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ultracapacitors and drivers

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Jul 25, 2012
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just wondering, has anyone used an ultracapacitor with a diode laser setup?

being capacitors, it doesnt seem like they shouldnt give any voltage surge when first connected, meaning they might be able to be connected directly without need for a driver, but i am not sure if the diodes internal resistance would be high enough to prevent it burning out

can anyone confirm any of this?

i plan on using 2 2.7v 3000F supercapacitors in series

if noone knows ill probably end up testing them myself, but the caps i bought wont arrive for at least a week or two. once i do the tests ill post the results here.
 
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The voltage drops with use just like it does with batteries.

It'll work fine assuming you're still using a driver. However, one of your supercaps has 11kJ of energy (assuming a driver can work down to 0V, which it can't, so it's probably closer to half of that) while an 18650 has up to 40kJ of energy, and is much, much smaller.

Supercaps are good for high current, not for high capacity. If you're not using them in a high current application, you might a well use a battery instead.
 
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The only real upside to them if you don't need high current is that they can charge full quickly.
 
Searching for a Wikipedia graph...

Found it!

350px-Supercapacitors_chart.svg.png


Edit: Whoops, not so good on a dark background. Second try:

58066255_1f0a08fb61.jpg
 
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The voltage drops with use just like it does with batteries.

It'll work fine assuming you're still using a driver. However, one of your supercaps has 11kJ of energy (assuming a driver can work down to 0V, which it can't, so it's probably closer to half of that) while an 18650 has up to 40kJ of energy, and is much, much smaller.

Supercaps are good for high current, not for high capacity. If you're not using them in a high current application, you might a well use a battery instead.

thanks for the input :)

i know the capacitors wont hold anywhere near as much energy, but the main issue is charge/discharge rates, the potential for not needing current drivers. (since i'd need one per diode, and i plan on using a single power supply for multiple diodes)

what the original post is about is the internal resistance of the diode. for an A-140 at 5.4 volts, to get a current of 1250mA, the diode needs a resistance of around 4.3 ohms. if the resistance is much lower than that, they may overheat too quickly, and i have no way of testing them until they arrive.

also, the capacitors allow storage at lower than the stated voltage, 2 in series would give 5.4v, but they can be charged to 4.5v and hold close to the same amount of charge, that is the final reason why i'm experimenting with them.

does anyone know the internal resistance of the A-140 445nm diodes?

tl;dr - do drivers regulate current and voltage? or just voltage?
 
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Laser diodes need regulated current, they will basically explode the uS you connect them to a supercap. Sometimes you can get away running them directly from batteries due to the battery's internal resistance, which ultracapacitors have very little.
 
^Even with zero internal resistance, as long as the voltage is between ~3.5 and 4.5V, it won't blow the diode. The current range within that narrow voltage range is very large though, so the output is unstable.

The supercaps only stay at 4.5V until you put a load on them, then the voltage drops because they are discharging.

Laser diode drivers regulate current only, although the voltage range is limited (you can't expect a 1A driver to supply 1A at 500V for example). You will need a driver even when using supercaps.
 
^Even with zero internal resistance, as long as the voltage is between ~3.5 and 4.5V, it won't blow the diode. The current range within that narrow voltage range is very large though, so the output is unstable.

The supercaps only stay at 4.5V until you put a load on them, then the voltage drops because they are discharging.

Laser diode drivers regulate current only, although the voltage range is limited (you can't expect a 1A driver to supply 1A at 500V for example). You will need a driver even when using supercaps.

exactly what i needed to know. thanks :)
 
Ideally, you'd really want to be using a boost driver designed specifically for this purpose. With super caps, you want to squeeze as much of the low voltage range from them as possible.

I know that a member here has actually been working on a relatively low voltage boost driver, and I think has a design that can boost from as low as 1V.

Does someone have the math handy, for how to equate the ratings of a supercap into something somewhat analogous with mAh ratings for li-ions? I've figured it out before, but if someone has been thinking on this ^ stuff for the last few hours and has the approach handy, let me know :)
 
Ideally, you'd really want to be using a boost driver designed specifically for this purpose. With super caps, you want to squeeze as much of the low voltage range from them as possible.

I know that a member here has actually been working on a relatively low voltage boost driver, and I think has a design that can boost from as low as 1V.

Does someone have the math handy, for how to equate the ratings of a supercap into something somewhat analogous with mAh ratings for li-ions? I've figured it out before, but if someone has been thinking on this ^ stuff for the last few hours and has the approach handy, let me know :)

there are 3.6 coulombs per mAh, and 1 farad is 1 coulomb per volt.

so the 3000 farad caps will be
2.7 x 3000 = 8100 coulombs
8100 / 3.6 = 2250 mAh
 
Yup^. Except the average voltage would be 1.3V instead of the 3.6V of a Li-ion. So it's comparable with a AA NiMH cell.

I prefer the energy approach E=CV²/2. If you can only discharge them to 1V, that's 9.4kJ (3000/2*2.7²-3000/2*1²) in comparison to 3000mAh 3.6V cell which is 40kJ (3*3.7*3600).

So apples-to-apples, it would have roughly the same energy as a 700mAh lithium cell.
 


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