Commodore PET 3016

[SiriusHardware]

Some further tests, all with power OFF.

We will now identify the ICs by their circuit part number / mainboard part number rather than their chip type number.

Set meter to 200 ohms resistance range, and check that the following connections are very low resistance / short-circuit.

The first set of measurements is to verify that the read-input connections between UC6 and UA7 are as they should be. For these measurements the exact resistance is not important as long as it is very, very low. You are just making sure the connection between these pairs of pins is present, rather than absent.

UC6 pin 14 to UA7 pin 3
UC6 Pin 15 to UA7 pin 5
UC6 Pin 16 to UA7 pin 11
UC6 pin 17 to UA7 pin 13

Still on 200 ohms range, check the resistance between the following pairs of pins of UC6. This is to find out if any of the read-input lines between UA7 and UC6 are shorted together (they should not be). With the meter on its 200 ohms range I would expect the reading to be high or infinite in each of the following three cases. If you see a very low resistance or short between any of these pairs of pins below, flag that up.

UC6 pins 14 and 15
UC6 pins 15 and 16
UC6 pins 16 and 17

Now move to the highest resistance range you meter has, and with the black meter probe on a 0V point, check the resistances between there and each of

UC6 pin 14
UC6 pin 15
UC6 pin 16
UC6 pin 17

I would expect those to be medium to high readings, the main thing is that all four should be roughly the same resistance. If any of them looks significantly different to the others, flag that up.

Still on the highest resistance range, move the red meter probe to UC6 pin 20, keep it there and measure the resistance from there to each of these pins again, this time moving the black probe from one pin to the next. Again, I would expect relatively high resistances, but all four roughly the same, whatever the value is.

UC6 pin 14
UC6 pin 15
UC6 pin 16
UC6 pin 17

Now let's move to the port buffers, starting with UA8 and UA7 which buffer the 8 data bits of the port. Change to the 20K resistance range and with the black probe on a 0V point, measure the resistance between each of the following pins and 0V by moving the red probe from one pin to the next. As before, all the resistance measurements seen here should be roughly similar. If they are all 'infinite' dial up to a higher resistance range and try again.

UA8 pin 2
UA8 pin 6
UA8 pin 10
UA8 pin 14
UA7 pin 2
UA7 pin 6
UA7 pin 10
UA7 pin 14

Move the red probe to UC6 pin 20 and measure the resistance from there to the same pins, this time moving the black probe from one pin to the next.

UA8 pin 2
UA8 pin 6
UA8 pin 10
UA8 pin 14
UA7 pin 2
UA7 pin 6
UA7 pin 10
UA7 pin 14

(Of the two groups above, it will be interesting to see if the 'D' line and possibly the 'C' line (UA7 pins 10 and 14) are noticeably different to the rest).

Now let's move to the remaining buffer, UA9. Staying on 20K and with the black probe on a 0V point, move the red probe from one pin to the next to measure the resistances between these pins and 0V. As before, if they all read 'infinite', dial up to a higher resistance range.

UA9 pin 2
UA9 pin 6
UA9 pin 10
UA9 pin 14

And, with the red probe held on UC6 pin 20, move the black robe from one pin to the next and measure the resistances to

UA9 pin 2
UA9 pin 6
UA9 pin 10
UA9 pin 14

That lot will keep you occupied for a while.

As mentioned previously this method of attack can prove useful, or it may not. If you find a very noticeable difference between one pin and the functionally identical other pins on a chip, then there is a good chance you have identified a fault. Sometimes, however, a fault can be buried so deep inside a chip as to cause no measurable difference on those parts exposed to the outside world. It's worth a try though.

There are a few other signals on the port which will also need checking. We'll get to those later.

[John Earland]

I’ll get onto it! Thank you.

[John Earland]

Originally Posted by SiriusHardware

Some further tests, all with power OFF.

We will now identify the ICs by their circuit part number / mainboard part number rather than their chip type number.

Set meter to 200 ohms resistance range, and check that the following connections are very low resistance / short-circuit.

The first set of measurements is to verify that the read-input connections between UC6 and UA7 are as they should be. For these measurements the exact resistance is not important as long as it is very, very low. You are just making sure the connection between these pairs of pins is present, rather than absent.

