# VLSI Homework 12 - Aidan Sharpe

## Problem 12.5
Estimate the minimum delay of a 10:1024 decoder driving an electrical effort of $H=20$.

A 10:1024 decoder will contain 1024 AND gates. It will have a branching effort of $B=1024/2=512$. Assuming $G\approx1$ for a 10-input AND gate, the path effort will be $F=GBH=40960$. The optimal number of stages will be, assuming the stage effort is $\rho=4$, $\hat{N} = \log_4(40960)=7.66$. We will use 8 stages in our solution, seen below.

![](decoder-six-stage.png)

### Using static CMOS gates
For a static CMOS design, $G = (4/3)^4 = 256/81$. Therefore, the actual stage effort is $F = GBH = 32363.5$. The parasitic delay is $P = 12$, so the total delay will be $D=NF^{1/N} + P = 41.3\tau$.

### Using footless domino gates
A footless domino design has $g=1/3$ for inverters and $g=2/3$ for NAND2. Therefore, $G=(1/3)^4(2/3)^4=16/6561$, and $F=GBH=24.97$. The parasitic delay for inverters is $p=2/3$, and $p=1$ for NAND2. Therefore, the path parasitic delay is $P=4 + 8/3 = 20/3$. Finally, the total delay will be $D=NF^{1/n} + P = 18.62\tau$.

## Problem 12.12
NAND ROMs will have a higher resistance than a comparable NOR ROM, thereby making them slower.