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Oct 19, 2019 at 2:01 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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S May 14, 2018 at 20:16 history suggested Dorian CC BY-SA 4.0
Corrected the schematic following explanation in the comments
May 14, 2018 at 10:02 answer added Dorian timeline score: 1
May 14, 2018 at 8:50 review Suggested edits
S May 14, 2018 at 20:16
May 13, 2018 at 22:27 comment added Hans For more in-depth background, www.pinrepair.com/6803/index.htm#opto will show the exact boards in question, and should answer most questions on how this board is used.
May 13, 2018 at 22:18 comment added Hans I can definitely share the use for this one. It's 4-switch opto board, used in the ball trough of a pinball machine. Detects how many balls are in the trough. Ties into, I believe, a 6521 or 6821 peripheral interface adapter (PIA) chip. All 4 circuits are identical. I could do the whole schematic as well, but the font is hard to read, so wanted to zoom in closer.
May 13, 2018 at 21:22 comment added jonk @Hans You say there are several similar sections, only one of which you are showing us. I can't tell from reading your writing what the original purpose of the circuit is. For all I know, it might be a paper-tape reader from 1970. Would you mind spending a little time talking about the "design I need to replicate, function wise?"
May 13, 2018 at 20:41 comment added D.A.S. Normally NPN bias current is sourced from the Vcc so reverse biased from Vcc to base. or many other ways too google.com/…
May 13, 2018 at 17:35 history edited Hans CC BY-SA 4.0
added 299 characters in body
May 13, 2018 at 17:32 comment added Hans No, definitely not 9v, it's a TTL logic source. Again, limitation of the schematic software, never used it before. Hopefully this works to clarify. I'll edit the image link into the main post as well. siegecraft.us/images/IMGP5973.JPG
May 13, 2018 at 17:31 comment added Dorian Are you shure there is a 9V source or it's just the voltage you measured there?
May 13, 2018 at 17:29 history edited Hans CC BY-SA 4.0
added 129 characters in body
May 13, 2018 at 17:28 comment added Dorian Something is really wrong in the first (also the second) schematic. Q1 will certainly burn in the first moment, the second circuit does nothing.Could you send pictures instead ? Of the original circuit.
May 13, 2018 at 17:27 comment added Hans So, essentially, just flip D2 around? Just put in a revised schematic to show that change.
May 13, 2018 at 17:23 comment added D.A.S. Yeah you’re almost right photo diode is backwards it’s a current source when reverse biased
May 13, 2018 at 17:19 comment added Hans Verified the schematic on the first one again. I agree, it really does look odd to me as well, but I did confirm it via both an original schematic and also measurements on a PCB. That's part of my confusion, the original is just not what my research has turned up for a photodiode circuit of any kind.
May 13, 2018 at 17:13 comment added Hans In this schematic, yes, I used photodiodes. I didn't see an available option for phototransistors or photodarlingtons. The original circuit left the base leg unconnected. I'll double check that first circuit, the original schematic is pretty difficult to read, and I can't post photos yet.
May 13, 2018 at 17:09 comment added Leon Heller D1 and D2 are photodiodes.
May 13, 2018 at 17:05 comment added Hearth Are you sure about that first one? It doesn't look right.
May 13, 2018 at 17:00 review First posts
May 13, 2018 at 17:08
May 13, 2018 at 16:58 history asked Hans CC BY-SA 4.0