Open circuit voltage - 812mV, Source impedance 3 ohms = matched load.
So maximum power out is 0.812*0.812/(3+3) = 110mW, 1/2 of which is
available in the load.
So you have a 400mV source that will give you 55mW.
Unfortunately 400mV is not enough to turn a silicon transistor on but
you might be able to sustain an oscillation once you got it going.
There are circuits aroound that boost a (dead) 1.5V cell to drive an
LED.
(eg <http://www.nifty-stuff.com/LED-boost-circuit-joule-thief.php>)

But I doubt you could do much useful with it.
RP


On 24/11/05, Olin Lathrop <olin_piclist@embedinc.com> wrote:
> Brooke Clarke wrote:
> > I got a Honeywell Q313 series "750 millivolt Powerpile Generator" and
> > have measured it's output when in one finger of a gas stove flame.
> > It puts out about half a volt at 100 ma with a 5 Ohm load (50 mw) and
> > probably would produce more power with a lower resistance load.
> > See: http://www.pacificsites.com/~brooke/batt.shtml#Heat
>
> Your figures for open circuit and 5 ohm load indicate the pile has about 3
> ohms resistance.
>
> > Is there a circuit that would transform this into say 3.3 volts at 10
> > ma that would be suitable to power a PIC?
>
> Not without some serious magic.  3.3V x 10mA = 330mW.  That's nearly 7 times
> what you measured with a 5 ohm load.  Where exactly do you expect this power
> to come from?
>
>
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