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Adding capacitor to wireless power transfer circuit. Need guidance.

ThunderBear

Feb 6, 2022
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Hello.

This is my first post in this forum, and I know very little about electronics.

I am currently using wireless power to light up a LED inside of a wooden sculpture.

IMG_4487-B.jpg

So far I have been using matching components from a kit:

https://www.aliexpress.com/item/32944450041.html

The sculpture has a stand with the transmitting coil inside. The receiving coil will usually be in the most favorable position to receive power from the transmitter.

To do get this I take apart the receiving coil and the LED, and reconnect the coil to the LED through thin wires.

The coil have a tiny capacitor attached to it that I have been told is for resonance with the transmitter. I have also learned that the transmitter sends AC power, and that this is no problem for the LED to handle as it will just block half of the pulses and use the rest.

IMG_20211214_080421d.jpg

What I want to achieve is for the LED to turn on slowly when put in it's stand, and fade out slowly when it is taken off the stand.

I have managed to create a circuit that gives me the wanted effect, but the question here is how to make it safe, and what is the best components to use.

This is what I have done so far:

C1. I want to use a super-capacitor to delay the light up, and to store the energy. This is because of the high amount of charging cycles before replacement is needed. In my working example I use a 2.7V 0.5F super-capacitor.

D2. In order to use a super capacitor I have to turn the AC from the coil into DC, right?, so the first component I have added right from the coil (after the tiny resonance capacitor) is a Shottky diode 1N5819.

This is my circuit so far:
LED-Circuit-1.png
R1. I was told that i need a resistor before the LED, so I found some formulas and found that i needed on at 3ohm. Anyways I have tested the function with and without resistor, and it works in any case, but the question here is to find out what is safe since the whole assembly will be put inside a wooden sculpture.

L1 is the receiving coil shown i the photo. C2 is the tiny capacitor shown in the same photo. D1 is the LED in the same photo. All these three parts was the original receiving assembly.

Now to what I am wondering here.

1. The super-capacitor is made for 2.7V, but the voltage in this circuit is much higher. When I try to measure the voltage of the AC directly from the coil i get around 2V, but I don't think this is a correct measurement, because the measurement i get after adding the diode is up to 28V. I have measured this directly with my multimeter, but also indirectly by charging a 100V capacitor and taking measurements from it after i take it out of the circuit. The LEDs are rated from 2.2V to 3.6V, so this does not make sense to me. Could i destroy the super capacitor by charging it at this voltage, or is it just mye readings that are wrong for some reason?

2. I don't really know how to choose a resistor for this circuit, or if i need it at all.

When i test this circuit on my breadboard it works just the way I want, but the readings make me concerned about the safety. Maybe the capacitor and the resistor can generate too much heat over time closed inside the wooden sculpture? What is going on here with the voltages?

I have tested on my breadboard a lot, and no component ever gets hot to an extent where i can sense any heat with my fingers. Still these are critical questions because i sell these sculptures to customers who may have them connected to power day and night over years.

Can anyone help me to understand what is going on here?
 

bertus

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Nov 8, 2019
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Hello,

What is the voltage you see on the supercapacitor?
As stated it may not go over 2.7 Volts.
What color led is used?
A white led may need 3 to 4 volts to be lit.

led color guide.png

Bertus
 

ThunderBear

Feb 6, 2022
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Hello,

What is the voltage you see on the supercapacitor?
As stated it may not go over 2.7 Volts.
What color led is used?
A white led may need 3 to 4 volts to be lit.

View attachment 54169

Bertus

The reading on across the super capacitor is max 2.750V, and so far I have not felt it getting hot in any way.

The current that can be drawn from the transmitter is max 750mAh, but it is dimentioned for at least 3 small receiver coils.

I have tested this circuit with orange, green, royal blue, and 30 000K white LEDs.

The white and the royal blue LEDs gets the shortest fading time. The green light fades slower, and the orange LED get the longest fading time. All these colors light up well when the capacitor is "filled up", so the LEDs seems to be quite flexible. I have seen this table before so I don't really understand why the white LED light up with only 2.7V.
 

Audioguru

Sep 24, 2016
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A car battery is rated in hundreds of Cold Cranking Amps because cold oil is very thick which causes the starter motor to draw many more amps. A Super Capacitor is not rated in Amps, its is rated in capacitance that discharges quickly.
 

