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Cbm 2001 Pet strange boot

Ok

I think you need to spend some time understanding the functions of logic chips

In this case the chip takes four signals on pins 20 to 23 on which a binary number from 0000 to 1111 can be sent by the processor (ie each line can be 0v or 5V representing a 0 or a 1). This means there are a possible 16 combinations of these four signals.

The chip then interprets this four bit binary number and selects the individual output that matches this number

so for 0101 (pins 23 & 21 high and pins 22 & 20 low), this would be a binary number of 5 in decimal, this would mean that output 5 (pin 6) would be active, but because it has a little circle on it, it means its an 'active low' so it would read 0V while every other output pin would read 5V.

Similarly pins 23,22,21 and 20 high would be a binary number of 1111, which is 15 in decimal which would mean pin 17 would read 0v.

Now as homework, read this datasheet and try to match how I have described it working to how the datasheet does
 
I think its important that, as Dave suggested, you take some time to learn how 74 logic chips operate and how to read the datasheets (I think we all assumed you could and were a little frustrated when what seemed like obvious logical steps weren't taken)

Also, I missed out in the description about about the G1 & G2 signals as in the PET implementation, they basically dont do anything, but they act as 'gates' so when they are at 5v, the chip doesn't let anything through, and all outputs sit at 5V, but when they are low (and in the PET, they are connected to 0V), the input signal affects the output (hence a 'gate'). In the datasheet I quoted, the G1 & G2 symbols have a line over them.

There are a few variations in symbols that take time to learn.

We have already seen the circle symbol in the PET drawing which indicates that these pins are 'active low' but sometimes the same is indicated with a line above the symbol (a bar) or a little sloping line , these all indicate that the particular signal is 'active low'

In this datasheet, you will also see a table, which shows what combinations of inputs give on the outputs which is helpful and can often make it easier to understand what the chip does.

Any questions ?

IEC symbols
 
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Gary C said:
but sometimes the same is indicated with a line above the symbol (a bar) or a little sloping line (well, not quite), these all indicate that the particular signal is 'active low'
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active low or high? Thanks!

Maybe bad Ud2 in this case?? Can i try to replace?
 
Do you get what I mean by 'active low' ?

It means that we think of the signal as binary 1 but the actual voltage on the board is 0v (which is normally thought of as binary 1)

SO back to the drawing for the PET

We know that the inputs are working and that pins 9,10,11,13,14,15,16 & 17 have waveforms that show brief blips down to 0v ?

Further looking at the drawing, shows that these pins are the ones used by the PET itself so if they are working, UD2 can remain.

Pins 1,2,3,4,5,6,7 & 8 should also be showing the same waveform but aren't used inside the PET so if one of them isn't working, it doesnt really matter.
 
Desperado said:
active low or high? Thanks!

Maybe bad Ud2 in this case?? Can i try to replace?
Click to expand...

I gave you the answer back in post #1,434...

Swapping UD2 should give you a working pin 4, but this is unlikely to fix any of your current problems. But it will rule out one very small probability.

Dave
 
Worth then thinking about what these outputs do.

So if the CPU puts binary 1100 (decimal 12) onto UD2, pins 20 & 21 are high at 5V and thus output 12 is 'active' which means that pin 14 is at 0v ('active low' outputs remember)

Which means that SEL C is now 'active'. Note that SEL C has a bar above it to indicate its an 'active low' signal. The line then has a number 4 on it indicating we go to sheet 4 to see whats done with it.

On sheet 4 we see that it goes to pin 20 of UD6 to an input CS1 with a bar above it. CS1 means chip select and the bar means its 'active low' so when this line (SEL C) is low, chip UD6 is selected to operate :)
 
Please, please, please

Take time to understand what I have said, please ask questions to clarify as its important to you (and our sanity) that you begin to be able to diagnose some of the issues yourself.
 
Gary C said:
Worth then thinking about what these outputs do.

So if the CPU puts binary 1100 (decimal 12) onto UD2, pins 20 & 21 are high at 5V and thus output 12 is 'active' which means that pin 14 is at 0v ('active low' outputs remember)

Which means that SEL C is now 'active'. Note that SEL C has a bar above it to indicate its an 'active low' signal. The line then has a number 4 on it indicating we go to sheet 4 to see whats done with it.

On sheet 4 we see that it goes to pin 20 of UD6 to an input CS1 with a bar above it. CS1 means chip select and the bar means its 'active low' so when this line (SEL C) is low, chip UD6 is selected to operate :)
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unfortunately I don't understand any of this ... it's a difficult topic and since I don't know the English language very well, everything becomes more difficult for me ... Maybe I have to think about abandoning this repair :(
 
Desperado said:
unfortunately I don't understand any of this ... it's a difficult topic and since I don't know the English language very well, everything becomes more difficult for me ... Maybe I have to think about abandoning this repair :(
Click to expand...

Ok, maybe too much

Did you get the point that UD2 takes four lines and selects one out of 16 lines ?
 
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