I've zipped up all the Ainslie NERD scopeshots I've been able to find. Here is the archive. I've also put it in the "downloads" section of the forum.
Note that I've taken her false "jpg" extensions and renamed them correctly to .bmp, then converted to actual .jpg files. It's all in there.

If anyone knows of an Ainslie/NERD scopeshot that I've not included, please let me know and I'll add it to the archive.

TK:

QuoteDo you mean if you take the Pisnt curve and take the "rms" of that? Yes, there is another "square" in that. But that is not how you get RMS Power, which is Irms x Vrms... and this is already an "average" power. What is its sign?

My choice of words could have been better in that posting.  Just some basics that everybody should know.  You have an RMS voltmeter and you can use it to measure the power dissipated in a resistive load, which is (Vrms-squared/R).  You can also determine the RMS current, which is Vrms/R.

The current is derived from the voltage and when you increase the voltage the current increases, which I am calling a "square factor."  So you incorporate the "square factor" into a single measurement only, typically voltage, by calculating the RMS voltage.  The higher the voltage, the higher it contributes to the average power and it is a non-linear relationship.

So you don't take Pinst over time and do the RMS on that.  Pinst already incorporates a "square factor" so all that you need to do is average it over your chosen time period.

There is no such thing as "RMS power," there is only average power.  i.e.; the RMS voltage-squared divided by your load resistance is the average power.  Again, the whole point to the "RMS" is to give higher voltages a higher weighting factor when you want to determine power or current for a given resistive load.

Irms x Vrms doesn't work either.  Both of those terms imply a weighted averaging over a certain period of time.  You don't know what I(t) and V(t) and even R(t) are, so you have lost the time dimension to correlate one variable to the other when you say Irms x Vrms.

I can lose track sometimes myself but I am pretty confident what I stated above is correct.

QuoteBOTH phases of the power get dissipated in the circuit ! When you look at the "negative" current swing of an AC signal, does this indicate that you are returning power to the wall? Of course not: that "negative" power is also dissipated in the same place that the positive power is dissipated. That is why you are charged a positive number of dollars on your electric bill, rather than zero or a negative number. The "negative" power and the "positive" power are coming from the utility (or the battery) and they are dissipating  in your load (the entire circuit).

Both phases of power get dissipated in a given purely resistive circuit when  you apply an AC mains signal.  Note however that the voltage and current are in phase in a case like this.

In the circuit we are looking at when the MOSFET drain-source capacitance starts to discharge back through the circuit, it's stored energy and the voltage and current are in "antiphase."     So it's not comparable to a typical AC mains power situation.

Ultimately this circuit is a "pulse circuit" that is based on a repeating pulse train, and the sloshing currents moving about and the associated voltages should be measurable.

300 MHz = 1 meter, so 2 MHz = 150 meters.  So all of the interconnect wires should be at an equipotential.  So that means when you are getting the "bounce back" when the current reverses direction and the capacitor starts discharging, you should be able to take a scope probe and go through the entire main loop of the circuit and find out where the voltage drops are.

My gut feel is that you will see most of the voltage drop across the battery itself, and some of the voltage drop across the load resistor.  Note I am keeping it simple on the first go round and ignoring any capacitive and inductive effects.

To be continued...

MileHigh

TK:

I am just on a little journey here, I am not asking you to do anything specific.

Going back to when the MOSFET capacitance does it's discharge with the current reversed, in the DSO capture I annotated the reverse current waveform which looked like a triangle wave.  So low reverse current -> high reverse current -> low reverse current.  That could be used as a timing reference for looking at the voltage drops around the main loop.  Obviously you know what the cap voltage is as you traverse through the reverse current waveform.

