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Author Topic: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011  (Read 741368 times)

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #405 on: March 27, 2011, 05:59:09 PM »
Sorry Rosemary . Another Senior Moment there! This business of one cycle every 3 Minutes is just so Gobsmacking that my subconscious keeps trying to tell me its 3 seconds . This makes it even easier to drive the microswitch at realistic frequencies for experiment with a pendulum .Although at one cycle/3 minutes you would need one hell of a pendulum . As I said a while back you could actually switch it manually with a stopwatch .

LOL  I also took a day or two to realise that it was minutes not seconds.  And then only because it was pointed out to me by one of our academics.  Interestingly there is NO drop in temperature over the resistor.  Also - if you disconnect - for 3 minutes then the temperature drops dramatically.  It's not in the material of the iron.  So?  Here's the question.  How come the batteries are 'retaining  charge' and yet dissipating energy?   You see why it is that our academics are closing there eyes here?  LOL.

So glad we've got you to deal with the practicalities of this design Neptune.  I have no skills here - at all. 

Kindest as ever,
Rosie

added
BTW Neptune.  The ideal would be to get it to 'trigger' into oscillation mode and then just leave it to do it's thing.  No need for any further switching - I'd have thought?  It's holding the temperature - so it's doing work - and it's not - apparently, losing any charge at the batteries.  It would be a really good test - if you guys - someone?  can set this up.  I'm going to give it at go at this end but, because I'll be relying on others to do your circuit - then it'll take time. 

neptune

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #406 on: March 27, 2011, 07:25:21 PM »
Hi Rosemary . Yes all points noted . If experiment shows that transients are necessary , and just switching on and off that 9 volt battery does not do the job , I would think the next step would be having two batteries each with its own pot and the hand operated microswitch idea . That way we have alternate pos and neg pulses on the gate , and we can adjust the voltages of the pos and neg pulses independently .When the oscillations start , just release the switch and it defaults to a steady neg condition . I am more of a practical guy than a theoretician  , but from reading the article about prevention of parasitic oscillation in parallel Mosfets , I get the impression that 2 is the minimum number of mosfets to use , But choose the value of your load resister so as not to draw more current than 2 mosfets can handle . Rose I understand you are waiting for someone to help set up experiments but I am sure we will get there soon.

Magluvin

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #407 on: March 27, 2011, 07:33:59 PM »
So?  Here's the question.  How come the batteries are 'retaining  charge' and yet dissipating energy?   You see why it is that our academics are closing there eyes here?  LOL.



Kindest as ever,
Rosie



Could it be,  hmmm,  could it be that the heat created in the resistor is not really a loss felt by the batteries, and it is just an artifact of current flow.   If we think about it, as the resistor heats up, the resistance will become higher as the heat becomes hotter.  As the resistance becomes higher, and current becomes lower, yet the heat is higher.   More heat for less energy?

This was just a blurt.  I have to think about this.

Mags

Bubba1

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #408 on: March 27, 2011, 07:59:08 PM »
Exactly my point Bubba.  vi dt.

vi dt is not power, it is energy. power is vi.

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #409 on: March 27, 2011, 08:12:42 PM »
Guys - there's someone contacted me who may be prepared to fund the required to take this to an application for LED's.  Neptune - if you're game - let me know - or anyone who wants to take this further.  It's nothing to do with me.  I'm just  go-between.  But if you're up for it. PM me and I'll send on the details.

Kindest regards,
Rosemary

Also possibly Mags, nul-points  - whoever.  It may be a way of getting this tested DIRECTLY onto a limited application.  And I'm sure that there'll be some compensation for the outlay.  Just let me know.  I'm back here later tonight or tomorrow morning early.  I like the thinking.  Just get it up and running and explain how it works after.  Something like that.  You'd be able to test the battery durations and the whole gamut. 


Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #410 on: March 27, 2011, 08:18:56 PM »
vi dt is not power, it is energy. power is vi.
Bubba - why are you going on about this?  Energy and power are generic terms.  Applied to electric energy then Power is still volts times amps x time.  And when time is factored in then its represented as Joules.  I just don't see your point.

Rosemary

alexandre

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #411 on: March 28, 2011, 12:19:09 AM »
Hello,
I was wondering how much would the batteries last, if they weren´t getting any charge. This is a very crude aproximation:

If you use 6 12v batteries to power a DC load dissipating 50 watts, the current is around 50/72 which is 0.7 amps .  Dividing the AH rating of the batteries, lets say 100 AH, by the current, we get the run time of 142.8 hours.

If the load is pulsed, it seems there will be extra run time, even more so with lead acid batteries.

What is needed is replication and continuous operation of the heater, this is better than the the measurements route. Any takers?

