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This chapter focuses on understanding the basic operations of CMOS circuits using inverters, analyzing their performance indices such as speed and power consumption, and optimizing the transistor sizing for speed and energy efficiency.
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Chapter 5 The Inverter April 10, 2003
Objective of This Chapter • Use Inverter to know basic CMOS Circuits Operations • Watch for performance Index such as • Speed (Delay calculation) • Optimal Transistor Sizing for speed and Energy • Power Consumption and Dissipation
V DD V V in out C L The CMOS Inverter: A First Glance
V DD CMOS Inverter N Well PMOS Contacts Out In Metal 1 Polysilicon NMOS GND
Two Inverters Share power and ground Abut cells Connect in Metal Vin Vout Vout Vin
V V DD DD R p V out V out R n V V V 0 = = in DD in CMOS InverterFirst-Order DC Analysis VOL = 0 VOH = VDD
t = f(R .C ) pHL on L = 0.69 R C on L CMOS Inverter: Transient Response V V DD DD R p V out V out C L C L R n ln(2)=0.69 V 0 V V = = in DD in (a) Low-to-high (b) High-to-low
I Dn V = V +V in DD GS,p I = - I D,n D,p V = V +V out DD DS,p 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 DS,p DS,p out V =-1 GSp V =-2.5 GSp V = V +V V = V +V in DD GSp out DD DSp I = - I Dn Dp (Vdd = 2.5V in 0.25um CMOS Process) (Vt = 0.4V as shown in Table 3-2) PMOS Load Lines
CMOS Inverter VTC VM: Vin = Vout Switching Threshold Voltage
Switching Threshold as a Function of Transistor Ratio NMOS and PMOS are in Saturation Modes For r = 1, and saturated velocity NMOS = 2 PMOS, Wp = 2Wn
(V) V Switching Threshold as a Function of Transistor Ratio 1.8 1.7 1.6 1.5 1.4 1.3 M 1.2 1.1 1 0.9 0.8 0 1 10 10 /W W p n
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 Good: Smaller oxide thickness, smaller L, higher W, smaller VT
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 loading capacitances (CL) small • Increase transistor sizes (add CMOS gain) • Watch out for self-loading (for the previous stage)! • Increase VDD (????) Power consumption??
NMOS/PMOS ratio tpHL tpLH tp b = Wp/Wn (See pp. 204) Fig. 5-18
Device Sizing (for fixed load) Self-loading effect: Intrinsic capacitances dominate (Fig. 5-19)
Inverter Chain In Out CL 1 f1 f2 • 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 tpLH = (ln 2) RPCL tpHL = (ln 2) RNCL Delay (D): 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 and Para. Cap. CP = 2Cunit Delay 2W Cint CL W 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 effective 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
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