Exploring the Inductive Resistor Heater

[lanenal]

Hi Greg,

I have watched your video, and I have a question for you about the measured voltage drop over SH3 when the circuit is hooked up. Since the voltage should be oscilating, what you measured must be some sort of average voltage, right? Is it RMS? or is it something else?

great job, and thanks for sharing.

lanenal

[lanenal]

@lanenal: You do realize that Gmeast is saying that the most current that can pass through a MOSFET gate to the drain or source is 100 nA, right? And that PW has explained from the circuit theory standpoint, and I have illustrated empirically in the video above, that the 100 nA claim is definitely not true for the kind of signal that is being applied to the gate by the gate driver which is capable of supplying 9 amps (if not restricted by the inline resistor.) Right?
Yet you see how he responds. He is refuted time after time but cannot deal with the refutations; instead he says "Fuck them all dead". That is his argument!

@TinselKoala: I am not a judge between you.

Greg in his first post already estimated the maximal energy contribution from the Gate Driver, which seems to be no big deal. As for the Gate current, Greg might be talking about gate leakage current, not the current to induce the voltage difference.

Transistors are gated by currents, while MOSFETS are gated by voltage difference, and to develop voltage difference there is a "hidden cap" (so to speak) at the Gate. The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap. The gate leakage current is due to the imperfect insulation inside the "hidden" cap.

regards,

lanenal

[picowatt]

Greg has done a great job in his calculations and very honest in his proof.


PW is doubtful and cautious, but he can not show any calculations, so there is no way to assess how significant is the deviation in energy calculation due to the factors listed by him. It could be that the conclusion from Greg's calculation still stands even if those factors are totally ignored.


It seems to me, the "increase in battery capacity under pulsed load" argument is conducing rather than deducing to Greg's calculations, because when a battery is increasing in its capacity while draining in its stored energy, then the voltage drop should be an exaggeration (because there is a bogus voltage drop due to increased capacity without change in stored energy). So under that observation, Greg's calculation of input to the original circuit (a pulsing load) is an exaggeration, and the OU is even more pronounced than what his calculations shows.

Lanenal,

I am not sure how you figure that an increase in capacity would somehow cause an exaggeration in Vdrop.  If the capacity of a battery is increased, it will hold a given voltage for a longer period of time under a given load.  There would be no "bogus voltage drop".  The larger capacity would just allow the battery to sustain a given load for a longer period of time before reaching a given stop voltage.

As the bulk of the OU is only measureable when comparing the battery discharge characteristics of a lead acid battery under a dynamic versus a static load profile, it would seem wise to devise an alternate measurement method to reconcile the difference between this "battery rundown" method and his direct measurements.

As well, a further investigation of any possible FET driver contribution to the circuit should be considered.  But, one must first understand that under dynamic conditions, the gate of a MOSFET can indeed pass more than nanoamperes into the circuit, before one can devise methods to quantify this. 

In my original post, I merely stated that as it seems that many of these "OU" pulsed circuits require a battery, and that a lead acid battery is, for the most part, the chemistry of choice.  The possibility exisits that the observed OU is moreso related to differences in battery capacity under static versus dynamic loading, than to some previously unobserved phenomenon related to inductors.  Desulphators have been around for some time, and as well, the effects of pulse plating in the world of electroplating are also well known.


PW

[lanenal]

Lanenal,

I am not sure how you figure that an increase in capacity would somehow cause an exaggeration in Vdrop.  If the capacity of a battery is increased, it will hold a given voltage for a longer period of time under a given load.  There would be no "bogus voltage drop".  The larger capacity would just allow the battery to sustain a given load for a longer period of time before reaching a given stop voltage.

As the bulk of the OU is only measureable when comparing the battery discharge characteristics of a lead acid battery under a dynamic versus a static load profile, it would seem wise to devise an alternate measurement method to reconcile the difference between this "battery rundown" method and his direct measurements.

As well, a further investigation of any possible FET driver contribution to the circuit should be considered.  But, one must first understand that under dynamic conditions, the gate of a MOSFET can indeed pass more than nanoamperes into the circuit, before one can devise methods to quantify this. 

In my original post, I merely stated that as it seems that many of these "OU" pulsed circuits require a battery, and that a lead acid battery is, for the most part, the chemistry of choice, the possibility exists that the observed OU is moreso related to differences in battery capacity under static versus dynamic loading.  Desulphators have been around for some time, and as well, the effects of pulse plating in the world of electroplating are also well known.


PW

PW, let me first explain the exaggeration thingy. If I understood you correctly, from what you have said, increased capacity would allow the battery to last longer for the same voltage drop under a given load. (BTW, this looks like an OU statement: if the battery increases its capacity under pulsing load of equivalent wattage, then it lasts longer, which means that the total energy output is larger as time X watt = output energy).


So if the stored energy remains constant (by the law of energy conservation), AND if the base voltage does not change, then the initial voltage must drop. In math:


Battery Capacity =  Energy Output / Delta Voltage.


If the battery starts from full charge to full drain:


Battery Capacity = Total Stored Energy / (Initial Voltage - Base Voltage)


Now if Battery Capacity increases, and Total Stored Energy and Base Voltage remains constant, then Initial Voltage must drop to keep the identity.


regards,
lanenal

Edit: I am not making a strict argument, as batteries not not linear. If we assume that the point-wise Battery Capacity (as a function of current voltage) is bloated up by a constant factor, then the above argument can be easily transformed into integrations, with the conclusion unchanged.

[picowatt]

PW, let me first explain the exaggeration thingy. If I understood you correctly, from what you have said, increased capacity would allow the battery to last longer for the same voltage drop under a given load. (BTW, this looks like an OU statement: if the battery increases its capacity under pulsing load of equivalent wattage, then it lasts longer, which means that the total energy output is larger as time X watt = output energy).


So if the stored energy remains constant (by the law of energy conservation), AND if the base voltage does not change, then the initial voltage must drop. In math:


Battery Capacity =  Energy Output / Delta Voltage.


If the battery starts from full charge to full drain:


Battery Capacity = Total Stored Energy / (Initial Voltage - Base Voltage)


Now if Battery Capacity increases, and Total Stored Energy and Base Voltage remains constant, then Initial Voltage must drop to keep the identity.


regards,
lanenal

Lanenal,

That is exactly my point.  You think that the example I gave appears to be OU.

If one connected a desulphator to a battery and applied a load and noted the discharge time to a given stop voltage, and then did the same using the same load but without the desulphator connected and noted a shorter discharge time, there would be those who would claim that the desulphator produces overunity.

But, who ever said that a battery is "unity" to begin with?  A battery produces waste heat during both charge and discharge.  Suppose the charge/discharge efficiency of a lead acid battery is 70%.  Suppose the use of a desuplhator can increase this to 85%.  Though better, still not OU.

Desulphators supposedly produce finer grained sulphate crystals effectively increasing plate area.  As well, pulse and reverse pulse plating similarly are known to produce finer grained structures.  Visualize large clumps of sulphate crystals forming at the plates as opposed to a much smoother, finer grained structure.  As well as an increase in surface area, less resistive losses occur over the finer grains.  This is, supposedly, how a desulphator functions, and the waveforms used to desulphate have much in common with the waveforms of these OU circuits.

Do a search for "battery desulphator".  There are both construction articles as well as commercial units available.

I am not stating with any certainty that these effects are the reason for the observed OU, but, as these effects have been observed regarding lead acid batterys when being pulsed or reverse pulsed, should they not, at the very least, be considerd and ruled out as the reason for the observed OU?

As well, has anyone ever demonstrated the ability to discharge more energy from a battery than was required to charge it to begin with?


PW