COP 17 Heater | Rosemary Ainslie

[witsend]

Clearly my error.
EDIT - SPOKE WAY TOO SOON. TOTAL DENIAL THAT ANY SUCH WAS THERE. JUST KEEPING THE POST FOR THE RECORD.

[ren]

Nice work Jet Love that little ring....

[Aaron]

tk

Now can we please have some power measurements?

TK said with a (cringe) that he may hook my timer circuit to his current test. At least he'll be able to do it with an oscillating mosfet now and supposedly he has established a baseline and anything over is over 1.0.

Also, Stefan at ou said he didn't block you. With your IP address, he can make sure it isn't in a bulk list of spamming IP addresses. If someone in your internet address range spammed, then it might be blacklisted.

If you don't mind, I can give him your IP so he can do that. Hopefully, you'll be able to see what is happening there.

[Joit]

@wistend
Maybe he will bring you the flowers and the cake once,
because, when i did not get this wrong, hes in Africa too.

Btw, can someone tell me, if the Shunt is good for anything else or just for measurements.
Is a Potentiometer a equal Replacement?
And where is the use of 24V instead only 12V. Thats about the better increasing Performance?

Right now i play with a Coil and a heat element in series around too, till i got my Shunts.
Got nice Spikes back,i should try it with some more Amps.

[Jetijs]

Thanks ren and Aaron
Joit, the resistor in series of the coil will make the circuit to consume less current, thus also less heat will be produced, so in order to get enough current through the circuit to get some good heat, you will have to increase the voltage.

[Joit]

Jetijs
Ok, sounds good.
Still thinking about, if i should go away from one 12V Batt with a Diode,
to load a Cap, and got the 12V for the TimerCircuit

I still think about it, if you dont move beside a balanced Point with a Pot around

[MileHigh]

I'm going to post a clarification for everyone about this circuit. I wince because you are not going to like it, but here goes.

Let's discuss the circuit without the diode. Everyone, please open up your tab to the simplified schematic diagram in Rosemary's paper.

When the MOSFET is on, the battery discharges through the coil-resistor, through the MOSFET itself, and then through the shunt resistor. Assume the MOSFET has been switched on long enough so the coil is fully charged and is now just acting like a resistance.

So now we have what looks like three resistors in series, everybody following!? It's a 10-ohm (coil-resistor) connected to a 0.01 ohm (MOSFET) connected to a 0.25 ohm (shunt). Assume a 12-volt battery you can calculate the current.

So the voltage drops as you go around the loop are roughly... 12 volts at the battery/coil-resistor node, about 0.26 volts at the coil-resistor/MOSFET node, and about 0.25 volts at the MOSFET/shunt node. I hope everybody is following that. As you go clockwise around the circuit starting at the battery positive, the voltage takes a drop as it flows through the three resistances.

Note that almost all of the voltage drops across the resistive part of the coil-resistor, and almost no voltage drops across the MOSFET, and a very small amout of voltage drops across the shunt resistor.

Then the MOSFET switches off and becomes a very high resistance, and we know that we get the spike from the coil. Imagine it like the coil now becoming part of the circuit. The coil was "asleep" and causing no voltage drop when the current had stabilized. Now the coil "wakes up" and appears in the circuit when the MOSFET switches off.

Let's keep it simple and say that the coil generates a 100-volt spike for a short period of time. We all know that the voltage across the coil reverses in these cases.

So now we are going to walk around the circuit clockwise just like we did above. We start at the battery positive at +12 volts. Now, in the previous case when the MOSFET was on discussed above, we know that the voltage went DOWN across the coil-resistor. Therefore in this case, the voltage has to increase and go UP. You go DOWN when the MOSFET is on. When the MOSFET switches off, you reverse the voltage and go UP.

That means we start at the battery at +12 volts, and we go UP another 100 volts because the coil has "woken up". Imagine the coil is connected to the +12 volts of the battery on one side, and the 10-ohm resistor on the other side. This means that the output at the coil part of the coil resistor is at +112 volts. Is everybody checking their schematic and following? You really should because it's important.

