Claimed OU circuit of Rosemary Ainslie

[TinselKoala]

The simple solution here is to simply put an ammeter in the supply path and run the circuit without oscillation and check the average current draw, then put it in oscillation and see if the current draw drops. If it does, then energy is being put back, if it doesn't then its not.

I have a cold beer in my fridge for anyone that can prove the current reduces instead of increases as the heater goes into oscillation.

You mean like I did last night? The average current indicated on the Simpson in series was about 600 mA just before onset and about 1 or 1.2 amps during, and could go as high as 2 amps. During the "oscillations" my current-viewing shunt, which is made from 4 ea. 1.0 ohm 1/2 Watt precision resistors in parallel, gets noticeably warm and the mosfet and load get quite hot quite quickly. The voltage drop across the shunt looks like a continuous 500 to 600 milliAmps (EDIT: should read milliVOLTS of course, for a current of I = 0.6 / 0.25 or about 2.4 Amps ) with a bit of ripple on top, so the current is nearly continuous, even though the mosfet is spiking.

So I guess you'll have to save that beer for someone Else. I'm sure Aaron will be claiming it soon.

Why is your mosfet now on a heatsink, Aaron? Did you discover that it gets hot, after all?

[Hoppy]

The simple solution here is to simply put an ammeter in the supply path and run the circuit without oscillation and check the average current draw, then put it in oscillation and see if the current draw drops. If it does, then energy is being put back, if it doesn't then its not.

I have a cold beer in my fridge for anyone that can prove the current reduces instead of increases as the heater goes into oscillation.

Harvey.

I've already done that and I'll wager that your beer will stay in the fridge!

Hoppy

[MileHigh]

Jibbguy:

The glass is more than half-full in my book.  No need to go into the "big picture" stuff yet.

If the effect is real, then forget about the "100 possible real-world applications", it would represent a paradigm shift for the world.  Free heat, you could do anything with that.  That's a big win!

Or, just figuring out what the circuit is doing, even if there is no free heat, is another win.  It would be a fun learning experience.

TK has reported a flaky 555 circuit producing a semblance of oscillation.  You can run with that and look for free heat if you want to.  Perhaps Aaron's observations are totally different.  He needs to check his 555 waveform.

Or suppose that is fixed, and you invoke the "real" MOSFET oscillation by playing with the gate resistance and the 555 waveform.  Then you can look for free heat there.  Then of course the non-oscillatory mode is also supposed to produce free heat.

Or you can focus on the battery spikes and run-down times, which is not what Rosemary's paper is really about but I am not in the driver's seat!

It's fun, don't get too serious yet!

Hoppy:  Thanks for the info!

.99:  he he

MileHigh

[MileHigh]

Hoppy:

I am reasonably confident in saying that any spikes getting back into the battery are not charging the battery to any useful extent.

For your setup I would agree with you also.  You mentioned that as you turn your gate trimpot you can see the spikes on the battery getting higher and lower.  So when you have the highest spikes, assume that your MOSFET is switching the fastest.

With that setup and then trying the diodes out as you did, you are enabling and disabling the feedback spikes to the battery.  When you check the battery voltage with your multimeter, you should definitely see the DC battery voltage creep up a bit if the spikes are there.  You indicated that you didn't, demonstrating that the return spikes have no affect on the standard voltage drop when the battery is powering the circuit.

Don't mind me I am just repeating what you said for the sake of completeness.  That would just confirm that the spikes are returning only a very small amount of power back into the battery.  This is in agreement with just eyeballing the spikes and performing the integration in your head.

Also, I acknowledge what .99 said about diodes and transients.  I will go out on a limb and assume that the spikes the diode would be blocking coming from the coil discharge reflection would be much larger than whatever could go through the diode junction capacitance.  They have to be if you are assuming the return spikes have any chance of "recharging" the battery, a.k.a lowering the discharge rate.

MileHigh

[Hoppy]

Hoppy:

For your setup I would agree with you also.  You mentioned that as you turn your gate trimpot you can see the spikes on the battery getting higher and lower.  So when you have the highest spikes, assume that your MOSFET is switching the fastest.

With that setup and then trying the diodes out as you did, you are enabling and disabling the feedback spikes to the battery.  When you check the battery voltage with your multimeter, you should definitely see the DC battery voltage creep up a bit if the spikes are there.  You indicated that you didn't, demonstrating that the return spikes have no affect on the standard voltage drop when the battery is powering the circuit.

Don't mind me I am just repeating what you said for the sake of completeness.  That would just confirm that the spikes are returning only a very small amount of power back into the battery.  This is in agreement with just eyeballing the spikes and performing the integration in your head.

Also, I acknowledge what .99 said about diodes and transients.  I will go out on a limb and assume that the spikes the diode would be blocking coming from the coil discharge reflection would be much larger than whatever could go through the diode junction capacitance.  They have to be if you are assuming the return spikes have any chance of "recharging" the battery, a.k.a lowering the discharge rate.

MileHigh

MH

I totally agree with your assessment.

Hoppy