Harald Kapp

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the measurement i get after adding the diode is up to 28V.
My guess: this was measured without load (LED) attached, right? The capacitor will charge up to the peak of the voltage that is produced by the coil without load. When you attach the LED plus resistor, the voltage you'll measure will be much, much lower.
2. I don't really know how to choose a resistor for this circuit, or if i need it at all.
In this circuit you'll probably not need a resistor (you've been rightly told that usually you'll need one). Here the power (current) delivered by the coil is limited by the power that can be transmitted by the transmitter. This is probably designed such that the current through the LED is limited by design.
The reading on across the super capacitor is max 2.750V, and so far I have not felt it getting hot in any way.
The capacitor won't get hot from a little overvoltage, but its lifetime can be drastically reduced. I recommend you use a 5 V supercapacitor. This will give you a good safety margin and longer lifetime.
The white and the royal blue LEDs gets the shortest fading time. The green light fades slower, and the orange LED get the longest fading time.
Of course: the LEDs will take longer to fade the lower the "turn-on" voltage is (but read the next paragraph for "turn-on" voltage) because it will take longer to discharge the capacitor to the lower voltage.
I have seen this table before so I don't really understand why the white LED light up with only 2.7V.
Note the "VfTYP" in the table header. This table represents typical values. Specific values vary from LED type to LED type and from manufacturer to manufacturer. Also n LED is not a binary element that is either on or off. An LED has a characteristic like any diode with an exponential current vs. voltage curve and accordingly light vs. voltage. Here's a concrete example:
upload_2022-2-9_13-8-44.png
As you can see forward current is at 1 mA for a voltage of 2.7 V and rises to 80 mA at 4 V (the graph is not exponential due to the logarithmic y-scale!).
 

ThunderBear

Feb 6, 2022
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Wow..thanks for your detailed answer.

My guess: this was measured without load (LED) attached, right? The capacitor will charge up to the peak of the voltage that is produced by the coil without load. When you attach the LED plus resistor, the voltage you'll measure will be much, much lower.

Yes, the reading of 28V is just the coil, the tiny original capacitor, and a schottky diode to rectify. No load or resistor.

Is it a coincidence that with the 30000K LED the voltage over the capacitor stops at exactly the max voltage for the capacitor at 2.74V ?

I wonder this because when i use the orange LED the voltage over the capacitor stops at 1.98V.

In this circuit you'll probably not need a resistor (you've been rightly told that usually you'll need one). Here the power (current) delivered by the coil is limited by the power that can be transmitted by the transmitter. This is probably designed such that the current through the LED is limited by design.

Ok, that makes sense.

The capacitor won't get hot from a little overvoltage, but its lifetime can be drastically reduced. I recommend you use a 5 V supercapacitor. This will give you a good safety margin and longer lifetime.

Thanks for this reassurance, because I had just ordered 5.5V super capacitors to make sure I am well within the limits.

Of course: the LEDs will take longer to fade the lower the "turn-on" voltage is (but read the next paragraph for "turn-on" voltage) because it will take longer to discharge the capacitor to the lower voltage.

Thank you. That is very useful information to understand the different LEDs i have. I just hope I don't need a different circuit for each color, but I can do it if I have to.

So, from what I can understand here it is important to avoid overvoltage into the capacitor, so it is better to go for a capacitor with a higher voltage rating. I will try with 5.5V capacitors.

My next concern then is what happens if any component burns out. As long as the LED is working the voltage is much lower as you say, but if the LED burns out, then the capacitor may be subject to the 28V directly after the schottky diode. This will probably kill the capacitor, right?

It is not a big problem if the capacitor have to be changed when the LED have to be changed, but the question is if there could be any fire hazard or other risky things going on when the LED dies.

What happens if the capacitor dies? Will the capacitor keep power from the LED then, or will the circuit just work as if the capacitor wasn't there?
 

