# VLSI GATE LEVEL DESIGN UNIT - III P.VIDYA SAGAR ( ASSOCIATE PROFESSOR) Department of Electronics and Communication Engineering, VBIT

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1 VLSI UNIT - III GATE LEVEL DESIGN P.VIDYA SAGAR ( ASSOCIATE PROFESSOR)

2 contents GATE LEVEL DESIGN : Logic Gates and Other complex gates, Switch logic, Alternate gate circuits, Time Delays, Driving large Capacitive Loads, Wiring Capacitances, Fan-in and fan-out, Choice of layers. 2

3 BASIC CIRCUIT CONCEPTS In VLSI design the wiring up (interconnection) of circuits takes place through the various conductive layers which are produced by the MOS processing. So, it is necessary to know the resistive and capacitive characteristics of each layer. Concepts such as resistance R S and a standard unit of capacitance c g which helps in evaluating the effects of wiring and input and output capacitances. The delays associated with wiring with inverters and with other circuitry evaluated in terms of a delay unit τ. Sheet Resistance R S The sheet resistance is a measure of resistance of thin films that have a uniform thickness. Consider a uniform slab of conducting material of resistivity ρ of width W, thickness t and length between faces A&B is L. as shown in figure. 3

4 Consider the resistance R AB between two opposite faces Where A is area of cross section Consider a case in which L = W. It means square of resistive material Where R S is ohm per square or sheet resistance Hence, the resistance of the MOS layers depend on the thickness and the resistivity of the material of the layer. Sheet resistance concept applied to MOS Transistors and Inverters: Consider the transistor structures by distinguish the actual diffusion (active) regions from the channel regions. The simple n-type pass transistor has a channel length L = 2λ and a channel width W=2λ. Here the length to width ratio denotes the impedance (Z) and is equal to 1:1. Consider another transistor has a channel length L = 8λ and width W = 2λ. Thus, channel resistance 4

5 Layer R S ohm per square 5µm Orbit Orbit 1.2µm Metal Diffusion Silicide Polysilicon n- channel X X 10 4 p-channel 2.5 X X X 10 4 Sheet resistance for Inverters L = 8λ; W = 2λ Z = L/W = 4 Sheet resistance R = Z.R S = 4 X 10 4 = 40 KΩ For pull down transistor the channel length 2λ and width 2λ, then the sheet resistance is R = Z.R S = 1 X 10 4 = 10 KΩ Hence Z p.u to Z p.d = 4:1 hence the ON resistance between V DD and V SS is the total series resistance i.e., R ON = 40 KΩ + 10 KΩ = 50 KΩ 5

6 Area capacitance of layers From the concept of the transistors, it is apparent that as gate is separated from the channel by gate oxide an insulating layer, it has capacitance. Similarly different interconnects run on the chip and each layer is separated by silicon dioxide. For any layer by knowing the dielectric thickness, we can calculate the area capacitance as follows Where, A is area of the plates, D is the thickness of Sio 2, Є 0 is the permittivity of the free space and Є ins is the relative permittivity of insulator (Sio 2 ). 6

7 Typical area capacitance values of MOS circuits Layer Value in pf X 10-4 / µm 2 (Relative values in brackets ) 5µm Orbit Orbit 1.2µm Gate to channel 4 (1.0) 8 (1.0) 16 (1.0) Diffusion 1 (0.25) 1.75 (0.22) 3.75 (0.23) Polysilicon to substrate 0.4 (0.1) 0.6 (0.075) 0.6 (0.038) Metal 1 to substrate 0.3 (0.075) 0.33 (0.04) 0.33 (0.02) Metal 2 to substrate 0.2 (0.05) 0.17 (0.02) 0.17 (0.01) Metal 2 to metal (0.1) 0.5 (0.06) 0.5 (0.03) Metal 1 to poly silicon 0.3(0.075) 0.3 (0.038) 0.3 (0.018) 7

