On Sat, Dec 17, 2016 at 6:34 PM, Adrian Graham <wit...@binarydinosaurs.co.uk> wrote: > On 17/12/2016 14:28, "Tony Duell" <ard.p850...@gmail.com> wrote: > >> On Sat, Dec 17, 2016 at 2:02 PM, Adrian Graham >> <wit...@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. >
Yes. I would agree with that. >> 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... 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