Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12EE201 Lecture 16 P. 1Capacitors- a new circuit elementA parallel plate capacitor. Between plates is a dielectric material.As plates become charged, a potential difference vC(t) appears across capacitor. The current through the circuit is proportional to rate of change in vC(t).ic(t) = C d vC(t) dt(16.1)where C is the capacitance.ic(t)vs+ + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - E_+ _+_vc(t)EE201 Lecture 16 P. 2Symbol:Units: Farads ( F) 1F = 1 Capacitance is a measure of a capacitor to store energy.The integral form is, +vc(t) _ amp-sec voltt0=t1Cvc (t0) +ic ()dt0Initial voltage across capacitor (MEMORY)1Ct- ic ()dvc(t)= 1C1C-ttoic ()d + ic ()d=Note: vc(-) = 0ic(t)EE201 Lecture 16 P. 3Example: For a source current is(t) = e-t A for t 0, compute vc(t) and vs(t) across the current source for t 0 .Assume vc(0) = 1V.+_ is(t)0.5 F2 vC(t)vR(t)+vs(t)+__Step 1: Calculate vC(t) using integral relationship.1C-tis (t)d =vc (t)1C-0is ()d1Cis ( )d +t0=vc (t)EE201 Lecture 16 P. 4vC(t) = {1 – 2(e-t – 1)} VStep 2: Calculate vR(t) using Ohm’s Law.vS(t) = 2e-t + 3 – 2e-t VvS(t) = vR(t) + vC(t) vC(t)= (3 – 2e-t) VvR(t) = 2e-t VStep 3: Calculate vS(t) from KVL.vS(t) = 3 VvC(t) = {1 + e- d}V 1 0.5 t0EE201 Lecture 16 P. 5Continuity Property of CapacitorsGiven a discontinuous (but bounded) current waveform, the voltage vC(t) across a capacitor is continuous.is(1 -) ≠ is(1 +) t iS(t) vC(t)tDiscontinuousvc(1 -) = vc(1 +) Continuous1 1EE201 Lecture 16 P. 6Principle of Conservation of Charge The total charge transferred into or out of a junction (node) is zero. Applying KCL to a junction with n capacitors, i1(t) + i2(t) + i3(t)+…+ in(t) = 0A consequence of Eq.(16.1) is, q(t) = C vc(t) Integrating KCL equation wrt time yields, C1 v1(t)+ C2 v2(t)+ C3 v3(t) + …+Cnvn(t) = 0because qj(t) = Cj vj(t).EE201 Lecture 16 P. 7Example: For the circuit find vc1(0+) .vc1(0-) = 1 V vc2(0-) = 0 Vvc1 +__vc2+0.75F0.25FConservation of charge:q1(0-)+ q2(0-) = q1(0+)+ q2(0+) q1(0-)- q1(0+) = q2(0+)- q2(0-) C1 [vc1(0-)- vc1(0+)]= C2 [vc2(0+)- vc2(0-)]0.75 [1- vc1(0+)] = 0.25[vc2(0+) –0 ]0.75 - 0.75 vc1(0+) - 0.25 vc1(0+) = 0 vc1(0+)= vc2(0+)= 0.75 Vt = 0EE201 Lecture 16 P. 8And the instantaneous stored energy in a capacitor is,Energy Storage in a CapacitorWc (t0, t1) =pc ()dWc (t0, t1) =vc () ic ()dt1t0t1t0Wc (t0, t1) =12C [vc2(t1)- vc2(t0)]Wc (t) =12C vc2(t)EE201 Lecture 16 P. 9Capacitors in series+_C1CeqC3C2++++____vc1vc2vc3vcicicFrom KVL: vC = vC1 + vC2 + vC3dvC1dt+ dvC2dt+ dvC3dtdvCdt= Taking the derivative gives,dvCndtBut,=iCnCnand,dvCdt=iCCeqCapacitors in series behave like resistors in parallel. For two capacitorsCeq=1 1/ C1+ 1/ C2 + …+1/ CnCeq= C1 C2C1 + C2Ceq=11/ C1+ 1/ C2 + 1/ C3 =iC1C1+iCCeqEE201 Lecture 16 P. 10iC2C2iC3C3+From KCL: iC = iC1 = iC2 = iC3=1C1+1Ceq1C21C3+= Cnn1For n capacitors in seriesEE201 Lecture 16 P. 11Capacitors in parallelC1+_+_Ceq+_vcC2+__Cn+icicicnic1ic2vc1vc2vcnFrom KCL: iC = iC1 + iC2 + … + iCndvC1dt+ dvC2dt+…+ dvCndtdvCdt= Using differential form of I-V relationship,But, from KVL: vC = vC1 = vC2 = vCnCeqC1C2CnCeq=N capacitors in parallel C1+ C2 + … + CnCapacitors in parallel behave similar to resistors in seriesEE201 Lecture 16 P. 12Series-Parallel CombinationsExample: Find the equivalent capacitors+_9 mF18 mF9 mF5.4 mF30 mFStep 1: Reduce 9mF//18 mF to equivalent 9mF + 18mF = 27mFStep 2: Combine series capacitances1/ Cs= 1/9 mF + 1/27 mF + 1/ 5.4 mF = 6/54mF + 2/54mF +10/54mF = 18/54 mF Cs = 3mFStep 3: Combine parallel capacitorsCeq = 30mF + 3mF = 33
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