EE201 Lecture 16 P 1 Capacitors a new circuit element ic t vs vc t E A 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 where C is the capacitance 16 1 EE201 Lecture 16 Symbol ic t vc t Units Farads F 1F 1 P 2 amp sec volt Capacitance is a measure of a capacitor to store energy The integral form is 1 t vc t ic d C 1 to ic d C 1 vc t0 C Note vc 0 1 t ic d C t0 t ic d t0 Initial voltage across capacitor MEMORY EE201 Lecture 16 P 3 Example 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 vs t vR t 2 0 5 F vC t Step 1 Calculate vC t using integral relationship 1 t vc t is t d C 1 0 1 t vc t is d is d C C 0 EE201 Lecture 16 P 4 t e d V vC t 1 0 5 0 vC t 1 2 e t 1 V 1 vC t 3 2e t V Step 2 Calculate vR t using Ohm s Law vR t 2e t V Step 3 Calculate vS t from KVL vS t vR t vC t vS t 2e t 3 2e t V vS t 3 V EE201 Lecture 16 P 5 Continuity Property of Capacitors Given a discontinuous but bounded current waveform the voltage vC t across a capacitor is continuous iS t vC t 1 is 1 is 1 Discontinuous t 1 t vc 1 vc 1 Continuous EE201 Lecture 16 P 6 Principle 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 0 A 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 0 because qj t Cj vj t EE201 Lecture 16 Example For the circuit find vc1 0 vc1 0 1 V P 7 t 0 vc1 0 25F 0 75F vc2 0 0 V Conservation 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 V vc2 EE201 Lecture 16 P 8 Energy Storage in a Capacitor Wc t0 t1 t1 pc d t0 t1 Wc t0 t1 vc ic d t0 Wc t0 t1 1 2 C vc2 t1 vc2 t0 And the instantaneous stored energy in a capacitor is Wc t 1 2 C vc2 t EE201 Lecture 16 P 9 Capacitors in series ic ic C1 vc1 C2 vc2 C3 vc3 vc Ceq From KVL vC vC1 vC2 vC3 Taking the derivative gives dvC dt But dvC3 dvC1 dvC2 dt dt dt dvCn iCn dt Cn dvC iC and dt Ceq EE201 iC Ceq Lecture 16 iC1 C1 iC2 C2 P 10 iC3 C3 From KCL iC iC1 iC2 iC3 1 Ceq 1 C1 Ceq 1 C2 1 C3 n 1 Cn 1 1 C1 1 C2 1 C3 Capacitors in series behave like resistors in parallel For n capacitors in series 1 Ceq 1 C1 1 C2 1 Cn For two capacitors Ceq C1 C2 C1 C2 EE201 Lecture 16 Capacitors in parallel ic vc1 ic1 C1 vc2 ic2 C2 vcn P 11 ic icn Cn vc Ceq From KCL iC iC1 iC2 iCn Using differential form of I V relationship dvC2 dvC dvCn dvC1 C Ceq C2 1 Cn dt dt dt dt But from KVL vC vC1 vC2 vCn N capacitors in parallel Ceq C1 C 2 C n Capacitors in parallel behave similar to resistors in series EE201 Lecture 16 P 12 Series Parallel Combinations Example Find the equivalent capacitors 9 mF 9 mF 30 mF 18 mF 5 4 mF Step 1 Reduce 9mF 18 mF to equivalent 9mF 18mF 27mF Step 2 Combine series capacitances 1 Cs 1 9 mF 1 27 mF 1 5 4 mF 6 54mF 2 54mF 10 54mF 18 54 mF Cs 3mF Step 3 Combine parallel capacitors C 30mF 3mF 33 mF
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