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Author Topic: Exploring the Inductive Resistor Heater  (Read 77172 times)

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #30 on: April 30, 2013, 09:01:45 PM »

This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V.  For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the  load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.


The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V.  19.78Wh / 25.28Wh = 0.78  So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER.  I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.


You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!


The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.


Take off your BLINDERS you guys.


Regards,


Greg

Greg,  I understand what you are saying with regard to your watt hour calculations, which was the next step beyond those I covered in my previous post.

But, what I was asking, to make sure I understood your methods, was, up to the points discussed in my previous post, did I follow your methods correctly and the measurements you made? (understanding, of course, that you did not use them for any in/out comparisons at that point)

PW

gmeast

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Re: Exploring the Inductive Resistor Heater
« Reply #31 on: April 30, 2013, 09:20:23 PM »
Using CSRs to measure power in certain types of pulsing circuits returns inaccurate and unreliable results.  It's fine for sine waves, nicely packaged square waves, saw-tooth waves, triangle waves, but not the types observed here.

gmeast

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Re: Exploring the Inductive Resistor Heater
« Reply #32 on: April 30, 2013, 09:22:00 PM »
Greg,  I understand what you are saying with regard to your watt hour calculations, which was the next step beyond those I covered in my previous post.

But, what I was asking, to make sure I understood your methods, was, up to the points discussed in my previous post, did I follow your methods correctly and the measurements you made? (understanding, of course, that you did not use them for any in/out comparisons at that point)

PW


Good ... then we're done.

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #33 on: April 30, 2013, 09:25:16 PM »
Using CSRs to measure power in certain types of pulsing circuits returns inaccurate and unreliable results.  It's fine for sine waves, nicely packaged square waves, saw-tooth waves, triangle waves, but not the types observed here.

Greg,

Please just answer yes or no as to whether I was following along correctly up to the points in my previous post.

You say "I don't get it", so I am trying.  Did I follow along correctly up to that point?



As to using CSR's with complex waveforms, it seems you relied on the voltage measured at SH3 to determine input power, if I have followed along correctly up to the point in my previous post.

PW

gmeast

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Re: Exploring the Inductive Resistor Heater
« Reply #34 on: April 30, 2013, 11:49:37 PM »
Greg,

Please just answer yes or no as to whether I was following along correctly up to the points in my previous post.

You say "I don't get it", so I am trying.  Did I follow along correctly up to that point?



As to using CSR's with complex waveforms, it seems you relied on the voltage measured at SH3 to determine input power, if I have followed along correctly up to the point in my previous post.

PW
Generally "yes" up until your last statements (3).  I said the draw-down test will give you 'some but not all' of the information you need to determine the input 'power' of the circuit. I did not say that this 1st draw-down test would give you THE input power.  Also you sited the 3.09, 3.099 values. Those only relate to the DC draw-down in which case the SH3 values are extremely accurate and reliable. So you were only correct in your understanding up until (3).


It's clear that you are still hung up on using real-time power values to determine efficiencies ... and you are wrong to assume that's correct with these types of circuits.


So, 'no', I'm afraid you still don't 'get it' and I've answered the "why" since that posting and posted jpegs from the slide show.


gme

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #35 on: May 01, 2013, 12:28:32 AM »
Generally "yes" up until your last statements (3).  I said the draw-down test will give you 'some but not all' of the information you need to determine the input 'power' of the circuit. I did not say that this 1st draw-down test would give you THE input power.  Also you sited the 3.09, 3.099 values. Those only relate to the DC draw-down in which case the SH3 values are extremely accurate and reliable. So you were only correct in your understanding up until (3).


It's clear that you are still hung up on using real-time power values to determine efficiencies ... and you are wrong to assume that's correct with these types of circuits.


So, 'no', I'm afraid you still don't 'get it' and I've answered the "why" since that posting and posted jpegs from the slide show.


gme

Greg,

So, all is well regarding my understanding of step 1 and 2.  You seem to be saying I am incorrect about step 3, but are you actually saying that my understanding of the tests and measurements are correct, but only that you do not want that data to be used to calculate efficiency?  At least let me know I am grasping your test methods correctly.

