Race conditions

Question

I'm currently stuck trying to understand two things related to race conditions.

Issue 1:

I have been presented with the following question:

We consider the digital circuit and the value of its inputs a,
and b as given below. For all logic gates, we assume that
there is a gate delay of exactly one time unit (i.e. the gate
delay equals the time between two dotted lines in the
diagram). Give the values of c, d, e, f in the digital circuit for
every point of time between 0 and 8.

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And the answer given is as follows:

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How exactly is this achieved? This is what I think so far:

c starts at 1 because a start at 0
d starts at 0 because b start at 1
When time is equal to 2 a becomes 1... there is a propogation delay of 1 for c to switch over to 0 hence it becomes 0 at 3 time units
Same logic applies to d
e and f are meant to be constant 1 or 0 but seem to be affected by something.

What's really going on here? Is it related to a boolean function or soemthing. If so what?

Issue 2:

Does anyone have a simple way or logical approach in which to produce a simple circuit (using XOR, AND, OR, NOT, NAND boolean functions with:

static race condition - when the value is meant to be constant
dynamic race condition - when a value is expected to change

Many thanks in advance!

@Hans haha it's revision for an Computer Science exam – methuselah Jun 12 '12 at 18:42 — methuselah, Jun 12 '12 at 18:42

Hans Z · Accepted Answer · 2012-06-12T19:06:13.900

Okay, so race conditions in asynchronous circuits happen when inputs change at different times for a gate. Let's say your logical function looks like this

λ = ab + ~b~a

the most straightforward way to implement this function with gates looks like

NOTE: I assume your basic building blocks are AND, OR, and NOT. Obviously in CMOS circuits, NAND, NOR and NOT is how you build circuits, but the general principal stays the same. I also assume AND, NOR, and NOT have the same delay when in reality NAND and NOR have different delays if the output goes form 0 to 1 or 1 to 0, and NOT is about 20% faster than NAND or NOR.

a ->| AND |-------->| OR | -> λ
b ->|  1  |         |    |
                    |    |
a ->| NOT |->|AND|->|    |
b ->| NOT |->| 2 |  |    |

Now, assume that AND and NOT both have a delay of 2ns. That means the OR gate sees the value at it's first position change 2 ns before it sees the value at its second position change.

Which means if a and b both go from 1 to 0, you would expect λ to stay the same, since the output of the first AND gate is going from 1 to 0, but the output of the AND gate goes from 0 to 1, meaning the OR condition stays true.

However, if you get the output from the second AND gate a little bit after the first AND gate, then your OR gate will momentarily see 0,0 at it's input while transitioning from 1,0 to 0,1. Which means λ will have a momentary dip and it'll look like

     __
   a   |___________
     __
   b   |___________
     ____
AND1     |_________
            _______
AND2 ______|
     ______   _____
   λ       |_|

If you look at the inputs of the OR gate right between when AND1 goes down and AND2 goes up, it propagates a 0 through the OR gate, and sure enough, there's a dip in the output 2ns later.

That's a general overview for how race conditions arise. Hope that helps you understand your question.

How would you read `λ = ab + ~b~a` - we've only been using AND, OR, NOT when naming our logical functions. — methuselah, Jun 14 '12 at 13:03
ab is equivalent to (a and b). The reasoning is if you take 0 to mean false and any other value to mean 1, then ab is 0 if either a and b is 0. a + b is equivalent to a or b. The reasoning is if either a or b are > 0, then a + b > 0. ~a is another way of writing not a. as is !a, -a, and a'. — Hans Z, Jun 14 '12 at 13:20

Race conditions

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