Full Adder Question

[he11fire]

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 ...

 

[ericgibbs]

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**

 

[birdman0_o]

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!

 

[he11fire]

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**

Sorry but I don't understand what do you mean by 000h to 3E8h ...is that in hex ?

 

[ericgibbs]

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

 

[he11fire]

Yes , thanks for your help

 

[he11fire]

Yes ,exactly ,...thanks

 

[stephen87]

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!