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1 Using 2-opr adder Carry-save adder Wallace Tree Dadda Tree Parallel Counters Multi-Operand Addition.

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Presentation on theme: "1 Using 2-opr adder Carry-save adder Wallace Tree Dadda Tree Parallel Counters Multi-Operand Addition."— Presentation transcript:

1 1 Using 2-opr adder Carry-save adder Wallace Tree Dadda Tree Parallel Counters Multi-Operand Addition

2 2 X 1 +X 2 + …+X k with n-bit each Sum= n +log 2 (k) bits Serial Addition  T add = (k–1) (T Adder + T Reg )  A add = A Adder + A Reg Two-Operand Adders

3 3 Serial Addition: Cascading  T add =  A add =  A Adder = bits adders Two-Operand Adders(2)

4 4 Parallel Addition: Binary tree  T add =  A add = Two-Operand Adders(3)

5 5 Add more than two numbers (says n) Carry not added (Carry save) 3 #  (3,2)  2# Carry-Save Adder

6 6 CPA in the last step 3 #  (3,2)  2#; T CPA +  log 3/2 k  T FA Carry-Save Adder(2)

7 7 4#  (4,2)  2#; T CPA + 2  log 2 k  T FA Carry-Save Adder(3)

8 8 7#  (7,3)  3#; T CPA + 2  log 7/3 k  T FA Carry-Save Adder(4)

9 9 Applying (3,2) FA & (2,2) HA with dot notation n=6&k=6  12 FA in 1st level Carry-Save Adder(5-1)

10 10 Applying (3,2) FA & (2,2) HA with dot notation n=6&k=6  5 FA & 2HA in 2nd level Carry-Save Adder(5-2)

11 11 Applying (3,2) FA & (2,2) HA with dot notation n=6&k=6  12HA in 2nd level  4 numbers (same)  Not so good Carry-Save Adder(5-2-1)

12 12 Applying (3,2) FA & (2,2) HA with dot notation n=6&k=6  5 FA & 1HA in 3rd level  7-bit CPA in last level Carry-Save Adder(5-3)

13 13 k-input Wallace Tree reduces to two (n+log 2 k –1)-bit outputs  h(k)=1+h(  2k/3  )  h(k): the smallest height of an k-input Wallace tree Wallace Tree

14 14 7-input Wallace Tree reduces to two (n+log 2 k –1)=(n+2)-bit outputs  Wallace Tree(2)

15 15 Reduce the number to the next lower number Dadda Tree

16 16 Reduce the number to the next lower number Ex1: k=8  8 (2CSA)  6 (2CSA)  4 (1CSA)  3(1CSA)  2  CPA Ex2: k=12  12 (3CSA)  9  6  4  2 Dadda Tree(2)

17 17 One column  (3,2) counter  at most 3 1’s  2 bits  (k,m) counter  at most k 1’s  m=  log 2 (k+1)  bits How about multi-column?  (k,k,m) counter: at most k 3’s  m=  log 2 (3k+1)  bits Parallel Counters

18 18 Ex: (5,5,m); m=  log 2 (3*5+1)  = 4 bits  Overlapped  m/(# of col.)  = 2 bits  CPA at last stage too Parallel Counters(2)

19 19 Ex: (5,5,5,m); m=  log 2 (7*5+1)  = 6 bits  Overlapped  m/(# of col.)  = 2 bits  CPA at last stage too Parallel Counters(3)


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