So you can imagine the battery set taking this "hit" every time the reverse current waveform happens.  And we know that the cap voltage is falling when this happens.  I suspect that you would see a significant voltage drop across the battery set when it gets hit.  For example, imagine something like 90 volts at the one side of the load resistor and 80 volts on the other side of the load resistor.  Then imagine some voltage lost in the interconnect wire so you see 79.5 volts on the battery positive terminal and 0.5 volts on the battery negative terminal.  Note I am talking about one instant in time during the cap discharge.  That huge voltage drop across the battery is NOT charging the battery - it's too fast for the battery - and it just plows through the battery like a small hurricane.  The battery ends up looking like a big dumb resistor to the cap discharge.  Perhaps 98% of the cap discharge through the battery becomes heat and only 2% charges the battery.

So with a lot of patience and a lot of work, you could characterize the cap discharge and know how much of a voltage drop the battery is sustaining during the actual one microsecond of the discharge.

Why do all of this crazy stuff even if it is just an intellectual exercise?  It's to explore what is really going on in the circuit and dismiss out of hand Rosemary's ridiculous claim that the battery is supplying "potential only" and that the binding material in the load resistor is responsible for the energy return.  As of course the zipons are part of the fabric of the binding material and the always want to seek a balanced state.

What's responsible for the current reversal and corresponding energy return is a lousy boring capacitor, not a zipon in sight.

MileHigh

Quote from: MileHigh on May 17, 2012, 08:17:23 PM
TK:

My choice of words could have been better in that posting.  Just some basics that everybody should know.  You have an RMS voltmeter and you can use it to measure the power dissipated in a resistive load, which is (Vrms-squared/R).  You can also determine the RMS current, which is Vrms/R.

The current is derived from the voltage and when you increase the voltage the current increases, which I am calling a "square factor."  So you incorporate the "square factor" into a single measurement only, typically voltage, by calculating the RMS voltage.  The higher the voltage, the higher it contributes to the average power and it is a non-linear relationship.

So you don't take Pinst over time and do the RMS on that.  Pinst already incorporates a "square factor" so all that you need to do is average it over your chosen time period.

There is no such thing as "RMS power," there is only average power.  i.e.; the RMS voltage-squared divided by your load resistance is the average power.  Again, the whole point to the "RMS" is to give higher voltages a higher weighting factor when you want to determine power or current for a given resistive load.

Irms x Vrms doesn't work either.  Both of those terms imply a weighted averaging over a certain period of time.  You don't know what I(t) and V(t) and even R(t) are, so you have lost the time dimension to correlate one variable to the other when you say Irms x Vrms.

I can lose track sometimes myself but I am pretty confident what I stated above is correct.

Both phases of power get dissipated in a given purely resistive circuit when  you apply an AC mains signal.  Note however that the voltage and current are in phase in a case like this.

In the circuit we are looking at when the MOSFET drain-source capacitance starts to discharge back through the circuit, it's stored energy and the voltage and current are in "antiphase."     So it's not comparable to a typical AC mains power situation.

Ultimately this circuit is a "pulse circuit" that is based on a repeating pulse train, and the sloshing currents moving about and the associated voltages should be measurable.

300 MHz = 1 meter, so 2 MHz = 150 meters.  So all of the interconnect wires should be at an equipotential.  So that means when you are getting the "bounce back" when the current reverses direction and the capacitor starts discharging, you should be able to take a scope probe and go through the entire main loop of the circuit and find out where the voltage drops are.

My gut feel is that you will see most of the voltage drop across the battery itself, and some of the voltage drop across the load resistor.  Note I am keeping it simple on the first go round and ignoring any capacitive and inductive effects.

To be continued...

MileHigh

Well, there's a lot in that post. Let's just see if we can get an "average power" value for that SCRN0235, please. I KNOW that the scope trace indicates a very large power value, DC, and this problem about the power in the AC portion of the signal is just exactly why I didn't want to discuss the AC part FOR NOW. But OK..... we are forced to.

I still seek some explanation for how the "average power" can come out to 50 Watts if the power in the DC portion of the signal is as high as it is. 360 W, for as long as it is on which is 0.45 x the period.  What is required for the oscillations, in order to bring down the "average" to 50 Watts across the entire envelope containing both the DC and the oscillations?

Can we at least agree that the DC power shown in that scope trace during the Gate HI period is 5 x 73 = 365 Watts and that the duty cycle shown is 45 percent ON, not including the ramp-up to the steady state?