8 Standard unit of capacitance c g It is defined as the gate to channel capacitance of a MOS transistor having W = L. i.e., standard square as shown in figure. The unit is denoted by c g. c g may be calculated for any MOS process as follows For 5µm MOS circuits Area/standard square = 5µm X 5µm = 25 µm 2 Capacitance value = 4 X 10-4 pf/ µm 2 Thus, standard value c g = 25 µm 2 X 4 X 10-4 pf/ µm 2 = 0.01 pf For 2µm MOS circuits Area/standard square = 2µm X 2µm = 4 µm 2 Capacitance value = 8 X 10-4 pf/ µm 2 Thus, standard value c g = 4 µm 2 X 8 X 10-4 pf/ µm 2 = 0.01 pf For 1.2µm MOS circuits Area/standard square = 1.2µm X 1.2µm = 1.44 µm 2 Capacitance value = 16 X 10-4 pf/ µm 2 Thus, standard value c g = 1.44 µm 2 X 16 X 10-4 pf/ µm 2 = pf 8

9 Calculation for capacitance value Consider the area defined as shown in figure of length 20λ and width 3λ Area relative to the standard gate Relative area = Area (L X W)/standard gate area = Consider the area in metal 1 Capacitance to substrate = relative area X relative C value (from table) =15 X c g =1.125 c g 2.consider the same area in polysilicon capacitance to substrate = 15 X 0.1 c g =1.5 c g 3.consider the same area in n- type diffusion capacitance to substrate = 15 X 0.25 c g =3.75 c g 9

10 10

11 We know that the transit time (τ sd ) from source to drain Here the V ds varies as C g charges from 0 volts to 63% of V dd in period τ. Thus the average value of V ds = 3V. For 5µm technology τ sd = 0.13 n sec, τ sd τ Similarly the transition point of an inverter or gate is 0.5 V DD which is approximately equal to 0.63 V DD (time constant). From this we can conclude that we can use the transit time and time constant interchangeably and stray capacitances are allowed for doubling the theoretical values calculated. Thus, τ is used as the fundamental time unit and all timings in a system can be assessed in relation to τ,hence for 5µm MOS technology τ = 0.3 nsec. for 2µm MOS technology τ = 0.2 nsec. for 1.2µm MOS technology τ = 0.1 nsec. 11

12 INVERTER DELAYS In general, the delay through a pair of similar nmos inverters is T d = (1 + Z p.u /Z p.d ) τ Assume that τ = 0.3 n sec. Then, Td = (1 + 4) 0.3= 5 τ Thus, the inverter pair delay for inverters having 4:1 ration is 5τ. Hence, a single 4:1 inverter exhibits undesirable asymmetric delays, since the delay in turning ON is τ and delay in turning OFF is 4τ. Formal estimation of CMOS inverter delay In CMOS inverter by the charging and discharging of a capacitive load C L, we can estimate the Rise time and fall time from the following simple analysis. 12

13 Rise time estimation In this analysis we assume that the p-device stays in saturation for the entire charging period of the load capacitor C L. Assume that t = τ r when V out = V DD then If V tp = 0.2V DD, then By making similar assumptions we can write for fall-time estimation, From the above two estimations we can deduce that 13 We know that So that the rise time is slower by a factor of 2.5 when using minimum size devices for both n & p

14 Cascaded inverters as drivers N cascaded inverters, each one of which is larger than the preceding by a width factor f. Now both f and N can be complementary. If f for each stage is large the number of stages N reduces but delay per stage increases. Therefore it becomes essential to optimize. 4: f 4: f 2 4:1 1:1 1: f 1: f 2 GND C L 14

15 Super buffer Use super buffer to reduce delay problems V in =1 : Q1, Q2 & Q4 Turned ON, Q3 pull down to 0 with small delay V in =0: gate of Q3 rise to Vdd, Q4 is off & Q3 made to conduct with Vdd That is twice the avg. voltage would applied if gate tied to source. 15

16 With 5μm tech driving load of 2pF with 5ns rise time 16

17 BiCMOS drivers Tr having g m & I/A char. That are greatly superior to MOS high I driving capabilities I increases exponentially in BJT (with low V be ) Effect of temp. V be depend on W B, N A, μ n & I c it is linearly depend on T T in initial time to charge B to E junction(2ns for BiCmos, 1ns for MOS & ps for GaAs) High current drive capabilities for small areas in silicon Working of bipolar transistor depends on two main timing components: 1) T in the time required to charge the base of the transistor which is large 2) T L the time take to charge output load capacitor which is less 17

