A capacitor consists of two conducting plates, separated by an insulator. Conduction plates: e.g., Aluminum foil Insulator: air, mica, ceramic, etc

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1 3//7 haper 6 apacors and Inducors Makng preparaon for dynamc crcus, whch hae far more applcaons han he sac crcus we hae learned so far. 6. apacors Sore energy n elecrc feld nsulaor onducng plaes A capacor consss of wo conducng plaes, separaed by an nsulaor. onducon plaes: e.g., Alumnum fol Insulaor: ar, mca, ceramc, ec q rcu q No drec exchange of charge beween he plaes Elecrons moe from one plae o anoher a he res of crcu Same amoun of pose charge and negae charge: q Basc Propery: The olage s proporonal o he charge q. The capacance: rcu Symbol Measured n Farads (F) q= =q/ Fxed capacors Varable capacor

2 3//7 apacors are passe elemens, assgned by passe sgn conenon Basc propery: Wha s he relaonshp beween and? Take derae on boh sde of, dq d = d = dq = urren s proporonal o he me derae of olage If curren s gen, d = d = () = ( ) Assume, () - d = () = ( ) Power delered: d p= = Energy sored: d w() = p = - - d = ( ()- ( )) () - w Snce =q/, q w=

3 3//7 Imporan hngs o remember:. A capacor s an open crcu under D conon Under D conon, = No curren hrough. The olage on a capacor canno change abruply. As a funcon of me, () mus be connuous () = ( ) () s fne, negraon mus be connuous 3. An deal capacor does no dsspae energy 4. A real, non deal capacor has a large leakage ressance e.g., MΩ In summary: d = ; () = ( ) Example: Gen = F, () = 6e 3 ma, ()=V, ()=? In compuaon, eeryhng should be n sandard un. = F= 6 F, () = 6e 3 3 A, () = ( ) -3-3 = -6 6e -3 3 = 3e -3 = -e -3-3 e ( ) 3 e V 3

4 3//7 Example: Fnd and under D conon 6A 3Ω Ω 5Ω 4Ω 6A 3Ω Ω 5Ω 4Ω Under D conon, capacors are open crcu. No curren hrough 5, no olage drop. same as olage across 4. Also, Ω 4Ω are n seres. By curren dson, =6 (3/9)=A, =4V, =8V 6.3 Seres and parallel capacors Fnd equalen capacance Parallel connecon: = eq d eq Wha s he relaonshp beween and? By KVL, same olage across, By propery of capacor: d d = ; = By KL: d = =( ) eq = 4

5 3//7 Seres connecon: By KL, same curren hrough, By propery of capacor: = eq d eq () = ( ) () = ( ) By KVL: ()= () ()= ( ) ( )( ) = eq = eq 6.4 Inducors sore energy n magnec feld Inducors are usually consruced as a cylndrcal col wh many urns of conducng wre around a core e.g., ferromagnec core Also a passe elemen: curren and olage assgned accordng o passe sgn conenon. 5

6 3//7 Basc propery: d = L Volage proporonal o he me derae of curren L : Inducance, measured n Henry (H) a consan dependng on he srucure and maeral I can be dered: () = ( ) L rcu Symbol: Energy sored: w = L Assume Varable nducor Imporan hngs o remember:. An nducor s a shor crcu under D conon Under D conon ()=cons, = / No olage drop. The curren on an nducor canno change abruply. As a funcon of me, () mus be connuous () = ( ) L () s fne, negraon mus be connuous 3. An deal nducor does no dsspae energy 4. A real, non deal nducor 6

7 3//7 6.5 Seres and parallel Inducors Seres connecon: By KL, same curren hrough L, L By propery of nducor: L L L L d d ()= ; ()= ; By KVL: d ()= () () = (L L) Equalen nducance: L eq =L L Parallel nducors: By KL: L L = L L L eq () = ( ) ( ) By KVL, same olage across L, L By propery of nducor: () = ( ) L ( )= ( ) L L L ( )( )( ) L = eq L L 7

8 3//7 Example: Fnd capacor olage and nducor curren under D conon Inducor shor 4Ω L D conon: apacor open 4Ω L 3V 6Ω 3V 6Ω = = =3/(46)=3A =6 =8V Example: Fnd c and L under D conon. A H _ 5mF Under D conon, Inducor = shor, capacor = open Use source ransformaon: 3 8 8V 7 8 _ A _ 8 7//

9 3//7 Applcaons: Power sysems dren by baery/supercapacor hybrd energy sorage deces, by Hoeguk Jung (Ph.D suden) Archecure of Elecrc Vehcle 9

10 3//7 Rpple reducon n wo sage nerer: expermen seup Rpple reducon n boos conerer nerer Mosfe6 g m D.8 g S. R RL.5RLg..5 R4 RLg Mosfe Mosfe5 g R3 m D S D g D.6.4 Mosfe3 m S Mosfe4 m S D V m g Mosfe m g S D D S

11 3//7 Pracce : For a capacor wh =F, gen () = 6sn4 A, ()=V. Fnd () for. Pracce : For an nducor wh L=.H, gen () = e 5 A. Fnd (), w(). Pracce 3: Gen = 4mF, ()=A, and () Fnd A and B. 5 V, < 6 A e Be, Pracce 4: Fnd c, L under D conon 4 L A.5F V.5H

12 3//7 Pracce 5: Fnd L,, R under D conon. mf 6V 4 6 A 4 R H L Pracce 6: Fnd and R under D conon. A R 3A H 5mF 4 4