| Electronics > Beginners |
| Will this work (optocoupler over shunt)? |
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| Zero999:
--- Quote from: spec on December 06, 2018, 03:05:56 pm --- --- Quote from: ArthurDent on December 06, 2018, 02:27:44 pm ---spec - "AD, is this what you had in mind >>" That is what I had in mind and I couldn't see why there would need to be 2 supplies. The only question I have with the circuit is the input connections on the unused comparator. I think connecting the unused inputs directly to ground is recommended as in section 8.1 of this application note. http://www.ti.com/lit/ds/symlink/lm393-n.pdf --- End quote --- Well well well- I have been doing it wrong for all these years! Thanks for the info, but I am not too happy with tying both unused inputs to 0V because, it seems to me, that then the comparator would be in an undefined state and likely to oscillate (the inputs are still active at 0V input). --- End quote --- Some hysteresis would also be a good idea. |
| spec:
I think you just made an oscillator ::) There is no real need for hysteresis in this application, because the inputs are never close to one another. |
| Zero999:
--- Quote from: spec on December 06, 2018, 03:35:48 pm ---I think you just made an oscillator ::) --- End quote --- How? Please take another look. --- Quote ---There is no real need for hysteresis in this application, because the inputs are never close to one another. --- End quote --- The inputs do get quite close, about 12mV between them, when the bulb is working and C5 makes it worse, as it increases the length of time when the comparator will be in its linear region. I do agree, hysteresis not essential, as the relay has some hysteresis built-in, there's no hysteresis on any of the discrete designs posted here, but the relay will last longer with it and it's only two extra resistors, so why not? |
| ArthurDent:
Hero999 - "Some hysteresis would also be a good idea." Although it probably isn't necessary, if I were adding hysteresis I would put a 1M resistor between pin1 (output of LM393) and pin3 (+ input) instead of the way the schematic shows it between pin2 (- input) and the collector of Q2, which is the noisiest point in the circuit. I also might go with .1uF for C5 and C6 with C6 moved to be across R15.. |
| Zero999:
--- Quote from: ArthurDent on December 06, 2018, 07:51:23 pm ---Hero999 - "Some hysteresis would also be a good idea." Although it probably isn't necessary, if I were adding hysteresis I would put a 1M resistor between pin1 (output of LM393) and pin3 (+ input) instead of the way the schematic shows it between pin2 (- input) and the collector of Q2, which is the noisiest point in the circuit. I also might go with .1uF for C5 and C6 with C6 moved to be across R15.. --- End quote --- Yes, that was my first instinct, but then I noticed Q2's base clamps the comparatot's output voltage to around 700mV, which would make it less effective, so I opted for the collector and pin 2 instead. How is Q2's collector noisy? It just goes to the relay coil? There's no active noise source. |
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