Review: Design Abstraction Levels SYSTEM MODULE + GATE CIRCUIT Vin Vout DEVICE G S n+ D n+ Review: The MOS Transistor Gate oxide Polysilicon W Gate Source n+ L Drain n+ Field-Oxide (SiO2) p substrate p+ stopper Bulk (Body) Design metrics for gate CMOS Inverter: A First Look VDD Vin Vout CL CMOS Inverter: Steady State Response VDD VDD VOL = 0 VOH = VDD VM = f(Rn, Rp) Rp Vout = 1 Vout = 0 Rn Vin = 0 Vin = V DD CMOS Properties • Full rail-to-rail swing high noise margins – Logic levels not dependent upon the relative device sizes transistors can be minimum size ratioless • Always a path to Vdd or GND in steady state low output impedance (output resistance in k range) large fan-out (albeit with degraded performance) • Extremely high input resistance (gate of MOS transistor is near perfect insulator) nearly zero steady-state input current • No direct path steady-state between power and ground no static power dissipation • Propagation delay function of load capacitance and resistance of transistors Review: Short Channel I-V Plot (NMOS) X 10-4 2.5 VGS = 2.5V 2 VGS = 2.0V 1.5 1 VGS = 1.5V 0.5 VGS = 1.0V 0 0 0.5 1 1.5 2 2.5 VDS (V) NMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = 0.4V Review: Short Channel I-V Plot (PMOS) l All polarities of all voltages and currents are reversed -2 VDS (V) -1 0 0 VGS = -1.0V -0.2 VGS = -1.5V -0.4 -0.6 VGS = -2.0V -0.8 VGS = -2.5V -1 X 10-4 PMOS transistor, 0.25um, Ld = 0.25um, W/L = 1.5, VDD = 2.5V, VT = -0.4V Transforming PMOS I-V Lines l Want common coordinate set Vin, Vout, and IDn IDSp = -IDSn VGSn = Vin ; VGSp = Vin - VDD VDSn = Vout ; VDSp = Vout - VDD IDn Vout Vin = 0 Vin = 0 Vin = 1.5 Vin = 1.5 VGSp = -1 VGSp = -2.5 Mirror around x-axis Vin = VDD + VGSp IDn = -IDp Horiz. shift over VDD Vout = VDD + VDSp CMOS Inverter Load Lines PMOS 2.5 NMOS X 10-4 Vin = 0V Vin = 2.5V 2 Vin = 0.5V Vin = 2.0V 1.5 Vin = 1.0V 1 Vin = 2V 0.5 Vin = 1V Vin = 1.5V Vin = 1.5V Vin = 0.5V Vin = 1.5V Vin = 1.0V Vin = 2.0V Vin = 0.5V 0 Vin = 2.5V 0 0.5 1 1.5 Vout (V) 2 2.5 Vin = 0V 0.25um, W/Ln = 1.5, W/Lp = 4.5, VDD = 2.5V, VTn = 0.4V, VTp = -0.4V CMOS Inverter VTC NMOS off PMOS res 2.5 NMOS sat PMOS res Vout (V) 2 1.5 NMOS sat PMOS sat 1 NMOS res PMOS sat 0.5 NMOS res PMOS off 0 0 0.5 1 1.5 Vin (V) 2 2.5 CMOS Inverter: Switch Model of Dynamic Behavior VDD VDD Rp Vout Vout CL Vin = 0 CL Rn Vin = V DD l Gate response time is determined by the time to charge CL through Rp (discharge CL through Rn) Relative Transistor Sizing • When designing static CMOS circuits, balance the driving strengths of the transistors by making the PMOS section wider than the NMOS section to – maximize the noise margins and – obtain symmetrical characteristics Switching Threshold • VM is defined as the point where Vin=Vout •Graphically it means an intersection of the VTC with the line given by Vin=Vout •In this region both PMOS and NMOS are saturated as VDS=VGS •We Solve for the case in which the supply voltage is high enough for the device to be velocity-saturated • Analytical expression for VM is obtained by equating the currents through the transistors Notice: We have ignored channel length modulation But NMOS PMOS Switching Threshold NMOS PMOS Switching Threshold Notice since same , therefore it is assumed here that t OX is For VDD>>VTp And VDD>>VDSATp r=1 For symmetrical VTC Sizing In general Sizing set to the desired Vm r Vm Wp And vice versa