Help wanted

Which 8-bit processor?

Maybe if you googled "multiple-precision arithmetic" you'd find some suggestions.

Or, since (2^45/Fclk) is invariant, what about just a LUT?

Good Luck! Rich

You will need to change it to 6 or 8 bytes from 4 bytes, but the theory is the same. Here is a code segment from a multimeter that uses the routine "Ratio" to divide two 32-bit numbers when calculating ohms. It is written for an 8-bit "AVR". Sorry it's not "C". if it's a homework problem, it's just a constant anyway ;-)

Regards,

Frank Raffaeli

Good idea, dear Sphero! Many thanks !!

You need an exact 48-bit answer, so with either an 8-bit, 16-bit or

First factor the Fclock (presumably fixed, right?) to

Fclock = 2^8 * 390625 = 2^8 * 0x5F5E1

Then you can write

Data = Fout* (2^40/0x5F5E1) = (Fout * ((2^40 * 2^k)/0x5F5E1))/2^k

Round and use the smallest value of k that gives you accuracy down to the last bit. You can reduce this to shifts and multiplication by a constant, which might be desirable since divide is typically quite slow.

Il Fri, 3 Mar 2006 16:16:05 +0100, nonostante la corrente di placca pericolosamente elevata, Roberto IZ5FCY, senza effettuare gli accordi, volle trasmettere :

.. ehmm... sorry ! Sephro ;-)