Building an Arithmetic Logic Unit (ALU) for MIPS: Logic Gates, Adders, and Subtraction, Slides of Computer Science

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Arithmetic II

CPSC 321

Today’s Menu

Arithmetic-Logic Units

Logic Design Revisited

Faster Addition

Multiplication (if time permits)

Logic Gates

• AND gate
• OR gate
• NOT gate

What are the truth tables?

Logic Gates

• NOR gate
• NAND gate
• XOR gate

What are the truth tables? Docsity.com

Full Adder

cin a b c (^) out s 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1

Give a Boolean formula for s

• s=cin xor a xor b

Give a Boolean formula for c (^) out

• cout =ab+cin (a xor b)

Design now a circuit using and, or, xor.

Full Adder

c in

a

b

c out

s

s=c in xor a xor b c out = ab+c in (a xor b)

Critical Path

c in

a

b

c out

s

Suppose that each gate has a unit delay. What is the critical path (= path with the longest delay)? Docsity.com

Ripple Carry Adders

• Each gates causes a delay
  • our example: 3 gates for carry generation
  • book has example with 2 gates
• Carry might ripple through all n adders
  • O(n) gates causing delay
  • intolerable delay if n is large
• Carry lookahead adders

Fast Adders

Iterate the idea, generate and propagate

ci+1 = g (^) i + pici

= g (^) i + pi(g (^) i-1 + pi-1 ci-1 ) = gi + pig (^) i-1 + pipi-1 ci- = g (^) i + pig (^) i-1 + pipi-1 g (^) i-2 +…+ p (^) ipi-1 …p 1 g (^0) +pipi-1 …p 1 p 0 c 0

Two level AND-OR circuit

Carry is known early!

Carry Lookahead Adders

• Based on the previous identity

Fast because critical path is shorter O(log n) gate delays [assuming 2-input gates] More complex to implement Design is less regular Layout of one bit adder cells depend on i

• Compromise

couple blocks of carry lookahead adders

Tailoring the ALU to the MIPS

• Need to support the set-on-less-than instruction (slt)
  • remember: slt is an arithmetic instruction
  • produces 1 if rs < rt and 0 otherwise
  • use subtraction: (a-b) < 0 implies a < b
• Need to support test for equality (beq $t5, $t6, $t7)
  • use subtraction: (a-b) = 0 implies a = b

SLT

• Determine a<b
• Calculate b-a
• If MSB equals
  • 1, then a<b
  • 0, then a>=b
• Changes?
  • Operation less than
  • Output of subtraction
  • Overflow

0

2

Result

Operation

a

1

CarryIn

CarryOut

0 1

Binvert

b

a31 0 Set

ALU0 Result

CarryIn

a

a1 0 Result

a2 0 Result

Operation

b

b

b

b

Result Overflow

Binvert

CarryIn

Less

CarryIn CarryOut

CarryInALU1Less CarryOut

CarryInALU2Less CarryOut

CarryInALU31Less

• LSB indicates

whether a<b

• 0 if false
• 1 if true

Test for equality

• Notice control lines:

000 = and 001 = or 010 = add 110 = subtract 111 = slt

• Note: zero is a 1 when the result is zero! a31 0 Set

a0 Result

a1 Result 0

a2 Result 0

Operation

b

b

b

b

Result Overflow

Bnegate

Zero

CarryIn ALU0Less CarryOut

CarryInALU1Less CarryOut

CarryInALU2Less CarryOut

CarryInALU31Less