It is not standard physics.  There must be some misunderstanding.

The oscilloscope probes are placed directly across the batteries that ground is at the source rail and the probe is at the drain rail.  Which is standard convention.  Then. During the period when the oscillation is greater than zero - in other words - when the battery is DISCHARGING - then it's voltage falls.  And it SERIOUSLY falls.  It goes from + 12 volts to + 0.5.  Given a  supply source of 6 batteries for example, then it goes from + 72 volts to + 3 volts.  At which point the oscillation reaches its peak positive voltage.  And this voltage increase is during the period when the applied signal at the gate of Q1, is negative.  WE KNOW that this FAR EXCEEDS THE BATTERY RATING.  In order for that battery to drop its voltage from + 12V to + 0.5V then it must have discharged A SERIOUS AMOUNT OF CURRENT.  Effectively it would have had to discharge virtually it's ENTIRE potential as this relates to its watt hour rating.  We EXPECT the battery voltage to fall during the discharge cycle.  But we CERTAINLY DO NOT expect it to fall to such a ridiculous level in such a small fraction of a moment AND SO REPEATEDLY - WITH EACH OSCILLATION.

The absolute worst case load that can be applied to the batteries is determined by the DC resistance of the load. This is because any AC present simply increases the over-all impedance. Therefore, with a load of 11 Ohms DC (this is the worst case), and a battery voltage (B+) of roughly 72VDC, the worst case (highest) current that can be drawn from the batteries is simply:

72VDC/11 Ohms = 6.5 Amperes.

With for example a 100 Amp-hour (A-h) battery, there would be roughly 15 hours of use available before the batteries were considered fully discharged. Out of interest, the power delivered by the batteries would amount to about 471 Watts.

So, if you were to take your load resistor and connect it directly to your battery array, this is approximately how long the batteries would last before they were considered "dead".

Your actual circuit however is one harboring a considerable amount of parasitic inductance throughout, especially in the long connecting wires to the battery array. As such, when the MOSFET bursts into its 1.5MHz oscillation, the circuit impedances become active and limit the net average current and power delivered to the load.

Taking this inductance and oscillation into consideration, it is not good practice to acquire battery voltage measurements at the "Drain Rail", because at this point there is an excessive inductive reactance between this point and the actual B+ terminal. As such, what will be observed is a large voltage swing, far in excess of the B+ voltage. Power measurements computed with this voltage measurement can only produce a "reactive" power result (Google "reactive power"). The unit for reactive power is "VAR", Volt-Amps-reactive.

I see this clearly in the simulations.

To obtain a "real" Battery power computation, the B+ must be measured directly between the battery posts, i.e. between B+ and B-. (Google "real power").

I'll get back to you later Poynty.  I'm finally beginning to feel tired.  What a pleasure.  Maybe I'll get some sleep.

R

PC,

Placing a DC meter or VOM directly across the battery terminals will not produce too much excitement I'm afraid. Why? Because the battery voltage (measured directly) doesn't actually dip that much. And for what little it does vary, the meter will average those small variations out and retain a fairly steady voltage reading.

A VOM wouldn't drop to 0.5v or is the 'recharge rate' really high or something?. I'm confused, I thought that was what the excitement was about!
PC
edit:changed frequency for recharge rate

OK Poynty - I think I've understood what you're saying here.

The absolute worst case load that can be applied to the batteries is determined by the DC resistance of the load. This is because any AC present simply increases the over-all impedance. Therefore, with a load of 11 Ohms DC (this is the worst case), and a battery voltage (B+) of roughly 72VDC, the worst case (highest) current that can be drawn from the batteries is simply:

72VDC/11 Ohms = 6.5 Amperes.

Indeed.  No problem with this.  Except I'll reserve comment related to 'This is because any AC present simply increases the over-all impedance.'  I take it that this increase to the impedance is related to the applied frequency.  I'm not sure that there's much difference in the computed AC amperage flow from energy applied from our grids to the energy applied from a battery with the load simply placed in series with that supply.  Much of a muchness.  It's at those higher frequencies that there's a reduced current flow due to higher impedance.  Which would, unquestionably increase the amount of relative resistance and REDUCE the rate of current flow - correspondingly. 

With for example a 100 Amp-hour (A-h) battery, there would be roughly 15 hours of use available before the batteries were considered fully discharged. Out of interest, the power delivered by the batteries would amount to about 471 Watts.

