Circuto LEDs AC230v <=> DC320v, download capacitor.
This solution or I do not understand it or I am not able to interpret the graph :(
I understand that you put an AC relay, so I don't see the solution.
I will try to explain how I have it, because I actually already use a RELAY
1 - switch, this switch is always ON
2 - once the AC power passes, the neutral goes to one side of the bridge and the phase passes through a RELAY controlled with Arduino, when Arduino gives the order, he opens the RELAY and feeds the other end of the bridge.
3 - the bridge feeds the capacitor and the capacitor removes the + and - from the circuit
and of course, sorry for writing it apart ..
4 - when Arduino cuts the RELAY it is when the LEDs continue to illuminate, slowly lowering the intensity until almost the time of effect is off, although as I said the RELAY continues with a more than high load close to 240v
Discharging the capacitor is unnecessary in addition to generating useless consumption whenever the circuit is energized.
The correct thing is to produce an interruption, one at the alternating input and the other at the continuous output with the same relay that has a double circuit (inverter).
In any case, the capacitor capacity value seems too much.
Hi Jose Miguel
Yes, that was my initial idea to solve the problem, but since I have no DC RELAY, I wanted to solve it without it.
I have not seen RELAYS that allow to do that, but it does not matter either since 2 can be used for AC and one for DC, although I should buy the one from DC.
I currently use the model '24V-380V 40A 250V SSR-40 DA' and of those I have 2 ... I have not tried to use it in DC but I also do not understand what system can have a simple RELE so that it does not serve in the same way for DC and I've never considered trying it.
How do you see that use the same relay to cut the DC, explode :)?
by the way, YES .. I understand with your comments that the capacitor is exaggeratedly excessive, but allow me a question about this.
What real problem can this have in my circuit? Energy inefficiency, damage to the circuit?
* I understand that the farads are simply a value of 'storage capacity', what I do not understand is if I want to get 1 bottle of milk a month and have a box of 24 bottles stored, what would be the problem, more than the date of expiration?
Greetings and thank you
The difference from one DC to another AC is the coil. The one that is AC is worth to DC if you do not energize it for a long time.
The coils that are designed to work in AC have an inductive reactance component that has been taken into account when calculating its working current and if you feed it continuously the coil current will be higher so it can be overheated.
What are the relay plates are indifferent if they control AC or DC. the only thing that the amps that these platinois can withstand is greater in AC than in DC.
Whenever you read on a relay that is for AC or DC keep in mind that it refers to its coil.
Thanks Jose Miguel
So that RELAY does not help me because the lights are usually on for many hours and even if they only transport 1A I suppose it will be broken, I will buy one for DC even if this means waiting to solve it.
By the way! checking I found an already used capacitor that marks '100uF 450v' when testing it gives me that it has 92uF.
The question I ask you is.
Having a new 3300MFD and an old one of 100uF that actually offers 92uF.
Do you think without hesitation that it would be better to use the old 100uF? and because?
The circuit consumes 1a and the frequency of my AC is 50hz
It is advisable to use between 1000 and 2000 microfarads per amp. You can use the AC relay by limiting the coil current with a resistor or with a zener so that it works with 30% less voltage.
Regarding your question about capacitor capacity. I will tell you that in principle there is no problem in that it is oversized except what is happening to you with excessive download time.
If the relay is a 24 volt coil, place a 8.2v 9.1v zener in series to reduce the intensity and no longer burn.
Thanks Jose
Yes, the 2000 Ampere had read it and so I bought one of 3300 thinking that nothing would happen to me, THE PROBLEM is that I did not notice that I was measuring with different measurements and of 3300 microFaradios that I thought I had bought, what I bought They were 3300000 microFaradios.
On the RELAY, if I have to buy a Zener diode and put it better I buy another DC RELAY and I am sure, unless I can fix it with what I already have.
1st delivery that I don't know what coil it takes, '24V-380V 40A 250V SSR-40 DA' the 24v that appears is part of a range that I suppose will be the AC feature it allows.
2º knowing how little I know of a Zener diode, could I not do something similar with a resistance of 1, 5 or 50ohm (they are the only ones I have) + a normal IN540 diode that is the one I use for the bridge?
I rectify, they are NOT megafaradios I went to see the pdf of the manufacturer and where it says MFD is exactly uF bone microfarad, so I also did not buy anything so disproportionate ... double what is necessary? well it's not so much :)
By the way, how do LEDs without capacitor suffer?
