Digital Root circuit using tree of 4-bit CLA's with Cout fed into Cin

Jul 16, 2003 3 Replies

Hi,



In order to generate hexadecimal Digital Root of a number (i.e, sum-of-hexdigits until it reduces to a single hex number - a nibble)



For e.g 0x123a = 0x1 + 0x2 +0x3 +0xa = 0x10 = 0x1 + 0x0 = 1 is the hex digital root of (0x123a)



I am planning to use tree of 4-bit Carry-Look-Ahead adders with the Cout fed into Cin of the same CLA.



Question: Can I directly fed back the Cout signal to Cin at each CLA to get the Digital Root . Will it not form a combinational loop? Due to feedback will it affect the next level CLA's o/p.



Thanks lrl


I believe that can be done. Consider how a ones complement adder works.

Well, I would think that you would need G and P in, but other than that it should be fine.

I thought about this some time ago wondering why anyone would build ones complement machines, such as CDC did. If they couldn't do a fast add that would be a big disadvantage. I then decided that CLA should work.

Now, it might be that there is a better way to do the problem you are asking about, but that is a different question.

-- glen

I believe that can be done. Consider how a ones complement adder works.

Well, I would think that you would need G and P in, but other than that it should be fine.

I thought about this some time ago wondering why anyone would build ones complement machines, such as CDC did. If they couldn't do a fast add that would be a big disadvantage. I then decided that CLA should work.

Now, it might be that there is a better way to do the problem you are asking about, but that is a different question.

-- glen

I believe that can be done. Consider how a ones complement adder works.

Well, I would think that you would need G and P in, but other than that it should be fine.

I thought about this some time ago wondering why anyone would build ones complement machines, such as CDC did. If they couldn't do a fast add that would be a big disadvantage. I then decided that CLA should work.

Now, it might be that there is a better way to do the problem you are asking about, but that is a different question.

-- glen

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