02.12 Adders

Addition of Two 1-Bit binary numbers - Half Adder

$$\begin{array}{lcc}& & x\\ +& & y\\ & c& s\end{array}$$

Here, we have $s$ the sum of the, and $c$ the carry

Let’s draw out the truth table of what happens :

$$\begin{array}{cccc}x& y& s& c\\ 0& 0& 0& 0\\ 0& 1& 1& 0\\ 1& 0& 1& 0\\ 1& 1& 1& 1\end{array}$$

We then can conclude that the logical expression representing addition is given by

$$\begin{array}{lccc}& & & \\ s& =& \bar{x}y+x\bar{y}& =x\oplus y\\ c& =& xy& \end{array}$$

Addition of Two N-Bit binary numbers - Full Adder

A binary $n$-bit adder has two operands $0\le x,y\le 2^n-1$ and a carry-in $c_{\text{in}}$ as inputs It produces a sum of $0\le s\le 2^n-1$ and a carry out of $c_{\text{out}}$

We have the sum of the $i$-th term, and the carry given by

$$\begin{array}{lcc}{s}_i& =x_i\oplus y_i\oplus c_i& \\ c_{i+1}& =(x_i\oplus y_i)c_i+x_i{y}_i& =c_{i+1}=x_i{y}_i+x_i{c}_i+y_i{c}_i\end{array}$$

In fact, we can construct it from two half adders:

In the optimized form, we get the circuit given by:

Basic Ripple-Carry Adder

If we daisy-chain the full adders together (with the carry-in wired to the carry-out of the previous adder), we can do the sum of $n$-bit numbers

Subtractors

Subtractors are basically the same thing as adders

N-Bit Unsigned Subtractor

We again start with only one circuit for two bits, instead of doing a truth table for the full circuit

We can then daisy-chain them, just as we did with adders to make a ripple-carry subtractor

In Two’s Complement

Adders don’t change, let’s do subtractors

Subtractors

The nice thing we can do is add a complemented form of the second number to do subtraction

$$\text{SUB}(X,Y)=\text{ADD}(X,\bar{Y},1)=X+\bar{Y}+1$$

Adder-Subtractor in TC

We have a control signal $m$ that determines the mode (whether we do addition or subtraction)

$$S(X,Y,m)=\{\begin{array}{ll}X+Y,& \text{if }m=0\\ X+\bar{Y}+1,& \text{if }m=1\end{array}$$

We can then have

$X$	$Y$	$m$	$S(X,Y,m)$
Any	Any	$0$	$X+Y$
Any	Any	$1$	$X+\bar{Y}+1$
Which gives this expression $S=\bar{m}(X+Y)+m(X+\bar{Y}+1)$

By simplifying the expression, we get

$$S=X+m\oplus Y+m$$

Fast Adders

We know that performance matters The system value is given by

$$\text{value}=\frac{\text{performance}}{\text{cost}}$$

For full ripple-carry adders, the “ripple” has to reach the last carry-out to get the correct result This is not good for performance

The Critical path is the slowest input-to-output path and determines when the result is ready