Could a Joule Thief Power A Peltier?

[Krb686]

I've been checking out some JT's on Youtube lately but I haven't actually built one, so I'm curious if a Joule Thief combined with a simple battery source (AA's) could power a small sized Peltier.

My application in case you're wondering is a Peltier that can cool approx. 20oz of water to about 12-15C below room temp. for an extended period of time in a portable size (a water bottle?)

Here are my (amateur) calculations. 

20 oz = 567 grams --> 575 for good measure
1 kCal cools 1kg of water by 1 degree Celsius.
575 calories will cool 20oz by 1 degree.
Room temperature water ~ 21 degrees Celsius.
Approx target = 12-15 degrees C subtraction(I know it's not what I posted above).
575 calories ~ 2400 Joules

2400 * 12 = 28,800 Joules
2400 * 15 = 36,000 Joules

Ok, so there's my range, 29 kJ to 26kJ


So then, 29-36kJ comes out to 8-10 Watt Hours.


Well obviously, I don't want to have to wait 8-10 hours to get cold water, besides the fact that it wouldn't work due to dissipation.

So, here I am trying to choose a peltier, and there are a few problems. The more powerful the peltier, the less portable it can be.  I'm looking at these peltiers for referencing, sorted by current draw.

http://www.tetech.com/Peltier-Thermoelectric-Cooler-Modules/Standard.html

I think the biggest limitation would obviously be the current draw because you sure as hell can't get batteries to handle the current necessary for some of the chips up there.

4 3.7v Lithiums in series would give 14.8V but only a max current draw of 1.5A at best that I've seen, or just  3 in parallel would give the necessary current of 4.5A but only 3.7v

Is there a way to raise the voltage (and by necessity the resistance) of 3 lithiums in parallel while keeping the current draw still within the ratings for the batteries?

[MrMag]

I'm not sure of your application but I really don't think a joule thief will cut it. A jt is more of a voltage device and it won't give you the required current. I am not sure why you cannot use a power adapter or the length of time you need it to operate but have you looked into boost caps.

They wouldn't take very long to charge up and would give you the required voltage and amperage for quite some time. Are you willing to give a little more information about your application?

[Krb686]

I changed some stuff in my above post.  I don't want to use a wall adapter because I would like for it to be portable =D

To be honest I think I am confused about some very basic principles of electricity, voltage boosting especially.

I guess I'm asking

1) Is there a way to boost voltage without boosting current by changing resistance? (Is this what a JT does?)

2) If #1 is possible, then say a 3.7V battery supply was boosted to 15v.  Does this affect the current output of the battery? Or does that still ONLY depend on the draw of whatever it is hooked to? AKA if the peltier DRAWS 3A at 15v, then 2 1.5A batteries should be able to handle it.

[sm0ky2]

The relationship between Voltage, Current, and Resistance
is defined by the equation:   V = IR
where V is the Voltage, I is the current, and R is the resistance.

So by raising the Voltage, you are in effect changing the current, and/or the resistance. (doesnt necessarily have to be both )

In a battery, the current the max current output cannot be easily overdriven.
But for example,
3.7v = 1.5A (R) ; (3.7) / (1.5) = 2.46~ Ohms (internal resistance)
the resistance in this case, is not a variable number, any changes made to the voltage, will result in a direct change in current.
up to the maximum current availability of the battery. which is why they are usually just connected in series to add up the voltage linearly when current required is over the limit.
V / (multiplier factor) = current; so 3.7V / 4 = 0.925A
so, if you boost one battery up to 15 Volts, the multiplier circuit can only drive 0.925 Amps, (you will have to adjust this by the resistance of the multiplier, PLUS the internal battery resistance... its been a long time and i dont remember wether that is added in once, or sucessively with each step of the multiplier.... sorry)

The Joule Thief circuit converts low voltage, high-current pulsed DC into higher voltage, lower current A/C, which we usually then rectify back into DC to get a steady higher-voltage, low current DC output.
This works great for lighting LEDs and flourescent (really any non-filament) bulbs.

Peltier devices are generally high-current applications.

The problem with the peltier device, is not its size... even the 86Watt units listed in the above link are about the size of your palm. Portability doesnt break down until you get into tens of kilowatts.
a 50KW peltier unit on a semi-truck is about as big around as a basketball. These are designed to operate in reverse (seebeck) and produce power for the truck electronics, but the unit itself is made up of the same peltier cooling devices, several of them in series/parallel.

if your target power is .. say 10 Watt-hrs,
and the target cooling time is 30 minutes.
then you are shooting for approx. 20 Watts, these things do not convert electricity to heat at 100% efficiency. So in reality,
its going to need probably a 22W unit, or two 12W units something like that...... (assuming an 80% conversion efficiency)
now, operating this peltier at Max voltage and current will probably burn it out.. so the Voltage and Current (max) ratings on the list are overpowering the device. Idealy you want to operate it at less than max voltage and current, equivalent to or less than the (max) power rating of the module.


Now, lets play with some numbers a little...
let's use rechargable nickel metal hydride, rated at 1000mA-hrs
@ 1.5v this is 1.5 Watt hrs. per battery.
you will need at least 10 AA batteries
you will have to run the numbers with your 3.7V batteries to see how many it will actually take.

If you use 2 batteries going up each side of the cup
if you run them in series this gives you  7.4V @ 3Amps
This may be enough "energy" to do the job, but i dont see any ~22W chips that would fit in that range.
using the above link as a reference, the TE-63 [1.0.2.0]unit (11.4W) might be perfect for this application. using two of em
each being operated by 2 Li batteries in series (1.5A each)

you will want to place a small resistor in the circuit, because preliminary math analysis tells me that the batteries will run dry just short of the 30min goal if you just wire it up directly with only a switch.
thats based on if this were done using 10 AA's instead of the lithium batteries, it will all boil down to the mA-hrs rating of the lithium ones.

it may not be the ideal arrangement, for using these devices, and a lot of it will have to do with the materials in the cup and their ability to transfer heat to the liquid.

But hopefully this will give you a point to start from. There are many mobile-peltier applications you could use as an example; ranging from mini-fridges to a CPU cooler on a laptop or remote sensors that must maintain a constant internal temperature.

[Krb686]

I know the peltier chips are small themselves, It just needs to run off of a relatively small number of batteries otherwise it becomes a hassle since it would take longer to charge/more to replace, and would be heavier and bulkier.


1000mAh is fairly small for a AA, there are some that go as high as 2850 mAh I think.  But I don't think AA's will cut it.  10 in series won't have the required current and 10 in parallel won't have the required voltage.

I might have to combine some series/parallel grouping