ECEN 607 (ESS) Liittin fr Op Ap due t input ignl TANSIENT ESPONSE.- Let u cnider vltge fllwer nd deterine it tep repne Then v in 0 v Thu fr i v (t) i u t r i t Tking the invere Lplce it yield t / v t e ; where ( rd) f t The riing tie t i defined the tie fr the utput v (t) t rech 90% f nd trting t 0%. t (ln( 0.9) ln(0.)). /( f t )
Let u cnider next ltge Gin fr the nn-inverting plifier Fr Ndl Anlyi v x nd v (t) v x v in x 0-0 v, 0 i () v in (t) i H H Let () in * eference: Pive nd Active Netwrk Anlyi nd Synthei, Ar Budk. ; v t v x t v in
3 i i i B G bece nd, Nw fr ) ( ; ) gnitude ( tep f then if i i The crrepnding tie-din exprein bece t t e t v e t v
TANSIENT ESPONSE EXPESSION DEIATIONS Finding dely tie τ D : 0.5K = K ( e τ Dω 3dB ) = e τ Dω 3dB ; τ D = 0.693 ω 3dB Finding riing tie τ = t t (t) K 0.9K 0.5K 0.K τ D τ t t t 0.K = K e t ω 3dB ; t = 0. ω 3dB 0.9K = K e t ω 3dB ; t =.3 ω 3dB Thu τ = t t =. =. = 0.35 ω 3dB πf 3dB f 3dB 4
x dv dt t plitude tep gin - bndwidth prduct Nte tht the Gin Effect h been cncelled fr the nn-inverting ce. In rder t keep the circuit perting in the liner regin the fllwing inequlity ut be tified. Slew te (S) where i expreed in r/. If thi cnditin i nt tified the tep repne i lewing-rte-liited. Furtherre, if we cnider inuidl input c t, n utput Then i. e. x f Aide, dv dt S then FPB 6. KHz fr p p bece the turtin full pwer bndwidth (FPB) SAT 3, S 0.50 6 S SAT SAT, then v ut be in r/ unit in t Fr the Inverting Aplifier v t = v x dv t dt e t + = + v Then the inequlity bece + v < S 5
SIMULATION AND EFICATION Unity Gin ltge Buffer Exple 6
() 0 Input Step in v 0.043 μ < S t v=0.0 τ =. =. π.55 0 6 = 0.6μ NOTE: OFFSET = τ D = 0.693 9.734 0 6 0.07μ 7
(b) 5 Input Squre Wve in iing Slpe: v 0.35 μ S t v=5 Flling Slpe: v 0.35 μ S t v=5 8
(c) 3dB Cut Off Frequency f 3dB = f t.55mhz ω 3dB = πf t π.55mhz = 9.739Mrd/ 9
EXAMPLE: NON-INETING AMPLIFIE An pertinl plifier with S = 0.35 /µ nd ω t = = 9.739M rd/, i ued t ipleent nn-inverting plifier f gin 0, which will liit the rte f rie f the utput, the lewing r the bndwidth when ) The input i =0 tep b) The input i = 0.6 tep c) The input i inuidl 0.5 in π 5k t d) The input i inuidl.5 in π 5k t 0
Slutin ) Deterin = 0.0 0.097/μ And S = 0.35/μ 9.739Mr = Since S i the lrger f the tw, the rte f the utput i liited by the BANDWIDTH b) In thi ce, = 0.6 5.84/μ 9.739Mr = Thu, S <. The rte f rie i liited by the lewing rte. Nte tht the utput will chnge linerly rther thn expnentilly.
Slutin c) Deterine xiu chnge in cled lp. ( i the pek vlue) ω in = + ω in = 0.5 0 π 5 0 3 = 0.57 μ < S Since ω in i ller thn the S, the rte f rie f the utput i liited by the BANDWIDTH
SIMULATION AND EFICATION Nn-inverting Aplifier f Gin = 0 S 0.35/μ = π.55mr/ 3
BW LIMIT () 0 Input Step x in = in - x v t v=0.0 0.057/μ < S τ = + (rd) =.03μ Clcultin. τ =.6μ Meured t ~3.4µ NOTE: OFFSET = 0.9 4
S LIMIT (b) 600 Input Step x in = in - x v t v=0.6 0.3/μ ~ S 5
BW LIMIT (c) 5 khz 0.5 Input Sin Wve x in SPECTUM OF OUTPUT WAE N Nticeble Hrnic & Ditrtin 6
(d) SAT @ 5 khz.5 Input Sin S LIMIT ω in =.5 0 π 5k = 0.47 μ > S = 0.35 μ SAT = 7.8 = 3.9 SPECTUM OF OUTPUT WAE Nticeble Hrnic & Ditrtin v t v=.5 0.35/μ SAT 7