18 T L - charge o/p load cap. Time is less for BJT by factor of h fe Delay of bicmos INV. is 18

19 Delay related to Rc, high Rc lead to long propagation delay. Buried subcollector region keeps Rc as low as possible. 19

20 Propagation delay Cascaded pass transistors: delay introduced when the logic signals have to pass through a chain of pass transistors 20

21 The transistors could pose a RC product delay Ex: the response at node V2 is given by the distance along the network from the input (x). C dv 2 /dt =(I 1 -I 2 )= [(V 1 -V 2 )-(V 2 -V 3 )]/R The propagation time τ p from a signal to propagate a distance x is τ p α X 2 if all sheet resistance, gate-to-channel capacitance R S and cg are lumped together R total = nr R s C total = nc C g overall delay for n sections τp = n 2 rc(τ) 21

22 Design of long polysilicon wires Long polysilicon wires contribute distributed series R and C and inconsequence signal propagation is slowed down For long polysilicon runs, use buffers 22 Effects :signal propagation is speeded up, reduction in sensitivity to noise, input voltage spends a relatively long time in the vicinity of V inv so that small disturbances.

23 Wiring capacitances (i)fringing fields For fine line metallization, the value of fringing field capacitance (C ff ) can be of the order of area capacitance. l = wire length t = thickness of wire d = wire to substrate separation. Then, total wire capacitance C w = C area + C ff 23

24 ii) Interlayer capacitances capacitance between the layers due to parallel plate effects. knowing these capacitances--the accuracy of circuit modeling and delay calculations will be improved. iii)peripheral capacitance p-diffusion regions & n-diffusion regions forms junctions with n-well & p-well resp. diffusion regions (form diode structure) Cp increases with reduction in source or drain area. (pf/unit length) Total diffusion capacitance is C diff = C area + C p 24

25 R C Low Low High Moderate Low Moderate Moderate High 25

26 Fan in and Fan-out: Fan-in: The number of inputs to a gate is called as fan - in. Fan-out: The maximum number of similar gates that a gate can drive while remaining within the guaranteed specifications is called as fan-out. Effects of Fan-in and Fan-out on propagation delay: An additional input to a CMOS logic gate requires an additional nmos and pmos i.e., two additional transistors, while in case of other MOS logic gates, it requires one additional transistor. In CMOS logic gates, due to these additional transistors, not only the chip area but also the total effective capacitance per gate also increased and hence propagation delay increases. Some of the increase in propagation delay time can becompensated by thesize-scaling method. By increasing the size of the device, its current driving capability can be preserved. Due to increase in both of inputs and devices size, the capacitance increases, Hence propagation delay will still increase with fan-in. An increase in the number of outputs of a logic gate directly adds to its load capacitances. Hence, the propagation delay increases with fan-out. 26

27 Choice of layers The following are the constraints which must be considered for the proper choice of layers. 1. Since the polysilicon layer has relatively high specific resistance (RS), it should not be used for routing VDD and VSS (GND) except for small distances. 2. VDD and GND (VSS) must be distributed only on metal layers, due to the consideration of Rs value. 3. The capacitive effects will also impose certain restrictions in the choice of layers as follows (i) Where fast signal lines are required, and in relation to signals on wiring which has relatively higher values of RS. (ii) The diffusion areas have higher values of capacitance to substrate and are harder to drive. 4. Over small equipotential regions, the signal on a wire can be treated as being identical at all points. 5. Within each region the propagation delay of the signal will comparably smaller than the gate delays and signal delays caused in a system connected by wires 27

28 Choice of layer 1) Vdd and Vss lines must be distributed on metal lines due to low Rs value 2) Long lengths of poly must be avoided because they have large Rs 3) The resistance effects of the transistors are much larger, hence wiring effects due to voltage divider effects b/w wiring and transistor resistances 4) Diffusion areas must be carefully handled because they have larger capacitance to substrate. 28

29 29 Thank you

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