Is it really a problem for the LED to have such a quick curl, 100Hz?
I have removed the capacitor to go testing and I have left it without it and because it was low light I removed an LED from the series and the truth is that I do not see anything wrong to go testing while the ideas discussed here.
nothing happens to the LEDs because they do not have a capacitor just to produce the strobe effect even though the eye almost does not appreciate it and another thing that happens to them is that they will illuminate something less.
Upload a photo of the relay so that the nomenclature is visible.
Play ..
As I said in this post the capacitor marks 3300MFD that actually seeing the PDF of the manufacturer the 'M' should actually be 'm' lowercase, so it is not millifaradio but micro, that is 3300uF I comment it just after.
As EVERYBODY insisted on the exaggeration of the capacitor, I interpreted that the capital M was thousands, but they are micro and seeing the size that the 'mFD' is already named more accurately, I doubt that there is one that is 'Mili', the capacitor measures 11 * 6cm and I repeat it is 3300uF, I don't think the rookie mistake is so big to give it so much ball
sorry .. Jose Miguel
I am new to the forum, I responded outside our conversation but I copy the important thing here.
https://i.ebayimg.com/images/g/2ZcAAOSw-RFaUtE1/s-l1600.jpg
Hi Juite and Servimat1
I may not be very focused on these issues and the values do not express them real.
1 - Yes, if I charge the capacitor directly (although I really don't know what you mean by direct) although maybe you say it about 3f and I went to look well at the capacitor to add the real value, the value that marks is 3300MFD 450VDC, I suppose for your answer that this value will not really be '3f' as I wrote, is it perhaps as servimat1 3.3uF said?
I apologize to all for that failure
2 - If servimat1, it is electrolytic and brand 3300MFD 450VDC ... is it useful for this application?
I leave a link for you to see another methodology of connection of the LEDs. Regards
http://www.circuitosimpresos.org/2011/12/04/lampara-led-a-220v/
thanks servimat1
I was looking at the first option you put, the resistance. I tried to interpret well how to do it
but first a question about what the capacitor is very large.
I thought that since the circuit had a consumption of 1a, it would take a large capacitor to remove the curl. Is this really a problem to remove it and use a smaller one? Is it dangerous to be using 20 times more uF than you propose? The truth is that it works perfectly except for the exposed problem.
and about solution 1 that you expose
Is it to put a resistor joining one end of this to the positive and the other to the negative of the capacitor and then simply continue the circuit as if nothing?
I ask you because this that I described that I understood, gives me a scary hair :)
Yes, yes ... servimat1
I look at everything slowly because I need to review everything slowly to understand it.
Link 2, I see that it is your 'solution 1' and add a polyester capacitor also with parallel resistance on the alternating side.
How did you put these 2 solutions, do you think your second option is better since it has the polyester capacitor or could I really avoid it without much trouble?
I do not need the light to go out immediately, I better not exceed 10min
good! Excuse me, I see that my questions bothered you and certainly you misunderstood it, although it doesn't matter if it's me who sees it that way.
I just asked and even without answering many of the questions that would help me to learn better, I thank you for your help, I press the 'best solution' button that I have NEVER doubted and not bothered anymore.
thanks for sharing
That is an electronic or solid state relay and has no coil. It suits you perfectly for what you are using. That relay can work in the same maneuver with alternating as with continuous but if you spit it with continuous it has polarity that is why it indicates it in its input.
What if you do not use to switch direct current because it is a triac relay.
I did not know very well what you say :(
First you say that if it is valid and then the final sentence, "What if it is not enough for you to switch direct current because it is a triac relay." It leaves me in doubt.
What I intend is to switch continuously or to cut off of letting current go by, is it called differently? :)
The case is that the manufacturer tells me that if I want to use it for continuous I should buy the DD model instead of DA
https://i.ebayimg.com/images/g/1uQAAOSwLahawytm/s-l1600.jpg
but of course, this brand up to 220VDC so it would also be a problem since I put DC330v
and good! doing tests, but EYE ... they are only tests of low intensity and short time.
Of course it works, I did not expect it to not work, what I hope is that it does not toast ...
I put the positive in 1 and 2 gives me the energy when I 'turn it on' with the 5v of the arduino, but the test was to use 24v delivering 1st and for only 10 minutes
