Ultracaps tested for excess energy

[MileHigh]

Double-check on Paul's battery Joule calculations:

Here we go, I'll use my 700mAh AAA NiMH battery instead.  That's only 2400 joules.

Do we assume 1.3 volts for the battery?

That gives 1.3 x .7 x 3600 = 3276 Joules.   What the deal?

I just realized what gadget was saying. He does not discharge the UC all the way because it's not as efficient charging something near zero volts. That makes sense.

I assume that we are talking about using a JT to charge here.  The JT will charge the ultracap with the same efficiency at any ultracap voltage.  The reason for this is that's the way discharging inductors work, they will raise or lower their voltage to whatever voltage is required to discharge their energy into a load.

MileHigh

[PaulLowrance]

I ended the experiment. AAA is 0.491V, UC is 1.313V. I'll have to wait till tomorrow to see what was wrong with it.

Paul

[PaulLowrance]

how are you measuring the frequency without a frequency meter  ?

Actually I used a frequency meter, but I think you're right. The input current was 40mA, but the output was only 11mA. I didn't plan it that way, but 40mA just so happens to be the current the JT wanted from the battery. I guess tomorrow I'll adjust the pot to get lower current.

Paul

[MileHigh]

I ended the experiment. AAA is 0.491V, UC is 1.313V. I'll have to wait till tomorrow to see what was wrong with it.

I am not an expert on Joule Thieves but I can make some generic comments that apply to JTs.

The real way to make sure your JT is running as efficiently as possible is to check the timing with a scope, it's all about the timing.  You don't want the main transistor to be on for too long.  If it's on for too long then you are dissipating energy in the JT firing coil for nothing, just producing heat energy that is lost forever.  You also want the base resistor to be as high as possible so that you don't waste any extra current going through the transistor for nothing.  By the same token you want enough base current going through the transistor to be sure that the transistor collector-emitter junction is saturated and the transistor is really on 100%.

I will assume that it is harder to control the running frequency and you have to start playing with the numbers of turns in the trigger coil, not so easy.  However for the purposes of charging an ultracapacitor, it doesn't really matter.  The higher the running frequency the better in fact.  The real critical issues are making sure the transistor is not on for too long and that the base resistor is the correct value.  That way you will transfer the maximum possible battery energy into the ultracapacitor and loose the minimum amount of battery energy as heat.

Oops one last thing.  The battery output impedance is a pain in the butt.  In theory you want a DC load on a battery to be as high an impedance as possible.  Jenna said it, "the faster you discharge a battery the less energy you get out of it."  That's because the higher the discharge current, the more energy lost in the battery itself due to its internal output impedance.  With a JT it's a pulse load with a fairly high pulse current.  It's hard to gauge how this issue affects the battery, since it's a pulse load, not a DC load.  It could get complicated though and if you were hard core you would investigate this.  One possibility would be to use a high inductance inductor and switch off the transistor before the maximum current is reached.  This implies a longer transistor ON time coupled with a bigger coil.  That way the battery doesn't dissipate too much energy internally because the current is lower.  There is another bonus here, the inductor will be more efficient, burning less energy as heat and having proportionally more energy available for the spike.

I almost forgot ha! ha!  There is something to keep in mind about the inductor.  If you make a larger inductor, then you use more wire, so there is a higher resistance in the wire -> more lost energy.  However, the inductance is proportional to the square of the number of turns in the wire.  So your increase in inductance rises faster than your resistance, which is good news.  If you wanted to "pimp out" your JT and avoid loosing energy in the battery, and in the firing coil, then you want a large inductor with a low resistance and you don't switch on the transistor for too long.  That implies the bigger the inductor and the thicker the wire the better.  I have a 4000 watt water-cooled power supply for my computer, and the whole computer is sitting in an aquarium filled with mineral oil!  IST built it for me.  lol

I also forgot to mention the toroid or ferrite core to increase your inductance - how could I forget that!  I think that different core materials have different properties optimized for different applications.  I am no expert here - but I would think that you would want the lowest loss possible ferrite core.  If you are going for a toroidal core as most are, I would guess that you can get laminated low-loss cores for transformers - the typical square form that you can use to build and wind your own transformer with - an off-the-shelf lowest-loss-as-possible laminated transformer core.  They may actually make special low-loss square/toroidal parts.  They would probably be expensive.

I think it is worth investigating cores because for sure some must be cheap and relatively lossy designed for applications where this is not a critical design issue.

MileHigh

[Vortex1]

From MileHigh

You will also notice that the rate of the voltage increase starts to slow down the higher the voltage on the capacitor.  Do you know why this is happening?

No one answered so I'll give it a go:

Since there is a finite amount of energy per pulse, it takes more pulses of energy to charge the cap from say 4 to 5 volts than it does from 1 to 2 volts

E(J) =1/2CV^2.