! Inverter Power. ! Dynamic Characteristics. " Delay ! P = I V. ! Tricky part: " Understanding I. " (pairing with correct V) ! Dynamic current flow:


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1 ESE 570: Digital Integrated ircuits and LSI Fundamentals Lecture Outline! Inverter Power! Dynamic haracteristics Lec 10: February 15, 2018 MOS Inverter: Dynamic haracteristics " Delay 3 Power Inverter Power! P = I! Tricky part: " Understanding I " pairing with correct ) 5 Static urrent Switching urrents! P = I static DD! Dynamic current flow:! If both transistor on: " urrent path from dd to Gnd " Short circuit current 6 7 1
2 urrents Summary! I changes over time! At least two components " I static no switching " I switch when switching " and I sc Switching Dynamic Power LK φ ramp_enable RAMP 8 9 Switching urrents harging! I total t) = I static t)i switch t)! I switch t) = I sc t) t)! t) why is it changing? " I ds = f ds, gs ) " and gs, ds changing I sc I static I DS ν sat OX GS T DSAT 2 ) " ) I DS = µ n OX L GS T ) DS 2, DS Switching Energy focus on t) Switching Energy focus on t) I sc I static E = Pt) = It) dd = dd It)
3 Switching Energy Switching Energy! Do we know what this is?! Do we know what this is? t) Q = t) E = Pt) = It) dd = dd It) E = Pt) = It) dd = dd It) Switching Energy Switching Energy! Do we know what this is?! Do we know what this is? Q = t) Q = t)! hat is Q? E = Pt) = It) dd = dd It)! hat is Q? E = Pt) = It) dd = dd It) Q = = It) Switching Energy Switching Power! Do we know what this is? Q =! hat is Q? E = Pt) = It) dd = dd It) t) Q = = E = dd 2 It) apacitor charging energy! Every time output switches 01 pay: " E = 2! P dyn = 01 trans) 2 / time! 01 trans = ½ of transitions! P dyn = trans) ½ 2 / time
4 Short ircuit Power Switching! Between TN and dd  TP " Both N and P devices conducting Short ircuit Power Short ircuit Power Peak urrent! Between TN and dd  TP " Both N and P devices conducting! Roughly:! I peak around dd /2 " If TN = TP and sized equal rise/fall I DS ν sat OX GS T DSAT 2 ) I sc in ddthp thn dd in ddthp thn dd time time dd Isc dd Isc out tsc tsc time 22 out tsc tsc time 23 Peak urrent Peak urrent! I peak around dd /2 " If TN = TP and sized equal rise/fall I DS ν sat OX GS T DSAT 2 ) It) I t 1 peak sc 2 ) in dd ddthp thn time dd Isc! I peak around dd /2 " If TN = TP and sized equal rise/fall I DS ν sat OX GS T DSAT 2 ) It) I t 1 peak sc 2 ) E = dd I peak t sc 1 in 2 dd ddthp thn time dd Isc out tsc tsc time 24 out tsc tsc time 25 4
5 Short ircuit Energy! Make it look like a capacitance, S " Q=I t " Q= " " E = dd I peak t sc 1 2 Short ircuit Energy! Every time switch 01 and 10) " Also dissipate shortcircuit energy: E = 2 " Different = sc " cs fake capacitance for accounting) E = dd Q S E = dd S dd ) = S 2 dd S = I peak t sc 2 dd Inverter Delay Dynamic haracteristics! aused by charging and discharging the capacitive " hat is the? Inverter Delay Inverter Delay
6 Inverter Delay Inverter Delay gb = gbn gbp = dbn dbp gdn gdp int gb Usually db >> gd sb >> gs dbn dbp int gb Inverter Delay Propogation Delay Definitions n = fanout 1 DD 0 t DD 0 dbn dbp int n gb = DD /2 35 Propogation Delay Definitions Propogation Delay Definitions t
