Rosemary:

TK does not have a "monumental lack of abilities related to power analysis."

It's a brazen LIE, and it's disgusting and it's time for this to stop.

YOU ARE LYING - and I could easily find another 50 instances like this if I reviewed the past month's postings.  IT MUST STOP.

YOU MUST STOP LYING.

Get yourself out of the gutter, stop compulsively lying, show some self-respect, and try to act like a decent human being.

MileHigh

Golly.

Rosemary,

I think most if not all are in disagreement with me on the impedance issue.

My extensive research and findings on this issue indicate that the correct value to use when computing average power, even in the presence of high voltage oscillations, is the DC resistance of the CSR (providing the voltage probes are as close as possible to the CSR body).

In short, the reason being is that the resistance is a static fixed value, just as the battery voltage is, and therefore it (they) can be averaged.

Quote from: poynt99 on May 16, 2012, 09:22:42 PM
I have no qualms with what the MOSFETs may be dissipating, I'm simply saying that the indications are that roughly 50W of average power is being delivered by the batteries in that shot. And yes, 50W will do some decent heating.

So then what is the power necessary during the LO portions required to bring the overall average over the entire period down to roughly 50 W? Since you now agree that there are indeed 5 amps at 73 volts during 45 percent of the total time, I hope.

And where IN ANY SCOPE TRACES WE HAVE SEEN is there that much power in the oscillation traces? There are plenty of blowups out there and every one shows that the negative current and the positive current in the oscillations are nearly exactly equal, and that is without any filtering.

If average power is 50 watts and the current and voltage during the ON time is 5 A and 73 V, and the duty cycle is 45 percent ON,  then:

((73 x 5) x 0.45) + ( Px x 0.55) = 50. Solve for Px.

(Px x 0.55) = 50 - (365 x 0.45)
Px = (50 - (365 x 0.45)) / 0.55
Px = -208 Watts

So there must be NEGATIVE 208 WATTS in the oscillation portion to bring the average down to +50 Watts, given the power level that is clearly happening during the gate HI time. Right? And that requires a current average of I=P/V or 208/73 or about 2.8 Amps average in the NEGATIVE direction during the gate LO, oscillations present, portion of the cycle.
Where is this current passing? Does a Negative current flow not dissipate heat in the elements it is flowing through? Toasters heat no matter which phase of the AC is going through the element at any given instant. 

Where have I gone wrong? Please show me, that I may understand.

Look at the high-res oscs below. This is at 62 volts, I believe. Where is the evidence for a negative current averaging close to 2 amps during the oscillation?

Quote from: poynt99 on May 16, 2012, 09:33:45 PM
Rosemary,

I think most if not all are in disagreement with me on the impedance issue.

My extensive research and findings on this issue indicate that the correct value to use when computing average power, even in the presence of high voltage oscillations, is the DC resistance of the CSR (providing the voltage probes are as close as possible to the CSR body).

In short, the reason being is that the resistance is a static fixed value, just as the battery voltage is, and therefore it (they) can be averaged.

I KNOW you depend on this argument Poynty.  It's another of those cards that we've not played yet. You're wrong - but nor am I going to argue it on this thread.

Rosie Pose
And POYNTY POINT.  WHEN in the proud history of science has the MAJORITY opinion every carried an argument?  Surely you know better than to rely on this to support your allegation?

R