On 24 Nov 2005 at 9:23, Richard Prosser wrote:

> 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
> 

	Maybe you could try a germanium transistor like those old AC187
running a "joule-thief" circuit.

	Mark Jordan


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