Ok let us leave the Arduino side because that may take us on another journey.The thing I don't know is how to play in the big league that is 3 phase o the Hot side of the optocouplers. Ok let me put it this way,you see that image I posted showing the single phase side connection, imagine I have 3 of those, ona 3 phase how do I go about it. First take an example if there is no nuetral second we look at it if the nuetral was there in these two how do I do the connections in such a way that when ever we are in the positive half the optocoupler gives a HIGH to Arduino pin.
What your post indicates is not actually what you need, what you are requesting is a power source to power a few electronic components, a source of potential difference
With no neutral using ground as the return of a medium voltage circuit would violate code in most countries but that is one source of potential difference, not one I would use. Another source is to connect components between phases, which takes us from dangerous to extremely dangerous. Using a ~400v primary to ~5v secondary (or thereabouts) transformer (or maybe three) is getting safer and greatly reduces the chance of electrocution. You can ground one leg of the secondary side and you have gnd/neutral.
My choice would be to buy the certified tool for the job, there are many options and they are relatively inexpensive. I have in my toolbox a Fluke T+ pro which does the job and is a decent all round tester. They run about $125 to $150 brand new. I don't know the condition but I saw one on ebay for $30 while looking a few minutes ago.
Please be aware I am thinking about your safety @derrick-pro and a 3 phase 400 volt system has killed a lot of "pros" so please take care.
Yes but I want it I this way, phase sequence is all about the order in which each phase reaches it's peak right? if so then I want to detect the order in which one phase comes first the other second and the third to their respective peaks. But that is not the point of interest here the point is how do I connect the optocouplers on the phase side or hot in order to give me a high when ever there is a positive zero cross that is it the Arduino part doesn't matter at the moment.
Reaches zero..
Like you would do with monophase. Three optocouplers phase to phase. But your diode doesn't make sense to me.
Optocoupler something like Vishay H11AA1 (or more recent alternatives).
Thank you sumguy and I must say I like the way you talk and organize your worlds very carefully I like that.But you see in our countries that is allot of money.Some people survive on one dolar a day, some even take months without out getting 30 dollars.So if you are not stubborn like some of us you don't get the money hope you get what am trying to say.
Be careful anyway. The kick from 400V is not comparable to 230V kiss... I tried when I was kid.
As the Vishay H11AA1 is bi-directional, the diode is necessary to detect the positive part of the wave.
Just connect them the way you did with 220V but with increased resistor values and power capacity. Under 220V. 2x47k draws 2.34mA and dissipates 0.25W each.
If you use four 39K resistor in series (instead of two 47K), current draw under 400V will be 400/(39000x4) = 2.56mA and dissipated power will be 400²/(39000x4) = 1.026W total or ~0.25W per resistor.
Connect one opto between phase 1 and phase 2 and one opto between phase 2 and phase 3 and the (programmed) logic should do the rest...
As for the logic, use the 3 sinewave drawing post #35 and note (using colors for example) when each opto is blocked or passing.
Yes, if that is what you are looking for.
Etienne _74 which angel has brought you, it was was as if I was thrown in a den of angry lions but any way that's just a joke. But this is what I've been looking from people to tell me.That if there is no neutral use say fr example a delta-wye transformer or RC circuit.Thank you Etienne_74 this what am looking for connect this connect that. The next question i was going to ask is if I have 3 single phase loads and want to connect them on these 3 phase lines how would i do it coz with a neutral present this would be similar.Now without the neutral can we say that is impossible, the virtual ground cannot hold the 3 leds?
Let me draw this on the paper I post you tell me if this is what you mean.
What kind of loads do you indent to connect ?
If these 3 loads are identical, you can connect them as a "Y" (all neutrals together), it will create a "virtual" ground. It is true with resistors but less true with 3 single phase electric motors as each motor will have a different impedance depending on the mechanical load applied to it.
If your country doesn't have very strict regulations, drive a copper stake into the ground, it should provide a decent ground.
Otherwise, there's no easy way to create a ground as it should absorb the current difference between the loads.
The loads are only 3 leds in the optocouplers no other.Here is the diagram is this what you meant in post #47? Please correct me if am wrong.
With no duplicate between phase 3 and phase 1 you will have 180 degree ambiguity! You don't know if you are looking at an old phase time or the current phase time.
You're right, that's what I mean.
If you want a weak virtual ground to connect 2.5mA LEDs to detect zero crossing, use 3 opto and connect the diode between pin 1 and 2 of each opto.
In other words, keep your drawing post #36 (including the 47k) but connect the 1N4007 between pins 1 & 2 (cathode or ring to pin 1 the PC817 so that the opto is conductive during positive half and the diode during the neg half of the sinewave). Connect all three N together and each L to one phase. As the loads (the 47K) are identical, the Y is balanced and you created a virtual ground (as long as you don't connect any other thing to it)
It would be very clear for me if you draw for me a simple diagram like this,it would make my life easy since am a slow learner.
Draw a simple diagram and show what you think is right I think in that way the puzzle will be easy solved.
Tho am not understanding how this works at the moment question is with only two optocouplers, how will I sense that this rising edge is from this phase A, B or C ? I need a clear drawing on paper may be am not getting it right and if someone draws this on paper then maybe my mind will be cleared.
Sorry, I have nothing to draw on right now.
You're almost right:
- remove the old diode, you only need the one between pin 1 & 2
- use only two 47k and one opto + one 1N4007, see your post #36
- replace "phase 2" with "N"
- add a third identical leg between "N" and "phase 2". This creates a 3 leg star also know as a "Y", each leg consist of two 47K resistors, one 1N4007 anti-parallel to one PC817.
The zero crossing time in milliseconds will give you the order of the phase in each of the three wires. After three measurements, reset to zero and measure again. You know the wire number for each of the three opto-couplers.
