Arithmetic Circuits – Addition, Subtraction, & Multiplication
The adder is another classic design example which we are obliged look at.
Simple decimal arithmetic is something which we rarely give a second thought
to, but it is useful to closely examine the way we do this before we think about
how to achieve it using logic devices and binary arithmetic.
Consider the following sum:
3 8
+ 514
----9 2
First we took the eight and the four and added them which produced a result
of 12, or in other words a sum of two was produced and a carry of 1. We
wrote down the 2 and put the carry in the next column to the left. Next we
added the five and the one to produce a sum of 9. This time no carry was
generated.
If we want to build a device to perform the addition it might look like this:
3
5
+
9
8
carry
4
+
2
The box to the right adds two numbers and produces two outputs, the sum
and the carry. The box to the left however, adds three numbers: the 3, the 5
and the carry from the right hand box.
We can generalise our design:
An Bn
Sn
A1 B1
A0 B0
S1
S0
Performing the addition in binary is much the same but a little more tedious.
1001 10
+ 1 1 0 1111 0
--------------10111 00
(38 in decimal)
(54 in decimal)
(92 in decimal)
If we now move on to consider what happens inside these black boxes it
should be apparent that the box on the right is simpler than the others
because it only ever adds two numbers. For reasons which (might!) become
clear later we call this box a ‘half adder’ and the others we call ‘full adders’.
As usual we can express the requirements in a truth table. Our half adder has
two inputs: A and B, and two outputs: S and C (really we have two truth tables
in one).
A
B
S
C
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
Hopefully you have spotted that S is merely an EX-OR operation and C is a
AND operation, so our half adder is simply:
A
B
S
C
With a little thought we can figure out how to build a full adder from two half
adders (hence the names). Remember that the full adder is adding three bits,
let’s call these X, Y, and Z, so the sum we need is S = X+Y+Z. It doesn’t
matter what order we add them in as X+Y+Z = (X+Y)+Z = X+(Y+Z), so let’s
take X and Y and add them first to produce an intermediate sum, R.
X+Y=R
We only have two bits (X and Y) so we can do this with a half adder and our
final result will be given by:
S=R+Z
We can do this with a half adder too, so we’re almost home and dry. Our full
adder is beginning to take shape:
Z
X
Y
S
R
Ca
Cb
We still have to deal with the carries. Our immediate thought might be to add
them both together using another EX-OR gate. This will certainly work but it
isn’t the simplest solution, nor is it one you will ever see in any textbook! Think
carefully about the difference between an Exclusive OR gate and a normal
(Inclusive) OR gate.
The only difference occurs when both inputs are 1 , the OR has an output of 1
and the EX-OR an output of 0. Returning to our half finished full adder, on
close examination it turns out that the two carries, Ca and Cb in the diagram,
can never both be 1, so it makes no difference whether we use an OR or an
EX-OR! OR gates are simpler than EX-OR gates so it makes sense to use
the easiest option and our completed full adder is:
Cin
A
B
S
Cout
Unfortunately this is a somewhat contrived example, in reality we wouldn’t
build full adders we’d buy a chip containing several of them, however it is
useful to understand how they work.
Subtraction
We could repeat this exercise for subtractor circuits but there isn’t much point.
We already have a technique whereby we can use the adder to perform
subtraction. Suppose you want to subtract B from A (A-B), this is exactly the
same as saying A+(-B). In other words, if we have a way of representing
‘minus B’ then all we need to do is add this to A to produce a result. We
already have such a representation in the twos complement scheme we
looked at earlier. To negate a number all we need to do is invert it and add 1.
A – B = A + (-B) = A+ /B + 1
This is easy to achieve using the adder. We can invert B using inverters and if
we use a full adder instead of a half adder we can add one by using the carryin of the first stage. With a little ingenuity we can use the same circuit to
perform addition and subtraction. The diagram illustrates a four bit
adder/subtractor designed around a four bit adder. The EX-OR gates will
invert B if the switch is set to 1 or do nothing if it is set to 0, check the truth
table for EX-OR to confirm this.