UC6 pin 14 to UA7 pin 3 = 00.8
UC6 Pin 15 to UA7 pin 5 = 00.8
UC6 Pin 16 to UA7 pin 11 = 00.9
UC6 pin 17 to UA7 pin 13 = 00.8

Still on 200 ohms range, check the resistance between the following pairs of pins of UC6. This is to find out if any of the read-input lines between UA7 and UC6 are shorted together (they should not be). With the meter on its 200 ohms range I would expect the reading to be high or infinite in each of the following three cases. If you see a very low resistance or short between any of these pairs of pins below, flag that up.

UC6 pins 14 and 15 = infinite (1)
UC6 pins 15 and 16 = infinite
UC6 pins 16 and 17 = infinite

Now move to the highest resistance range you meter has, and with the black meter probe on a 0V point, check the resistances between there and each of
2000k

UC6 pin 14 = infinite
UC6 pin 15 = infinite
UC6 pin 16 = infinite
UC6 pin 17 = infinite

I would expect those to be medium to high readings, the main thing is that all four should be roughly the same resistance. If any of them looks significantly different to the others, flag that up.

Still on the highest resistance range, move the red meter probe to UC6 pin 20, keep it there and measure the resistance from there to each of these pins again, this time moving the black probe from one pin to the next. Again, I would expect relatively high resistances, but all four roughly the same, whatever the value is.

UC6 pin 14 = 689
UC6 pin 15 = 686
UC6 pin 16 = 687
UC6 pin 17 = 697

Now let's move to the port buffers, starting with UA8 and UA7 which buffer the 8 data bits of the port. Change to the 20K resistance range and with the black probe on a 0V point, measure the resistance between each of the following pins and 0V by moving the red probe from one pin to the next. As before, all the resistance measurements seen here should be roughly similar. If they are all 'infinite' dial up to a higher resistance range and try again.

UA8 pin 2 = 1.85
UA8 pin 6 = 1.86
UA8 pin 10 = 1.87
UA8 pin 14 = 1.87
UA7 pin 2 = 1.89
UA7 pin 6 = 1.89
UA7 pin 10 = 1.91
UA7 pin 14 = 3.00

Move the red probe to UC6 pin 20 and measure the resistance from there to the same pins, this time moving the black probe from one pin to the next.

UA8 pin 2 = 1.71
UA8 pin 6 = 1.72
UA8 pin 10 = 1.73
UA8 pin 14 = 1.73
UA7 pin 2 = 1.74
UA7 pin 6 = 1.75
UA7 pin 10 = 1.77
UA7 pin 14 = 2.52

(Of the two groups above, it will be interesting to see if the 'D' line and possibly the 'C' line (UA7 pins 10 and 14) are noticeably different to the rest). Yes the C line ([in 14) is different!

Now let's move to the remaining buffer, UA9. Staying on 20K and with the black probe on a 0V point, move the red probe from one pin to the next to measure the resistances between these pins and 0V. As before, if they all read 'infinite', dial up to a higher resistance range.

UA9 pin 2 = 2.01
UA9 pin 6 = 2.04
UA9 pin 10 = 2.03
UA9 pin 14 = 0.38

And, with the red probe held on UC6 pin 20, move the black probe from one pin to the next and measure the resistances to

UA9 pin 2 = 2.46
UA9 pin 6 = 2.49
UA9 pin 10 = 2.48
UA9 pin 14 = 0.83

That lot will keep you occupied for a while.

As mentioned previously this method of attack can prove useful, or it may not. If you find a very noticeable difference between one pin and the functionally identical other pins on a chip, then there is a good chance you have identified a fault. Sometimes, however, a fault can be buried so deep inside a chip as to cause no measurable difference on those parts exposed to the outside world. It's worth a try though.

There are a few other signals on the port which will also need checking. We'll get to those later.

I have completed these checks with some interesting anomalies! Connections to the buffers seem okay to me. However, I think these results are suggesting that UA7 and UA9 buffers are faulty?

[SiriusHardware]

Those are very good results: Or bad, depending on your point of view I suppose. You have done your usual forensic recording job, and the results are very informative. I agree with your analysis that pin 14 of UA7 and pin 14 of UA9 are highly suspect.

