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Commodore 8032 repair.

daver2

10k Member
Joined
Jun 19, 2012
Messages
15,343
Location
UK - Worcester
Finally started repairing my 8032 with an 8032030 80 column main board.

Electrical safety checks - OK.

Transformer AC supply checks - OK.

DC supply checks (black smoke time) - OK.

Reset signal (CPU pin 40) - OK.

Phi 2 clock (CPU pin 39) - NO.

Chased the clock (or lack of) all the way back to UE4 (74S04).

Pin 1 is HIGH.

Pins 2/3 - undefined logic level.

Pins 4/5 - LOW.

Pin 6 - HIGH.

I forced pin 1 to a logic LOW.

Pins 2/3 went LOW.

Pin 4 stayed LOW.

I forced pin 1 to a logic HIGH (via a 47 Ohm resistor).

Pins 2/3 went HIGH.

Pin 4 stayed LOW.

All the other pins of UE4 appear to operate as inverters (as expected).

All of the above indications made with a logic probe.

I am suspecting UE4 (74S04) gates A and/or B being duff.

The change in state of pins 2/3 being induced by the presence of resistors R30 and/or R29.

Any further thoughts before I exchange UE4?

Dave
 
Hi daver2,
I don't want to jinx you, but I hope there are more problems than the clock! This thread would make a great 'How To' on troubleshooting a PET with you leading the effort.
-Dave
 
Who knows until I get in there?!

I see from the underside of the board that a few ICs have been repaired in the past. I will clean the flux off and finish the job off for them!

I have a disk drive and a printer as well. Never played around with either of these, so I will be doing some learning myself with this one...

Another business trip coming up, so it will be a while before I get around to this.

Dave
 
Just one thing, if you are using a testing methodology where you force a pin high or low, it pays to do it only briefly with a low value current limiting resistor maybe 33R, or similar.

TTL's are very good at sinking current to ground but poor at sourcing it. Most pins (inputs) you are forcing high or low will be connected to another TTL output, or at least a tie resistor. 74S series can sink more current as I recall than 74LS, but it pays to look that up in the book. Anyway if you do force an input "against the will" of an output it is connected to, it is kinder to the IC's output stage to do it only briefly and with a current limiting resistor helps a little.
 
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