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Question

How to find the gate delay of carry out \$C_{n}\$ and sum \$S_{n}\$ in ripple carry adder?

I encountered this doubt while going through the book by Carl Hamacher.

It is written as:

Using the implementation indicated in figure (I have attached), \$C_{n−1}\$ is available in \$2(n − 1)\$ gate delays, and \$S_{n−1}\$ is correct one \$XOR\,\,\$ gate delay later. The final carry-out, \$C_{n}\$, is available after \$2n\$ gate delays.

enter image description here

I am completely stuck, how gate delay of carry out \$C_{n}\$ and \$S_{n}\$ is \$2{n}\$ despite carry is using total of 4 gates ans sum only 1 gate?

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  • \$\begingroup\$ Looking at it quick he appears to mean 2 gate delays times n stages (n is however many bits wide are being added). The AND, OR, and XOR gates only depict 1 bit adder, but block (FA) is a multiple bit adder. \$\endgroup\$ – Entrepreneur Jun 5 '17 at 16:18
  • \$\begingroup\$ @Entrepreneur let it be \$n\$ bit adder consisting of obviously \$n\$ block.My question is that each block will take \$1\$gate delay ,then it must be total of \$n\$ delay ,how \$2n\$ is coming? \$\endgroup\$ – laura Jun 5 '17 at 16:21
  • \$\begingroup\$ I am looking at the 3 AND gates feeding the OR gate. A carry-in (Ci) enters an AND gate then passes from the AND gate output through the OR (2 gates total traversed) before the Ci+1 output propagates to the next stage. \$\endgroup\$ – Entrepreneur Jun 5 '17 at 16:26
  • \$\begingroup\$ okk,i too assumed the same but got confused in case of sum .why in case of sum , \$2n\$ gatedelay is used ?it is using a single \$3\$ input XOR gate.shouldn't be it \$n\$? \$\endgroup\$ – laura Jun 5 '17 at 16:29
  • \$\begingroup\$ Each stage has two functions. 1) The XOR creates the new value of the output bit for the current stage bit result. 2) The AND-OR group creates the carry-out to the next stage. All of the carry bits must ripple from the lowest to highest stage to have a complete addition (like adding 1 to 99999 changes all higher digits). So the carry logic (can) delay the guaranteed result. \$\endgroup\$ – Entrepreneur Jun 5 '17 at 16:36
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The figure on the right of your image is the carry of one bit position. Given a wordlength of \$n\$ bits and position \$i\$, the carry \$c_{i+1}\$ goes to the exact same circuit, just with inputs \$x_{i+1}\$, \$y_{i+1}\$ and \$c_{i+1}\$. So to calculate the carry of the total summation \$c_{n+1}\$ the signal takes \$2 \cdot t_{Gate-Delay}\$ times the number of bits: \$2 \cdot n \cdot t_{Gate-Delay}\$

Since the sum \$s_i\$ is calculated using two \$XOR\$-Gates it needs \$2 \cdot t_{Gate-Delay}\$ as cascading \$XOR\$-Gates is done by put them one ofter another, so calculating \$s_n\$ also takes \$2 \cdot n \cdot t_{Gate-Delay}\$.

xor cascading

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The carry output is made up of "3 AND gates" and "one OR gate" as can be seen from the figure in the right side. The inputs to the "3 AND gates" is provided simultaneously that's why the effective delay from "3 AND gates" is "1 only" and not 3, there is another delay caused by the "OR" gate, so to generate a carry at initial position 2 gate delays are required thus to generate carry at nth position 2n gate delays are required because each carry-out dependent on the previous carry-in.

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