I have trouble understanding what's critical path delay of n-bit Ripple Carry Adder. In the book I read, given N-bit Ripple Carry Adder formed from N single 1-bit full adder:
the critical path delay is 2N
I know that for each single 1-bit full adder, the Cout(Carry-out)-delay is 2 and the Sum-delay
is 3, from this observation why the critical path delay is not 2(N-1)+3 = 2N+1 (I mean the last one is calculated with Sum-delay instead of Cout-delay)? So assume that these 1-bit full adders are denoted by A(0) to A(N-1), the critical path delay is from Cin(A0) to Cout(A(N-1))? So I should not substitute the last one with Sum-delay?
the delay to get the sum (Sum delay) is 2N+1
I can understand this one, since now 2(N-1)+3 = 2N+1 should work, but isn't that this implicitly assumed that the last 1-bit full adder will be the slowest one? (You may skip this one if you want since my main focus is 1.)
Edit: I found a YouTube video explains why I mean, but I don't what whether he mean critical path delay.