So,again, is my understanding correct regarding your methods and measurements in step 1 thru 3?

If I am correct, but you are actually just "hung up" on not using any of that data "as is" to measure efficiency, fine, just say so and I can move on to step 4.

So, as I understand it, in step 4 you use the step 3 drawdown data (using the rheostat set for 5.6mv across SH3) and note the time it takes the battery to discharge to the step 1 end voltage (27.44V I think it was).  The step 3 drawdown reached 27.44V in less time than the step 1 test, so you use the shorter time interval of that drawdown to calculate the watt hours used in the step 1 BH circuit run.

Is this correct?

PW

gmeast

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Re: Exploring the Inductive Resistor Heater
« Reply #36 on: May 01, 2013, 01:16:16 AM »
Another important note here regarding the Gate Driver: The power the gate driver is drawing either in-circuit or on its own support battery is NOT supplementing the HEATING of RL. Aside from its overhead from 'just being there', as it does its 'driver thing' the increase in power draw is from doing what it has to do to source and sink the current required in charging and discharging the gate capacitance in order to maintain the required gate charge voltage whether turning it 'ON' or 'OFF'.  This 'sourcing and sinking' is isolated from the rest of the MOSFET except for a leakage of 100nA ... 100 billionths of an Amp ... the published data for the UCC2732x drivers.


So, this is the reason why the gate driver's power is not considered in Exploring the Inductive Resistor heater. The ultimate goal is to have the components self-oscillate instead of relying on controlling circuitry such as a PWM and Driver arrangement ... something I've almost figured out how to do ... but not quite.


Everything is contained in that slide show, and in enough detail, to answer any and all questions. That's the reason I made it.  No more 'on the witness stand' "yes" or "no" questions ... capiche?


Regards,




Greg

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #37 on: May 01, 2013, 01:21:37 AM »
Greg,

Is my understanding correct up to and inclusive of step 4 in my previous post?

PW

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #38 on: May 01, 2013, 02:00:56 AM »
Another important note here regarding the Gate Driver: The power the gate driver is drawing either in-circuit or on its own support battery is NOT supplementing the HEATING of RL. Aside from its overhead from 'just being there', as it does its 'driver thing' the increase in power draw is from doing what it has to do to source and sink the current required in charging and discharging the gate capacitance in order to maintain the required gate charge voltage whether turning it 'ON' or 'OFF'.  This 'sourcing and sinking' is isolated from the rest of the MOSFET except for a leakage of 100nA ... 100 billionths of an Amp ... the published data for the UCC2732x drivers.


So, this is the reason why the gate driver's power is not considered in Exploring the Inductive Resistor heater. The ultimate goal is to have the components self-oscillate instead of relying on controlling circuitry such as a PWM and Driver arrangement ... something I've almost figured out how to do ... but not quite.


Everything is contained in that slide show, and in enough detail, to answer any and all questions. That's the reason I made it.  No more 'on the witness stand' "yes" or "no" questions ... capiche?


Regards,




Greg

Greg,

Your comments regarding the operation of a FET fall short of the true picture.  While it is true that under static DC conditions the gate leakage is typically very low, in the pico or nanoamp range, the dynamic conditions during turn on and turn off are quite different.

As you say, to turn the FET on or off the gate capacitance(s) must be charged or discharged.  To do this quickly requires a significant amount of current.  The driver you are using is capable of sourcing/sinking 4 amps for this purpose, but there are gate drivers capable of 20-40amps also available.

The most significant capacitances at the gate are the gate to source capacitance, Cgs, and the gate to drain capacitance Cgd.  Cgs is almost always the largest, and it is equivalent to a  capacitor connected between the gate and source of the FET.  Cgd is similarly equivalent to a capacitor connected between the gate and drain of the FET.  To charge Cgs from zero to 12 volts, the charging current must flow between the gate and source.  Simultaneously, to charge Cgd, charging current must flow between the gate and drain.  During hi speed switching, these charging currents can have very large peak currents.

