14.45 For a CMOS inverter fabricated in a 0.13-m process with VDD = 1.2V, Vtn =

Question

14.45

For a CMOS inverter fabricated in a 0.13-m process with VDD = 1.2V, Vtn = Vtp = 0.4 V, kn = 4kp =430 A/V2 , and having (W/L)n = 1.5 and (W/L)p = 3.

A) Find tPHL, tPLH , and tP when the equivalent load capacitance C = 10 fF. Use the method of average currents.

B)  Insert a 40 nH inductor in series with the capacitor and Determine the shape (equation) for the waveform using a similar method as described in class with the addition of the inductor and calculate the new propagation times. (RC circuit is now a RLC circuit)

C) Design a NOR gate using transistor sizes to get equivalent timing performance as the inverter.

Explanation / Answer

Using average current methods, Iavg(HL) = 0.5*[(ic(Vin=VOH,Vout=VOH) + ic(Vin=VOH,Vout=V50%)] where PMOS is off. NMOS is on.

Where Vin=VOH =VDD=1.2V and V50% =0.6V,Kn =430micro

ic(Vin=VOH,Vout=VOH) = 0.5*Kn (VOH -Vtn)2 = 0.5*430micro*(1.2-0.4)2 = 0.136mA

ic(Vin=VOH,Vout=V50%) =0.5*Kn [2(VOH -Vt)V50% - V50%2 ] = 0.129mA

Therefore Iavg(HL) = 0.5(0.129m +0.136m) = 0.132mA.

Therefore tPHL = Cload *(VOH - V50% )/Iavg(HL) = 10fF*(1.2-0.6)/0.132m = 45.4psec.

Now,Iavg(PLH) = 0.5(Ic(VOL ,VOL ) + Ic(VOL ,V50% )),VOL =-1.2V ,Kp = 430micro/4=Kn /4

Ic(VOL ,VOL ) = 0.5*Kp (VOL -Vtp)2 = 34.4microA

Ic(VOL ,V50% ) = 0.5*Kp [2(VOL -mod(Vt))V50% - V50%2 ] = 32.25microA

Icavg(PLH) = 33.33microA

tPLH = Cload*(V50%- VOL )/Icavg(PLH) = 18nsec

Therefore tP = 0.5(tPHL + tPLH) = 11nsec

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