Switching Threshold • VM where Vin = Vout (both PMOS and NMOS in saturation since VDS = VGS) VM rVDD/(1 + r) where r = kpVDSATp/knVDSATn • Switching threshold set by the ratio r, which compares the relative driving strengths of the PMOS and NMOS transistors • Want VM = VDD/2 (to have comparable high and low noise margins), so want r 1 (W/L)p kn’VDSATn(VM-VTn-VDSATn/2) = (W/L)n kp’VDSATp(VDD-VM+VTp+VDSATp/2) Switch Threshold Example • In our generic 0.25 micron CMOS process, using the process parameters , VDD = 2.5V, and a minimum size NMOS device ((W/L)n of 1.5) NMOS PMOS VT0(V) 0.43 -0.4 (V0.5) 0.4 -0.4 VDSAT(V) 0.63 -1 k’(A/V2) 115 x 10-6 -30 x 10-6 (V-1) 0.06 -0.1 (W/L)p 115 x 10-6 0.63 (1.25 – 0.43 – 0.63/2) = (W/L)n x -30 x 10-6 x -1.0 (1.25 – 0.4 – 1.0/2) (W/L)p = 3.5 x 1.5 = 5.25 for a VM of 1.25V = 3.5 Simulated Inverter VM VM is relatively insensitive to variations in device ratio 1.5 1.4 1.3 setting the ratio to 3, 2.5 and 2 gives VM’s of 1.22V, 1.18V, and 1.13V 1.2 1.1 Increasing the width of the PMOS moves VM towards VDD 1 0.9 0.8 0.1 1 (W/L)p/(W/L)n Note: x-axis is semilog ~3.4 10 Increasing the width of the NMOS moves VM toward GND Noise Margin approximation Noise Margins Determining VIH and VIL 3 VOH = VDD 2 VM 1 B VOL = GND0 A VIL C Vin VIH A piece-wise linear approximation of VTC Noise Margins Determining VIH and VIL By definition, VIH and VIL are where dVout/dVin = -1 (= gain) 3 VOH = VDD NMH = VDD - VIH NML = VIL - GND 2 VM Approximating: VIH = VM - VM /g VIL = VM + (VDD - VM )/g 1 VOL = GND0 VIL Vin VIH A piece-wise linear approximation of VTC So high gain in the transition region is very desirable Gain Now Rearranging Since Idn Device is functioning in saturation Idp Gain Idn Device is functioning in saturation gain Technology parameters Idp Gain Determinates Vin 0 0 -2 -4 -6 -8 0.5 1 1.5 2 Gain is a strong function of the slopes of the currents in the saturation region, for Vin = VM (1+r) g ---------------------------------(VM-VTn-VDSATn/2)(n - p ) -10 -12 -14 -16 -18 Determined by technology parameters, especially channel length modulation (). Only designer influence through supply voltage and VM (transistor sizing). CMOS Inverter VTC from Simulation 0.25um, (W/L)p/(W/L)n = 3.4 (W/L)n = 1.5 (min size) VDD = 2.5V VM 1.25V, g = -27.5 VIL = 1.2V, VIH = 1.3V NML = NMH = 1.2 (actual values are VIL = 1.03V, VIH = 1.45V NML = 1.03V & NMH = 1.05V) Impact of Process Variation on VTC Curve 2.5 Good PMOS Bad NMOS Vout (V) 2 1.5 Behavior remains unchanged Nominal 1 Bad PMOS Good NMOS 0.5 For good one 0 0 0.5 1 1.5 2 2.5 Vin (V) lProcess variations (mostly) cause a shift in the switching threshold Scaling the Supply Voltage Vm ɡ & Scaling the Supply Voltage 2.5 0.2 2 Vout (V) Vout (V) 0.15 1.5 1 0.1 0.05 0.5 Gain=-1 0 0 0 0.5 1 1.5 Vin (V) Device threshold voltages are kept (virtually) constant 2 2.5 0 0.05 0.1 0.15 Vin (V) Device threshold voltages are kept (virtually) constant 0.2 Performance of CMOS inverter The dynamic Behavior CL has to be as small as possible for high performance Major component of CL Gate-Drain capacitance Assumption: Vin is driven by an ideal voltage source with zero rise and fall timesWe will account for capacitances connected to the output node Assumption: All capacitance are lumped together into a single capacitor CL, located between Vout and GND. Diffusion Capacitance Unfortunately very non-linear So