Pin 14 of UA7 is actually the 'D' line and if you remember, when you did Tony's output voltage tests that line, alone of all the eight, was outputting a 'high' voltage of around 5V when the other seven lines were outputting around 3+ volts.

It's not looking good for the 'D' line of UA7, but then, this is exactly the sort of thing we were hoping to find: An explanation for the drive operating problem and some way of narrowing it down to a particular device or devices.

Pin 14 of UA9, the _ATN control line, looks even worse. I think it is reasonable to assume, for now, that those two devices are partly damaged.

Well done, good job! What with this and managing to remove the old ULA from the ZX81, you are on a bit of a roll!

[John Earland]

Originally Posted by SiriusHardware

Those are very good results: Or bad, depending on your point of view I suppose. You have done your usual forensic recording job, and the results are very informative. I agree with your analysis that pin 14 of UA7 and pin 14 of UA9 are highly suspect.

Pin 14 of UA7 is actually the 'D' line and if you remember, when you did Tony's output voltage tests that line, alone of all the eight, was outputting a 'high' voltage of around 5V when the other seven lines were outputting around 3+ volts.

It's not looking good for the 'D' line of UA7, but then, this is exactly the sort of thing we were hoping to find: An explanation for the drive operating problem and some way of narrowing it down to a particular device or devices.

Pin 14 of UA9, the _ATN control line, looks even worse. I think it is reasonable to assume, for now, that those two devices are partly damaged.

Well done, good job! What with this and managing to remove the old ULA from the ZX81, you are on a bit of a roll!

That's good news - at least we're on the right track (for now!). Could it be third time lucky and I try to solder the socket into the ZX81 We'll see.

I came across a really lovely program for the PET - it creates a maze randomly forever - I love it:

10 Print "{CLR/HOME}"
20 Print CHR$ (205.5+RND(1));
30 Goto 20

Simple but lovely to look at I think! I'll try it on the ZX81 tomorrow but will have to track down a different CHR$!

[julie_m]

If you are reusing ICs that have previously been soldered, you probably will have better luck with the cheaper double-wipe type of IC sockets than with the better-quality turned pin ones. The former type seem to be more forgiving of pins with some residual tinning of solder on them. Alternatively, it may be possible to cut the pins below the "knee", close to the board, and solder the remains to the top of a turned-pin socket. It won't win too many beauty awards but ask yourself, do you want a looker or a worker?

[John Earland]

Originally Posted by julie_m

If you are reusing ICs that have previously been soldered, you probably will have better luck with the cheaper double-wipe type of IC sockets than with the better-quality turned pin ones. The former type seem to be more forgiving of pins with some residual tinning of solder on them. Alternatively, it may be possible to cut the pins below the "knee", close to the board, and solder the remains to the top of a turned-pin socket. It won't win too many beauty awards but ask yourself, do you want a looker or a worker?

Thank you for this. Someone mentioned something similar to me with using new ICs. instead of taking out the old ICs put the new ones on top matching up the pins in a piggy-back style? Interesting concept. He says it's worked for him when he's done it - something to think about perhaps? I'd rather take the old ones out though myself but it is worth thinking about.

[SiriusHardware]

No, absolutely not.

The 'bad' ICs will still exert influence over the signal lines unless they are physically removed from them. Remove and replace them, preferably put the new ones in sockets. As Julie suggested, you can plug desoldered ICs into 'normal' IC sockets as long as the pins of the desoldered ICs are clean and clear, even if they are a bit short.

Anyway, I would have thought that for someone who managed to take a 40 pin IC out of a double sided PCB using a manual solder sucker, removing three much smaller ICs would be child's play.

[John Earland]

Originally Posted by SiriusHardware

No, absolutely not.

The 'bad' ICs will still exert influence over the signal lines unless they are physically removed from them. Remove and replace them, preferably put the new ones in sockets. As Julie suggested, you can plug desoldered ICs into 'normal' IC sockets as long as the pins of the desoldered ICs are clean and clear, even if they are a bit short.

Anyway, I would have thought that for someone who managed to take a 40 pin IC out of a double sided PCB using a manual solder sucker, removing three much smaller ICs would be child's play.

