Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[poynt99]

Here is a crazy thought if we really want to be pedantic .What is magic about the battery terminals . These are just the point at which the lead "wires" [bus bars] inside the battery change to copper wires outside the battery .Should we not really put our probes INSIDE the battery on the actual plates?

It is unfortunate that you perceive me as being pedantic.

The issue of where the battery voltage is measured is paramount to obtaining a valid battery power computation by the scope.

How was your post helpful in any way?

.99

[hartiberlin]

Actually no, that's not really what I am trying to say. My point was to show that the two measurements are not the same. If the two measurements are not the same, one of them must be incorrect. I wasn't expecting anyone else to answer the question, as I had hoped Rose herself would be allowed to see the point I am trying to make.

Well the second is wrong, but I don´t understand what your point is here...what you want to say with it..
hmm...a bit confusing..

You will see the actual voltage across the battery, which is the required goal for obtaining the correct PVbat. A small ripple is precisely what you should see. You seem to be thinking contrary to this....why?

No, I don´t think contrary.
The voltage will be almost constant, just only a small ripple, i.e. rising of the voltage,
when the battery current is negative and a bit falling, when the current is positive.
But the differences  will be only in the MilliVolts range, as the internal
resistance of the batteries is pretty low.

The battery works here as a big capacitor, where the voltage can not jump on it,
so only a small ripple will be seen on the DC supply voltage.

This is why we can neglect the ripple voltage and can calculate with
a "constant" battery supply voltage and just
observe the current on the shunt.

When the current trace area below the ground line , which I painted green
in the posted scope shots on the shunt, is bigger than the
area above the ground line(painted red), then we can already see, if energy
is flowing back into the battery or energy is flowing out of the
battery. Then we don´t need the battery voltage.

Hope this helps.
Regards, Stefan.

[poynt99]

Stefan,

One of the two battery measurements is incorrect. That is problem number one that needs attention.

Second, now that you bring up the current measurement, that too has it's problems.

Referring to your post; your area fill-in of the csr voltage is going to come out very close to equal when comparing both halves. First inaccuracy is that cycle mean is being used, and there are multiple cycles displayed. This is not the intended way to use cycle mean. Furthermore, the scope can not know what constitutes one cycle (it simply looks for zero-crossings), and therefore it completely missed that fact that one full switching cycle includes the portion of the cycle I highlighted.
http://www.overunity.com/index.php?topic=10407.msg279211#msg279211

Second (see scope shot below), realize that you are only looking at about half of the cycle. The other half clearly shows that there is positive current sourced from the battery. I have highlighted this in a red elipse.

In summary; the shunt voltage mean value shown is of no use, and does not reflect what the real average current is for that measurement.

.99

[cHeeseburger]

Now, Stefan and Poynt, we are converging on the truth finally.  The battery is a fixed DC potential with millivolts of actual ripple due to its internal resistance.  The battery voltage does not actually have the 150VAC 1.5MHz signal Rosemary is feeding the 'scope.

The importance is now properly focused on the shunt and the actual current flow there.

This demonstration shows the difference between the waveforms obtained across the inductive shunt and the resistive portion of same shunt.  The scale factors are identical (1V per division) on both traces.  Notice three super-important  things:

1)  The amplitude when we include the inductance is way higher and does not agree at all with the actual current measured just across the resistor.  The inductance allows a much larger voltage swing, fooling us into thinking the current is much larger than it really is.

2)  Look at the areas above and below zero.  In the larger (inductance included) trace, by eyeball, it looks like the areas are close to even or maybe even slightly more negative.  But in the real current trace it is clear that the area above zero is easily greater than that below zero.

3)  There is significant phase skew between the two waveforms and this will ruin the accuracy of any multiplied samples.  The true current (across just the resistive part of the shunt) does not peaqk at the same time as the false, inductor-polluted "current" trace and is in fact not always the same polarity at a given instant in time.  Notice the inductive shunt trace is approximately at its peak at the zero-crossings of the real current:  almost 90 degrees phase shift.  Basic fundamentls when the L vastly dominates the R of shunt!

So, the amplitude, waveshape and phase angle of the "current" signal Rosemary is feeding into the scope is by no means an accurate picture of the true instantaneous current flowing in the circuit. When the "battery" voltage also has an enormous misrepresentation due to series inductance inside the measuring points, and we multiply the data samples point by point, the numbers are so far from any believable reality that it boggles the mind and the results could come out anywhere and are totally meaningless, sorry to say.

Here is a challenge for Rosemary:  Submit this post to your favorite Tektronix Applications Engineer.  He or she is a certified oscilloscope measurement expert and is called on all the time to sort out these kinds of measurement questions.  Ask him or her to write a paragraph about it, agreeing or disagreeing with what I have wriiten here, attach his or her name to it, and publish it here for us.

Cheeseburger

[Rosemary Ainslie]

Right, Part 2 then.

What will happen to our PVbat calculation if rather than using this agreed upon equation:

PVbat = V(P1 - P4) x V(P3 - P4)/0.25

we make a small change and use this instead:

PVbat = V(P2 - P4) x V(P3 -P4)/0.25

What happens to the PVbat calculation?

.99

Poynty - Still not called off your dogs?  Shame on you.

Now.  Regarding that equation.  P never, to the best of my knowledge - is represented in any of those equations that you've put forward.  Power is ALWAYS vi dt.  Or Volts x amps x time.  THAT's it.  You can try and argue this till the cows come home Poynty.  This is the fundamental requirement for wattage analysis and this over time = POWER.  NOTHING ELSE.

SO.  Take that example that you've given us.  I'm looking at your schematic.  The amount of current discharged from your batteries will be determined by the amount of resistance in the path of that current.  Therefore resistance will be R1 + R2 + R3.  IF R1 = 1 Ohm and R2 = 1000 Ohms and R3 = 0.25 - then the total resistance determined by that circuit will be 1001.25 Ohm. 

Let us further assume that VBatt = 24 volts.  Therefore the current discharged by that supply source will be 24/1001.25 = 0.024 amps.  THEREFORE  vi dt = 24 (vbatt) x 0.024 (amps) = 0.575 watts x (say) 5 minutes would be 0.575 x 60 x 5 = 172.5 Joules.  THAT'S IT.  The ONLY correct way to determine that power.

So.  To get back to your question.  The Ohmage in the path of that P value that you refer to cannot be considered in isolation to the power over the entire circuit which will be distributed according to the resistance over the whole circuit.  You CANNOT look at one isolated part of the equation and expect it to represent a true value.

Now.  To get back to that same circuit that you drew and REPLACE R2 with a whole pile of MOSFETs in parallel.    Then.  Replace the R1 @ 1 Ohm with R = 10.86 Ohm.  NOW.  Apply a switch that the battery is ONLY connected during 20% of the time and for 80% it is disconnected and THEREFORE NO POWER IS DELIVERED. 

P = vi dt.  THEREFORE.  10.86 (R1) and 0.25 (R3) + 0 resistance at the MOSFET.  Therefore R = 11.11 Ohms.  THEREFORE if Vbatt = 24 then 24/11.11 = 2.16 amps.  24 (v) x 2.16 (i) = 51.85 watts.  Assume a 5 minute run time.  Therefore 51.85 x 60 x 5 = 15.552KJ. 

The actual question here is what happens during the period when the switch is open and the battery APPARENTLY is not able to discharge any current flow.  Hopefully you're looking at this.

Rosemary