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Author Topic: Exploring the Inductive Resistor Heater  (Read 77176 times)

lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #60 on: May 04, 2013, 09:29:08 AM »
According to you, and to Gmeast, this is impossible. So I want your explanation as to how I did it.
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #61 on: May 04, 2013, 09:39:37 AM »
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]

DID YOU WATCH THE VIDEO? Apparently not.

You really have no idea what you are talking about. I am showing MUCH MORE CURRENT than "gate current" or "leakage current" flowing through the mosfet from GATE to SOURCE and from GATE to DRAIN. Enough to make a 12 volt light bulb light up brilliantly.

Explain how this is done, since you claim that the mosfet CANNOT pass more than the leakage current. You might also get off your high horse and watch the video where  I switch the mosfet using only the charge from my fingers, illustrating that only a tiny current, providing a sufficient CHARGE, is needed to switch the mosfet. So what is your explanation for the hundreds of milliAmps of current that I show passing through the mosfet's GATE-SOURCE and GATE-DRAIN capacitances?

Come on, Mosfet master, I want to hear your explanation that explains how ONLY 100 nA can pass, yet the bulb will light nevertheless.

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #62 on: May 04, 2013, 09:47:10 AM »
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[size=78%]This clearly shows you didn't understand. You were not even able to discern gate current and leakage current.[/size]
I DID NOT MEASURE, NOR DID I ATTEMPT TO MEASURE, either of those currents. What I DID DO, is to show that the Gate-Drain and Gate-Source capacitances are essentially transparent to currents oscillating at the frequencies being used, and that they will easily pass FAR MORE than the 100 nA leakage current.
Your posts are showing more and more that you do not understand the operation of mosfets. "Gate current" applies to bipolar transistors and is different from the gate CHARGE that must be applied at the mosfet gate to turn it on. Gate leakage, for a mosfet, is just that, and in DC conditions the 100 nA figure is reasonable. The current supplied by a highspeed mosfet driver is there to charge the gate fully at high speed in spite of the gate capacitances, which is not properly "gate current" in the same sense as for a bipolar transistor.
It is definitely the case that power from the gate drive circuit can make its way to the load by way of the gate-drain and gate-source capacitances, at far higher currents than 100 nA, and this is what I showed in my video. It's really too bad that you continue to deny what is being demonstrated right before your eyes. Or would be, if you'd actually watch the video.

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #63 on: May 04, 2013, 09:52:07 AM »
Does this sound familiar?
Quote
This circuit should be the easiest to tune as well (Reason: when the pulse is at 6V, all transistors should be open switch, so logically, there will never be battery shorts). The pulse voltage source V5 can be replaced by a 555 astable circuit. V6 can be obtained by using an equal voltage divider (two 5K resisters, for example) sharing the same power as the 555 circuit. All the JBT transistors can be replaced with equiv. MOSFETs. I used voltage dependent current source and a resistor to mimick the behavior of a transformer. You can use 4 transformers instead in your implementation (MOSFETs are probably better than JBTs, I am affraid, as MOSFETs have no gate current). I used no diodes for switching purpose in this circuit. LTspice simulation is performed and the plot of the current over the load R3 is given below (I omitted the two capacitors at each end of the load R3 for the sake of simulation, in your implementation, you should include them).

Do they, or don't they? Why is a 9 amp driver needed then? Because of the 100 nA leakage?

poynt99

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Re: Exploring the Inductive Resistor Heater
« Reply #64 on: May 04, 2013, 03:03:59 PM »
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
What happens when the "hidden cap" is getting charged? Is the "hidden cap" between the Gate and GND, or between the Gate and Source and the Gate and Drain?

picowatt

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Re: Exploring the Inductive Resistor Heater
« Reply #65 on: May 04, 2013, 04:22:00 PM »

PW, you don't seem to understand what I am talking about. Your very argument about battery capacity is against the law of energy conservation. Your lengthy argument only seem to mislead and misinform people.


lanenal

Lanenal,

If I understand you correctly, you wish to treat a lead acid battery as if it were an ideal capacitor.

There are a lot of EV companies out there that wish this were true!  But, sadly, it is not.

There are many factors associated with most battery chemistries that affect the total energy they can deliver, or their efficiency during the charging or discharging process. 

To treat them as ideal capacitors falls very short of depicting their true nature.

