Ohm's and kirchhoff's law - have I misunderstood? - Page 1
Forum section: electronics / beginners

author: FriedMule
author_rank: Frequent Contributor
timestamp: July 09, 2022, 06:25:20 am
content: So just to try to learn more do I often find schematics and try to find out how they work, often can I find the logic behind, but this time do I have to give up!!What are R20 and R21 doing there?Would the optocoupler not just burn up due to too many voltage and current?Will the normal LED even turn on?

author: m k
author_rank: Super Contributor
timestamp: July 09, 2022, 09:17:16 am
content: The artist has clearly some problems with par vs. ser and V vs. I.

author: FriedMule
author_rank: Frequent Contributor
timestamp: July 09, 2022, 10:38:30 am
content: Yes I thought so, but what is the thought behind these two resistors?R20 is bleeding down to Q8 if the voltage should get too low, but this does I not imagine would happen.I imagine they can adjust the voltage, but do not touch the current in any way?Would the two resistors R20 and R21 not be better placed in series with the two LEDs?

author: magic
author_rank: Super Contributor
timestamp: July 09, 2022, 02:46:09 pm
content: I don't understand what's the point of this circuit and it doesn't make much sense.700mA current source is too much for the combination of LED and optocoupler.The general rule is that if the LED isn't passing 700mA then the rest of the 700mA current will flow through the opto into Q8 gate and turn it on, activating the LED. So most of the current passes through the LED, but the LED is rated for 20mA only.Where did you find this schematic?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 09, 2022, 03:17:57 pm
content: I gave that circuit about 30 seconds before I gave up.It's just not worth spending time on.The quality of "random" stuff from the 'net varies a lot, and part of the art of electronics is to identify the garbage quickly and dismiss it.

author: FriedMule
author_rank: Frequent Contributor
timestamp: July 09, 2022, 03:41:31 pm
content: But would this not only happen if the voltage is high enough AND not too high?The LED is 3.5V, so does that not only begin to use current at about that voltage level?Would the opto not be burned off long before hat would ever happen?Why would you use a current source instead of a voltage source?I found it by clicking randomly around after searching for simple latching push button, it led me out on a journey to a site that does not use the same type of letter we do here, so I have no idea of what it is, but it looks strange! :-)

author: FriedMule
author_rank: Frequent Contributor
timestamp: July 09, 2022, 03:43:33 pm
content: LOL and I gave it 30 minutes, then 2-3 hours again and ended up asking here. Call me naive but if a person posts a circuit, do I stupidly assume it has some function. :-)

author: m k
author_rank: Super Contributor
timestamp: July 09, 2022, 04:05:21 pm
content: My guess is that R21 is limiting the LED.That then leads to the assumption that the rest are flying too high.

author: fourfathom
author_rank: Super Contributor
timestamp: July 09, 2022, 04:28:26 pm
content: All circuits have value, even if only to serve as a bad example. This one is an excellent example of what not to do.

author: pcprogrammer
author_rank: Super Contributor
timestamp: July 09, 2022, 04:38:35 pm
content: I'm not very good with this kind of electronics, but I would say that R21 only takes a little bit of the current. The LED is a limiting factor and if this is actually driven by a current source that can supply a high enough voltage to get the 700mA flowing the whole circuit will burn.To get 700mA in a 10K resistor you need 7000VBut the LED it is parallel with will limit the voltage to ~3.5V because that is what a LED does, so the current can't flow through the 10K resistor. The LED will only take so much above the 20mA before it pops.Then the optocoupler. a) what is the use, because nothing is connected to the transistor of it, and b) here the same applies as for the LED, only the voltage seems to be 1.2V with 100mA forward current. So the 1M resistor won't do a lot there. The 1K resistor is the one that limits the optocouplers led current since it is in series. The gate of the FET has a very high resistance so no current there.The question is if the FET will turn off when the voltage across the 1K resistor (R22 and diode) drops below the turn on voltage of the FET.It is a very dubious circuit indeed.

author: rstofer
author_rank: Super Contributor
timestamp: July 09, 2022, 05:20:04 pm
content: Quote from: FriedMule on July 09, 2022, 03:43:33 pmYou might get some insight by putting this circuit in LTspice. Perhaps even Falstad but I haven't used that app.

author: pqass
author_rank: Super Contributor
timestamp: July 09, 2022, 05:52:22 pm
content: That's a lot of parts for just lighting a high power LED!?EDIT: Unless, you don't assume an unlimited compliance voltage.ie. It maxs-out at 12V and the optocoupler notifies when the LED is burnt-out? Still too many parts for that.Simulation here:http://www.falstad.com/circuit/circuitjs.html?ctz=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-YHgwT9TmqNETRsAM6LPHrS5OOqNSt0EFgt77XPgDZAAMOSFET beta changed to 6 using Rds(on)=0.018R, Vgs=10V, Vgs(th)=0.75 as perhttps://www.falstad.com/circuit/mosfet-beta.htmlDatasheet:https://datasheetspdf.com/pdf-file/589863/Alpha&OmegaSemiconductors/AO6408/1

author: pcprogrammer
author_rank: Super Contributor
timestamp: July 09, 2022, 06:06:48 pm
content: Nicely done. In real life it would go poof the LED is now sticking to the ceiling

author: FriedMule
author_rank: Frequent Contributor
timestamp: July 09, 2022, 06:40:40 pm
content: Quote from: m k on July 09, 2022, 04:05:21 pmBy limiting, do you mean reducing the voltage or do you mean splitting the current in some ratio?Quote from: pcprogrammer on July 09, 2022, 04:38:35 pmWhat if we just pretend the circuit had some strange "black box" function, and we had to correct it without moving the resistorsThe R20 does to me look like a bleeder?Am I wrong in that R21 looks like it is reducing the voltage that FD6 without touching the current in FD6What if we reduced the current to 120mAQuote from: pcprogrammer on July 09, 2022, 04:38:35 pmI am trying to absorb everything you have written above and really understand it, I think you are very great at explaining so I really gets it, it does just take time. I am looking at the 1k or R22 as it is also known as. :-)

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 09, 2022, 06:52:00 pm
content: We are not seeing what is behind this "current source".Since ideal current sources (that have virtually no voltage compliance) exist only in fantasy (and in Spice), here it sure has a compliance voltage, so a current source basically acts as a current limiter.With that said, we can't really figure out the purpose (if there is any) of this piece of schematic without seeing the entire schematic. And maybe it's indeed some random sh*t. Just keep the above in mind just in case.

author: m k
author_rank: Super Contributor
timestamp: July 10, 2022, 12:41:52 pm
content: Quote from: FriedMule on July 09, 2022, 06:40:40 pmOnly artist knows.But one guess anyway.First low voltage is starting FET through R21.Then voltage rises a bit and opto replaces R20.Then voltage rises some more and LED starts turning on.Then voltage rises even more and FET turns off.