So, if you were to take your load resistor and connect it directly to your battery array, this is approximately how long the batteries would last before they were considered "dead".

So.  This is still in line with that standard model. 

Your actual circuit however is one harboring a considerable amount of parasitic inductance throughout, especially in the long connecting wires to the battery array. As such, when the MOSFET bursts into its 1.5MHz oscillation, the circuit impedances become active and limit the net average current and power delivered to the load.

Absolutely.

Taking this inductance and oscillation into consideration, it is not good practice to acquire battery voltage measurements at the "Drain Rail", because at this point there is an excessive inductive reactance between this point and the actual B+ terminal. As such, what will be observed is a large voltage swing, far in excess of the B+ voltage. Power measurements computed with this voltage measurement can only produce a "reactive" power result (Google "reactive power"). The unit for reactive power is "VAR", Volt-Amps-reactive.

Not actually.  That rather MONSTROUS voltage swing is never apparent on a standard switching circuit.  All that one sees there is the very high spiking that is managed at each switch.  The Spike itself - may exceed the battery supply voltage - but the battery voltage stays on track - more or less.

To obtain a "real" Battery power computation, the B+ must be measured directly between the battery posts, i.e. between B+ and B-. (Google "real power").

I keep telling you this.  We have done this ENTIRE TEST with a FULL OSCILLATION with the scope probe directly ON the positive battery terminal and the probe ground directly ON the negative terminal.  You asked us to do this.  NO INFLUENCE WHATSOEVER from those leads.  THAT SWING IS ALWAYS EVIDENT.

I am reasonably satisfied that there is absolutely no way that the measured voltage over any of the circuit components is ever WRONG.  If the scope shows 12 volts then indeed it's measuring 12 volts.  If the scope shows 0.5 volts then it's measuring 0.5 volts.  24 volts and it's measuring 24 - and so on.  That voltage measurement is SPOT ON.  ALWAYS.  That's our guarantee from the oscilloscope manufacturers.  And those rather zut instruments that we are privileged to access, can read those voltages in real time EASILY.  It is well able to adjust to the applied frequency.  What MAY vary is the current to be determined by that voltage reading.  That can vary if there's a phase shift - which is NOT applicable in our own waveforms.  Or it can vary in line with the impedance.  But that simply needs to be factored in.  It most certainly does NOT make that voltage reading across the battery incorrect.  Indeed, the scopes that we use are PRECISELY ACCURATE to within the smallest and most irrelevant margin of error.  So.  IF it is giving a measurement - you can take that measurement to the bank.  It is PRECISELY CORRECT.  Which also means that IF it is measuring 0.5 volts then THAT'S WHAT IT IS FOLKS.  The LeCroy and the Tektronix manufacturers have staked their reputations on it.  They give us all kinds of guarantees to this effect.

It may be advisable to check that there's no phase shifts.  And it would be advisable in the computation of the AMPERAGE DELIVERED - that one also factors in the impedance.  But that has NOTHING to do with the voltage reading across the battery.  It's SPOT ON. 

Kindest regards,
Rosemary
Changed I to It

A VOM wouldn't drop to 0.5v or is the 'recharge rate' really high or something?. I'm confused, I thought that was what the excitement was about!
PC

It's virtually impossible to pull down a single somewhat charged 12V battery to 0.5V, much less 6 batteries in series!

The voltage measurement is not taken on the battery posts, which is the reason the scope shows a large voltage swing, and this is due to the impedance of the wire between the load and the batteries.

Rosemary,

The battery voltage measurement is essentially the "meat of the matter" for my argument against the validity of your power measurements. If you will not argue this, then it would seem we have very little left to discuss.

I NEVER SAID THAT I WON'T ARGUE IT.  I said that you really need to stop saying that the scopes are picking up the wrong voltage.  OF COURSE I need to argue why that voltage swings.  I have argued it.  I'll try it again.  But right now - what I AM saying is that the scope meter is not wrong.  Those measurements are NOT ERRONEOUS.  IF THEY WERE I'd have cause to quarrel with LeCroy - and that would be RIDICULOUS.

I'll try that argument again.  Meanwhile could I impose on you to JUST READ what I've already tried by way of an explanation?  If it's not understandable then tell me where?  That might help.

Kindest regards,
Rosemary