If you are using the PG50, Cgs and Cgd are fairly large values.  From the data sheet you will see that Cgs and Cgd vary with the drain to source voltage, Vds, and when the Vds is near zero volts (as it is when the FET is turned on in your circuit for example), these capacitances are at their maximum value.  As the gate driver's waveforms typically contain a large amount of high frequency content (i.e., fast rise and fall times), the reactance of Cgs and Cgd can be quite low at these high frequencies causing large peak currents (many amps) to flow from the gate to both source and drain.  This is why a gate driver is typically used, to provide the amps of current needed to charge these capacitances rapidly.

In your circuit, however, you are limiting the gate drive current by using the 100 ohm resistor between the driver and the gate.  Assuming there is only small amount of stray capacitance between the driver output and the gate, whose reactance would effectively be in parallel with the 100 ohm gate resistor, your 100 ohm resistor limits the peak current available to charge the gate capacitances and hence greatly slows your on/off switching times as compared to what your driver is capable of.


It is a shame you are unwilling to discuss your circuit further.

PW




gmeast

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Re: Exploring the Inductive Resistor Heater
« Reply #39 on: May 01, 2013, 05:17:14 AM »
Greg,

Your comments regarding the operation of a FET fall short of the true picture.  While it is true that under static DC conditions the gate leakage is typically very low, in the pico or nanoamp range, the dynamic conditions during turn on and turn off are quite different.

As you say, to turn the FET on or off the gate capacitance(s) must be charged or discharged.  To do this quickly requires a significant amount of current.  The driver you are using is capable of sourcing/sinking 4 amps for this purpose, but there are gate drivers capable of 20-40amps also available.

The most significant capacitances at the gate are the gate to source capacitance, Cgs, and the gate to drain capacitance Cgd.  Cgs is almost always the largest, and it is equivalent to a  capacitor connected between the gate and source of the FET.  Cgd is similarly equivalent to a capacitor connected between the gate and drain of the FET.  To charge Cgs from zero to 12 volts, the charging current must flow between the gate and source.  Simultaneously, to charge Cgd, charging current must flow between the gate and drain.  During hi speed switching, these charging currents can have very large peak currents.

If you are using the PG50, Cgs and Cgd are fairly large values.  From the data sheet you will see that Cgs and Cgd vary with the drain to source voltage, Vds, and when the Vds is near zero volts (as it is when the FET is turned on in your circuit for example), these capacitances are at their maximum value.  As the gate driver's waveforms typically contain a large amount of high frequency content (i.e., fast rise and fall times), the reactance of Cgs and Cgd can be quite low at these high frequencies causing large peak currents (many amps) to flow from the gate to both source and drain.  This is why a gate driver is typically used, to provide the amps of current needed to charge these capacitances rapidly.

In your circuit, however, you are limiting the gate drive current by using the 100 ohm resistor between the driver and the gate.  Assuming there is only small amount of stray capacitance between the driver output and the gate, whose reactance would effectively be in parallel with the 100 ohm gate resistor, your 100 ohm resistor limits the peak current available to charge the gate capacitances and hence greatly slows your on/off switching times as compared to what your driver is capable of.


It is a shame you are unwilling to discuss your circuit further.

PW


The driver I'm using supplies as much as 9Amps required to negotiate the challenges of the Muiller Plateau. The current is needed as I said it is and does NOT pass through to the drain. As I said the PG50 can only leak 100nA. My assessment does not fall short. I have had several good conversations with TI on the subject of the 321 firing the PG50 and you're all wet on this one. It's amazing how you clowns try and muddy the water when it comes to this Inductive Resistor Heater.


It's amazing how I start this thread and ass holes like you hikack it. FUCK YOU AND YOU TOO HARTMAN FOR LETTING THIS SHIT CONTINUE.

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #40 on: May 01, 2013, 05:37:18 AM »

This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V.  For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the  load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.


The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V.  19.78Wh / 25.28Wh = 0.78  So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER.  I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.


You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!


The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.


Take off your BLINDERS you guys.


Regards,


Greg

Greg,

I have reread the above again, and I believe I now understand your method.  It is actually quite brilliant.