Multiplication and Division
It would be nice if we could design similar circuits for multiplication and
division, however it isn’t easy to design such circuits and they tend to be very
complex and expensive. In actual fact, most microprocessors do not even
have multiplier circuits, they cheat!
It is very easy to multiply a decimal number by 10 or a binary number by 2. If
we multiply a decimal number by 10 we simply add a zero to the right hand
side eg 28 x 10 = 280. Similarly, adding a zero to the right hand side of a
binary number is equivalent to multiplying the number by 2. Adding a zero to
the right hand side of a hexadecimal number is equivalent to multiplying by 16
etc. etc.
Consider the binary number
Add an extra zero:
1011 (eleven in decimal)
10110 (twenty two in decimal)
A more general way of thinking about this, particularly if fractions are involved
is ‘shifting the point’. Each time we shift the decimal point to the right we
multiply the number by 10, or if we shift it to the left we divide by 10. Because
of the way binary numbers are stored in the logic circuits of computers it
actually makes a lot more sense to shift the number rather than the (binary)
point. Consider the following example:
128
64
32
16
8
4
2
1
0
0
0
0
0
1
1
0
1
1
1
1
1
1
0
0
1
1
1
0
1
1
1
0
------ a
------ b
------ c
Beginning with a (59 decimal) we shift one position to the right to obtain b (29
decimal), and one place to the left to obtain c (118 decimal). There is a slight
complication if negative numbers are involved; in this example a zero was
shifted in on the left hand side of b, and a one was lost from the right hand
side. Suppose our starting number was 10000000 (-128 in twos complement
notation). If we applied the same procedure we’d get 01000000 (+64) by
shifting it right one place. If we’d shifted a one into the left hand side we would
have 11000000 (-64) which is the correct answer. The solution to our problem
is to shift a zero in when the number is positive (the msb is zero) and a one
when it’s negative (the msb is one). You will encounter this later when we look
at microprocessors in more detail.
Now we can think about multiplying by any number, rather than just ‘two’.
Once again it is useful to think about decimal arithmetic; let’s suppose we
want to multiply the two decimal numbers 425 and 218. Nowadays most of us
would simply reach for an electronic calculator to do this and remembering
how to do it without one might be a challenge!
425
218
--------85000
4250
3400
--------92650
------ 1
------ 2
------ 3
In step 1 we began by writing two zeroes and then multiplied 425 by 2.
In step 2 we wrote one zero and multiplied 425 by 1.
In step 3 we multiplied 425 by 8.
Finally, adding the three numbers together produced our result: 92650
Let’s do this again, but this time, instead of 218 we’ll use 101 (one hundred
and one) as the multiplier.
425
101
--------42500
------ 1
0000
------ 2
425
------ 3
--------42925
Spot the difference? This time it was a lot easier because our multiplier only
contains 1s and 0s.
In step 1 we wrote down the two zeroes and then multiplied 425 by 1. Of
course there is nothing tricky about multiplying any number by 1!
In step 2 we multiplied 425 by 0 which is fairly trivial too!
Let’s try once more, this time we’ll replace the 425 by 111 (one hundred and
eleven)
111
101
--------11100
------ 1
0000
------ 2
111
------ 3
--------11211
There shouldn’t be any surprises so far. What if 111 and 101 were actually
binary numbers (7 and 5 respectively in decimal)? Steps 1,2, and 3 will be
exactly the same! We just have to remember we’re working in binary when we
add everything together:
111
101
--------11100
0000
111
--------100011
------ 1
------ 2
------ 3
The result 100011 is what we’d expect (7 x 5 = 35) and we’ve succeeded in
multiplying our numbers without really doing any multiplication at all! All that
we’ve done is shifted the 111 two places to the left in step 1. Done nothing in
step 2, then written down 111 in step 3, and finished off by adding everything
up.
