Serial Binary Subtracter

A serial subtracter can be obtained by converting the serial adder using the 2's complement system..could anyone explain that with this corresponding figure.

I don't think the subtractor is using two's complement, it looks more like one's complement. The way ones complement arithmetic works is to do an end around carry(borrow). I see the Flip-Flop for holding the carry state but I don't know if the end around carry happens on step four or if there is an extra step. I also don't know what the inputs to the subtractor are after four steps.

Papabravo said:

I don't think the subtractor is using two's complement, it looks more like one's complement. The way ones complement arithmetic works is to do an end around carry(borrow). I see the Flip-Flop for holding the carry state but I don't know if the end around carry happens on step four or if there is an extra step. I also don't know what the inputs to the subtractor are after four steps.

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I think that also, it's more like one's complement than 2's complement but that's maybe because Cout is fed back to Cin so when Cout is "1" then Cin="1" so it would be added with X & Y .........

BTW, what did you mean by step four?!

thanks

End-around carry would require feeding the entire result back into the input to propagate the carry. Otherwise, a sequence of low-order 1's won't be converted to a sequence of 0's.

To get two's complement, instead of adding 1 after complementing, immediately add the 1 by setting the carry bit to 1. The complementing is still in the data path preceding the adder, so effectively the complementing "happens" before adding 1.

In short, the circuit will perform two's complement subtraction, provided the carry flag is set to 1 before clocking the circuit.

I think tkbits is correct in his analysis of adding 1 by setting the carry. Step 4 is the one that subtracts the most significant bits of the two operands. Step 1 would subtract the least significant bits and the initial carry.