Full Adder Question

he11fire

New Member
How many 74LS283 adders would be required to add two binary numbers each representing decimal numbers up through 1000 (in decimal) ?

I'm a little confused with the part "up through 1000" ...maybe my english is not that good ...
 
How many 74LS283 adders would be required to add two binary numbers each representing decimal numbers up through 1000 (in decimal) ?

I'm a little confused with the part "up through 1000" ...maybe my english is not that good ...

hi,
The way I read it is:
numbers 1 and 2 , can both have a range value of 000h thru 3E8h

Is this what you are asking.?

**broken link removed**
 
Last edited:
The question is basically, "How many bits in 1000" and then how many adders to add that many bits.
1000 = 2^n implies that n = log(1000)/log(2)... n = 9.9(etc) bits so 10 bits.

I assume each adder is a 4 bits, so we would need 3 of them!
 
Sorry but I don't understand what do you mean by 000h to 3E8h ...is that in hex ?
hi,
Yes.
0011 1110 1000 its 10 bits wide for the input values and 11 bits wide for the output values
 
you need a 10bit adress since the most near total addresses are 1024 (2^10). so the 74LS283 is not able to address 1000, because the maximum addresses is of 2^4 = 16 address.
Solution is to put 3 of them and will end up to a 12 bit address!
 
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