This is an extremely common method of performing multiplication. So
common in fact, that it has its own, rather unimaginative name, ‘the shift and
add algorithm’!
If you’re wondering how on earth you would draw Karnaugh maps to describe
this operation and build a circuit then don’t worry. You can’t. This algorithm is
sequential, there are several steps involved in producing an answer and the
logic design methods that we’ve looked at so far simply can’t do this.
Microprocessors are very good at this kind of thing and even very simple
devices often have ‘Multiply’ instructions. Usually, when such an instruction is
executed, a lot of shifting and adding happens behind the scenes.
Division is another story, although it is somewhat similar. Just as we can live
without dedicated circuits to perform multiplication, we can get by without
divider circuits too. We won’t explore division here but you might not be too
surprised that a ‘shift and subtract’ algorithm is usually used.
Lateral Thinking
Is the method outlined above, the only way in which we can multiply two
numbers together? Perhaps in your primary school days you used ‘table
cards’?
A table card is a very simple device for avoiding arithmetic (although they are
usually intended to encourage mental arithmetic!)
Consider the table below:
1
2
3
4
5
6
7
8
9
10
11
12
1
1
2
3
4
5
6
7
8
9
10
11
12
2
2
4
6
8
10
12
14
16
18
20
22
24
3
4
3
4
6
8
9
12
12
16
15
20
18
24
21
28
24
32
27
36
30
40
33
44
36
48
5
5
10
15
20
25
30
35
40
45
50
55
60
6
7
6
7
12
14
18
21
24
28
30
35
36
42
42
49
48
56
54
63
60
70
66
77
72
84
8
8
16
24
32
40
48
56
64
72
80
88
96
9
10
9
10
18
20
27
30
36
40
45
50
54
60
63
70
72
80
81
90
90
100
99
110
108
120
11
11
22
33
44
55
66
77
88
99
110
121
132
12
12
24
36
48
60
72
84
96
108 120 132
Using this table we can perform multiplication by simply ‘looking up’ the
answer. For example, in the table above we can see that the product of 7 and
9 is 63.
144
Could we use such a device in a computer using a ‘lookup table’ in the
computer’s memory? When we multiply two numbers we call one of them the
‘multiplier’ and the other the ‘multiplicand’. Suppose that our multiplier and
multiplicand are both 8-bit integers, this means that the product may be a 16bit integer. It also means that our table will be big; 256 rows and 256 columns
(including zero). This means that the table will require 65,536 16-bit words
(128 kilobytes!) of storage. That’s a lot of memory for a 8-bit multiplication!
There is a way around this. Consider the following:
(a + b)2 = a 2 + b2 + 2ab
(a – b) 2 = a2 + b2 - 2ab
---------- (1)
---------- (2)
Subtract (2) from (1):
(a + b)2 - (a – b) 2 = 4ab
Rearranging this:
ab = (a + b)2 - (a – b) 2
4
So we can multiply two numbers by finding their sum and differences.
Squaring these. Subtract one from the other, and then dividing the result by
four. At first this seems like a perfectly idiotic way of doing things but let’s
check that it does actually work: Let a=9 and b=7 to match the example
above.
9x7 = (9 + 7 )2 – (9 – 7) 2
4
9x7 = 162 – 2 2
4
9x7 = 252
4
9x7 = 63 It works!
How does this help us? We can now use a lookup technique in this approach
because only a table of squares is required. Following on from the example
above, our table will be much smaller and look like:
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
n2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
196
289
324
361
400
441
484
529
576
625
So, in our example, to ‘calculate’ 9x7 we find the sum and difference; 16 & 2
Now look up their squares; 196 & 4. Then 196 – 4 = 192.
Divide 192 by 4 to give the result of 63 (we could have missed this stage by
using a table which is already divided by 4).
However crazy this might seem, it can be significantly faster than a ‘shift and
add’ algorithm.