However, does this not presuppose that the batteries have the same amp hour rating, and hence draw down characteristics, with a purely resistive load versus a pulsed load? (as per my original concern regarding load profiles versus battery capacity)

As an analogy, suppose you have a flooded lead acid battery on a motorized table that gently rocks the battery so that the electrolyte is being stirred constantly.  A load resistor is applied and the voltage and current is monitored and its draw down from a start and stop voltage is noted over a measured time period.  The same test with the same load is again performed but this time the "stirring table" is turned off.  Would you expect the battery to necessarily measure the same capacity in both tests?

As well, suppose a pulsing desulpator is connected to a lead acid battery driving a resistive load so that sulphate crystals formed during discharge are maintained at a smaller size, and as well, the effects of pulse plating produce a finer grain structure, that is, a greater conductive area, during discharge, so that the battery appears to have a larger amp hour rating than it does when similarly loaded without the desulphator connected.  Would this prove that the desulphator produces overunity or would it only prove that the capacity of a lead acid battery can be increased by pulsing the battery during discharge?

I bring this up because for many years there have been various claims of overunity with pulsed circuits, but for some reason the "overunity" always requires a battery.  And a lead acid battery appears to be, for the most part, the most popular battery chemistry used.

This is why I asked if you had ever attempted to operate your circuit using only a well filtered supply to determine if the circuit itself is truly overunity or if the observed effect is moreso related to the battery having a different capacity under different load profiles.

You say that your circuit will not oscillate when operating from the DC supply.  This could be further investigated by installing a network between the supply and battery that models the measured equivalent series resistance, inductance, and capacitance of your batteries.  If necessary, the supply can be isolated at AC by installing inductors between the supply and network.  Doing so might allow you to operate your circuit from the supply and make your input measurements at DC.  This would assist in determining if the observed OU is due to the operation of the circuit, or moreso, to the increase in battery capacity under a pulsed load profile.


As an aside, you seem to have a certain disdain for modern test equipment regarding the ability of any equipment being able to measure the voltage at SH3 because of its "complex" waveform.  The waveform there is not all that complex nor particularly fast, and direct measurement there, given a very low inductance CSR (due to your use of .05ohms), can be accurately performed.  I believe you stated that you measured the SH3 voltage using both a scope and .99's multimeter approach and had close agreement with both methods.  So why then, do you dismiss that measurement out of hand as inaccurate? 

In any event, as it appears that your input power determined by direct measurement and by use of the drawdown method are in significant disagreement with each other, would you not at least agree that a third method is in order to determine which method of measurement is more accurate?

Thank-you for your patience... us old guy's are slow on the uptake (just wait till you get there!)

PW   

   



   

 

MileHigh

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Re: Exploring the Inductive Resistor Heater
« Reply #41 on: May 01, 2013, 05:38:03 AM »
Gmeast:

i belive PW is a seasoned retired analog electrical design engineer with perhaps 40 years of design and test experience under his belt.  That's just my gut feel I am not asking him to divulge any details about himself.  He has more knowledge about this stuff on the tip of his pinky then you will ever have in your entire life.  He could figure skate circles around you backwards and blindfolded.  You are an insignificant pea in a darkened pod and PW is a mountain under the stars.

PW is being very polite to you and you act like a rude, vulgar, ignorant and stupid wanker.  It's pathetic.

Go ask Rosie, even she knows about AC current coupling through the gate-source capacitance of a MOSFET.  You can have a very nice and very uncomfortable conversation with her.  She will stroke you and tell you how brilliant you are.

Jackass.

MileHigh

P.S.:  Everybody I apologize for my strong comments and I won't repeat them again.  I just wanted to set the record straight.

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #42 on: May 01, 2013, 06:29:02 AM »

The driver I'm using supplies as much as 9Amps required to negotiate the challenges of the Muiller Plateau. The current is needed as I said it is and does NOT pass through to the drain. As I said the PG50 can only leak 100nA. My assessment does not fall short. I have had several good conversations with TI on the subject of the 321 firing the PG50 and you're all wet on this one. It's amazing how you clowns try and muddy the water when it comes to this Inductive Resistor Heater.


It's amazing how I start this thread and ass holes like you hikack it. FUCK YOU AND YOU TOO HARTMAN FOR LETTING THIS SHIT CONTINUE.

