Physics 214 INTRODUCTION TO LABORATORY ELECTRONICS Lecture 3 Topics: capacitors, transients in RC (resistor + capacitor) circuits

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1 Physics 4 INTRODUTION TO LABORATORY ELETRONIS Lecure 3 Topics: capaciors, ransiens in R (resisor + capacior) circuis As adverised, we now urn o devices ha have frequency dependen characerisics. A capacior is a wo erminal device ha sores charge. The amoun of charge he device sores depends on is capaciance,, (which depends on he size, geomery and maerial of he device) and he volage across he erminals: Q= V APAITANE is measured in FARADS, (F): F = / V = / J usually encouner micofarads (µf = -6 F) or picofarads or puff ( pf= - F). Occasionally we ll encouner nanofarads (nf= -9 F). APAITORS: ommon Geomeries: For PARALLEL PLATES, of area A, disance beween conducing plaes d, A = ε d. ε is he perimiiviy of free space = 885. / J m. Example:.5cm by.5cm plae wih disance beween plaes of.mm has a capaciance of 885. / J m cm / J m. 5. 5m = = 3. mm. m =. / J =. pf The capaciance can be increased if he space beween plaes is filled wih a dielecric maerial. Tha is, an insulaing maerial such as ceramic, plasic, meal oxide, wax paper. The capaciance hen changes by he dielecric consan or permiiviy, ε, of he insulaor: εε A = d Typical values of ε are: wax paper=.5, cellulose aceae=5, ianae ceramics=5-. For YLINDRIAL capaciors wih inner conducing cylinder of radius a and ouer conducing cylinder of radius b and lengh εε = π ln( b/ a) Here ε is he dielecric consan of he maerial beween he wo cylinders. PRATIAL APAITORS: In pracice, need also beware of exceeding he dielecric srengh of he maerial: i.e. if he volage across he erminals is oo high, here will be a breakdown hrough he dielecric resuling in a shored and useless capacior. Pracical compac capaciors can be creaed by rolling up wo shees of foil separaed by an insulaor (microf o millif) hese have breakdown volages of up o 5V; or evaporaing elecrodes ono a ianae ceramic gives capaciances beween pf and.microf wih volage raings up o kv. Also, you ll encouner elecrolyic capaciors which are no used for A circuis. These have srips of Al in elecrolye of aluminum hydroxide. The srips are one elecrode (+) and he elecrolye is he oher elecrode (-). Applying a volage causes an oxide layer o build up on he Al srips which form he dielecric. d is very small so is large. The polariy is marked: negaive sign or black band indicaes he negaive polariy. If polariy is reversed, he oxide layer is removed and he capacior shors ou.

2 STORED ENERGY The energy sored in any capacior is given by inegraing he energy of moving charge dq from one elecrode o he oher hrough he poenial, V: Q zq q W = Vdq = dq = = Q z V. This sored poenial energy can be reurned hen discharging he capacior. Noe ha afer a large D volage has been applied o a capacior, he capacior should be discharged for safey reasons. An IMPORTANT equaion: dq() q v dv () () = () = d d i dv () () = curren d ime deriviive of volage across capacior If he volage across a µf capacior changes a a rae of V/s, hen he curren hrough he capacior is µa. apaciors in Parallel: i () = i() + i() dv () dv dv () () = + d d d dv() dv() dv() bu v() = v() = v() so = = d d d = +

3 apaciors in Series: i () = i() = i() dv () dv dv () () = = d d d dv() dv() dv() dv() = and = d d d d bu v () = v() + v() dv() dv() dv() so = + d d d dv() dv() dv() or = + = + d d d = +

4 R Transiens: harging up a apacior hrough a Resisor We can charge up a capacior by hrowing he swich in he following circui. Once ha swich is hrown, we can jus wrie down Kirchoff s Loop law: dv () V R i () v () = bu i () = d V R dv () v() = his s order differenial equaion d Boundary condiions:, v ( ) < he volages have o be =, v ( = ) = we' re charging i up from scrach OK so now les solve i firs by rearranging he erms: dv () d R v V + () =...() R B G we sar by leing v () = Ae + Fe + K Bu he v ( ) < condiion c F= so and d B v () B v () = Ae + K = ABe d puing his ino (): B ABe + + ) = R Ae B K V ( R equaing coefficiens of exponenial AB + A= B= R R equaing ime independen erms: K = V hus: / R v () = Ae + V now applying he second boundary condiion: v c ( = ) = = A+ V A= V / R v() = V (- e ) he ime consan is R.

5 Discharging a apacior hrough a Resisor Similarly, we can discharge a capacior which sars wih a charge Q and so volage V =Q/, hrough a resisor, R. vr() + v() = R i() = v () R dq () R dv [ ()] = = v () d d R dv c() dvc() = v d v R d () = () z dv () z = v R d () K / ln( v( )) = / R+ K v( ) = e e / =, v() = V v () = V e he ime consan is R. R