17 Dec
2016
17 Dec
'16
6:49 p.m.
On Sat, Dec 17, 2016 at 6:34 PM, Adrian Graham
<witchy at binarydinosaurs.co.uk> wrote:
It's not a 7493 as the resets don't match up. I would guess at 74x92 at least
for the moment. See if the rest of the circuit makes any sense.
-tony
On 17/12/2016 14:28, "Tony Duell"
<ard.p850ug1 at gmail.com> wrote:
Yes. I would agree with that.
On Sat, Dec 17, 2016 at 2:02 PM, Adrian Graham
<witchy at binarydinosaurs.co.uk> wrote:
> Thanks all, the pinouts are matching
the LS9x counters so I just need to
> trace more lines to hopefully narrow it down. Pins 6 and 7 are definitely
> inputs so you're right Tony, the reset must come from elsewhere. One of the
> outputs is confusing though since it appears to come FROM 5V via a resistor,
What value resistor?
It's a 5-band red-red-black-black-violet so either 220R or 70k? Based on
what Pete said about the Z80 I'm going for 220R without pulling it out of
circuit.
It is possible
that whatever it is driving needs a swing to +5V,
rather than just
a TTL high level. Adding a pull-up resistor is a way to kludge this.
I thought that but it doesn't appear to go anywhere else. I'll keep looking,
its only a 2 layer board so there's nothing hidden.
The 7490 has 2 sets of reset inputs (4 pins
total), one pair to reset
it to 0, the
other pair to reset it to 9. Since these inputs are active high, they
can't be left
floating, they must be conneccted to something (even if directly to ground).
So
you could see if those pins go anywhwere, if not, then you can eliminate the
7490
One problem I have is that I've already found a few chips with dead outputs
so I've no idea if these will be any different. The pinouts I have match the
LS92 since pins 2/3/4/13 are NC. All testing so far has been done with a DMM
and cheap logic analyser. Since one of the possibly-LS92s is out of circuit
I'll build a little test circuit to see if it does actually count given a
clock source...