another small breakthrough on our NERD technology.

[Rosemary Ainslie]

Rosemary:

I hate to interrupt, but I noticed your reply #273 directing people to your blog.  I read something on your blog that I could hardly believe.  From November 17, 2010:

"...This would certainly account for current flow.  But the problem is this.  Our scientists know the speed at which one valence electron would influence another valence electron.  And it would take up to half an hour  for it to travel through the average two meters of circuit wire before it would reach the light to light it or to reach the kettle to heat it.  There would be a required delay between the switching of the switch and the lighting of the light to get that process started...."

Do you believe that standard theory says a light bulb should take up to half an hour to turn on?  I have managed to live my whole life (up to now) without hearing that one.
Seriously, a half hour?

Bubba1

Yes Bubba.  The rate at which one valence electron would influence another valence electron takes a certain KNOWN quotient of time.  Therefore, IF current flow comprises the flow of VALENCE ELECTRONS - given some required length of wiring between the plug and the appliance - then it would take about twenty minutes before the light would light - or the kettle start to cook.  That's not my math Bubba.  That's standard physics.  I mention it because - I like you - find that when I switch on an electric appliance that current flow is also that dynamic that it's instantaneous.  Which means that it is possibly not entirely valid to claim that current flow is the transfer of energy via valence electrons - is my point.

Regards,
Rosemary

edited

[Bubba1]

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

[Rosemary Ainslie]

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

Bubba - if you've read that much then read on.  There are many 'optional' proposals to account for the flow of current to be the flow of electrons.  But the one contradicts the other.  If you take the trouble to speak to a physicist as opposed to an engineer - then you'll find that the purists ONLY refer to current flow as the flow of charge.  The imposition of the 'electron' according to Dyson in his 'conceptual physics' was simply to model the concept for purposes of 'understanding' the transfer of energy.  But the concept has been used for so long now that everyone refers to the flow of current as the flow of electrons - assuming that it carries the FULL weight of scientific endorsement.  It does not.  There are huge gaps in our knowledge.  It's presumed that all is known.  Far from it.  And I assure you - that example is only one of MANY concepts that are intrinsically contradictory.  I've listed some of them.   

Regards,
Rosemary

[poynt99]

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").

[PhiChaser]

Hey guys, I seriously enjoy reading your discussions!!! My question would be why you don't have VOMs connected directly to the batteries while running your tests? Sorry to interrupt but it seems like if you're measuring batteries you would want some sort of meter bettween the + and - (B+ and B-?), or at the ends of the battery bank or whatever, especially if they are discharching so quickly. Would be fun to watch an analog VOM drop that fast eh? I'm a noob here so please forgive my intrusiveness...

PC