Greg,

You only gave your driver part number as a UCC2732x.  As you are limiting the gate current via a 100R gate resistor, I just picked the 4 amp version as I thought the 9 amp version would just be even further overkill.

In any event, even with the 9 amp version, that 100 ohm resistor limits gate drive to 120ma at 12volts.  So either way, 4 amp or 9amp, with the 100R in series to the gate, the driver never approaches anywhere near its drive current rating nor does the FET's switching speed ever come close to the driver's switching speed.  It would take 100 volts across that 100R at the gate just to produce 1 amp of gate drive current.  So, just because the specs say your driver can switch in nanoseconds, your FET is switching way, way slower than that.

Irregardless of how fast the driver switches, it is the 100R and the gate capacitance time constant that is determining the FET's switchng speed.  However, that 100R likely allows for, in concert with Cgs and Cgd, a bit of positive feedback making the circuit less stable and allowing it to ring as it does.

But feel free to ask TI what your FET's switching speed would be if you use their driver and whatever FET you are using with a 100R in series between the driver output and the FET gate.  They can tell you how slow it would be, but will probably ask you why you even need a high current driver when such a low gate drive current results from using that 100R. 

PW


TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #43 on: May 01, 2013, 07:54:34 AM »
Wouldn't you love to see Gmeast standing next to his demonstration at a poster session at a physics conference? With his attitude, he'd get arrested before the first coffee break. Or punched in the nose.

Using a 100R resistor in series with a gate driver chip is just laughable, especially with a mosfet with such high internal capacitances as the PG50. It's like driving a Ferrari with the throttle limited to 10 percent by sticking a brick under the accelerator pedal.

Well, maybe it can't go very fast... .but it's a FERRARI, stupid, so fuck you and where are the moderators, I want to whine and cry instead of facing the issues. And instead of demonstrating that different discharge schedules produce different "total energy content" values for used lead-acid batteries -- something that has been known for some time --  the results presented MUST indicate OU performance.... because that is what GMeast set out to +prove+. 

But of course the Red Queen has defined Gmeast's "work" as a replication of her claims...... even though the components are different, the circuit is different, the waveforms are different, the operation is different, the behaviour is different, the testing is different, and even the claims made are different.

While at the same time, true replications that use the same components as Ainslie, in the same (several different) circuits claimed by Ainslie, with the same waveforms and behaviour as Ainslie's circuit, tested by the same methods that Ainslie mis-uses, and that give the _same objective results_ as Ainslie's...... are dismissed with false statements and bogus insults by Ainslie.... because they clearly show that her own claims are wrong and are full of lies and misrepresentations.

It's amazing, isn't it? Proper use of "Pin, Pout" measurements made with inline CVRs have been used for many years and have resulted in the ability to design things like robot spacecraft that hit their targets and send pictures home from a billion miles away, things like digital storage oscilloscopes, home computers, and even things like efficient electric automobiles. But now we learn that this method isn't valid. And what's the evidence for that claim of invalidity? Why.... simply that the method doesn't give the results that GMeast desires. The "Pin, Pout" measurement shows ordinary underunity performance... so therefore it's invalid, because we already know this system is OU.



picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #44 on: May 02, 2013, 07:57:04 PM »
Quote from Greg "over there":


Quote
I mean what I said about those jack-offs over on OU. I've invested too much to shelve all of this. I will continue to test what I have and post anything significant here as I will on EF and Heredical since I have moderator privileges there at H. I am so surprised that Stefan Hartman is such a dolt as to allow the TKs and the poynty-heads and the picowatts to hijack all of OU. I wonder who's paying these pricks to sit on that site? Picowatt is so self absorbed in thinking he knows everything about MOSFETS ... it's amazing.  I've talked to the engineers at TI about the UCC27321 MOSFET Gate driver driving the gate of the IRFPG50 mosfet, and they assure me that the only thing that can get through and add heating to RL is the 100nA as published in the PG50's data sheet (gate to drain leakage). They (at TI) will not answer questions (officially) about stuff like this, but will get back to you 'unofficially' in time ... like a month. picowatt has got (loser) TK convinced of his authority on these topics and it's laughable.  I have found out that they contacted many (so-called) circuit experts to attest that the gate driver is adding heating to the 'element'. They just kept fishing until one agreed to attest to it. One that wouldn't attest to it contacted me today (of all times). He wishes NOT to get embroiled in any of this and I respect that.

