🚀 go-pugleaf

In article <192@nikhefk.UUCP> henkp@nikhefk.UUCP (Henk Peek) writes:
+---------------
| >In article <18007@amdcad.AMD.COM> rpw3@amdcad.UUCP (Rob Warnock) writes:
| > |Notice that any resistor value less than "V_max_0/(I_max_0 * #of_inputs)"
| > |will be fine, as it's no worse than a true gate output would be. For a
| > |7400 TTL single "standard load" input, that's 0.4/.016 = 25 ohms or less.
| >The (low) input curent of a standard single load 7400 TTL input is 1.6 mA.
| >The maxium serie R for a single load standard input is 0.4/0.0016 = 250 ohms.
| >henk peek  ..!seismo!mcvax!nikhefk!henkp.UUCP
+---------------

Yup! I blew it! Will the net ever forgive?!?!? Mea culpa, folks...
(That's what I get for not doing the whole thing in millivolts and milliamps.)

But the part about using a 1/10-watt resistor for pulldowns and the guy in
manufacturing who wanted to occasionally zap the pulldowns from a bed-of-nails
tester and needing to use short pulses was right, I just forgot what value
we ended up using for a pulldown.

Let's see if I can do simple arithmetic tonight:

I'm sure it was 1/10-watt (they were *small*), so using R = (V^2)/P, if
the tester pulled up to +5 the resistor would have had to been less than
250 ohm to burn up. A 100 ohm resistor would dissipate 1/10 watt if held
high at 3.2 V (and a bit more if held at 3.6), so maybe I was just off by
a factor of 10 and we were using 100 ohm 1/10-watt pulldowns. (Crosscheck:
100 ohms could pull down 4 mA at 0.4 V, which is two 74S' loads. Yes, I know
an 'S low is 0.5 V...)

So. I dropped a power of 10, but was otherwise reporting true.

Sorry again for the error.


Rob Warnock
Systems Architecture Consultant

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