EE477 - s05 solution to study problem 5

A. Parker EE 477L Spring 2005

Study Problem 5 Solution

5. An inverter drives a long wire (5 mm) on a special kind of metal1 minimum width

(3 lambda) to the input of another inverter. Both these inverters have identical

size transistors. In order to speed up the circuit, Ernie Engineer inserts an

identical inverter in the middle of the 5 mm wire. Ernie’s boss wants Ernie to

insert an inverter that has transistors 3 times as wide as the other inverters in the

middle of the wire.

a) Compute the capacitance of each 2.5 mm section of the wire, assuming the

oxide under the wire is two layers of oxide thick. Assume the special kind

of metal1 has resistance .01 ohms per square.

C = ε₀ε_SiO2 A/t

A is 2500*.375 microns sq. =937.5 microns sq.=

.9375 × 10^-05 centimeters squared

Cint = 8.85x10^-14 F/cm*3.9*.9375 x 10^-5 cm²/(5000 x 10_-8cm)= 64.7 fF

Rint = .01 ohms/square * 2500/.375 square = 66.67 ohms

b) Show an RC ladder network for the original circuit that can be used to

compute Elmore delay. Assume Cg(n or p) = 100 ff, Rchn = 500 ohms, Rchp =

1000 ohms and Cd (drain or source) = 40 ff for the original inverters.

I only showed the rise time equivalent circuit here because it is the slowest. I also lumped Cg and Cint since they are essentially in parallel.

c) Use equivalent circuits and the RC time constants to show Ernie’s boss

whether Ernie’s design is better than the boss’s design or not.

Here the equivalent circuits have the same appearance as in part b. The only differences are that

a) Rint and Cint are half the values shown in part b, since we are cutting the wire in half

b) We must show a rising output followed by a falling output or vice versa.

c) For the Boss's design, the first inverter drives 3 Cg(n+p) and the second inverter possesses Rchboss=Rchoriginal/3, and Cdboss= Cd (n+p)*3

We break the equivalent circuit into two components and sum the RC delays.

RC total = RchpCd(n+p) + (Rchp + Rint) (Cint + Cg (n+p)) first to second inverter rising

+ RchnCd(n+p) + (Rchp + Rint) (Cint + Cg (n+p)) second to third inverter falling

For the boss we get RCtotal = 1000*80 + 1067 *(64.7 + 200*3)

+ 500/3 *80 * 3 + (234)(64.7 + 200) = 891 ns.

If we had reversed the order from falling to rising, we would have gotten .603 ns.

For Ernie we get RCtotal = .552 nsec. We get the same answer whether we look at the rising-falling or falling-rising combination.

So Ernie's design is faster than his boss's. Without the extra inverter the delay would have been .452 ns.

We could have tested whether to insert an inverter or not using the approach I did in Lec. 18, but the test is somewhat different because Rchn does not = Rchp