OK, so looking at SCRN0150 again, if we take the roughly -40 mVVs value from the "minimum integral" box and divide by 0.25 Ohms for the uncorrected value of the CVR we get -160mAVs, in other words, -160 milliWattSeconds, or Joules.

This is correct, units-wise which is encouraging. So if the LeCroy is integrating across the whole screen, then this value should be at the extreme right, if the integral trace was actually displayed. Like in the Tek shot from Tar Baby: the integral decreases to the right so the minimum value is at the right edge.

SO that means 160 milliJoules NET TOTAL went past the measuring point (the CVR) per the 5 milliseconds of the entire screen (10 divisions at 500 us per division).  So that gives us  0.160 J / 0.005 sec == 32 Watts, in the "negative" direction, overall average power.
Or if they use 1 Ohm for the inductive reactance at 1.5 MHz this becomes 8 Watts negative. And so on, scaled by whatever the impedance you use for the CVR.


Ok, fine. That is presumably the scope's calculation over the entire waveform, oscs and not oscs, across the whole screen.

But the DC power has already been shown to be POSITIVE 110 Watts or so for its 16 percent of the duty cycle.
So for the average power to be -32 Watts, the following equation using the duty cycle must be satisfied:
(110 Watts x 16 percent ) + (OSCpower x 84 percent ) = -32 Watts.

Solving for OSCpower:
(OSCpower x 0.84) = -32 - (110 W x 0.16)
OSCpower = (-32 -(110 x 0.16)) / 0.84
OSCpower needed to average -32 Watts overall = a bit over -59 Watts. (edited to correct the earlier miscalc, sorry)

Or, if a 1 Ohm impedance for the CVR at 1.5 MHz is used, the overall average power "-32" becomes "-8" and the equation becomes
OSCpower = (-8 -(110 x 0.16))/0.84
Oscpower needed to average -8 Watts overall = about -30.5 Watts.

So we need a fair amount of NEGATIVE power in the oscs to offset the POSITIVE power in order to give us an average power in Watts that agrees with the scope's displayed value in Joules for the Integral of the Math Trace.

I think. It's been a long day. Please go over this math and comment or correct as needed. Anyone's comments will be helpful IF THEY SHOW THEIR WORK AND REFERENCES TO COMMON KNOWLEDGE BASES.

So now the question I am asking is simply this:

What is the correct power dissipated by the circuit?

By analogy to the toaster, where the power in BOTH phases of the AC signal are dissipated in the circuit, is this "negative" power actually delivered and dissipated to the "toaster" or is it delivered to the "wall"?

Is the correct power the average shown by using the integral?
Or is it the DC power + the AC power, or is it the DC power - the AC power, or some other value?

I think it's simply the average power as computed using the integral properly as I hope I have done here. Negative 32 Watts  (or negative whatever, scaled by the total impedance of the shunt). 

SO.... IF THIS NEGATIVE WATTAGE IS NOT DUE TO THE MEASUREMENT ERRORS we have been discussing, which it most certainly is, I still maintain that it must be dissipated in the circuit: the load and the mosfet(s).
Simply recall the toaster: half of the power it is using to toast your bagel is "negative" by the definition we are using here, isn't it? Yet the toaster does not cool off during that part of the cycle, nor does the utility company deduct anything from your bill. And all it would mean in the present case is that the current "through the load" is being reversed part of the time by the clever..... inverting AC power supply..... kludged together by the NERDs. Does your inverter charge up the battery it's running from while it's providing AC power to your toaster? NO.... the AC goes into the load, not the battery.

The Truth is "Out There"....

Way out there.



Let me recap.
"He" is of course .99, and "he" is here being accused of holding back HIS data, when in fact it was she who either was oblivious to or was deliberately concealing her Mosfet Mistake.

But perhaps even more hilarious than that: She says his schematic is WRONG, then in the next paragraph admits that she has neither the skill nor the knowledge to DRAW an accurate schematic.  A rather obvious question arises: How does she know his yayhoo is wrong, if she can't even find her own yayhoo in the dark? So to speak.

And I really wonder what happened to that government interest and that of the "highly respectable" English institution. After all this post was made on the 30th of April.....2011. Yet I don't think we have heard another word about either one of these things. Have we?

TK,

Regarding SCRN0150, assuming zero NET power during the oscillation phase, and therefore computing only the power while the FG is HI, I get the following rough average power from the batteries:

49.5 x 0.5/0.25 x 1.5/7 = +21.21W = P_AVG

That's VBat x V_CSR/R_CSR x duty-cycle

Quote from: poynt99 on May 18, 2012, 06:58:36 PM
TK,

Regarding SCRN0150, assuming zero NET power during the oscillation phase, and therefore computing only the power while the FG is HI, I get the following rough average power from the batteries:

49.5 x 0.5/0.25 x 1.5/7 = +21.21W = P_AVG

That's VBat x V_CSR/R_CSR x duty-cycle

OK...  I am NOT including the ramp-up to steady state but am conservatively using 105 microseconds as the ON time, and the total period seems to me to be 675 microseconds.  But perhaps it is closer to 690 microseconds.  I see you are using 150 microseconds and 700 for the period. And I _initially_ estimated the voltage drop across the CVR as 0.6 V.
I still think it's a little higher than 0.5 V.
However, we are both using the same calculation and the same method. If I use your numbers I get your answer and if you use my numbers you get my answer. So we are both right ! 
So using my more "precise" readings I get
49.5 x (0.53/0.25) x (105/690) =  +15.96 W = P_avg which does not include the ramp-up time.
If I include the ramp up I need to reduce the "average" voltage slightly, so I might use 130 microsec for the ON time and 0.51 V for the voltage drop.
49.5 x (0.51/0.25) x (130/690) =  +19.02 W = P_avg

So it appears the major discrepancy between your calculation and mine is our different readings of the duty cycle. You are reading the ON time as 150 microseconds, and I can't see it being that long.

Here's the blowup I did again, where I used my most careful and optimistic reads of the trace values and obtained +20.4 Watts. (REVISION B)

I am measuring these things against a blowup on a 24 inch monitor and I'm using fine dividers against the screen scale. Since I know that I am always getting screwed up by the LeCroy's crazy minor division scheme: 5 per major div horizontally but only 4 per major div vertically, I try to check several times and in several places on the screen. I'd really like to meet the designer who came up with that little joke.

At any rate, using the revised numbers we still have the instantaneous power of 49.5 x (0.55/0.25) = 109 W acting during the DC portion of 19 percent ON. So... what is the instantaneous power required in the OSCS portion of 81 percent of the time to bring the overall average down to the -30 or -40 or whatever is claimed for this shot?