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Recall x y + x y = x y

Cout = CAB + CAB + CAB + CAB

= AB(C + C)

+ C(AB+AB)

= AB + C(A B)

= AB + C(A B)

= AB C(A B)

Adder- Recall truth table for binary addition

Cin Ai Bi Si Cout

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

So Si = (A B) C

Basic Ripple Carry Adder

- An n-bit unsigned integer ripple carry adder can be build from n Full Adders

Full Adder

Basic Ripple Carry Adder (RCA)

- Slow due to the ripple propagation of the carries

Overflow Detector

- Overflow occurs whenever the carry-in into the sign bit is different from the carry-out from that sign bit
- So we need an XOR at the output to indicate whether or not overflow has occurred

Ripple Carry Subtractor

- We can subtract using addition
- Form the 2’s comp of the subtrahend then add
- So by adding hardware, we can do both addition and subtraction
- We need a “mode” bit to tell us whether we are adding or subtracting

Bitwise controlled complementer

- If mode m = 0, indicate addition, so

S = A + B

- If mode m = 1, indicate a subtraction, so

S = A - B = A + B2 (my notation for 2’s comp)

- Recall B2 = (bn-1 bn-2 … b0) + 1
- Neglect the “+1” for now and try to form B1

Bitwise controlled complementer

bn-1 bn-2 … b0

Controlled Complimentor

m

b’n-1 b’n-2 … b’0

Where bi’ = bi, m=0

bi, m=1

0 0 0

0 1 1

1 0 1

1 1 0

Bitwise controlled complementer- This can be done with XOR
- For the “+1”, consider the following:
- if m=0, we are adding
- no complement
- no “+1”
- if m=1, we are subtracting
- compliment
- “+1”
- Why not just add m?
- Moreover, we can add m at the LSB carry-in!

RCA

- The RCS’s are slow
- very long “worst case” path
- assumes a carry may have to propagate from the LSB FA all the way to the MSB FA
- Why not compute each carry directly instead of having it propagate along?
- Time more important than gate space
- Gate space still important

Parallel adder

- Recall a 4-bit sum:

C4C3C2C1C0

A3A2A1A0

+ B3B2B1B0

C4S3S2S1S0

Recall Si = (Ai Bi) Ci

Ci = Ai-1Bi-1 Ci-1(Ai-1 Bi-1)

Conditional Sum Adder (CSA)

- Let the true augend, addend, and sum bits be designated by Ai, Bi, and Si
- Carries are indicated by Ci
- Subscripts indicate bit position
- Let superscripts indicate the assumption of the carry-in to bit position i:
- Si1 = “1” carried into sum bit position i
- Si0 = “0” carried into sum bit position i

CSA

- S0(k) and S1(k) denote 2 provisional sums
- each consists of multiple sections
- k addend/augend bits per section
- Then there are (n/k) sections for an n-bit addition
- C0(k) and C1(k) are provisional carry sequences formed by carries out of all the sections in S0(k) and S1(k), respectively

CSA

- Simultaneous addition are performed on all sections independently
- Compute all of the S0, C0, S1, and C1 for each section independently
- Select whether to use (Si0, Ci0) by the actual Ci-1
- Let k = 2j-1 for the jth step
- Take log2n steps

CSA: Stage 1 bits 1 and 2

Notice C20(2) = C20

C21(2) = C21

Also notice that

we don’t need C1

to decide (S20(2), C30(2))

or (S21(2), C31(2))

So we can compute these

before we know C1

- If C1 == 0:
- (S10(2), C20(2)) = (S10, C20)
- If C20(2) == 0:
- (S20(2), C30(2)) = (S20, C30)
- Else [C20(2) == 1]:
- (S20(2), C30(2)) = (S21, C31)
- If C1 == 1:
- (S11(2), C21(2)) = (S11, C21)
- If C21(2) == 0:
- (S21(2), C31(2)) = (S20, C30)
- Else [C21(2) == 1]:
- (S21(2), C31(2)) = (S21, C31)

CSA: Stage 1 bits 3 and 4

Notice C40(2) = C40

C41(2) = C41

Also notice that

we don’t need C3

to decide (S40(2), C50(2))

or (S41(2), C51(2))

So we can compute these

before we know C1

- If C3 == 0:
- (S30(2), C40(2)) = (S30, C40)
- If C40(2) == 0:
- (S40(2), C50(2)) = (S40, C50)
- Else [C40(2) == 1]:
- (S40(2), C50(2)) = (S41, C51)
- If C3 == 1:
- (S31(2), C41(2)) = (S31, C41)
- If C41(2) == 0:
- (S41(2), C51(2)) = (S40, C50)
- Else [C41(2) == 1]:
- (S41(2), C51(2)) = (S41, C51)

CSA: Stage 1 bits 2 and 3 (simultaneously with bits 0, 1)

- C20 and C21 both select new possibilities for (S20(2), C30(2)) and (S21(2), C31(2))
- If C20=0, we will select (S20(2), C30(2)) = (S20(1), C30(1))
- If C20=1, we will select (S20(2), C30(2)) = (S21(1), C31(1))
- If C21=0, we will select (S21(2), C31(2)) = (S20(1), C30(1))
- If C21=1, we will select (S21(2), C31(2)) = (S21(1), C31(1))

CSA: Stage 1 bits 2 and 3 continued

- So we still have 4 values:

S20(2), C30(2), S21(2), C31(2)

- The choice will depend on whether C1 = 0 or 1
- Next stage will choose which one based on C1

CSA: Stage 1 following bit pairs

- The CSA will work just like bits 2 and 3 for all subsequent bit pairs

CSA: Stage 2 bits 0-3

- The C1 output from stage 1 will select S1 = S10 or S11.
- The C1 output from stage 1 will also select S2 = S20(2) or S21(2)
- The C1 output from stage 1 will also select C3 = C30(2) or C31(2)
- This C3 output will be used in stage 3 to select outputs for higher order bits

CSA: Higher stages, higher bits

- We continue this setup for the higher-order bits

0 0 0 0 0

0 1 0 0 1

1 0 0 0 1

1 1 0 1 0

0 0 1 0 1

0 1 1 1 0

1 0 1 1 0

1 1 1 1 1

CSA: Conditional Cell- We want to produce the conditional sums
- Output does not depend on Ci

CSA: MPX

- MPX’s for the CSA have n pairs of inputs
- One line from each pair is selected by a single address line

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