Post by thread:[EE] Using pilot Flame Thermopiles as PIC Power So

Hi: 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 Is there a circuit that would transform this into say 3.3 volts at 10 ma that would be suitable to power a PIC? Have Fun, Brooke Clarke, N6GCE -- w/Java http://www.PRC68.com w/o Java www.pacificsites.com/~brooke/PRC68COM.shtml http://www.precisionclock.com

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? ****************************************************************** Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

> 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 > > Is there a circuit that would transform this into say 3.3 volts at > 10 ma that would be suitable to power a PIC? Yes. But if you are trying to use the heat rather than trying to use the specific device you may be better off using a Peltier 'cooling' module. These produce substantially more voltage (due to having more thermopiles in series)(and possibly the pile you have uses lower voltage but more robust thermopiles). Using the present module will be a challenge due to the low voltage. Not hard, but easy to be inefficient. Several boost converter topologies will work. A simple flyback converter would be an easy start. Inductor in the drain of a FET. Pulse FET on and then off and coil "rings" to a higher voltage. Coil resistance needs to be lowish. FET needs to be low Rdson, or if a bipolar transistor is used it needs good low saturation voltage. Care needed with all tracks etc to minimise losses. The circuit will almost certainly need to be operated from a higher voltage supply - probably from the output that it supplies. Starting a circuit using silicon components on 0.5 volts is approximately impossible. However, on no load the pile probably makes at least the claimed 750 mV and may make more. Even that is challenging and using an external higher voltage for starting makes the job much easier. A self oscillating converter with a couple of transistors could probably be implemented but an IC based one would be easier. You could use one of the single-cell converter ICs but a package of CMOS inverters would probably do as well. RM

At 12:44 PM 11/23/2005, Olin Lathrop 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? I'm probably being obtuse today, but 3.3V @ 10mA sure looks like 33mW to me. You might consider putting a pair of those thermopiles in series and then using one of the Maxim single-cell switchers to convert the resulting voltage to your desired level. Might just work . . . dwayne -- Dwayne Reid <spam_OUTdwaynerTakeThisOuTplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 21 years of Engineering Innovation (1984 - 2005) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email.

Olin Lathrop wrote: > Not without some serious magic. 3.3V x 10mA = 330mW. Oops. Looks like I slipped a decimal point. 3.3V x 10mA = 33mW, which is at least possible to get from 50mW. While it is theoretically possible, it's going to be difficult converting from such a low voltage. If you can somehow bootstrap, maybe with the use of a battery, to the 3.3V or 5V then you have enough to run a switcher to sustain the higher voltage. Getting there from only 500mV will be tough. Two of these in series would do it. That would allow a simple analog circuit to make 5V at low current and medium efficiency, which then allows more complex circuit to run that makes the final output voltage much more efficiently from the 500mV in. ****************************************************************** Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

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_piclistKILLspam@spam@embedinc.com> wrote: {Quote hidden}> 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? > > > ****************************************************************** > Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC > consultant in 2004 program year. http://www.embedinc.com/products > -

At 11:04 AM 11/23/2005 -0800, you wrote: >Hi: > >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 > >Is there a circuit that would transform this into say 3.3 volts at 10 ma >that would be suitable to power a PIC? Although it's not a complete solution, you may find this IEEE paper of interest: http://power.ece.uiuc.edu/chapman/papers/LVpaper.pdf Startup is the difficult issue with such low voltage circuits, after the circuit is started it can bootstrap enough voltage to run. Note the novel unijunction transistor method used in the paper, and also the mechanical switch idea. Although it's not as elegant, you could consider using button cells (perhaps you need a real-time clock anyway) for the startup and then run off of the thermopile when power is available. >Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" speffKILLspaminterlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com ->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff

Hi Brooke, You might try a variation on this: http://bobblick.com/3cell/ I forget exactly what voltage it starts with, but you should be able to play with resistors here or there and get it to start. Moving Q2's emitter to the input voltage is probably required too. Cheers, Bob On 23 Nov 2005 at 11:04, 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: www.pacificsites.com/~brooke/batt.shtml#Heat > > Is there a circuit that would transform this into say 3.3 volts at 10 ma > that would be suitable to power a PIC?

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

On Wed, 23 Nov 2005, Bob Blick wrote: > Hi Brooke, > > You might try a variation on this: > > http://bobblick.com/3cell/ > > I forget exactly what voltage it starts with, but you should be able to > play with resistors here or there and get it to start. > > Moving Q2's emitter to the input voltage is probably required too. Would a JFET based hf oscillator work to produce an initial bias ? A JFET may have gain at zero Vg. There are also 'programmable base' mosfets afaik. Could one of these be used ? e.g.: http://www.aldinc.com/ald_epadfaq.htm can produce parts programmed for Vg=0.2V. Last, a vibrator <g> circuit could be used to convert the bootstrap voltage, using a depletion mode mosfet to turn it off once the main voltage is present. Peter