So you are in the middle of the spike, and the end of the battery + coil is at +112 volts. In effect the coil is for a very short time acting like a second 100-volt battery in series with the real 12-volt battery, giving you +112 volts.

The load across this temporary +112V voltage source is, going clockwise, a 10-ohm resistor (the resistive part of the coil-resistor), a 10 Kohm resistor (the switched off MOSFET) and a 0.25 ohm resistor (the shunt). The swictched off MOSFET has the highest resistance by far therefore 99.9% of the power supplied by the temporary 112-volt battery is burned off in the MOSFET over a short period of time. This short burst of power x time gets turned into a fraction of a Joule of heat inside the MOSFET.

Now, if you have been following this here is the key point that you are not going to like: In this case, whatever "remains from the spike energy" after buring off in the MOSFET open switch does NOT go back to the source battery. Repeat: No energy is going back to the source battery.

In fact the source battery is DISCHARGING some of it's energy when the spike happens. The reason is simple: When the inductor discharges it's spike, the source battery and the "temporary coil battery" TEAM UP and add their voltages together. It's the same as putting two 12-volt batteries in series to make 24-volts and drive your Bedini motor faster. In this case the circuit puts the real battery in series with the coil-voltage-spike "temporary battery". They add up and give you more voltage to push through whatever load is across the terminals. Unfortunately, the poor MOSFET is the bearer of the brunt when the "battery-coil tag-team" decide to smack it with a nasty pulse of energy repeatedly. The poor MOSFET is hoping he can take the abuse and dissipate away heat fast enough, because if he can't take it, the continuous pounding of coil discharges will burn him up.

Anyway, the key point is that the spikes generated when the diode is removed from the circuit do nothing to charge the battery. In fact the battery discharges a tiny bit, and actually helps the coil do it's nasty MOSFET pounding.

MileHigh

[Aaron]

shunt

@wistend
Maybe he will bring you the flowers and the cake once,
because, when i did not get this wrong, hes in Africa too.

Btw, can someone tell me, if the Shunt is good for anything else or just for measurements.
Is a Potentiometer a equal Replacement?
And where is the use of 24V instead only 12V. Thats about the better increasing Performance?

Right now i play with a Coil and a heat element in series around too, till i got my Shunts.
Got nice Spikes back,i should try it with some more Amps.

Joit,

The shunt resistor should be low so it doesn't change what the circuit does. My new shunt is 0.05 ohms and it is calibrated - which is for measuring. You want it calibrated ideally so you know it is really that that ohms so you can do your measurements on a known resistance. I couldn't find anything less than 10 ohms at my local store. I had to get one from Peter since he had the good ones for measurement that he and Bedini used on many tests.

If the resistor isn't calibrated, you can see that most are not easily. Buy a pack of whatever ohm resistors and measure them. There are always differences. Anyway, no big deal, just use a small one. If you measure it and the measurement stays the same thru the test, I believe you're in good shape.

My original one was 10 ohms which is NOT what you want. I just wanted to put something there as a reminder of what I needed. It limited all the current almost so therefore, it did affect my circuit performance because I wanted the current pulses at the inductive resistor.

Potmeter there wouldn't be ideal because it isn't really fixed and the contact isn't perfect.

What is your 24v question? You mean batteries? Rosemary's test used 2 batteries in series. Each was 12v 20ah to my recollection.

[Aaron]

test

Thanks ren and Aaron
Joit, the resistor in series of the coil will make the circuit to consume less current, thus also less heat will be produced, so in order to get enough current through the circuit to get some good heat, you will have to increase the voltage.

Jetijs,

That would be a cool (maybe hot ) test to compare the higher voltage less current version compared to higher current lower voltage.

[Joit]

What is your 24v question? You mean batteries? Rosemary's test used 2 batteries in series. Each was 12v 20ah to my recollection.

I think Jetijs did answer it.
I understand it as, better put a bigger Pack into the Circuit, to get more circulation and Heat.
I was thinking about, if only a 12V isnt enough there, because i dont have a Ton from them, for now
Thanks.

