*** M E M O R U M ***

the LS138's at 7C and 7D decode the lowest 8 I/O ports 7C is the IN ports, 7D for OUT. So A0 is connected to both devices. If they're shorting, one or both of those could be holding A0 low.

Well, whatever is shorting A0, it's killed 7D. The plastic package has broken out!

I'll whip them both out and replace them I think They've probably both been toasted. I'll also remove the ROM C socket, which I soldered back in this morning after removing it and checking things. I can't see any shorts but you never know.

It's definitely a short to 0V somewhere as it has negligible resistance.

With all the chips in place, the resistance on the A0 line was about 15 ohms.

With 7D removed it went up to 31 ohms. With 7C removed, 71 ohms.

I removed 6D (LS157), the ROMs, the CPU and clipped out the to LS138 chips meaning that there shouldn'd be anything connected to A0.

Adding back chips one at a time, the LS157 was the only one to bring the resistance down. I have a spare, so I'll swap that in and see if it's the same.

After replacing the LS157 there's a high resistance. So it looks like it was actually three dead chips. The LS157 is actually one of the chips I'd already replaced, so I wonder what killed it.

A0 also goes to the CTC

If you have multiple chips all connected to the same line that have died, especially if some only have inputs to this line, that indicates that the line has experienced a severe over voltage or a negative voltage. A short circuit to 0V / GND or to the +5V rail will not harm or destroy most chips where the line is an input.

So the possibilities are:

• Something resulted in A0 becoming connected to the +12V rail,
• Something resulted in A0 becoming connected to the -5V rail,
• Something resulted in A0 becoming connected to the unregulated supply that feeds the 7805 and associated power supply circuitry.

In any event, it’s worthwhile testing on resistance between the above supply rails and the A0 line.

Mark

I've not got time to check tonight as I'm off to see Endgame, so I'll do it tomorrow.

In the meantime I've ordered practically a complete set of LS chips and sockets just in case. I didn't get a chance to get a set of full part numbers for the voltage regulators and capacitors so I'll have to order those later.

I've had a little time to check the board out. With all the chips connected to the low address lines removed I can find no shorts or leakage at all. It was the A3 line pin on 7D which looked "burnt" after I removed the chip.

Turning on, all the voltages look normal, though the 4070B chip is getting hot. I'll have to order a replacement just in case.

Putting just a "new" Z80 into the circuit intermittently got RAS/CAS running, as in some power-ups I got it and others I didn't, which suggests whatever initialises the Z80 is iffy. I think that goes through the 4070B. I think that the original Z80 is probably fried.

Nothing “initialises” the Z80 as such, but the Z80 does need /reset held low for a minimum time at startup, one of the 4070 XOR gates is in the reset circuit. I suppose you could call that beng “initialised”.

Again, the circuit diagram is in the manual, or fully described here http://www.primrosebank.net/computers/m ... _reset.htm

That's what I meant by "initialising".

The circuit diagram at the back of the manual is my bible at the moment.