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Digital Integrated Circuits A Design Perspective. Jan M. Rabaey Anantha Chandrakasan Borivoje Nikolic. The Inverter. July 30, 2002. The CMOS Inverter: A First Glance. V. DD. PMOS. 2 l. Out. In. Metal1. Polysilicon. NMOS. GND. CMOS Inverters. CMOS Inverter First-Order DC Analysis.
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Digital Integrated CircuitsA Design Perspective Jan M. Rabaey Anantha Chandrakasan Borivoje Nikolic The Inverter July 30, 2002
V DD PMOS 2l Out In Metal1 Polysilicon NMOS GND CMOS Inverters
CMOS InverterFirst-Order DC Analysis V V DD DD Rp VOL = 0 VOH = VDD VM = f(Rn, Rp) Vout Vin = 0 Vout Vin = 1 Rn
I Dn V = V +V in DD GSp I = - I Dn Dp V = V +V out DD DSp V out I I I Dp Dn Dn V =0 V =0 in in V =1.5 V =1.5 in in V V V DSp DSp out V =-1 GSp V =-2.5 GSp V = V +V V = V +V in DD GSp out DD DSp I = - I Dn Dp PMOS Load Lines
1.8 1.7 1.6 1.5 1.4 (V) 1.3 M V 1.2 1.1 1 0.9 0.8 0 1 10 10 /W W p n Switching Threshold as a function of Transistor Ratio
V out V OH V M V in V OL V V IL IH Determining VIH and VIL A simplified approach
Gain=-1 Gain as a function of VDD
2.5 2 Good PMOS Bad NMOS 1.5 Nominal (V) out Good NMOS Bad PMOS V 1 0.5 0 0 0.5 1 1.5 2 2.5 V (V) in Impact of Process Variations
V DD PMOS Metal1 Polysilicon NMOS CMOS Inverters 1.2 m m =2l Out In GND
Transient Response ? tp = 0.69 CL (Reqn+Reqp)/2 tpHL tpLH
Design for Performance • Keep capacitances small • Increase transistor sizes • watch out for self-loading! • Increase VDD (????)
Device Sizing (for fixed load) Self-loading effect: Intrinsic capacitances dominate
NMOS/PMOS ratio tpHL tpLH tp b = Wp/Wn
Inverter Chain In Out CL • If CL is given: • How many stages are needed to minimize the delay? • How to size the inverters? • May need some additional constraints.
Inverter Delay • Minimum length devices, L=0.25mm • Assume that for WP = 2WN =2W • same pull-up and pull-down currents • approx. equal resistances RN = RP • approx. equal rise tpLH and fall tpHL delays • Analyze as an RC network 2W W tpHL = (ln 2) RNCL Delay (D): tpLH = (ln 2) RPCL Load for the next stage:
Inverter with Load Delay RW CL RW Load (CL) tp = kRWCL k is a constant, equal to 0.69 Assumptions: no load -> zero delay Wunit = 1
Inverter with Load CP = 2Cunit Delay 2W W Cint CL Load CN = Cunit Delay = kRW(Cint + CL) = kRWCint + kRWCL = kRW Cint(1+ CL /Cint) = Delay (Internal) + Delay (Load)
Delay Formula Cint = gCgin withg 1 f = CL/Cgin- effective fanout R = Runit/W ; Cint =WCunit tp0 = 0.69RunitCunit
Apply to Inverter Chain In Out CL 1 2 N tp = tp1 + tp2 + …+ tpN
Optimal Tapering for Given N • Delay equation has N - 1 unknowns, Cgin,2 – Cgin,N • Minimize the delay, find N - 1 partial derivatives • Result: Cgin,j+1/Cgin,j = Cgin,j/Cgin,j-1 • Size of each stage is the geometric mean of two neighbors • each stage has the same effective fanout (Cout/Cin) • each stage has the same delay
Optimum Delay and Number of Stages When each stage is sized by f and has same eff. fanout f: Effective fanout of each stage: Minimum path delay
Example In Out CL= 8 C1 1 f f2 C1 CL/C1 has to be evenly distributed across N = 3 stages:
Optimum Number of Stages For a given load, CL and given input capacitance Cin Find optimal sizing f For g = 0, f = e, N = lnF
Optimum Effective Fanout f Optimum f for given process defined by g fopt = 3.6 forg=1
Impact of Self-Loading on tp No Self-Loading, g=0 With Self-Loading g=1
Buffer Design N f tp 1 64 65 2 8 18 3 4 15 4 2.8 15.3 1 64 1 8 64 1 4 64 16 1 64 22.6 8 2.8
Vdd Vin Vout C L Dynamic Power Dissipation 2 Energy/transition = C * V L dd 2 Power = Energy/transition * f = C * V * f L dd Not a function of transistor sizes! Need to reduce C , V , and f to reduce power. L dd
Adiabatic Charging 2 2 2
Transistor Sizing for Minimum Energy • Goal: Minimize Energy of whole circuit • Design parameters: f and VDD • tp tpref of circuit with f=1 and VDD =Vref
Transistor Sizing (2) • Performance Constraint (g=1) • Energy for single Transition
Transistor Sizing (3) VDD=f(f) E/Eref=f(f) F=1 2 5 10 20