[poynt99]

Here is what I found late last night, which is interesting. Can anyone else verify this?

.99, your simulation showed return to the battery was more without the diode? That was without oscillation in the mosfet right?

There may be something to that.

What I found was that when the mosfet was NOT in oscillation and was truly triggered by the 555 signal, the negative spike on the shunt was larger without the diode. That seems to corroborate with your simulation finding.

HOWEVER, with the diode there WITH oscillation in the mosfet, the negative spike on the shunt was BIGGER.

In both cases, the negative spike recharge is there with or without the diode.

So without oscillation, bigger spike on the shunt without diode.

With oscillation, bigger spike on the shunt with diode.

Yes correct on your questions.

Could you please try your circuit without that 1.1k input resistor present? I'd like to see if you still get your oscillation.

Also, see if you can get the MOSFET to oscillate this way: try placing about a 1k resistor from it's Gate to it's Source (or just Gate to gnd). This will cut down the peak voltage hitting the Gate. I think if the goal was to get that FET into oscillation, we would bias the drive voltage up to about 1V or 1.5VDC and ride the pulse input on that at a variable amplitude. This would put the MOSFET GATE right in the region of it's VGS(th) voltage, not turning it on too hard ON or OFF. This of course flies in the face of proper MOSFET drive practice, but for educational purposes, it might be interesting to see what happens if we try to purposely make it oscillate.

.99

[RAMSET]

TK responds to Aaron

Aaron to TK
You know exactly what my reference to you using a 10 ohm inductive resistor means. In the first video you showed using a 10 ohm resistor. If you changed it - you are deceitful in your ways - as you would have people believe that I am claiming your new shunt is the original 10 ohm one you used.

Do your followers actually buy this?



There was no need to post what resistor was on the 555's battery as it is irrelevant. You can make any variation of a 555 that will give you 50-99% with frequency control.

--------------

.99 - TK has clearly linked any triggering flaw to NON OSCILLATION in the mosfet, which of course is a bogus analysis since he didn't even know to verify the shunt to see if the mosfet was indeed in oscillation or not.

@ TK RESpONCE

That entire statement is inaccurate Aaron.

First, I used a 10 ohm "shunt" because that's what you used. Unlike you, I actually measure inductances and you will be surprised to know that that cement 10 ohm resistor has almost exactly the same inductance as your carbonfilm or metalfilm 10 ohmer.
And you can also see that your timer is "grounded" to the wrong side of your "shunt"... but OK.

You are the one who used that resistor on your timer power.

Your statement to .99 is also inaccurate. I have linked your triggering flaw to your induced oscillations in the mosfet, which are primarily caused by your improper signal driving the mosfet. I have indeed looked everywhere in that circuit, and your implication that I "don't know" to look somewhere is just ridiculous.

You are the one who is misusing your scope to "SHOW" what you are trying to prove. I am using mine rationally to see what is really there. I challenge you ONCE AGAIN: do a side-by-side comparison with your circuit using a decent oscilloscope properly operated, and you operating your scope like you do. Or compare my circuit with yours on your oscilloscope or mine or a third one.

You are seeing false triggering; I am reproducing your signal exactly --after all, I am using the SAME COMPONENTS AND THE SAME LAYOUT that you are using.

And I am in NO WAY apologizing to you, or "admitting" anything of the sort that you are implying. I still maintain that you are wrong; that the mosfet is oscillating because of "bad" driving signal and/or poor construction technique and/or damaged components; that you do not know how to use an oscilloscope properly, and/or there is something wrong with your scope, and/or it is just not "up to snuff" with a complex signal; and that you are not focussing on the real issue at hand: the many discrepancies in what Ainslie is telling you.

Now, by following your excellent suggestions, that is, by regressing to 10th grade electronics, and perhaps by using a roasted mosfet and a purpose-built chaotic 555 circuit, I have been able to show classic parasitic oscillations on top of the mosfet trace. Hooking up the probes incorrectly like you did helped too, thanks.

