CPC 6128 - white screen, black border

[McArti0]

I am not sure I do understand what you mean by stopping GA, GA will only stop if you stop the clock.

Take a look at the illustration in post no. 100. GA stops itself due to the Lo signal on pin 1 of 244. This freezes the clock generator. I'll never believe you could be as blind as Bryce It's not like you.

[SerErris]

A very cheap option would be to desolder the rom and put in a socket and a diagnostic eeprom.

That will tell you which ram chip is the problem here.

For all the RAM chips of bank1 check each output pin with the oscilloscope. This may show a stuck bit. However it might be also a problem with addressing. For instance a ram chip might have a singled dead or shorted address line. That would deliver wrong results upon reading as well.

But now with an oscilloscope you can start looking att the 244 enable line

What you are looking for is a square wave of some kinds when CPU wants to write to gate array. At best you solder a button from reset line to gnd to be able to reset the machine while having the oscillator running.

You can also check the other signals from gate array like crtc clock , ras and cas signals. They all need to form nice square waves with 5 volts. Any of those signals going to anything other than 0v or 5V is an indicator for a problem. Also any of those lines permanently stuck to 5V or 0v is as well a problem.

[SerErris]

I am not sure I do understand what you mean by stopping GA, GA will only stop if you stop the clock.

Take a look at the illustration in post no. 100. GA stops itself due to the Lo signal on pin 1 of 244. This freezes the clock generator. I'll never believe you could be as blind as Bryce It's not like you.

I am not sure what excuse you have to offend people here.

Neither is Bryce blind, nor is me. You might be a genius that we have been missing here all the time,  but from what you are writing, it is not the case.

The signal on pin one of 244 is normally generate by the gate array, which means that pin23 of the gate array is an output. It will not react to whatever you do to the line. The worst case if you ground that line is, that you will create a direct short and damage the output buffer in the GA.

The gate array does not stop if you pull pin 23 low. It will actually not result in any change of the ga operation whatever you do to pin 23 of the gate array or pin1 of the 244.

You may want to read some of the available documentation.

[McArti0]

Neither is Bryce blind, nor is me. You might be a genius that we have been missing here all the time,  but from what you are writing, it is not the case.

The signal on pin one of 244 is normally generate by the gate array, which means that pin23 of the gate array is an output. It will not react to whatever you do to the line.

Genius McArti for BLINDs.
You cannot view this attachment.

[Bryce]

I'm starting to think that this is some sort of never ending April fools joke..

Then prove to him that his memory is faulty. So far I have found out that the 3rd byte of the Rom has the value 61 instead of 89. If at all he did it well with 3 people. For now, it turns out that z80 writes to frames. And that the ram has the correct contents. Maybe you could stop being mean and start helping.

I don't have proof, I have experience. I've repaired 100's of CPC's over many, many years and despite owning an armada of expensive test equipment, if I get a start screen that looks like that I swap the RAM after just a few basic voltage checks. 99% of the time it fixes the problem. Time is more expensive than a set of 4164's.
To put it another way: If it looks like a duck and quacks like a duck, it's most likely a duck.

I am being helpful. I'm passing my experience on, and my experience says swap the RAMs.

Bryce.

[McArti0]

And the client wants proof without premonitions, because it's NOT HIS CPC.
He doesn't have the tools, so I help him the way he is capable of doing.
And don't insult my work.

To put it another way: If it looks like a duck and quacks like a duck, it's most likely a duck.

If it looks like a duck and quacks like a duck, sometimes it's not a duck.

[eto]

. Maybe you could stop being mean and start helping.

He did. He said very early in this process that it's most likely the RAM. Even if it's not the RAM there are only 6 more ICs that can be the culprit. Even if he would have swapped all relevant ICs it would have been maybe €20 and max 2h of work in total.

[McArti0]

Even if he would have swapped all relevant ICs it would have been maybe €20 and max 2h of work in total.

Ask mmldrm how many times he has desoldered an integrated circuit in other equipment in his life.

[Gryzor]

@McArti0 , I really don't care whether you're right or wrong. Be careful before insulting people next time.

[McArti0]

Be careful before insulting people next time.

I'm starting to think that this is some sort of never ending April fools joke.

[Gryzor]

Way to keep it up.

[mmldrm]

Even if he would have swapped all relevant ICs it would have been maybe €20 and max 2h of work in total.

If you have sources for the ICs (UK would be best for me) then I would love to know them.

The 64Kb of RAM that I bought was around £10, and so far I only saw a 40031 for £18 on eBay.

It would be good to know the best websites for CPC parts.

[eto]

If you have sources for the ICs (UK would be best for me) then I would love to know them.

The 64Kb of RAM that I bought was around £10, and so far I only saw a 40031 for £18 on eBay.

It would be good to know the best websites for CPC parts.

I did not factor in the HAL as it is almost certainly not your problem. Especially since you already bridged it (after message #6 in this thread) and got no change. So even if it would be damaged, something else would be too. But just to be sure: bridge the HAL and replace the RAM, then you can be 100% sure that it's not the cause and once your computer works (don't forget to change the bridge so you can test the second RAM bank too), put in the HAL again and see if it works. 

In case this HAL is the problem, you don't need the original. You can replace it by an ATF16V8. Not sure what it costs in the UK but here I can get it for about 2€ per piece in my favorite shop - or 5 pieces for 4-6€ from China. Programming is easy if you have an Eprom programmer or access to one. Even if not, I would be surprised if there is not someone on the forum who could program it for you. 

