t=0 t>0: + vr - i dvc Continuation
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1 hapr Ga Dlay and rcus onnuaon s rcu Equaon >: S S Ths dffrnal quaon, oghr wh h nal condon, fully spcfs bhaor of crcu afr swch closs Our n challng: larn how o sol such quaons TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr s Soluon c s ( c s Proof Usng Dffrnaon Show ha hs dffrnal quaon has hs soluon c s c s s s dncal s s c s.5 s c s c c s TUE/EE 57 nwrk analys 4/5 NdM frs ordr 3 qd TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 4 3. Ealua ngrals, us d ln K ad a K TUE/EE 57 nwrk analys 4/5 NdM Soluon by Ingraon. Sar wh Dffrnal Eq.. Spara arabls. Ingra LHS and HS 4. Absorb K s no K 3 K K c s c planaon c c s c c s ln( c s K K ln( c s K frs ordr 5 4. from prous sp 5. ponna LHS and HS 6. rsul 7. wr (dfn K K 3 TUE/EE 57 nwrk analys 4/5 NdM Soluon connud ln( s K 3 K 3 c s s K 8. Drmn K from nal alu: c ( s K s K K s 9. Fnal Soluon: s s ( s 3 frs ordr 6
2 Unknown arabl as a funcon of m. fnal alu of h arabl nal alu of h arabl TUE/EE 57 nwrk analys 4/5 NdM Gnral Soluon c ald for any frs ordr τ ( crcu wh D sourcs ( ( ( ( sady sa (saonar fnal alu of h arabl ransn [ (m of swchng ] m consan 3 frs ordr 7 Soluon (almos by Inspcon ( ( ( (. Idnfy sa arabl: capacor olag. alcula fnal alu 3. Drmn nal alu of sa arabl 4. alcula m consan TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 8 Eampl Sp : Idnfy Sa arabl Drmn ( for TUE/EE 57 nwrk analys 4/5 NdM Swch opns a Drmn ( for 3 frs ordr 9 Sa arabl s ofn obous: Sa mans ha hr s som mmory In crcus, mmory s prodd by capacor chargs ould us charg on capacor as sa arabl Bu ohr lmns ha olag as characrsc quany Br us capacor olag as sa arabl (Q/: smplr quaons (mayb unlss quson asks for charg as a funcon of m TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr Sp : alcula Fnal alu olag s no changng anymor Sady sa or fnal alu of capacor olag: lm c lm c c Sp : alcula Fnal alu D sady sa qualn modl of capacor s opn crcu draw: swch n fnal poson and rplacd by opn crcu! In h lm for, currn hrough capacor s zro plac capacor by opn crcu ( D sady sa qualn modl of capacor s an opn crcu Q: c (? TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr
3 Sp 3: alcula Inal alu In many cass, c ( s gn plcly Somms, mus b calculad. Ofn, h problm sas somhng lk Bfor, swch s s opn / closd for a long m Thn, nal condon s sady sa wh swch n h sar poson. draw: swch n sar poson and rplacd by opn crcu! Gn: Swch opns a, was closd for a long m Q: c (? ( Sp 4: alcula Tm onsan rcu afr swch has opnd (> Thr s only on and on nold, hus: Tm consan follows by nspcon: τ TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 3 TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 4 Eampl Sold Eampl Swch opns a losd for a long m bfor Drmn ( for Swch closs a Opn for a long m bfor Drmn ( for ( /τ ( ( ( ( ( Inal and fnal condons ar rrsd: c ( c ( c ( c ( τ TUE/EE 57 nwrk analys 4/5 NdM (( /τ /τ c ( 3 frs ordr 5 Equaln crcu for compung τ TUE/EE 57 nwrk analys 4/5 NdM τ s drmnd by wo s and on How? 3 frs ordr 6. KL: How can w drmn τ? ( ' ' TUE/EE 57 nwrk analys 4/5 NdM. Afr w ha nroducd Thénn/Noron 3 frs ordr 7 τ ' W can compu τ 3. I s no acly by nspcon W can mpro our mhod From compuaon Inror olag Transfr haracrsc ou f n < l ou f n > h.7,.8 TUE/EE 57 nwrk analys 4/5 NdM ou l h n 3 frs ordr 8 3
4 Inrr Par Swchng Spd How long dos ak for an nrr o swch from o l? ( c( ( ( ( swng / τpd ( ( / τ pd l 5% 63% % 9% / τ pd l 9% / τ l pd ln pd τ pd ln l m.69τ.τ.τ.3τ Whr Dos Powr Go n MOS Dynamc Powr onsumpon: s.9 hargng and dschargng capacors Shor rcu urrns Shor crcu pah bwn supply rals durng swchng Lakag Lakng dods and ranssors May b mporan for baryoprad qupmn.9 TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 9 TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr TUE/EE 57 nwrk analys 4/5 NdM Dynamc Powr Dynamc Powr E nrgy of swchng n (o frs ordr ndpndn of swchng spd dpnds on procss, layou Powr Enrgy/Tm P E T E PowrDlayProduc PD mporan qualy masur EnrgyDlayProduc ED combns powr*spd prformanc 3 frs ordr LowoHgh Transon Enrgy ( TUE/EE 57 nwrk analys 4/5 NdM ( E Enrgy sord on E ( Equaln crcu for lowohgh ranson No: full swng assumd. Book shows ha paral swng dsspas narly as much. ( c 3 frs ordr LowoHgh Transon Enrgy LowoHgh Transon Enrgy ( ( ( ( E Enrgy dlrd by supply E ( E Ec E c E Q: whr s h rs? Dsspad n ranssor! Edss Enrgy dsspad n ranssor Edss ( E E c TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 3 TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 4 4
5 HghoLow Transon Enrgy Equaln crcu s S Summary ( o s (o s ransn rspons D sady sa rspons oal rspons Show ha h nrgy ha s dsspad n h ranssor upon dschargng from o quals E dss ½ Unknown arabl as a funcon of m nal alu of h arabl fnal alu of h arabl fnal alu of h arabl [ (m of swchng ] m consan TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 5 TUE/EE 57 nwrk analys 4/5 NdM 3 frs ordr 6 5
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har 9 Transn sons har 9: Ouln N F n F Frs-Ordr Transns Frs-Ordr rcus Frs ordr crcus: rcus conan onl on nducor or on caacor gornd b frs-ordr dffrnal quaons. Zro-nu rsons: h crcu has no ald sourc afr a cran
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