Now that I can maybe induce the same oscillations as you (After all, you ADMIT that mine are the same as yours, in the above posts) it will be much easier to show that there is no free energy coming from the Ainslie system, even when the mosfet is oscillating.

[gotoluc]

Hi everyone,

last night I decided to Test the (gotoluc) circuit with just batteries and here are the results to date.

The supply battery is one 12 volt 5 Amp/hr Sealed Lead Acid Battery and an identical battery is in series on the flyback return side instead of the resistor or bulb. The 555 PWM that controls the IRF840 is set at 5KHz with a 30% duty cycle.

Test started Tuesday 21st at 10pm

Supply Battery voltage 12.95

Charge Battery voltage 12.82

Test stopped Wednesday 22nd at 9pm and measured after 2 hours to allow stabilization.

Supply Battery voltage 12.91

Charge Battery voltage 12.90


The Batteries will be re-tested again tonight but I will exchange their position to establish if the charged battery has been charge with usable power.

I will post the results tomorrow evening at about the same time.

Stay tuned.

Luc

[poynt99]

Some additional information

Regarding Asron's question about there being oscillation in the spikes I saw in the simulation, and regarding MH's clarification in his post #1058 above, I'd like to say that the "spikes" going back into the battery are indeed different with and without the diode.

I've mentioned this a few times and it's importance has been overlooked, but here goes again, just in case anyone wants to know how these "spikes" are showing up at all. It's because of the parasitic capacitance inherent in the MOSFET. If they were not there, the spikes would be gone.

Another overlooked device parasitic of sorts is the junction capacitance of the flyback diode. It too has a marked effect on the appearance of these spikes.

First without the diode, indeed MH's description is mostly correct. However, upon flyback of the coil there really is a very short duration of very high frequency ringdown, even though at first inspection it looks merely lilke a single negative spike, which it is not. This ringdown current does appear in the battery, but because there is an equal number of undulations of positive and negative current, the net charging effect in the battery is zero. These undulations occur on the trailing edge. So Rosemary, indeed the current during the ringdown is AC and fluctuates in both directions at about 2MHz.

The electrical path for these high frequency undulations while the MOSFET is "OFF", is mainly provided via the D-S, D-G, and G-S parasitic capacitors. Also we should not forget the Gate drive path, as this provides a path for the D-G capacitor. In general, there are several paths to allow current to flow during these undulations, but again their net average is zero.

When the flyback diode IS present, there is a slightly different scenario that plays out. Due to the junction capacitance of the flyback diode, there is a extremely short impulse of current that goes into the battery, and this on the leading edge! This is unexpected but it in fact results from the drive pulse on the Gate. The leading edge current path is through G-D, then through the flyback diode into the battery. We have in effect the MOSFET driver charging (very very very meagerly) the battery.

Fun fun fun

.99

[RAMSET]

TK answers Aaron

Besides, this oscillation thing is Yet another Red Herring.

Anyone with eyeballs can see that I have already reproduces essentially the heating profile that Ainslie claimed to get, and I'm using a known fast risetime clean 3 percent dutycycle with no parasitic oscillations. And my input power figures are nearly the same as Ainslie's.

The appearance of battery recharging and the accounting of the power flows through the circuit have been analyzed by Henieck and MileHigh and others, and this phenomenon also does not depend on the mosfet oscillation--as it can and has been observed in many other pulsed charging systems, that even Aaron can probably cite.

So, if the oscillation isn't necessary for the heat, and it isn't necessary for the appearance of battery recharging, what's it there for?

It is there to obfuscate the issue. Please tell me how the duty cycle figures cited in Ainslie's paper are compatible with the oscillations on Aaron's scope. Take single shots all you want...they will still result in regular traces that can be easily computed, when done properly.

(Don't forget, I have these 2 digital sampling storage oscilloscopes sitting here next to my analog ones. It is just possible that I do know whereof I speak, in spite of Aaron's enlightenment.)