There are no websites for CPC parts unfortunately. But that is no surprise as almost all ICs are standard ICs that can be sourced easily on Ebay or AliExpress for cheap. E.g. all the other 6 ICs that can be the culprit in your case are usually below 1€ per piece. 

 

[eto]

@McArti0 

It's obvious that you have thought about what you suggest - however it's not clear what exactly you want to achieve with your ideas - and if they are correct or not. It would be good if you could not only write what test to perform, but also what purpose this test has. Without that context it's impossible to distinguish if your test is useless - or genius.

E.g. what do you achieve with the connection to CE of IC115? 

[McArti0]

@eto 

1) iorq to Ready/wait (better by diode) 

2) start cpc
        Z80 executes LD BC,#7F89. OUT(C),C
        OUT Produces iorq which does an eternal wait.

3) GA reads the number 89h a million times a second.

4)  When the first clock inverter is blocking by HI or Lo.           The clock stops working.
         Attention. The old mainboard has the first inverter on pin 5 of the 74hcu04 but the new one on pin 13.

5) When connected  pin 1. 244 to pin 5 (13) of the first clock inverter, the clock will stand and GA will be stopped in its tracks. If we draw Lo on pin 1 244, we have stopped CPC from 89h on the z80 side and gate 244 is open. (Of course, the diode helps connect only the LO state to the clock)

6) You can easily check whether the number 89h is also on the GA side on the date lines.

[eto]

1) iorq to Ready/wait (better by diode) 

what will that do?

[McArti0]

@eto
During the wait line test, the TW z80 encounters a low state from IORQ in each clock. Therefore, the write execution of the first OUT instruction never completes.

[McArti0]

How to test 74LS373 without desoldering.

The pictures show the procedure for stopping CPC on a single read from the internal RAM memory. at the end we have the state of open 74LS373. RAMRD and READY are low.

ORG #0638
LD HL,#B100
LD DE,#B101
LD BC,#07F9
LD (HL),A
.L0642
LDIR    <----- here we are stoping CPC

By setting the data pins of the internal bus, you can check whether these states pass to the external bus side. RAM pin 14 set to Lo (grounded) should show the same on pin 2.

If you are surprised by the polarization of diodes, know that I am surprised too.

(WRONG IMAGES REMOVED UPON REQUEST)

[McArti0]

Where is catode?

my multimeter works according to alternative physics. It's strange that the NPN transistor shows correctly.

[McArti0]

Back to real physics, the previous pictures have incorrectly drawn diodes.

[Bryce]

How to test 74LS373 without desoldering.

The pictures show the procedure for stopping CPC on a single read from the internal RAM memory. at the end we have the state of open 74LS373. RAMRD and READY are low.

ORG #0638
LD HL,#B100
LD DE,#B101
LD BC,#07F9
LD (HL),A
.L0642
LDIR    <----- here we are stoping CPC

By setting the data pins of the internal bus, you can check whether these states pass to the external bus side. RAM pin 14 set to Lo (grounded) should show the same on pin 2.

If you are surprised by the polarization of diodes, know that I am surprised too.

Just out of interest. Have you actually tried to do this setup in reality and confirmed it works? Or is this all based on theory?

The reason I ask, is that connecting anything at all directly to a crystal pin will usually dis-balance the capacitance enough so that
the clock will either not start at all or will be unstable.

Bryce.

[McArti0]

Just out of interest. Have you actually tried to do this setup in reality and confirmed it works? Or is this all based on theory?

I did exactly that yesterday. Everything I write in this thread is tested (on real CPC16128)

The reason I ask, is that connecting anything at all directly to a crystal pin will usually dis-balance the capacitance enough so that
the clock will either not start at all or will be unstable.

And that's how it should work. GA is supposed to permanently turn off the clock and thereby freeze itself in a specific state.

[Bryce]

Just out of interest. Have you actually tried to do this setup in reality and confirmed it works? Or is this all based on theory?

I did exactly that yesterday. Everything I write in this thread is tested (on real CPC16128)

The reason I ask, is that connecting anything at all directly to a crystal pin will usually dis-balance the capacitance enough so that
the clock will either not start at all or will be unstable.

And that's how it should work. GA is supposed to permanently turn off the clock and thereby freeze itself in a specific state.

Well that's at least good to know. However, there's a reason why those capacitors and the 74 are extremely close to the crystal with very short traces. Just the additional capacitance and inductance of the extra wire is enough to disturb the clock. If the jumper wire is lying across the PCB and (as is the case here), the user doesn't have a scope to confirm the clock signal, then they have no way of knowing if their test is running as expected or not and the results can't be trusted.

Bryce. 

[mmldrm]

If the jumper wire is lying across the PCB and (as is the case here), the user doesn't have a scope to confirm the clock signal, then they have no way of knowing if their test is running as expected or not and the results can't be trusted.

I do have an oscilloscope now! And pin grabbers/clips should arrive today.

I know my kit was woefully inadequate initially (and still isn't ideal now) but hourly these will provide more reliable test results.

Thanks again for putting up with this situation.

[McArti0]

@Bryce.
Notice that the clock inverter gets a heavy zero at the input - it has no chance of ticking even once. The freeze is sure and strong.On the address lines there is &B100 or &B101. I use rectifier diodes so they have a 60ns propagation time, they may not stop at the first LDIR address.