1445 For a CMOS inverter fabricated in a 013m process with V

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.

Solution

Using average current methods, Iavg(HL) = 0.5*[(i_c(V_in=V_OH,V_out=V_OH) + i_c(V_in=V_OH,V_out=V_50%)] where PMOS is off. NMOS is on.

Where V_in=V_OH =V_DD=1.2V and V_50% =0.6V,K_n =430micro

i_c(V_in=V_OH,V_out=V_OH) = 0.5*K_n (V_OH -V_tn)² = 0.5*430micro*(1.2-0.4)² = 0.136mA

i_c(V_in=V_OH,V_out=V_50%) =0.5*K_n [2(V_OH -V_t)V_50% - V_50%² ] = 0.129mA

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

Therefore t_PHL = C_load *(V_OH - V_50% )/Iavg(HL) = 10fF*(1.2-0.6)/0.132m = 45.4psec.

Now,Iavg(PLH) = 0.5(Ic(V_OL ,V_OL ) + Ic(V_OL ,V_50% )),V_OL =-1.2V ,K_p = 430micro/4=K_n /4

Ic(V_OL ,V_OL ) = 0.5*K_p (V_OL -V_tp)² = 34.4microA

Ic(V_OL ,V_50% ) = 0.5*K_p [2(V_OL -mod(V_t))V_50% - V_50%² ] = 32.25microA

Icavg(PLH) = 33.33microA

tPLH = Cload*(V_50%- V_OL )/Icavg(PLH) = 18nsec

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