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Division Circuits

Jan 2013

Shmuel WimerBar Ilan University, Engineering Faculty

Technion, EE Faculty

2Jan 2013

Sequential DivisionLike sequential multiplication of k-bit operands, yielding a 2k-bit product, the division of 2k-bit dividend by k-bit divisor can be realized in k cycles of shift and subtraction.

2 1 2 2 1 0

1 2 1 0

1 2 1 0

1 2 1 0

: Dividend: Divisor: Quotient

: , Reminder

k k

k k

k k

k k

z z z z zd d d d dq q q q q

s z q d s s s s s d

3Jan 2013

2k bitsk bits

First subtraction must consider k+1 MSBs for non negative difference. Considering k bits would mean that quotient is 2k at least, which requires k+1 bits at least.

4Jan 2013

Since for unsigned division we have q<2k and s<d, to

avoid overflow we must have

z<(2k-1)d+d=2kd

d must be strictly greater than the k MSBs of z, as

otherwise the quotient would have more than k bit.

This is called overflow detection and done prior to

division and can be used also to detect division by

zero.

5Jan 2013

Fractional Division

2 22 2 2 2k k k kz d q s z d q s

Unsigned integer and unsigned fractional can be converted to each other as follows

Since z is 2k-bit and d, q, and s are k-bit, we obtain

frac frac frac frac2 kz d q s

We can therefore divide fractions just as integers, except that the reminder is right shifted by k bits.

Overflow detects that zfrac < dfrac , as otherwise the quotient would be no smaller than 1.

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Sequential Bit-at-a-Time Division

Jan 2013

It is performed by initializing s(0)=z, and successively

subtracting from it the properly shifted terms qk-jd.

Rather than shifting qk-jd rightwards, we shift the

partial quotient leftward, leading to left-shift division

algorithm.

1 02 2 with and 2j j kk kk js s q d s z s s

shift left

subtract

7Jan 2013

1 02 2 with and 2j j kk kk js s q d s z s s

shift left

subtract

2k pre multiplies d to unsure proper alignment. After k iterations the above recursion turn into

02 2 = 2 2k k k k ks s q d z q d s

1 0frac frac fracfrac frac frac frac2 with and 2j j k k

js s q d s z s s

In the fractional version the dividend is aligned to the radix point. The bits of the quotient begin from -1.

multiply divisor by 2k

initialize reminderleft-shiftsubtract

left-shiftsubtract

left-shiftsubtract

left-shiftsubtract

Jan 2013 8

9Jan 2013

1fracfrac frac

0fracfrac

fracfrac

2

with

and 2

j jj

k k

s s q d

s z

s s

In fractional division the dividend is aligned to the radix point. The bits of the quotient begin from -1.

10Jan 2013

Unlike multiplication that can be done by right and

left shifts, division can be done only by left shift.

This follows from the bit of the quotient that are

known progressively, starting from MSB. In

multiplication, then multiplier bits are known in

the outset.

11Jan 2013

The partial reminder is left shifted and its MSB is loaded into a special FF.

A trial difference2s(j-1)-qk-j(2kd) is computed.

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