A simple PIC microcontroller + drive transistor could set the gate voltage and provide a safety circuit breaker function.

Best,
Alex

cHeeseburger

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #412 on: March 28, 2011, 12:44:24 AM »
For anyone dubious about the idea that a shunt containing an inductance whose reactance is actually larger than the shunt resistance at the frequencies of interest could actually distort the direction of current flow and change the areas under the positive and negative portions of the trace, here is another even more astounding and revealing demonstration:

Here we have a current generator set up to ramp up and down linearly from zero to two Amperes.  The true current flow is always positive as shown by the trace taken across the pure shunt resistance.  It never once goes below zero.

Yet the voltage as taken across the R+L shows huge amounts of reverse (negative) “current flow”.  If we believed that, that is.

You see, the whole idea of a shunt is based on using a pure resistance to obtain the analog of the current flowing through it by looking at the voltage drop and applying Ohm’s law E=IR.  A resistor’s voltage drop across its terminals is a pure function of the instantaneous current flowing through the resistor.

The relationship between voltage drop and current is entirely different in inductors and capacitors.  In an inductor, the voltage measured across its terminals depends ONLY on how fast the current is changing and whether it is rising or falling and not at all on the actual amount or the polarity of current.  Inversely, in a capacitor, the current flow through it depends ONLY on how fast the voltage across its terminals is changing and not at all on the value of that voltage.

So, a pure inductor will have no voltage drop (zero) across its terminals no matter what the current flowing as long as that current is not changing.  If the current is changing at a steady rate (di/dt is a constant slope), the voltage across the inductor will be a non-zero value and will also remain steady.  If the current is rising, the voltage will be positive.  If the current is declining, the voltage will be negative.

If the rate of change of current (di/dt) suddenly changes from one slope to another, there will be a “spike” of voltage produced.  This is (di/dt)/dt, the rate of change of the rate of change.  That is why we see the spikes each time the slope of the current instantly changes on our trapezoid waveform, shown here.  If there were no resistance in series to damp these spikes, they would be infinite in amplitude if the change in slope was instantaneous NO MATTER HOW LARGE OR SMALL THE ACTUAL LEVEL OF CURRENT WAS.

So using a shunt that has internal inductance with a reactance far larger than its resistance at the frequencies of interest will primarily show the rate of change of the current and not the current’s actual value.

The scope traces here and in my sim of Rose’s circuit shown earlier prove the point that an inductive shunt can and does show negative voltages any time the true current is declining EVEN WHEN THE CURRENT ITSELF IS ALWAYS A POSITIVE NUMBER.

Humbugger

Bubba1

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #413 on: March 28, 2011, 01:38:57 AM »
Bubba - why are you going on about this?  Energy and power are generic terms.  Applied to electric energy then Power is still volts times amps x time.  And when time is factored in then its represented as Joules.  I just don't see your point.

Rosemary

Yeah, I see that.  Maybe it doesn't matter.

twinbeard

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #414 on: March 28, 2011, 09:45:21 AM »
Hi Rosemary,

So back to applications.  In your opinion, would it be possible to drive the existing resistive heating element in an off the shelf home water heater appliance from this circuit?  If not, what is necessary to retrofit an existing heater with a replacement element suitable for use with the circuit?  I think it is a easier path to mass implementation to provide a modular kit suitable to upgrade the existing devices in use, as opposed to the larger manufacturing requirements and hence consumer investment required
to replace existing systems altogether.

Cheers,
Twinbeard

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #415 on: March 28, 2011, 03:49:56 PM »
Guys,  It seems that OUR.com have banned me from viewing their forum.  Not sure of the thinking here.  Poynty himself HOWLED when Harti separated him from his work.  Somehow he feels this is due to me.  Apparently they want a free field where I can do nothing to stop all that slander.  I think this is justified because?  Actually I'm not sure why.  I thought I was knee deep in a friendly discussion with Poynty.  How wrong can one be. 

I've now reported their abuse EVERYWHERE.  I may as well add it here.  Mookie has put out a general appeal for all to 'feel free' to come and comment.  But i think they're only accepting ADVERSE comment.  Never seen a more blatant example of bigotry and more sanctioned intentions to indulge in 'hate speech'.  But Poynty apparently justifies it under the banner 'opinion'.  I see NO opinions.  Unless they do some retrospective editing.  Poor Mookie is trying to advance the general impression that I'm always in my pyjamas - and that I live in a dusty little hole where I weave my fantasies at whim.  Again.  Not sure of the relevance - but could someone perhaps advise him that my living quarters are ample and really well serviced.