we linearize it! Wiring Capacitance • Depends on the length, width of the connecting wire, • And also on the number of fan-out gates and their distance from the driving gate • Gaining Importance! Gate capacitance CL CL from High-low and low-high are approx same Propagation Delay: 1st order Analysis Intractable, so we can try switch model Vout (1 e t / )2.5 1 (1 e t / ) 2 1 e t / 2 Propagation Delay: 1st order Analysis To Increase performance Thin Oxide capacitance nearly constant over the years! Load capacitance Fanout Gate Capicitance Fanout capacitance Self Capacitance Self Capacitance I leave the calculation of delay for a chain of 4 inverters for you ! Propagation Delay (Design perspective) • 3-3.5 sizing although ensures symmetrical VTC and similar tpHL and tpLH (Reqn=Reqp) but =/=> tp(min) • (reason) widening of PMOS although improves tpLH but degrades tpHL (larger parasitic capacitance) • need to optimize (W / L) p (W / L)n ASSUMPTION: PMOS β times larger than NMOS Cdp1=βCdn1; Cgp2=βCgn2 ; R’eqp=Reqp/β Background maths Optimal performance (sizing) Sizing Inverter for performance Assume symmetrical Inverter Sizing Sizing Inverter for performance Gate Sizing for Optimal path Delay Inverter Chain Delay Optimization Intrinsic time constant Delay of a single stage In order to incorporate intrinsic delay The delay through the stages can be computed as Overall fan-out f1=f2=f3……….=fN f1f2f3……….fN=fN=CL/C1 =F 1/ N F N &vice versa tp can be optimized dt p dN 0 Sizing chain of inverters Both are proportional to gate sizing Overall fan-out f1=f2=f3……….=fN f1f2f3……….fN=fN=CL/Cg,1 =F Sizing chain of inverters for 1/ N F N &vice versa tp can be optimized dt p dN 0 Optimizing N Power Consideration 1. Dynamic Power Consumption 2. Dissipation due to direct-path current 3. Static Consumption Dynamic Power Consumption • Let us first consider low-to-high transition • Assume that the input waveform has zero rise and fall time. In other words NMOS and PMOS are never on simultaneously. Therefore low to high transition can be represented by following fig. This stored energy is dissipated through NMOS during H-L transition. Example As tP decreases with the advances of technology, higher value of f01 results Clock Rate:500MHz Average load capacitance of :15fF/gate Supply voltage: VDD=2.5V Per gate For 1 million transistor on chip Fortunately not all gates will switch at 500MHz rate Example Switching activity for complex gates In reality the switching activity is a function of nature and statistics of the input signals which in turn depends on network topology and the function being implemented Dissipation Due to Direct-Path Current Impact of load capacitance on Pdp Static Consumption Static CMOS: extension to static CMOS Inverter Complementary CMOS • Transistor can be thought of as a switch, controlled by gate • PDN NMOS, and PUN PMOS • Rules: NMOS series=AND, NMOS parallel=OR (PMOS series=NOR, PMOS parallel=NAND PDN PUN • De Morgan’s theorem is applied (series parallel and vice versa) •CMOS is naturally inverting NAND, NOR, XNOR are naturally implemented Static properties CMOS Strong pull-up Stronger VTn Propagation Delay LH A=B=10 A0, B=1 or A=1, B0 Worst case HL A=B=01 PMOS: RN=Rp/2 width should be twice NMOS: Ratioed Logic •Reduced number of transistor. •VOL is not equal to zero •Static power dissipation. Differential Cascode Voltage switch Logic (DCVSL)