PW

 


lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #66 on: May 05, 2013, 07:29:41 AM »

Explain how this is done, since you claim that the mosfet CANNOT pass more than the leakage current.


Can you show me where did I claim that? You probably misunderstood me somehow. I was only saying that Greg's 100nA could be referring to leakage current. I also said that gate current depends on quite a few factors.

lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #67 on: May 05, 2013, 07:35:42 AM »
Lanenal,

If I understand you correctly, you wish to treat a lead acid battery as if it were an ideal capacitor.

There are a lot of EV companies out there that wish this were true!  But, sadly, it is not.

There are many factors associated with most battery chemistries that affect the total energy they can deliver, or their efficiency during the charging or discharging process. 

To treat them as ideal capacitors falls very short of depicting their true nature.

PW


PW,


My initial layout of math is linear, in my PS I only briefly touched on how to treat the nonlinear case. In fact, there I put forth a capacity expansion ratio. And that's still a simplification. In its most general form, the ratio should also change with voltage, and the conclusion still holds even in this most general form as long as the ratio is always greater than 1, which is the mathematical equivalent of saying: the capacity expands with pulsing load.


lanenal

lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #68 on: May 05, 2013, 07:44:13 AM »
What happens when the "hidden cap" is getting charged? Is the "hidden cap" between the Gate and GND, or between the Gate and Source and the Gate and Drain?


Good question, and my answer was:


The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap.

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #69 on: May 05, 2013, 12:07:29 PM »

Can you show me where did I claim that? You probably misunderstood me somehow. I was only saying that Greg's 100nA could be referring to leakage current. I also said that gate current depends on quite a few factors.

Did I misunderstand, or did you? This is what Gmeast is claiming and what you apparently are defending:

Quote
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.

Yet as PW has described and I have shown empirically, this is manifestly NOT TRUE. The Gate-Source and Gate-Drain capacitances of the mosfet act like any other capacitances: they pass AC or pulsed DC currents. A substantial part of the power at 1.5 MHz applied to the Gate by the gate driver CAN INDEED make it "through" the gate to be dissipated in the load, or in the mosfet itself as heat.

Gmeast seems to be basing his claim that the gate driver power can be neglected as input power, on the 100 nA DC leakage current in the datasheet. He does not appear actually to have measured, to see if his assumption is correct or not, and I don't believe he could have described the situation accurately to his "consultants" who told him that no more than 100 nA could possibly be conducted across the Gate-Source or Gate-Drain capacitances.

Hopefully by now you will agree, lanenal, that this is not proper. It would seem that you must either agree that a lot of current can indeed pass, as I show in the video, and therefore Gmeast is wrong, OR.... if Gmeast is right, you need to explain how I did what I did in the video.

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #70 on: May 05, 2013, 12:11:53 PM »
Oh, and Gmeast: If you want your circuit to self-oscillate, that is easy. Just throw out all your nice construction techniques, and do what Rosemary Ainslie did: Use lots of excess wiring, especially on the Gate wiring, and use loose connections like clipleads. I know you don't care about my experience... even though it's a lot greater than yours... but that's exactly how I got Tar Baby to self-oscillate, after a proper tight layout would not. I looked at Ainslie's photographs and added the "rat's nest" of wires. Then the circuit self-oscillated without difficulty.
The magic oscillations you are trying for are nothing more than feedback.

lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #71 on: May 05, 2013, 03:31:03 PM »
Did I misunderstand, or did you? This is what Gmeast is claiming and what you apparently are defending:


TK, please read my post carefully. I didn't say that gate current can only get to 100nA. What I was saying is that that 100nA could be referring to the leakage current. Is that clear?


BTW, Greg already measured the a loose upper bound to the possible energy injection through the gate signalling. Even if he was wrong on the gate current, it does not invalidate his result.

poynt99

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Re: Exploring the Inductive Resistor Heater
« Reply #72 on: May 05, 2013, 03:57:51 PM »

Good question, and my answer was:


The current depends on the frequency, max voltage difference, the current limiting resistor, and the capacity of the hidden cap.
Well, read my two questions carefully; you didn't actually answer the two questions that I asked.

TinselKoala

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Re: Exploring the Inductive Resistor Heater
« Reply #73 on: May 06, 2013, 12:20:24 AM »

TK, please read my post carefully. I didn't say that gate current can only get to 100nA. What I was saying is that that 100nA could be referring to the leakage current. Is that clear?