Perhaps calling it "rms power" is where the confusion lies. It's a term used frequently for sinusoidal waveforms and it typically means Pavg (or Prms) = (Irms^2) x X where X is the inductive reactance of the load (which acts like "ohms" in the equation) or simply the R if the load is resistive.

I'm pretty confident that _this_ is correct: (the more complicated analytic form accounts for all phase variations and waveform complexities, while the strict Irms x Vrms applies to sinusoidal waveforms).
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powerac.html#c3

Quote from: MileHigh on May 17, 2012, 08:17:23 PM
My choice of words could have been better in that posting.  Just some basics that everybody should know.  You have an RMS voltmeter and you can use it to measure the power dissipated in a resistive load, which is (Vrms-squared/R).  You can also determine the RMS current, which is Vrms/R.

The current is derived from the voltage and when you increase the voltage the current increases, which I am calling a "square factor."  So you incorporate the "square factor" into a single measurement only, typically voltage, by calculating the RMS voltage.  The higher the voltage, the higher it contributes to the average power and it is a non-linear relationship.

So you don't take Pinst over time and do the RMS on that.  Pinst already incorporates a "square factor" so all that you need to do is average it over your chosen time period.

There is no such thing as "RMS power," there is only average power.  i.e.; the RMS voltage-squared divided by your load resistance is the average power.  Again, the whole point to the "RMS" is to give higher voltages a higher weighting factor when you want to determine power or current for a given resistive load.

Irms x Vrms doesn't work either.  Both of those terms imply a weighted averaging over a certain period of time.  You don't know what I(t) and V(t) and even R(t) are, so you have lost the time dimension to correlate one variable to the other when you say Irms x Vrms.

I can lose track sometimes myself but I am pretty confident what I stated above is correct.

Both phases of power get dissipated in a given purely resistive circuit when  you apply an AC mains signal.  Note however that the voltage and current are in phase in a case like this.

In the circuit we are looking at when the MOSFET drain-source capacitance starts to discharge back through the circuit, it's stored energy and the voltage and current are in "antiphase."     So it's not comparable to a typical AC mains power situation.

Ultimately this circuit is a "pulse circuit" that is based on a repeating pulse train, and the sloshing currents moving about and the associated voltages should be measurable.

300 MHz = 1 meter, so 2 MHz = 150 meters.  So all of the interconnect wires should be at an equipotential.  So that means when you are getting the "bounce back" when the current reverses direction and the capacitor starts discharging, you should be able to take a scope probe and go through the entire main loop of the circuit and find out where the voltage drops are.

My gut feel is that you will see most of the voltage drop across the battery itself, and some of the voltage drop across the load resistor.  Note I am keeping it simple on the first go round and ignoring any capacitive and inductive effects.

To be continued...

MileHigh

My dear MileHigh,

I LOVE this post - AND THE FOLLOWING.  I shall print them out tomorrow and then frame them and then hang them here by my computer.  And if and when that dim light of your 'real truth' EVER again dares intrude on my waking consciousness then I'll simply re-read these two posts of yours.  Because every re-read will remind me of your nascent comic genius which I initially mistook as an earnest dissertation.

What a boon to science.  A member with the ability to take logic by the nose - twist that nose - and thereby bring pure reason to it's knees.  WHAT a marvelous muddle.

Rosie Pose

LOL by the way (BTW) Guys, - I see now why our little TK depends on MileHigh's support.  He needs that precise scientific mind that MileHigh lacks in copious quantities - to elucidate the argument that TK lacks in any context at all.  It's a match made in heaven.  For my money I'd say that the greater skill is MileHigh's because he can avoid the issue with utter irreverence.  Whereas TK still reaches for some level of plausibility. 

They're a remarkably well matched team.  Two of the Marx Brothers.  But still looking for Groucho to make it perfect.  LOL. I think this is where sean will need to come to their rescue.  He - at least - doesn't underestimate our readers' intelligence.  And he has that caricatured commitment to a really bad mood.

Again.
R