7 Rise/Fall Times MOS Inverter Dynamic Performance! ANALYSIS OR SIMULATION): For a given MOS inverter schematic and, estimate or measure) the propagation delays! DESIGN: For given specs for the propagation delays and, determine the MOS inverter schematic METHODS: 1 Average urrent Model Δ HL = OH 50 Δ LH I avg,lh = 50 OL I avg,lh Assume in ideal MOS Inverter Dynamic Performance MOS Inverter Dynamic Performance! ANALYSIS OR SIMULATION): For a given MOS inverter schematic and, estimate or measure) the propagation delays! DESIGN: For given specs for the propagation delays and, determine the MOS inverter schematic! ANALYSIS OR SIMULATION): For a given MOS inverter schematic and, estimate or measure) the propagation delays! DESIGN: For given specs for the propagation delays and, determine the MOS inverter schematic METHODS: 2 Differential Equation Model METHODS: 3 1 st Order R delay Model i = d out d = out i 069 R n or Assume in ideal 069 R p Assume in ideal alculation of Propagation Delays Method 1 Average urrent Model Δ HL = OH 50 Δ LH I avg,lh = 50 OL I avg,lh 43 7
8 alculation of Propagation Delays alculation of Propagation Delays Δ HL = OH 50 Δ HL = OH 50 Δ LH I avg,lh = 50 OL I avg,lh Δ LH I avg,lh = 50 OL I avg,lh alculation of Rise/Fall Times alculation of Rise/Fall Times τ fall Δ I avg,90 10 = I avg,90 10 τ fall Δ I avg,90 10 = I avg,90 10 τ rise Δ I avg,10 90 = I avg,10 90 τ rise Δ I avg,10 90 = I avg, alculation of Rise/Fall Times τ fall Δ I avg,90 10 = I avg,90 10 τ rise Δ I avg,10 90 = I avg,10 90 Method 2 Differential Equation Model 48 8
9 alculating Propagation Delays ase 1: in Abruptly Rises  Assume in is an ideal pulseinput Two ases 1 in abruptly rises => out falls => 2 in abruptly falls => out rises => i DP  i Dn ase 1: in Abruptly Rises  ase 1: in Abruptly Rises ase 1: in Abruptly Rises  ase 1: in Abruptly Rises  out = DD T0n
10 ase 1: in Abruptly Rises  ase 1: in Abruptly Rises  d out i Dn d out i Dn d = out i Dn = t=t 50 out= DD /2 = t=t 0 out= DD 1 ) d out i Dn = 1 DD /2 1 ) d out ) i Dn DD T 0 n i Dn DD T 0 n DD d out out = DD T0n d = out i Dn = t=t 50 out= DD /2 = t=t 0 out= DD 1 ) d out i Dn = 1 DD /2 1 ) d out ) i Dn DD T 0 n i Dn DD T 0 n DD d out out = DD T0n 56 t 0 t 1 t 1 t ase 1: in Abruptly Rises  ase 1: in Abruptly Rises  saturation linear saturation linear t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn d out out = DD T0n t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn d out out = DD T0n saturation: i Dn = 2 in )2,sat = DD T 0 n DD " 1 2 DD )2 DD T 0 n d out,sat = d 2 out DD DD )2 2,sat = ) 2 58 ) linear: i Dn = 2 ) 2 in out out " DD /2 1,lin = DD T 0 n 2 2 ) 2 DD out out ),lin = 2 " DD /2 1 k DD n 2 ) out 2 T 0 n out ),lin = ) ln " out 2 ) out ) d out d out out= DD /2 out= DD T 0 n 59 ase 1: in Abruptly Rises  ase 1: in Abruptly Rises ,lin = ) ln out 2 ) out ),lin = ) ln 2 ) DD 2 2 out= DD /2 out= DD T 0 n saturation linear t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn d out out = DD T0n 2,sat = τ ) 2 PHL,lin = ) ln " 2 ) DD 2 DD 2 2 = ) 2 ) ln " 2 ) DD 2 DD
11 ase 1: in Abruptly Rises  ase 1: in Abruptly Rises  saturation linear t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn 2,sat = ) 2 d out,lin = ) ln " 2 ) 2 DD DD DD 2 out = DD T0n 1 ) 2 = ) ) ln 2 ), 2  Recall from static MOS Inverter: th = 1 DD ) 2 = ) 2 ) ln " 2 ) 2 DD DD DD 2 R n DESIGN: 1) th k R ; 2) ; 3) k R 1 ) 2 = ) ) ln 2 ), k R 1 1 k R k R = 2 " DD th th ase 1: in Abruptly Rises  Differential Model Approximation 1 ) 2 = ) ) ln 2 ), DD 2  Recall from static MOS Inverter: th = 1 DD ) k R 1 1 k R k R = 2 " DD th th DESIGN: 1) th k R ; 2) ; 3) k R 1) th k R ; 2) ; 3) k R 64 1 ) 2 = ) ) ln 2 ), 2  saturation linear t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn Δ is less than d out Approximate by assuming in saturation: DD /2 1 DD /2 DD i d out = Dn,sat DD 2 DD )2 DD ) R n 2 d out Differential Model Approximation Example 1: 1 ) 2 = ) ) ln 2 ), 2  Δ is less than 10 saturation linear t 0 t 1 t 1 t 50 DD T 0 n " 1 DD /2 " 1 = d out DD i Dn DD T 0 n i Dn d out Approximate by assuming in velocity saturation: DD /2 1 DD i d out Dn,vsat dsat Lv sat µ n DD /2 = v sat OX DD d DD  out dsat, 2 / DD 2v sat OX DD R n dsat 2 66! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1 and =625uA/ 2! Assume in is an ideal step pulse with instant rise/ fall times alculate the delay time necessary for the inverter output to fall from its initial value of 5 to 25 1 ) 2 = ) ) ln 2 ),
12 Example 1: Example 1:! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1 and =625uA/ 2! Assume in is an ideal step pulse with instant rise/ fall times alculate the delay time necessary for the inverter output to fall from its initial value of 5 to 25 1 ) 2 = ) ) ln 2 ), 2  = ) ), ln )5 1) 5 1) 25  = 052ns 1 ) 2 = ) ) ln 2 ), DD 2  = ) ), ln )5 1) 5 1) 25  = 052ns Approximate by assuming in saturation: DD ) R n Example 1: Example 2: 1 ) 2 = ) ) ln 2 ), DD 2  = ) ), ln )5 1) 5 1) 25  = 052ns Approximate by assuming in saturation:! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the average current method to calculate the fall time Assume OH = DD, and OL =0 DD ) R n ) ) = 05ns 2 Δ < 3 Usefull equations: Δ τ fall = I = k n D,sat 2 L GS ) 2 I avg,90 10 I avg,90 10 I D,lin = k n 2 L 2 GS ) DS 2 DS ) Example 2: Example 2:! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the average current method to calculate the fall time Assume OH = DD, and OL =0! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the average current method to calculate the fall time Assume OH = DD, and OL =0 = 1 [ 2 i c in = OH, out = 90 ) i c in = OH, out = 10 )] = 1 [ 2 i c in = OH, out = 90 ) i c in = OH, out = 10 )] = 1 2 [ i = 5, = 45 ) i = 5, = 05 )] c in out c in out = 1 2 [ i = 5, = 45 ) i = 5, = 05 )] c in out c in out = 1 " k n 2 2 L in ) 2 k n 2 L 2 in ) out 2 out ) = 1 " ) 5 1) ) 2 5 1)05) 05) 2 ) = 0988mA = 1 " k n 2 2 L in ) 2 k n 2 L 2 in ) out 2 out ) = 1 " ) 5 1) ) 2 5 1)05) 05) 2 ) = 0988mA
13 Example 2: Example 2:! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the average current method to calculate the fall time Assume OH = DD, and OL =0! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the differential equation method to calculate the fall time Assume OH = DD, and OL =0 = 0988mA Usefull equations: τ fall Δ I avg,90 10 = I avg, τ fall 1 10 = 4ns i = d out i d,sat = 1 2 k n i d,lin = 1 2 k n d = out i L GS ) 2 ) L 2 GS ) DS 2 DS Example 2: Example 2:! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, k n =20uA/ 2, and /L=10! Use the differential equation method to calculate the fall time Assume OH = DD, and OL =0 d i = out = 1 2 k n = k n L 2 L in ) 2 ) DD ) d 2 out = t=t sat out =40 ) 5 1) d = 2 out d out = d out = 0313ns t=t 90 out =45 76! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, =20uA/ 2, and / L=10! Use the differential equation method to calculate the fall time Assume OH = DD, and OL =0 d i = out = 1 2 k n L 2 ) 2 in out out ) 2 = k n L) 2 ) out 2 out ) d out t=t 10 = 2 out=05 1 t=t sat k n L) 2 ) out 2 out ) d out out=4 t=t 10 = k n L) 1 ln 2 in ) 10 ) in 10 t=t sat = ) ) 05 ln ) = 339ns t=t 10 t=t sat t 90 t sat = 0313ns 77 Example 2: ase 2: in Abruptly Falls ! onsider a MOS inverter with =1pF and DD =5 The nmos transistor has T0n =1, =20uA/ 2, and / L=10! Use the differential equation method to calculate the fall time Assume OH = DD, and OL =0 d i = out = 1 2 k n L 2 ) 2 in out out ) t 90 t sat = 0313ns saturation linear t 0 t 1 t 1 t 50! 1 50! 1 = 0 " i d out Dp " i Dp d out 2 = k n L) 2 ) out 2 out ) d out t=t 10 = 2 out=05 1 t=t sat k n L) 2 ) out 2 out ) d out out=4 t=t 10 = k n L) 1 ln 2 ) in 10 ) in 10 t=t sat = ) ) 05 ln ) = 339ns t=t 10 t=t sat t 90 t 10 = 37ns Avg method: 4ns 78 1 ) 2 = ) ) ln 2 ),