I hate "free-energy"!

Greg

Such a nice young man, and what a load of crap regarding the operation of his MOSFET (and some paranoid BS about contacting "circuit engineers")

For static DC conditions, it is true that only picoamps or nanoamps of current typically flow thru the gate.  This is the gate's DC leakage current value.  The amount of gate leakage current that flows is both voltage and temperature dependent (and given in data sheets). 

However, the gate of a FET/MOSFET appears to the outside world as a capacitor that must be charged or discharged to turn the FET on or off.  There are several "capacitances" related to the gate capacitance, but the primary gate capacitances appear as two variable capacitors connected between the gate and source (Cgs) and the gate and drain (Cgd).  The value of these two variable capacitors varies as the voltage between the drain and source (Vds) varies.  Typically the values of Cgs and Cgd increase as Vds decreases.  As well, the gate to source capacitance, Cgs, is typically the largest of the capacitances appearing at (connected to) the gate.

To turn the FET on or off requires charging or discharging these intrinsic capacitors to a voltage value above or below the gate turn on voltage.  Because the ends of these "capacitors" opposite their gate connections are connected to the drain and source, the current required to charge or discharge these capacitors must flow between both the gate and source and the gate and drain. 

The amount of current that flows thru these gate capacitances while charging or discharging them (turning the FET on or off), depends upon the value of the gate capacitance(s) and the rate at which these gate capacitances are being charged or discharged (how fast the MOSFET is being switched on/off). 

To turn the MOSFET on and/or off very quickly, that is, to charge or discharge the gate capacitance very quickly, requires a large amount of current, and this current flows between the gate and both the source and drain.  Because of its typically larger capacitance value, the greatest current flow during dynamic conditions is between the gate and source.  However, during hi speed transitions, i.e., when high frequencies are present at the gate, very significant currents can flow thru both Cgs and Cgd.

A gate driver IC is used to drive the gate when fast switchng is desired.  A gate driver IC can typically sink/source several amps to tens of amps to charge and discharge the gate capacitances very quickly.  A gate driver IC also typically has a very low output impedance to reduce losses during high peak currents at its output.  The current the driver IC drives the gate with during dynamic conditions flows between the gate and source, and as well, between the gate and drain.  Large currents only flow during dynamic conditions, wherein the voltage at the gate is changing and are rate dependent.  These currents can be tens of amps being dependent upon the desired/designed switching speed, the MOSFET gate capacitance, any resistance or reactance in series with the driver output and gate, and the driver's output currentcapability.  Once a stable DC voltage is reached at the gate, gate current, i.e., gate leakage current, reduces to a very small value, typically being picoamps or nanoamps depending upon the device.

Greg is using a 100 ohm resistor between his driver and gate.  Assuming the 100R resistor is low inductance and that there is little stray capacitance between the driver output and the MOSFET gate (whose reactance would be in parallel with the 100R resistor), irregardless of the driver IC's 9 amp current capability and very fast switching times, at 12volts, the 100R resistor limits his peak gate current to 120ma.  This limited current available to drive the gate (i.e., charge/discharge the gate) greatly reduces the rate (speed) at which his MOSFET can switch.  His diriver may be transitioning very quickly, but the rate at which the gate voltage can charge or discharge the gate capacitnce is dependent upon the time constant created by the 100R resistor in concert with the value of the gate capacitance. 

It is also very likely that, in concert with Cgs and Cgd, the 100R resistor allows for a bit of positive feedback to the gate which assists in causing the circuit to ring as it does.



Greg, please feel free to ask any EE, at TI or elsewhere, if anything I have written above in my somewhat "simplistic" description of MOSFET operation is untrue.  I am quite confident you will be unable to find any EE that would disagree with anything I have written above.

Possibly, by adjusting your attitude a bit and realizing that it may actually be YOU who is acting like the "know it all", you might learn a bit more than what your apparently closed mind is allowing.

Good luck...

PW

   
« Last Edit: May 03, 2013, 04:51:48 AM by picowatt »