Thank you and everyone included in this thread for this explanation! After wiring everything I realized that this is happening with my RAM, a whole day of trying to debug it I was exhausted. This saved my day and now I can finally continue and write my first program :)
Ok I have a question about the Schottky diode part of the solution. When the clock goes high the capacitor is charged. when the clock goes low how does the capacitor discharge with the Schottky diode in the way of current flow?
Note that on the other side of the capacitor from the diode is a connection to ground through a resistor. That pulls the voltage down on the capacitor side when the clock goes low. It is in fact a “pull down resistor”.
But I realize you mean what discharges the diode side of the capacitor. Admittedly I don’t have a good answer on how the electrons flow there. I can tell you (based on oscilloscope) the voltage does drop.
@@MichaelKamprath thanks, Doesn’t it matter if the input of the ram ICs is inverted? I use the 74LS219 instead of the 189 . Can I invert the write pulse there too?
@@labamichnetvoll5911 It matters insomuch that you have the sense of the signals right to the bus. The original design uses inverters to make the RAM out put active high because the RAM is active low. If you use active high RAM, the you should get rid of those inverters.
Amazing job figuring this out and providing a solution!
Thank you and everyone included in this thread for this explanation! After wiring everything I realized that this is happening with my RAM, a whole day of trying to debug it I was exhausted. This saved my day and now I can finally continue and write my first program :)
Glad it helped you!
Thank you very much!!!! It helped me a lot in solving the problem🙏👍
Thank you for your detailed explanation, very useful for a better understanding on what's under the hood.
Glad you found it useful!
A more general fix would be to latch the outputs of the 74LS157, though that would add another chip.
How would you do that when the 74LS157 output here is not intended to be synced with a clock and instead have immediate effect?
Thank you,you are my life saver!I deal with it all day and find this video!
and here is another question，if i change the address , sometimes when i switch back to the origin address, the data will be changed.
Happy it was helpful for you!
never mind,i fixed it
Thanks, I had exactly the same problem and this fixed it. Great explanation!
Glad it helped!
Nice rundown. I was having a similar problem myself.
Thank you for the explanation, great video
Glad you liked it
Could you tell me the measurements of the whole computer? These are eight breadboards one each side, but some power rails are taking off, right?
Ok I have a question about the Schottky diode part of the solution. When the clock goes high the capacitor is charged. when the clock goes low how does the capacitor discharge with the Schottky diode in the way of current flow?
Note that on the other side of the capacitor from the diode is a connection to ground through a resistor. That pulls the voltage down on the capacitor side when the clock goes low. It is in fact a “pull down resistor”.
But I realize you mean what discharges the diode side of the capacitor. Admittedly I don’t have a good answer on how the electrons flow there. I can tell you (based on oscilloscope) the voltage does drop.
Thank you.
You're welcome!
What did you use exactly for a schottky diode?
A BAT43 diode.
@@MichaelKamprath thanks,
Doesn’t it matter if the input of the ram ICs is inverted? I use the 74LS219 instead of the 189 . Can I invert the write pulse there too?
@@labamichnetvoll5911 It matters insomuch that you have the sense of the signals right to the bus. The original design uses inverters to make the RAM out put active high because the RAM is active low. If you use active high RAM, the you should get rid of those inverters.