14 ase 2: in Abruptly Falls  Differential Equation Model saturation linear 1 ) 2 = ) ) ln 2 ), 2  t 0 t 1 t 1 t 50! 1 50! 1 = 0 " i d out Dp " i Dp d out 1 ) 2 = ) ) ln 2 ), 21 ) 2 = ) ) ln 2 ), DD 2  DD ) 2 R p R p ONDITIONS for Balanced MOS Propagation Delays, ie! = µ! n " L µ p " L p n Delay Observations Delay Design Equations 1 ) 2 = ) ) ln 2 ), DD 21 ) 2 = ) ) ln 2 ), DD 21 ) 2 = ) ) ln 2 ), 21 ) 2 = ) ) ln 2 ), Delay Design Equations w/ Saturation Approximation DD ) 2 Design for Delays with More Realistic Model for i i dbn dbp int gb! DD " L µ n ox ) 2 n DD ) 2! DD " L µ p ox ) 2 p i i dbn n ) dbp p ) int gb
15 Design for Delays with More Realistic Model for Design for Delays with More Realistic Model for dbn n ) = [ n Y x j )] j0n K eqn n 2Y) jswn K eqn sw) dbp p ) = [ p Y x j )] j0p K eqp p 2Y) jswp K eqp sw) dbn n ) = [ n Y x j )] j0n K eqn n 2Y) jswn K eqn sw) dbp p ) = [ p Y x j )] j0p K eqp p 2Y) jswp K eqp sw) = α 0 α n n α p p α 0 = 2Y jswn K eqn 2Y jswp K eqp int gb 86 α n = Y x j ) j0n K eqn jswn K eqn α p = Y x j ) j0p K eqp jswp K eqp 87 Design for Delays with More Realistic Model for 1 ) 2 = ) ) ln 2 ), DD 21 ) 2 = ) ) ln 2 ), 2  Design for Delays with More Realistic Model for 1 ) 2 = ) ) ln 2 ), DD 21 ) 2 = ) ) ln 2 ), 2  = Γ n n and = Γ p p = Γ n n and = Γ p p Γ n and Γ P are set largely by process parameters and DD const Design for Delays with More Realistic Model for Design for Delays with More Realistic Model for = Γ n n = Γ p p = Γ n n = Γ p p = α 0 α n n α p p = α 0 α n n α p p / n ) n = α 0 [α n α p R] n = α 0 α n n α p p = α 0 α n n / p ) p α p p = α 0 [α n /R α p ] p = α 0 α n n α p p = α 0 α n n α p p / n ) n = α 0 [α n α p R] n = α 0 α n n α p p = α 0 α n n / p ) p α p p = α 0 [α n /R α p ] p where R = p / n = constant where R = p / n = constant Recall: th = 1 = µ p p when L p =L n ) k R µ n n τ α 0 [α n α p R] n PHL = Γ n n = Γ p α 0 [α n /R α p ] p p
16 Design for Delays with More Realistic Model for Design for Delays with More Realistic Model for τ α 0 [α n α p R] n PHL = Γ n n = Γ p α 0 [α n /R α p ] p p where R constant) = aspect ratio = p / n Hence increasing n and p will have diminishing influence on and as they become large, ie = limit = Γ n [α n α p R] n large = limit = Γ p [α n /R α p ] p large absolute minimum delays Design for Delays with More Realistic Model for Taking Into Account NonIdeal Input aveform ideal in nonideal in out to ideal in out to nonideal in MOS Inverter Dynamic Performance Idea! ANALYSIS OR SIMULATION): For a given MOS inverter schematic and, estimate or measure) the propagation delays! DESIGN: For given specs for the propagation delays and, determine the MOS inverter schematic METHODS: 1 Average urrent Model Δ HL = OH 50 2 Differential Equation Model d i = out d = out or 3 1 st Order R delay Model i Assume in ideal! P tot = P static P dyn P sc " an t ignore Static Power aka Leakage power)! Propogation Delay " Average urrent Model " Differential Equation Model " 1 st Order Model " More next time 069 R n
17 Admin! H 4 due today! H 5 posted today " Due 3/1! Quiz next week Thursday 2/
ESE 570: Digital Integrated Circuits and VLSI Fundamentals
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