I appreciate your confidence in me! But on my little precious PET Motherboard!!?

I could have a go, however, on the ZX81 I had cut out the offending IC - I wouldn't know what to do trying to remove all the pins without cutting them. How would you do that?

[SiriusHardware]

Ah, you didn't let on that you had cut the IC out (which is exactly what we would have told you to do anyway).

There is quite an art to getting an IC out of a double sided PCB intact without causing damage to the PCB - you will be able to do that one day in the near future, but I suggest you don't practice on the PET. This is one of those times when patience is definitely a virtue.

[julie_m]

Piggybacking a good IC on top of a bad one sometimes works for a quick test, but can't really be recommended as a long term solution.

What you are effectively doing is connecting the inputs and outputs of each gate within the good IC in parallel with its counterpart in the bad IC. But how successful this is really depends on the nature of the fault; and to some extent, the surrounding circuitry.

If you have an IC where one of the output devices has failed open-circuit -- say the bottom one is fine, so it can drive its zeros low, but the top one is weak and it can't drive its ones high enough -- then piggybacking a good one should help, because they both see the same inputs and the top transistor in the good IC will be conducting whenever the top transistor in the bad IC should be conducting.

But if the fault is different -- say the top transistor has failed short-circuit and now the IC's output is always high -- then piggybacking a good IC will not help, because the low resistance from the output to +5V will just drag up the output from the good IC even when it is trying to output a zero. If you are holding the new chip in place with your fingers, you might feel it getting warmer in time to shut off the power and avoid ruining it .....

(Incidentally, a similar warning applies to capacitors. Simple loss of capacitance can indeed be compensated by connecting an additional capacitor in parallel. But leakage, which is electrically equivalent to connecting a resistor in parallel with the capacitor, will not be affected by another parallel capacitor.)

A further complicating factor with connecting ICs in parallel is something called "fan-out". This is a property of digital logic ICs relating to how many inputs you can drive from a single output. In the real world, there are no perfect conductors -- or if there are, they are too few and far between for us ordinary people. The output transistors of a logic gate have a certain resistance when turned "on", the copper tracks that make up the wiring on board have resistance and the input of a logic gate has series resistance, parallel resistances to both supply rails and parallel capacitance to 0V. The circuit board is actually the least culpable in all of this; the intrinsic resistances and capacitances in the ICs usually dominate (though bad layout can certainly worsen an already-marginal situation).

At any rate, the overall result is that if you connect too many inputs to one output, it may not be able to drive its ones high enough or its zeros low enough. Too much capacitance is bad in fast logic circuits, because the voltage across a capacitor cannot change instantaneously; this means that it takes some time for a signal to change from a zero to a one or back, and if that time happens to be longer than half a clock cycle (inputs are presented to gates on one edge of the clock, the outputs are allowed half the clock period to stabilise and then they had better be ready to be looked at on the opposite edge), you're in trouble.

So another problem with piggybacking ICs is that by adding extra inputs, you may end up exceeding the fan-out rating of an earlier gate. And in some early microcomputer designs, some things were already pushing the margins a little. Which means even if the original IC had failed in a way that was amenable to piggybacking, connecting the new IC in parallel with it -- remember there is nothing wrong with the input circuits, which are still loading the output of the upstream circuitry -- might be enough adversely to affect the circuit operation.

I wouldn't say never, because there evidently are documented situations where piggybacking a known good part into parallel with a faulty one does work -- it's fine in the case of decoupling capacitors which have lost capacitance and funnily enough, ZX Spectrum RAM chips often fail in the right way for piggybacking to show up a duff one. But in general, it can't be expected to work; so don't expect it to, unless someone else has already tried exactly the same thing successfully or you understand the circuit well enough to think it viable. Even then, there are failure modes which are not amenable to the technique, and it's not inconceivable that a really faulty chip could kill a good one piggybacked onto it. If you were having a really bad day, a new chip could develop a failure mode that would ruin a good chip onto which it was piggybacked, leading to a trail of carnage. Ultimately, despite occasional successes, piggybacking is never going to be more than a bodge.

