Filter Analysis and Rectifier Circuits in Electronics Principles I - Prof. Hugh R. Grinold, Study notes of Electrical and Electronics Engineering

Download Filter Analysis and Rectifier Circuits in Electronics Principles I - Prof. Hugh R. Grinold and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! Lecture 9 ECE 331 Electronics Principles I Grinolds FA08 Rectification (cont) Filter Analysis iL = vo/RL iD = C dvo/dt + iL = C d(vi – VDo)/dt + iL + - Cvi iD RL iL iC vo = C dvi/dt + iL ECE 331 Electronics Principles I Grinolds FA08 When vo > vi , vo = Vomax e -t/CR L Vomax = VI if VDo << VI v i = VI sin(ωt) Then ∆Vo ~ VI - VI e -T/CR L ~ VI (T/CRL) e-T/CRL ~ 1 – T/CRL for T<< CRL Vr = VI (T/CRL) Half-Wave Rectifier Analysis V rms = 12.6 (60 Hz), R = 15 Ω, Vr < 0.75V What is the minimum for C and what are the specs for the diode? RCVrms vo i Vo = 12.6 √2 = Vp = 17.8 volts I = Vo/R = 17.8/15 = 1.19 amps C > VPT / R Vr = 17.8/0.75 *1/(60*15) ECE 331 Electronics Principles I Grinolds FA08 Chap 3 -5 > 0.03 F = 30,000 µF iDavg = I (1 + π√2VP/Vr) = 26.9 amps over the time ∆t = √2Vr/VP / 2πf ~ 0.8 mS iDmax ~ 2 iDavg = 53.8 amps A constant voltage drop model for the diode means: Vo = VP – VDo Diode Power Dissipation PDavg ~ VDo ∆T/T IP/2 ~ VDo IL Thus 1 * 1.19 = 1.19 watt ECE 331 Electronics Principles I Grinolds FA08 Chap 3 -6 But PDavg due to series resistance in the diode can be significant. PDavg = 1/T ∫ iD 2 (t)Rs dt = 4TIL 2Rs Where iD ~ ∆T IP/2 3∆T With Rs = 0.2Ω PDavg = 7.9 W Full Wave Rectifiers ECE 331 Electronics Principles I Grinolds FA08 + - + - ideal ideal ideal RD