another small breakthrough on our NERD technology.

[Rosemary Ainslie]

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

[poynt99]

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.

[PhiChaser]

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

[Rosemary Ainslie]

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

Hi PhiChaser -
Nice to see your post.  Actually I think it's a really good proposal.  We actually tried this but the problem is that they only operate at slower frequencies.  I can't remember what was stipulated.  What it probably DOES manage is to oscillate from current from grid supplies.  So it's frequency tolerance is possibly 50 Hertz or thereby.  (Sorry.  I wrote 220H - probably thinking of the supply voltage.  I'm getting seriously old) But at the range of frequencies that we apply it just sits tight - full center - with nowhere to go.  Just can't respond quick enough.  But I agree.  They're a neat means of actually seeing that current reversal number - as a rule.  Otherwise our only proof is our scope displays.

And you're right of course.  We're trying to explain that rather drunk swing of the voltage from an oscillation that during one half of the cycle falls to 0.5v's per battery and on the other half - climbs to a little under 24 volts.  In fact.  The upswing can be more than double.  And we've got record of the down swing that falls below zero.  So the peaks at each half of each oscillation go  WAY past the battery's capacity and rating.  And to argue that amount of discharge - recharge - we'd need to find CONSIDERABLY more power than is reasonable - and from somewhere that's NOT from that battery supply.

Anyway.  Welcome to the discussion. And feel free to ask questions.  We all need to.  It's the healthiest possible way to learn anything at all.  God knows.  I've got a fair share of my own.


Kindest regards,
Rosemary
edited for emphasis
SORRY.  I deleted the first as I took the wrong download.  This one may be clearer.  There's a 48 volt supply.  And the PINK trace is the battery voltage.  Channel 2.  Note that the battery voltage is nearly 3 x's the supply.  The mean battery average should be there in the display.  Sorry I forgot to check.  Anyway.  About 48 volts or thereby

[Rosemary Ainslie]

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