! 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 dd-thp thn dd in dd-thp 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 dd-thp 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 dd-thp 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 short-circuit 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 = fan-out 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 pulse-input 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 ), 2-1 ) 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 2-1 ) 2 = ) ) ln 2 ), DD 2-1 ) 2 = ) ) ln 2 ), 2-1 ) 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 2-1 ) 2 = ) ) ln 2 ), 2 - Design for Delays with More Realistic Model for 1 ) 2 = ) ) ln 2 ), DD 2-1 ) 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 Non-Ideal Input aveform ideal in non-ideal in out to ideal in out to non-ideal 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/