Ok. I have had a little time to look at this again. I'm still a bit stuck, but still willing to try and progress this.
Thank you for the drawing and the PCB photos. I think it would be worth focusing on the main connector, JP1, and U3 and U6, as they are quite visible, and located close to the connector.
One thing to point out is that although there are individual tracks on the PCB, all the other space is covered with copper too (you should be able to see that for yourself). This copper area is usually connected to ground, and called a ground plane. This means that a single pin (or group of pins) on JP1 for ground are connected to the ground plane, then any of the components that need to connect to ground don't need a track back to the ground pin on JP1, they just need to connect to the ground plane. You can see this most obviously with the capacitors. The capacitors have two connections, one at each end. You will see that a track goes to one end, and maybe continues somewhere else. This track is most likely the positive power connection (from JP1). The other end of the capacitor will seem to go nowhere, but it's connected to the ground plane, so there is a connection back to the ground pin on JP1. You will also no doubt have noticed that sometimes a track ends in a circular pad, and there is another pad directly in line with it on the other side of the board. This is called a via, and it basically connects the track through the board so that it can continue on the other side.
So, take a close look at JP1, remembering that the little arrow next to the gap in the connector on the back side of the board is pin 1, and that the pin order is reversed when we look at the top of the board. I think we have established that you are well aware of that, but it is easy to get confused when looking at the board from either side.
Unfortunately I can't really zoom in on your photos. They get very blocky, and it's hard to discern tracks and pins. Starting from basic assumptions, and some of your preceding information, can you confirm that JP1 pins 12, 13, and 14 are +5V? Certainly pin 13 seems to go to one side of C4 and in to pin 16 of U3. What do JP1 pins 12 and 14 do?
Similarly, JP1 pins 1, 3, 5, 7, 9, and 11, all appeared to be ground. Do you still think that is the case? It's not unusual to have several ground pins like this. In fact, if this is the case then it means that the ribbon cable will have alternating signal pins and ground pins, which helps to physically separate the wires in the ribbon cable, and therefore reduce interference and crosstalk.
Now, looking at U3 and U6. We don't care about pins 3, 4, 5, 6, 11, 12, 13, 14 as these are the inputs. They will be connected to the switches and resistor networks (RNx). We don't really care about pin 16 and 8 as they are power and ground and must be connected to power and ground or they won't work. We also don't care about pin 15. This is the ~CE pin, or Chip Enable. It must be pulled to ground to work. I don't think this pin is externally controlled. I think it is connected to ground for all the chips, so they are all enabled all the time. Again, you can look at pin 15 on all the shift registers and confirm this.
That basically leaves pins 1, 2, 9, and 10 on the shift registers. These are the important ones, and we really need to know where they go.
Pin 1 is ~PL, which loads data from the inputs into the shift register. I think that pin 1 on all U1 to U9 are connected together and go to a single pin on JP1. Similarly, pin 2 is CP, which clocks the data out of the shift register. I think all of these are connected together too, and go to a single pin on JP1. Again, these are assumptions. Don't take my word for it. Since you have the hardware in your hand you can verify it, or tell me what's different. On your drawing, pin 1 of U3 is connected to pin 1 of U6, so it's a good sign.
Finally pins 9 and 10, which are Q7 (shift register output) and DS (shift register input) respectively. These form a "daisy chain" across the PCB, where the output of one chip is wired to the input of the next, until you get to the end of the chain. Then the output of the last chip is connected to JP1, where it can be read externally. The input of the first chip is probably connected to ground.
I'm still not sure what U10 does. I know what it is and how it works, but not what its purpose is in this design.
Sorry if I am asking you to repeat what you have done already, but I'm really trying not to assume anything, and go on proof instead on supposition.
Thank you for the drawing and the PCB photos. I think it would be worth focusing on the main connector, JP1, and U3 and U6, as they are quite visible, and located close to the connector.
One thing to point out is that although there are individual tracks on the PCB, all the other space is covered with copper too (you should be able to see that for yourself). This copper area is usually connected to ground, and called a ground plane. This means that a single pin (or group of pins) on JP1 for ground are connected to the ground plane, then any of the components that need to connect to ground don't need a track back to the ground pin on JP1, they just need to connect to the ground plane. You can see this most obviously with the capacitors. The capacitors have two connections, one at each end. You will see that a track goes to one end, and maybe continues somewhere else. This track is most likely the positive power connection (from JP1). The other end of the capacitor will seem to go nowhere, but it's connected to the ground plane, so there is a connection back to the ground pin on JP1. You will also no doubt have noticed that sometimes a track ends in a circular pad, and there is another pad directly in line with it on the other side of the board. This is called a via, and it basically connects the track through the board so that it can continue on the other side.
So, take a close look at JP1, remembering that the little arrow next to the gap in the connector on the back side of the board is pin 1, and that the pin order is reversed when we look at the top of the board. I think we have established that you are well aware of that, but it is easy to get confused when looking at the board from either side.
Unfortunately I can't really zoom in on your photos. They get very blocky, and it's hard to discern tracks and pins. Starting from basic assumptions, and some of your preceding information, can you confirm that JP1 pins 12, 13, and 14 are +5V? Certainly pin 13 seems to go to one side of C4 and in to pin 16 of U3. What do JP1 pins 12 and 14 do?
Similarly, JP1 pins 1, 3, 5, 7, 9, and 11, all appeared to be ground. Do you still think that is the case? It's not unusual to have several ground pins like this. In fact, if this is the case then it means that the ribbon cable will have alternating signal pins and ground pins, which helps to physically separate the wires in the ribbon cable, and therefore reduce interference and crosstalk.
Now, looking at U3 and U6. We don't care about pins 3, 4, 5, 6, 11, 12, 13, 14 as these are the inputs. They will be connected to the switches and resistor networks (RNx). We don't really care about pin 16 and 8 as they are power and ground and must be connected to power and ground or they won't work. We also don't care about pin 15. This is the ~CE pin, or Chip Enable. It must be pulled to ground to work. I don't think this pin is externally controlled. I think it is connected to ground for all the chips, so they are all enabled all the time. Again, you can look at pin 15 on all the shift registers and confirm this.
That basically leaves pins 1, 2, 9, and 10 on the shift registers. These are the important ones, and we really need to know where they go.
Pin 1 is ~PL, which loads data from the inputs into the shift register. I think that pin 1 on all U1 to U9 are connected together and go to a single pin on JP1. Similarly, pin 2 is CP, which clocks the data out of the shift register. I think all of these are connected together too, and go to a single pin on JP1. Again, these are assumptions. Don't take my word for it. Since you have the hardware in your hand you can verify it, or tell me what's different. On your drawing, pin 1 of U3 is connected to pin 1 of U6, so it's a good sign.
Finally pins 9 and 10, which are Q7 (shift register output) and DS (shift register input) respectively. These form a "daisy chain" across the PCB, where the output of one chip is wired to the input of the next, until you get to the end of the chain. Then the output of the last chip is connected to JP1, where it can be read externally. The input of the first chip is probably connected to ground.
I'm still not sure what U10 does. I know what it is and how it works, but not what its purpose is in this design.
Sorry if I am asking you to repeat what you have done already, but I'm really trying not to assume anything, and go on proof instead on supposition.
Statistics: Posted by ame — Thu Jun 20, 2024 5:22 am