I've been searching for a long time for a way to use a capacitor bank as a relatively constant power source.

As we all know, voltage decreases as a capacitor discharges, and thus can't be used as a DC power source. The tolerance of my powered circuit is: 30-36 V dc. There are DC-DC converters that can be set to different input- and output voltages, but I can't find a circuit, or a basic schedule of a circuit which can take the steadily decreasing voltage of a capacitor and turn react to it to give an output voltage which does not exceed the limits of the powered circuit.

One would imagine that there could be a circuit which (when properly calculated) would react to any given input voltage, to cause a switching transformer circuit to output a voltage within the given voltage range (here 30-36 V). However no amount of research has given any results. Could you help me?

I need a cheap, reliable power source to power a 100W LED for about 30 seconds. If I build several of them I can set up an area to be lit up on a signal. It's for wildlife control hunting in the North. Batteries freeze here, freezing proof batteries are expensive so I'm looking at this both as a challenge and for practical reasons. I only need 30 seconds (3 kJ of energy after the transformer) of light because after the first shot, the herd of whatever you're after will run away. If you can't get a 'dead' centre shot off in 30 seconds, you shouldn't be hunting.

 

3kJ is a massive amount of energy, and such caps won´t be small nor cheap. You would need for example three or four of these http://uk.farnell.com/avx/sccy62b307vsb/super-capacitor-300f-2-7v-radial/dp/2696623 and a step-up converter circuit to get it done.

 

Getting a capacitor bank to provide 3+ amps for 30 seconds is a pretty tall order.

I would consider using batteries, and burying them below the frost line.

Three 12 volt batteries in series and a CC LED driver should do the trick.

And I can only assume you will also need power for the "signal", as well as keeping the caps charged?

 

Also capacitors self-discharge a lot. How do you plan to charge them in the first place?

 

capacitors are not batteries

 

While it may be a challenge I don't see this working out using a capacitor or bank of capacitors. Now as to batteries with lead acid type in mind, the below may be useful:

At a 40% state of charge, electrolyte will freeze if the temperature drops to approximately -16 degrees F. When a battery is fully charged the electrolyte will not freeze until the temperature drops to approximately -92 degrees F.

With that in mind and coupled with the suggestion of ElectricSpidey:

I would consider using batteries, and burying them below the frost line.

If surface temps are too cold you could consider burying the batteries. Also while here in NE Ohio it is not the frozen tundra batteries are used for many road signs and they are maintained using solar panels, a solar panel with a southern exposure (north of the equator) may be a viable solution to think about.

Ron

 

What are the actual temps? How long must a battery be there? One night?

 

Delta V = 6 V
I (avg) = 3 A
t = 30 seconds

First order approximation, E x C = i x t
6 x C = 3 x 30
C = 15 farads - that's a lot.

AND, to deliver 3 A, affordable supercaps will not work because the internal series resistance is too high. The supercaps in hybrid buses are *not* normal.

Consider an oversized battery, and use some of its charge to power a small heating pad bonded to its side.

ak

 

Put battery, termoregulator and heater into styrofoam box and keep temperature inside box a little above zero Celsius.

 

I've been searching for a long time for a way to use a capacitor bank as a relatively constant power source.

As we all know, voltage decreases as a capacitor discharges, and thus can't be used as a DC power source. The tolerance of my powered circuit is: 30-36 V dc. There are DC-DC converters that can be set to different input- and output voltages, but I can't find a circuit, or a basic schedule of a circuit which can take the steadily decreasing voltage of a capacitor and turn react to it to give an output voltage which does not exceed the limits of the powered circuit.

One would imagine that there could be a circuit which (when properly calculated) would react to any given input voltage, to cause a switching transformer circuit to output a voltage within the given voltage range (here 30-36 V). However no amount of research has given any results. Could you help me?

I don't follow this. This should be the easiest of the problems to solve.

DigiKey has about a quarter million different DC-DC converters. Narrowing it down to single output voltages in the 30 V to 36 V range and at least 100 W still left the list with over ten thousand entries.

You haven't given any hint as to what the input voltage range is that you are looking for, so it's hard to go much further.

Another source to look far are "battery eliminator" circuits used for RC applications. Many of these have wide input voltage ranges and can deliver a couple hundred watts.

I need a cheap, reliable power source to power a 100W LED for about 30 seconds. If I build several of them I can set up an area to be lit up on a signal. It's for wildlife control hunting in the North. Batteries freeze here, freezing proof batteries are expensive so I'm looking at this both as a challenge and for practical reasons. I only need 30 seconds (3 kJ of energy after the transformer) of light because after the first shot, the herd of whatever you're after will run away. If you can't get a 'dead' centre shot off in 30 seconds, you shouldn't be hunting.

A location of "the North" doesn't give us much to go on. WHAT temperatures are we talking about? There's a HUGE difference between -10°C and -60°C when it comes to designing an electronic system?

As others have said, explore your other alternatives. It shouldn't be too hard to use an insulated case, perhaps an ice chest or a cheap Styrofoam container or just a box lined with insulation to keep a battery from freezing overnight. You can put a heat pad in it or perhaps even just throw in some of the disposable handwarmers or MRE heaters to keep it warm enough through the night.

I'd do everything reasonable to use a battery over a capacitor bank. You will need farads and farads of low ESR capacitors which are not cheap. But 3 A for 30 seconds is only 25 mAh of charge and that is nothing for most batteries (meaning that a LOT of capacity is left to keep the battery warm).

 

Since you're hunting at night - what about a night scope? That way you don't have to carry daylight in a bucket. By that I mean you're carrying an energy source, be it battery or capacitors and some sort of flood light. You want 30 seconds of illumination. WOW! Sit and count out 30 seconds. If you can't get a shot off in five seconds then why are you out there? Light would surely distract your prey. If you're waiting for them to move into a clearing, shining a light - since I'm not a hunter I can't imagine prey behaving normally at night when a sudden flood of light comes their way.

Infrared comes to mind as well. Sure, I understand you want to know how to do this project according to the parameters you've set up - and if we have enough information we can probably help you solve the problems. But that's like taking a self propelled lawnmower and trying to turn it into a go-cart. Yeah, it'll go. And you might even make a few bucks cutting lawns, but it's not going to be fast. Sounds like you're way over-engineering this. Consider the alternatives.