Adders
RIPPLE CARRY ADDER
Introduction:
Ripple carry adder is designed in a way that mimics human’s way of adding two numbers.
Consider two decimal numbers, $A=5473$ and $B=2568$. The following explains how these two numbers are added.
Step $1$: The least significant digit of both numbers are first added. If we get a two digit number as answer, we retain one digit as sum and carry other digit to next position.
Step $2$: Then the next digit of both numbers are added with the carry of previous digit.
Step $3$: Repeat step $2$ until most significant digit is reached.
Binary Addition:
The method of adding two binary numbers is same as decimal addition, hence
Why the name???
The above type of addition method is termed as Ripple carry addition.
Why the name??
Ripple Effect: As in the above image, a small droplet causing small disturbance propagates/spreads outward to make a huge wave is called as Ripple effect.
In the above addition process, carry that is generated in the first stage propagates to the next position, then to next, next next… like the ripple effect.
That’s why the name ‘Ripple Carry Addition’
Implementation of Ripple Carry Addition:
As we can see, Decimal addition, Hexadecimal addition, Binary addition, etc., follows the same procedure. Hence we can generalize it as follows,
Whatever be the size of the number, we always add it bit by bit. So, to realize the above addition operation, we need $1$-bit adders. As we discussed earlier, half-adder and full-adder are $1$-bit adders.
Half-Adder: adds two $1$-bit numbers $$\{ C_{out}, Sum\}=A+B$$
Full-Adder: adds three $1$-bit numbers $$\{ C_{out}, Sum\}=(A+B)+C_{in}$$
Implementation of Ripple carry addition is explained bit by bit,
Stage $1$ $(A_0,\;B_0)$ :
Addition of LSB of $A$ and $B$ involves adding two bits $A_0$ and $B_0$ and getting Sum$(S_0)$ and Carry $(C_0)$ as output. This stage addition process can be realized using an Half-Adder.
Stage $2$ $(A_1,\;B_1)$ :
Next step in adding $A$ and $B$ involves adding two bits $A_1$ and $B_1$ and incoming carry $C_0$, which gives us Sum$(S_1)$ and Carry $(C_1)$ as output. This stage addition process can be realized using a Full-Adder.
Stage $3$ $(A_2,\;B_2)$ :
This stage addition process also can be realized using a Full-Adder.
Stage $4$ $(A_3,\;B_3)$ :
This stage addition process also can be realized using a Full-Adder.
The above addition process can be realized using a Full-Adder.
Combining the above four adders, we will get a $4$-bit Ripple Carry Adder.
First adder stage can be implemented with half-adder (as shown above) or with full-adder (as shown below),
Practice Question
How many EXOR gates are needed to implement a four bit ripple carry adder?
Correct Answer: Option A
To implement a four bit ripple carry adder, we need $3$ full adders and $1$ half adder.
One full adder has $2$ EXOR gates and one half adder has $1$ EXOR gate.
No of EXOR gates needed $=(3×2)+1=7$