BTW, Greg already measured the a loose upper bound to the possible energy injection through the gate signalling. Even if he was wrong on the gate current, it does not invalidate his result.

This is a direct quote from YOU. Please read your post carefully, including the part that I have highlighted in RED:
Quote
This circuit should be the easiest to tune as well (Reason: when the pulse is at 6V, all transistors should be open switch, so logically, there will never be battery shorts). The pulse voltage source V5 can be replaced by a 555 astable circuit. V6 can be obtained by using an equal voltage divider (two 5K resisters, for example) sharing the same power as the 555 circuit. All the JBT transistors can be replaced with equiv. MOSFETs. I used voltage dependent current source and a resistor to mimick the behavior of a transformer. You can use 4 transformers instead in your implementation (MOSFETs are probably better than JBTs, I am affraid, as MOSFETs have no gate current). I used no diodes for switching purpose in this circuit. LTspice simulation is performed and the plot of the current over the load R3 is given below (I omitted the two capacitors at each end of the load R3 for the sake of simulation, in your implementation, you should include them).
http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706

You here state that Mosfets HAVE NO GATE CURRENT. That is what the words "MOSFETs have no gate current" which you wrote in the post above mean in English. IS THAT CLEAR?

Now, once again, please explain how that can be true, at the same time that I can light up a light bulb on current passing through the Gate-Source and Gate-Drain capacitances of a mosfet.

The 100 nA cited in the mosfet data sheet as leakage current is leakage current, nobody has ever said that it wasn't. The issue is that both YOU (as shown above in the direct quote from YOU) and Gmeast both seem to think that is the only current that can pass between the gate and the other pins of the mosfet. Now, you are waffling about, since you have been presented with incontrovertible proof that your claims are false, now you are trying to pretend that you didn't make them. But as the DIRECT QUOTATION above proves.... you did indeed claim that "MOSFETs have no gate current." Is that clear?

What did you mean by that statement?  Why does the lightbulb probe light up in my video? Is there current flowing through my mosfet's gate-drain and gate-source capacitances, or not? Is this the current required to charge the gate and switch the mosfet, or is it in excess of that? 

Quote
Even if he was wrong on the gate current, it does not invalidate his result.
Of course it does. Crippling the gate driver by using the large inline resistor helps his cause by limiting the current here, but he is still wrong not to include the contribution of the driver to the circuit's overall power dissipation.

lanenal

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Re: Exploring the Inductive Resistor Heater
« Reply #74 on: May 06, 2013, 06:43:06 AM »
This is a direct quote from YOU. Please read your post carefully, including the part that I have highlighted in RED:http://www.overunity.com/6793/simplest-solid-state-tesla-switch/msg156706/#msg156706

You here state that Mosfets HAVE NO GATE CURRENT. That is what the words "MOSFETs have no gate current" which you wrote in the post above mean in English. IS THAT CLEAR?

Now, once again, please explain how that can be true, at the same time that I can light up a light bulb on current passing through the Gate-Source and Gate-Drain capacitances of a mosfet.

The 100 nA cited in the mosfet data sheet as leakage current is leakage current, nobody has ever said that it wasn't. The issue is that both YOU (as shown above in the direct quote from YOU) and Gmeast both seem to think that is the only current that can pass between the gate and the other pins of the mosfet. Now, you are waffling about, since you have been presented with incontrovertible proof that your claims are false, now you are trying to pretend that you didn't make them. But as the DIRECT QUOTATION above proves.... you did indeed claim that "MOSFETs have no gate current." Is that clear?

What did you mean by that statement?  Why does the lightbulb probe light up in my video? Is there current flowing through my mosfet's gate-drain and gate-source capacitances, or not? Is this the current required to charge the gate and switch the mosfet, or is it in excess of that? 
Of course it does. Crippling the gate driver by using the large inline resistor helps his cause by limiting the current here, but he is still wrong not to include the contribution of the driver to the circuit's overall power dissipation.


TK, what a great find -- I wonder what tool you have used, looks like you've got CIA team behind you :).


In that post, I was talking about the time average current (cause that's what matters in that case) while ignoring the leakage current, in that case, it is obvious that the time averaged gate current is zero.


lanenal