Timeline for How to use coordinate strip electrically?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2019 at 13:48 | vote | accept | xakepp35 | ||
| Feb 11, 2019 at 5:09 | comment | added | Dave Tweed | @jsotola: Yes, I know. That's the OP's term, and I was just quoting it back to him. | |
| Feb 10, 2019 at 18:15 | comment | added | jsotola |
@DaveTweed, optical slot sensor is a more accurate term than optoisolator
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| Feb 9, 2019 at 18:40 | comment | added | TimWescott | As far as limiting current, etc. -- there are a lot of different ways that quadrature encoded strips can be read; there's no telling the details of how yours work. Powering up the thing with the board from the printer and looking at the signals is a good first step. The most likely connection is one or two LEDs for illumination, and two phototransistors (with open-collector outputs) for readout. The read head may have its own signal conditioning, in which case the readout would be typical 0-5V or 0-3.3V CMOS. You need to do some poking and prodding to find out. | |
| Feb 9, 2019 at 18:37 | comment | added | TimWescott | Reading a quadrature position signal is not a patented algorithm. It's dirt-simple and dirt-cheap. Have you done a web search? | |
| Feb 9, 2019 at 17:09 | comment | added | xakepp35 | In practics, regardless to politics and policies, i want to ask - "do I need to limit a current, or is just a DC pin voltage is suffient?" and "Does it have a common ground, or is it optoisolated and actually 2 grounds? (3 grounds - there are 2 sensors+emitter, and each is 2pin device)" | |
| Feb 9, 2019 at 17:08 | comment | added | xakepp35 | Liked your reply, but "look at (reverse engineer) the original printer board" is some "company secret", which would lead only to "dumping+disassembling firmware". I am Russian, and so is very falimiar with it, but dont want to bother with USA. I want to go my own way from the grounds, ignoring their patented algorithms. I dislike that policy for being "anti-scientific" or "anti-educational" (human="stupid user") - "no datasheets for own assemblies provided, no cartrige refills", so on.. So we should ban that country worldwide, in theory, we not accept it. | |
| Feb 9, 2019 at 15:18 | comment | added | Dave Tweed | You'll have to look up the specific part, or make your own measurements, or take a look at (reverse engineer) the original printer board. That's why I said "more or less" -- some signal conditioning may be required, but in any case, it's a binary signal with just two states. | |
| Feb 9, 2019 at 15:05 | comment | added | Hearth | @xakepp35 Depends on how the output is configured; some of these have open-collector outputs instead of push-pull. | |
| Feb 9, 2019 at 15:00 | comment | added | xakepp35 |
So, theoretically that should have 4 legs - a common ground, a DC input +5VDC for powering light source (without any current limiting), and outputs for channels A and B. And they can output low(<0.6VDC) or high(>2.5VDC) voltages, relative to common ground. And I should just read A and B signals and do some "quadrature.." DSP with those two. Am I Right with that rough guess?
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| Feb 9, 2019 at 14:51 | history | answered | Dave Tweed | CC BY-SA 4.0 |