC s1 1000 j1000 s 2 1000 j1000 Vc k1e s1t k 2 e s2t For finding K1 and K2 we need 2 initial conditions. 1‐in t=0 we have Vc(0)=10 2‐ in t=0 we have dVc(0)/dt=0 Notice that 2 is true otherwise there would be a current in t=+0 in circuit and so we would have a voltage on R and that violates the KVL(since the voltage on C is still 10v in t=+0) Then we have: k1 5(1 j ) k 2 5(1 j ) And also for IL we have: dVc I L I c (C ) C ( s1 )k1e s1t ( s2 )k 2 e s2t dt The response would be underdamped and results are shown bellow. Prob 8) Here we have these two conditions for voltage and current in all times: 1)Vc VR VL 2) I c I R I L And also we know: Ic C dVc dI Vc ,IR ,VL L L dt R dt dVc Vc dt R Form 2 we have: I L I c I R C So we have: Vc V L L dV V d d d2 Ld I L L C c c LC 2 Vc Vc dt dt dt R R dt dt LCs 2 L s 1 0 R We can find s as: s 1 500 j 3122 s 2 500 j 3122 Vc k1e s1t k 2 e s2t Again like the prob7 we need 2 initial(or boundary) condition for finding K1 and K2 In t=+0s we know that C has the voltage equal to 60v so we have: Vc(0)=60 so K1+K2=60. And also we know that in t=+0 the inductance (L) don’t have any current(why?), sine if it has a current d I would be infinite and that’s not possible since VL Vc finite then its dt L So in t=+0 we have current only in C and R and I c I R or C From it we have: k1=30‐9603i k2=30+9603i d Vc Vc R dt And we have V c k1e s1 t k 2e s2t and IL 1 k1 s1t k 2 s2t (e e) L s1 s2 Then we can find and plot Vc and IL And as we can see the response in underdamped. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online