And of course if an IC is faulty, you're going to have to whip it out anyway. So what I'd recommend is, refer to the wiring diagram and your measurements, take more measurements if necessary, decide on the most likely culprits and desolder them. You'll probably have to cut them out but, as alluded to above, they may yet be salvageable. Turned-pin IC sockets make excellent connections with each other, so don't worry about putting what from below just looks like a socket into another socket.

Also worth mentioning: There are some "universal EPROM programmers" available which are essentially a bunch of general-purpose I/O ports connected to a quick-release socket; as such, they can also function as logic IC testers. If you are doing a lot of work with old computers, it might be worth investigating acquiring one, depending how easy it is to create custom test schedules for new devices. Of course it can also be used to burn custom ROMs

[John Earland]

That’s really comprehensive, thank you so much. Just to say I am printing all of these off for my own reference and I have quite a fantastic amount of solid guidance and advice. This is such a great experience. Thank you.

[John Earland]

Hi folks, saw this on YouTube:
https://youtu.be/XQVjwPsVFd8 Looked really neat

[SiriusHardware]

Indeed, but the key factor appears to be the electrically heated and electric pump-assisted desoldering iron he is using.

They certainly do make the job a lot easier, but they aren't cheap - at least they never used to be. Mine and the one I use at work were hundreds of pounds each originally.

It may well be that you can get something for a quarter of the price from China nowadays.

I should warn you that desoldering tools like that tend to be very high maintenance, you have to be very diligent about keeping them clear and maintained in good condition and in particular you have to be very careful not to accidentally desolder a small piece of component lead, otherwise once something like that gets jammed in the narrow channel in the nozzle, there's no getting it out.

Desoldering tips for the OKI /Metcal I use at work are about £26 a go at the time of writing this, so it's not something I'd want to have to replace every five minutes.

[Kyle__B]

I wish I had the pictures I took of the first time I tried desoldering a chip from an Atari ST. I didn't lift the tracks but I didn't get the chip out either, and it looked like a horrible unrecognizable mess.

Until recently I had a heated solder sucker, which was a good bit of equipment and didn't cost anywhere near as much as that electric suction pump in the video. It's a bit like using a hand powered drill though.

[SiriusHardware]

Funny you should mention that, here is a desoldering job I did on an Atari STe a few years ago, removal of the socket for the 68000 CPU. The images were originally taken to illustrate the non-standard (staggered) pin layout of the PLCC sockets used in the Atari ST / Ste range. The third image shows a standard socket with the same number of pins for comparison. Of course I didn't realise the sockets were different until the moment I tried to fit the replacement socket into the holes.

(I used the aforementioned OKI / Metcal electric desoldering iron to do this).

[John Earland]

Mmm-I think I’ll leave it until I am able to meet my friend once more! Thanks for the advice.

[John Earland]

Really good photos-those holes look like they’ve never been used!

[John Earland]

Hi guys,

Back to the drawing board for me! Horror of horrors my PET Monitor has stopped working!

I switched it on this morning and all was great. Ran a few pieces of software etc. for a couple of hours using the Tapuino. I tried to install a Chess game and the machine frozen - it does this occassionally, especially when loading, so I switched it off and now the monitor doesn't work! I did see a tiny pixel in the middle of the screen (I often do after it's been on a while) when I switched it off and it seemed to slide across the screen a little before fading away.

I've taken a look at the back of the monitor and the cathode is glowing and there is at least some power getting to the board because my little sound mod squeaks as usual when it powers down. And I can hear the 'hum' fom the transformer. I have swapped the 901447-10 IC with ones I know work but nothing on the monitor.

My main concern is the tube itself of course. I shall do some digging around.

[SiriusHardware]

Your tube heater is glowing, and there isn't much else on a tube which can go from working to not working in the space of a heartbeat. You can probably rule the tube out for now.

I recall that you said the display was occasionally glitching / bouncing, which makes me think that whatever was causing that has finally failed properly.

To help you out with this we'll need to identify the circuit diagram for the monitor chassis used in your specimen.

This is one possibility.

Layout? Does your monitor PCB look like this?

http://www.zimmers.net/anonftp/pub/c...01N/321446.gif

and possible circuit diagram:

http://www.zimmers.net/anonftp/pub/c...01N/321445.gif