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Re: Rosemary Ainslie circuit demonstration on Saturday March 12th 2011
« Reply #725 on: Today at 10:10:50 PM »QuoteModifyNote: This message is awaiting approval by a moderator.
I have already shown several times how a false negative voltage can appear across an inductive shunt. Now, with the revelation that at least part of any positive current is being diverted through the "upside down" group of MOSFETs and never appearing in the shunt at all and none of any negative current is thusly diverted, this further explains how a net average negative current could easily be displayed across the shunt when in fact the actual net current may well be quite positive.
Furthermore, there is a recurring theme here coming from Rosemary that it is somehow remarkable and anomalous that the drain waveform or so-called "battery voltage" is 180 degrees out of phase with the shunt waveform. This is absolutely normal and would occur in all cases and always does! This is true in MOSFET circuits, junction transistor circuits, tube circuits, switching circuits, linear amplifiers and linear oscillators of all kinds. Obviously, as the current in the D-S channel increases, the shunt voltage goes up and the drain voltage goes down! There is nothing at all surprising or anomalous there and, in fact, it would be virtually impossible to have it any other way.
This completely normal and expected behavior also easily explains why the V*V trace always shows a negative value in Rose's circuit. The so-called "battery voltage" is always positive but varies in a way that it is at its high point when the shunt voltage is at its lowest (most negative) point. So, when the two numbers are multiplied by the scope math, any negative voltage on the shunt is multiplied by a much larger number than during times when the shunt voltage is positive and the so-called "battery voltage" is at its lowest point.
Even if the negative voltage peak on the shunt was only, say, 1/4 of the positive voltage peak, as long as the so-called "battery voltage" multiplier was more than 4x larger at that moment than it is at its low point, you would see a net negative V*V product. All of this nonsense has nothing to do with the actual input power from the battery!
It is a completely normal, expected result of not actually using the real DC battery voltage (which is virtually constant) as a multiplier and instead, feeding in a huge ac voltage that actually comes from the wiring inductance and not the batteries and in truth represents only the di/dt (rate of change of current) and not the battery voltage at all.
When the wiring inductance of the battery stack was cut in half by measuring at the battery terminals and she observed an exactly corresponding reduction in half of the AC part of the waveform this instantly proved that all of the AC voltage on the battery measurement is due to the wiring inductance and that there was still about half of that inductance remaining inside the measurement path because of the long wires used to interconnect the battery stack (the inductance of which was never eliminated from the measurement loop).
Finally, the position of the shunt, being within the gate drive loop (and now also within the source path of 4/5 of the MOSFETs) is not by any means a true and exclusive measure of the battery current.
A very simple way to address all of these problems (and other problems I have pointed out before) is to place the shunt directly on the negative terminal of the battery stack instead of inside the gate/source drive loop as it now is. In addition, the actual DC battery voltage (without the AC effects of the inter-battery wiring inductance) can be easily had by placing the probe across only the most negative battery of the stack.
These simple changes will have no effect on the circuit operation itself (all the same waveforms and oscillations and phase relationships will remain exactly intact) but will provide the actual measure of power flow into/out of the battery.
Simply take the V*V mean, multiply it by 4 (for the shunt being 1/4 ohm) and then multiply that by the number of identical batteries. For absolute accuracy, the shunt inductance can easily be compensated out of the measurement (while leaving it in the circuit itself) by adding an RC time constant equal to the LR time constant of the inductive shunt; RC=L/R. The diagram below shows the existing shunt X'd out, but there is no problem with leaving it there so as to assure no effect on circuit operation. You just can't measure there and expect to see only the battery current!
If Rose is still confused about any of these observations and the necessary procedures to make true input power measurements without making any changes to the operating characteristics of the circuit itself, once again, I urge her to consult with her local Tektronix or LeCroy Applications engineer. Short of that, I will be more than happy to answer questions or give advice. Here's to proper measurements in the future!
Humbugger
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Topic: The Rosemary Ainslie Circuit (Read 477263 times)