Rosemary

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #416 on: March 28, 2011, 04:10:46 PM »
Hello,
I was wondering how much would the batteries last, if they weren´t getting any charge. This is a very crude aproximation:

If you use 6 12v batteries to power a DC load dissipating 50 watts, the current is around 50/72 which is 0.7 amps .  Dividing the AH rating of the batteries, lets say 100 AH, by the current, we get the run time of 142.8 hours.

If the load is pulsed, it seems there will be extra run time, even more so with lead acid batteries.

What is needed is replication and continuous operation of the heater, this is better than the the measurements route. Any takers?

A simple PIC microcontroller + drive transistor could set the gate voltage and provide a safety circuit breaker function.

Best,
Alex

Alex I don't suppose that the batteries are more than 100 ah's but I'll check.  And we've certainly run them for longer than 142 hours.  But that's 142 x 8 and even then - it's assuming a .7 amp discharge.  We've never managed any evidence of discharge.  Nor have we seen any depletion of battery voltage.  But it's not an argument that we'll win.  Ever.  And I, for one won't try it again.  But it MAY be relevant if any of you guys do these tests on smaller battery capacities. 

Not so keen on replications.  It would be nice to try a small application. But either way - it's your choice.

Kindest regards,
Rosemary

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #417 on: March 28, 2011, 04:14:26 PM »
Hi Rosemary,

So back to applications.  In your opinion, would it be possible to drive the existing resistive heating element in an off the shelf home water heater appliance from this circuit?  If not, what is necessary to retrofit an existing heater with a replacement element suitable for use with the circuit?  I think it is a easier path to mass implementation to provide a modular kit suitable to upgrade the existing devices in use, as opposed to the larger manufacturing requirements and hence consumer investment required
to replace existing systems altogether.

Cheers,
Twinbeard

Hi Twin.  That's more or less what we were considering.  When we chose that element it was simply to see how far from 'standard' we'd need to move.  The surprise was that we didn't need to move away at all.  Those elements that the guys are showing seem good.  But small Ohmage may be preferred - just so that you can keep the battery voltage low.  We could up the voltage because we had a generous donation of all those batteries.

Nice thinking.

Rosemary

alexandre

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #418 on: March 28, 2011, 04:31:06 PM »
Hello Rosemary,

I have been reading about it and I believe I have underestimated the capacity of the batteries. As for battery voltage, as pointed out already, it doesn´t represent the remaining charge. Especially when there is pulsing going on.

No consense on the mesurements either. IMO, more work is needed on this experiment. I would like to see continuous operation.

I hope you take this as constructive criticism.

Best
Alex

Rosemary Ainslie

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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #419 on: March 28, 2011, 04:47:54 PM »
Hello Rosemary,

I have been reading about it and I believe I have underestimated the capacity of the batteries. As for battery voltage, as pointed out already, it doesn´t represent the remaining charge. Especially when there is pulsing going on.

No consense on the mesurements either. IMO, more work is needed on this experiment. I would like to see continuous operation.

I hope you take this as constructive criticism.

Best
Alex

You all would.  And with good reason.  But I don't have the funds to get a continuous test going with the required constant supervision.   So. If even three experts stated that such a test would be definitive then that's another story.  I'd bend over backwards to get it going.  You see Alex - I'm anticipating a really slow but steady discharge from the battery.  Certainly, from previous experience - that's what was evident.  So.  I'd first run a control - say discharging 0.7 amps.  Then I'd need to run my own test.  The control would run for about 100 amp hours /0 .7 amps = about 142 hours x 6 batteries = 857 hours or a staggering 35 days.  Not too much of a problem because we'd be able to put that on a data logger.  Now comes our test.  Now we'd need constant supervision because it has a tendancy to trend into that heavy duty output mode which is hazardous.  So.  We'd first need to recharge those batteries then run it for the same period - another 35 odd days  before we got any kind of proof at all.  Then - to satisfy the picky complainers I'd probably have to run it for a further 35 days or to its point of absolute depletion - assuming it's depleting at all.  Then the argument will be to RERUN the control and the test because - you see -  the rate of charge at the start of both tests may have skewed the result.  That would take a further 34 - 35 days  each.  And at the end of it?  I'll be told that I'd simply fudged those results. 

I assure you - battery durations will never cut it as an argument.  And as I keep trying to remind you all - this is absolutely NOT the entire argument.  There is nothing - in prinicple or in fact - that prevents these applications on AC supplies.  What we've done should be more than enough as it depends on standard measurement protocols.

Kindest again,
Rosemary 

And I do take it as constructive suggestion.  I'm not sure that it's any kind of criticism at all.  It's just that what you're actually asking is the continually supervised for more than a month and possibly to be done twice.  I simply cannot afford it.

EDITED
I've had to change those numbers.  I multiplied instead of dividing.