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I want to run 3 LEDs off of a super capacitor, help? Answered

Alright, so I have a little project going on and basically, I am going to have a 5.5 volt .33ff supercapacitor to power 3 3mm white leds, at 3 volts each, each takes 30 miliamps. I need the led's to be in series. How can I do this? What resistor would I need to throw in the make it work out, or would I not need one? I want them all to run at full brightess, but how long will it last? I could modity the plans for two LEDs.

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rickharris
rickharris

9 years ago

I put a red LED directly across this capacitor - 10 farads at 2.5.volts  -not caring if it blew out.

It remained visibly on when viewed in darkness for over 8 hours - although very dim at the end.

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XOIIO
XOIIO

Answer 9 years ago

Nice, although I want axial capacitors, I'll keep this in mind.

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XOIIO
XOIIO

Answer 9 years ago

I mean horizontal radial

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XOIIO
XOIIO

9 years ago

I considered switching to one led but it really won't looks as good, but it does give me more room, I was wondering, is there a simple way I can have a small red LED (3mm again) fade up to full brightness as the capacitor gets to full charge? I think I've seen something like thois using transistors, not sure though.

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Jack A Lopez
Jack A Lopez

Answer 9 years ago

Perhaps you could use the Joule-Thief circuit described here:
http://www.evilmadscientist.com/article.php/joulethief
Except, instead of one LED, use a two, or three, of them in series, such that the series combination has a higher forward voltage than the highest voltage on your supercapacitor, which is 5.5V.  For example 4 red LEDs in a series gives 4*1.8 = 7.2 > 5.5.  Or two blue LEDs gives 2*3.6 = 7.2 > 5.5. 

The reason I am thinking you would need that rule is that otherwise you would just get a steady DC current  through that winding in series with the LED, if Vin were greater than VLED.  Also note that the usual Joule-Thief is intended to drive a single LED from a single battery cell, and the same rule applies;  e.g. 1.8 V, 3.6V, etc > 1.5V

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XOIIO
XOIIO

Answer 9 years ago

Yes, I wouldn't know how to adjust the resistor for 5.5 volts though, for optimal performance to boost it for the LEDs, but if I found a small toroid ring or other inductor then that might work, alebeight be a tad unsightly.

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XOIIO
XOIIO

Answer 9 years ago

This is a concept thing I just moded quickly, only the main board since I don't have female header pins in CAD yet, but bsically there would be one or more modules with a super capacitor you throw on the pinds, let charge, and then stick to the case or something metal with a magnet, so the led fader would be on the secondary board.

USB Tester Concept.bmp
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iceng
iceng

9 years ago

You didn't register that Jack used a two Farad capacitor only 20 minutes at 4 ma.

You have one third of a Farad and want to run threeLEDs at 90 ma total ! ! !

Do simple ratio math to determine your leds won't even light 2/10 of a second.

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XOIIO
XOIIO

Answer 9 years ago

I changed the cap actually, got a one farad one, couldn't find a two farad one. Still, doesn't sound good for a design with more than one LED :(

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iceng
iceng

Answer 9 years ago

Ahhh, a reasonable man....   Stored Energy = C × V × V ÷ 2

Put 6 capacitors in series, charge to 30VDC and regulate down
to 3 VDC and further second regulate to 10 ma...
You will have a constantly bright LED until the LDO input capacitor voltage
falls to 4.5 VDC

Hint, put more caps in parallel use a switching regulator and you can get hours of light ;-)

A

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XOIIO
XOIIO

Answer 9 years ago

If I make an exclusively mad eUSB supercaqp light I would definitely do that

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XOIIO
XOIIO

9 years ago

could I not have them in parallel, but woth one resistor, because I got this on one led calculator site

+----|>|---/\/\/----+ R = 100 ohms
+----|>|---/\/\/----+ R = 100 ohms
+----|>|---/\/\/----+ R = 100 ohms

but since all the positives will be conected, could one resistor on the other side not do it?

|+----|>|---
--/\/\/--|+----|>|---
|+----|>|---


with the resistor being 100 ohms?

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Jack A Lopez
Jack A Lopez

9 years ago

I actually built something similar to this, about 10 years ago.  It was a circuit driving a single LED, powered by 2 Farads of supercapacitor charged initially to about 5 volts.  If I can believe my notes, I got it to run for about 40 minutes when set up  for 2 mA of LED current, and about 20 minutes at 4 mA of LED current.

The driver IC I was using to do this was Zetec's ZXSC310.  At the time, I think I actually bought that IC from Digikey(r), but since then I have seen it, or something similar, cheaper, here:
http://www.dealextreme.com/p/3-6v-9v-800ma-regulated-ic-circuit-board-for-cree-and-ssc-leds-4-pack-3256
when built into a complete module.  Also if you buy the module, you get the inductor, a Schottkey diode, and a little board for it too.

The way you choose/set the current limit with the ZXSC310 is by way of a current-sensing resistor.  That is the regulator watches the voltage on the pin where that resistor is attached, and turns off the current when a certain voltage Vlimit  is reached, so that the current throught that resistor is Ilimit = Vlimit/Rsense  , and that's a trick that is familiar, if you've used this kind of current regulator IC before.

Anyway, some old pictures of that thing, plus a circuit diagram, are attached.

Regarding the circuit, this is an inverting style switching topology.  Not buck.  Not boost. The LED gets its current when the transistor turns off, and this circuit can run at input voltages above, and below, the forward voltage of the LED, which was 3.6 V for the blue LED shown, and I think the voltage limits for the ZXSC310 IC itself, were 2V min, and 8V max.

Link to big version of the circuit diagram here:
https://www.instructables.com/files/orig/FE8/DQD1/H0468C4M/FE8DQD1H0468C4M.png

LED-board-1.jpgLED-board-2.jpgLED-board-3.jpgLED-board-circuit.png
0
XOIIO
XOIIO

Answer 9 years ago

It also needs to be cheap, and bulk parts are the cheapest thing.

0
XOIIO
XOIIO

Answer 9 years ago

Ahh, see the problem is I don't want any microcontrollers or anything like that, mainly because I only started using CAD yesterday,I don't have the patience for microcontrollers, and this is designed to have the least complex parts possible.

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steveastrouk
steveastrouk

9 years ago

You can't, unless you use a boost mode LED driver - Maxim, Linear Technology and National, to name but three do such beasties. Look for "White LED boost drivers. They will work MUCH more efficiently than any home made circuit.

Kick me by PM next week, and I might get round to posting the Instructable I was writing on them.

Steve

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XOIIO
XOIIO

Answer 9 years ago

Well the thing is this is going to fit on a thmb drive sized board along with other stuff. I've seen supercaps with simmilar specs drive 40 led's with a resistor on each one, would I have to do it parallel or something like that?

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steveastrouk
steveastrouk

Answer 9 years ago

Oh, my circiuit would fit on a thumbsized board.
Otherwise, yes, you need PARALLEL circuits, and a lot of resistors - and you'd be blowing ~50% of your expensively stored energy as heat in the resistors

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XOIIO
XOIIO

Answer 9 years ago

Any idea how small it couuld be? This is the board design, do you think it could fit in the back without making it too long? It also needs to be fairly simple since I am using a single sided board.

USB Tester Adv.bmp
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XOIIO
XOIIO

Answer 9 years ago

Alright, cool. If you could draw up a rough schematic and send it to me I can try and make a board layout and test it out, I can even render it for your instructable. I'mjust anxious to see how small I can get it on there.

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steveastrouk
steveastrouk

Answer 9 years ago

http://www.linear.com/product/LT1932