Harald Kapp

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Is it a coincidence that with the 30000K LED the voltage over the capacitor stops at exactly the max voltage for the capacitor at 2.74V ?
It is a bare coincidence, depending on the LEd, as you observed by using another LED.
This will probably kill the capacitor, right?
Right.
What happens if the capacitor dies? Will the capacitor keep power from the LED then, or will the circuit just work as if the capacitor wasn't there?
If the capacitor goes short circuit, the LED will be off. Try this out by short circuiting the capacitor on the breadboard with a short piece of wire.
If the capacitor goes open circuit (loses capacitance), only the fading will be affected. Try this out by removing the capacitor from the breadboard.
he question is if there could be any fire hazard or other risky things going on when the LED dies.
Unlikely but not impossible. You could put a zener diode and another LED across the original LED. Thus in case the original LED goes open circuit the additional LED will light up and indicate a problem:
upload_2022-2-10_6-26-42.png
 

ThunderBear

Feb 6, 2022
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Unlikely but not impossible. You could put a zener diode and another LED across the original LED. Thus in case the original LED goes open circuit the additional LED will light up and indicate a problem:
View attachment 54193

Ok, thanks. I will try this out.

A couple of questions though:

Can the capacitor get harmed by too high voltage no matter how low the current is?

Is there any such thing as a voltage fuse?
 

Harald Kapp

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Can the capacitor get harmed by too high voltage no matter how low the current is?
yes.
Is there any such thing as a voltage fuse?
Afaik not in the sense you seem to have in mind like a single component.
I think you worry to much about the supercap. Brand supercaps are safe, see e.g. here. Which is not to say that the possibility of a hazard is 0. But it is very, very low.

As an extreme precaution you could D2 in my schematic from post #9 and replace it by the LED of an optocoupler, then use the optocoupler to turn off the transmitter.
Imho overkill.
 

ThunderBear

Feb 6, 2022
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As an extreme precaution you could D2 in my schematic from post #9 and replace it by the LED of an optocoupler, then use the optocoupler to turn off the transmitter.
Imho overkill.

I am still trying to understand how all this works. I don't really have a way to see a second warning diode from inside the sculpture. Is there a component that could make a sound instead?

Apart from that issue I am trying to understand the measurements i get.

If I use the 2.7V super capacitor in this circuit without a security system and without any resistors i never get more than 2.7V charge in the super capacitor as long as all components are working, but if i att a 100ohm resistor before the LED the super capacitor charges up to 3.6V. What is happening here?

In order to find max woltage from the coil i recently got some 50V 4.7mF capacitors. If I put only this capacitor in the circuit it charges up to 27V. If I add the LED in parallell I get between 2.4V and 3.5V depending of what color the LED is. What is happening here? How can max voltage in the circuit be so different with and without the LED?

There are probably very zero level newbie questions, but at this point I am very puzzeled about these things....
 

ThunderBear

Feb 6, 2022
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The capacitor can if you over volt it or get it’s polarity wrong..But this is probably not what you mean or want!:p

Martin
Right. These are products that will for the most part be turned on basically unattended, so this is not the solution I am looking for. Thanks anyways for your contribution to some humour in these dark times.
 

Harald Kapp

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Seeing how lithium batteries with much , very much higher potential for disaster are all over the place, I think you worry way too much about the capacitor.
Make sure it doesn't get connected to more than its rated voltage and you are good.
 

ThunderBear

Feb 6, 2022
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Seeing how lithium batteries with much , very much higher potential for disaster are all over the place, I think you worry way too much about the capacitor.
Make sure it doesn't get connected to more than its rated voltage and you are good.

Thanks, but the voltage is the exact problem here.

When i charge up a standard capacitor rated for 50V and 4700uF this capacitor charges up to 27V.

Still this same system is meant for LEDs rated at 3.2V - 3.6V.

Since I don't understand electricity very well, I don't know what that means for my circuit.

My super capacitors are rated for 2.7V, and i have some new super-capacitors coming rated at 5.5V to match the LEDs max voltage.

The max power the transmitter is rated to transfer is 750mA current. One small receiver can't pick up all that power, but probably about half of it, since the transmitter i meant for more than one receiver. Each receiver coil has one LED attached to it with a tiny capacitor in between. (for resonance i guess)

When I have a working LED in parallell with the standard capacitor (4700uF/50V) the capacitor reaches the same voltage as the LED is consuming. So as long as the LED is working there does not seem to be a problem.

My worry is that something hazardous might happen if the LED burns out after a few thousand hours, and a super capacitor rated for just a few volts is given up to 27V, even if the current is low, at probably 300mA.

These parts are to be installed inside wooden objects, so heat could be a serious issue.

I am planning to sell these sculptures.....
 
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