Claimed OU circuit of Rosemary Ainslie

[poynt99]

How can this be the case if all scope probe grounds are point connected at the power supply source?

Hoppy

Are they?


.99

[MileHigh]

.99:

I think that you are really close and your comments about the load resistor acting like a delay line and/or transmission line got me thinking last night and I hope what I say here is going to advance the cause.

One thing that I was forgetting is that there is a fairly long piece of wire required to make the load resistor, whether home made or commercial.  Let's keep it simple and say that there is three meters of wire in the load resistor.

We know that a signal propagates 30 centimeters per nanosecond in free space.  But inside the load resistor the permittivity is much higher, and we are going to take a wild guess and assume that a signal inside the load resistor will travel one-fifth as fast, or 6 centimeters per nanosecond.

Therefore we are guessing that it will take (300/6) = 50 nanoseconds for a signal to cross the load resistor.  It could be more, it could be less...

Now I am going to be bold and also throw in the fact that this is an inductor, and the distributed capacitance and inductance act as a passive delay line also, further slowing down the signal by let's say... one half.   So now we are going to assume that it takes 100 nanoseconds for a signal to cross the length of the load resistor.

Now comes the fun part.  The big high voltage spike is about 100 nanoseconds wide, and it takes about 100 nanoseconds to cross the load resistor, so we are in transmission line territory.  The equivalent-time length of the conductor in the load resistor is comaparable to the time width of the big spike.   This makes all the difference in the world.

Let's simply say that relative to the MOSFT drain pin, the closer end of the load resistor is a bit "upstream" and the farther end of the load resistor is much further upstream, about 100 nanoseconds further upstream.

So, when the MOSFET shuts off, is does not get whacked right away by the voltage spike generated by the inductor at all.  We see this in the DSO captures.

Here is the key:  when the MOSFET switches off, the inductor has to discharge its stored energy, there is no two ways about it.  So, let's make a very very simple model.  Let's say that the tangible energy in the inductor has to start from somewhere, and it has to go somewhere, those are givens.

It starts upstream from the MOSFET inside the load resistor itself.  The inductive energy gets pumped into the distributed capacitance inside the body of the load resistor.  Then it travels downstream out of the load resistor at 1/10th of "c" - the speed of light in a vacuum.

Lets turn this into a very simple visualization:  At the instant of time that the MOSFET shuts off, the bulk of the spike is sitting in the load resistor itself.  The part of the load resistor closest to the MOSFET is already at about 100 volts.  However, the center of the load resistor is at about 500-600 volts, the peak of the spike.

In other words, the main peak of the voltage spike is upstream from the MOSFET and barrelling down the coiled wire transmission line at 1/10th "c" on a collision course with the MOSFET drain pin like a bat out of hell.

When the peak voltage spike hits the the MOSFET, the MOSFET has already been shut off for about 70 nanoseconds.

I want to emphasize also that this is not a "voltage spike", it is really an energy spike.  You can't forget this, the inductor that is embedded in the load resistor stores energy through moving current.  If you stop the current flow, this stored energy becomes an energy spike - high voltage across the distributed capacitance inside the load resistor.

Back to the action, when the MOSFET switches off, the load resistor generates an energy spike that starts upstream and travels downstream, and about 70 nanoseconds after the switch-off the MOSFET is being hit by the peak of the spike.

Now back to transmission lines - the spike slams into the already-switched-off MOSFET so it is modeled as an open circuit at the end of the transmission line.  We know that when a voltage spike travelling down a transmission line hits an open circuit termination it is reflected back without being inverted in polarity (that happens for a short-circuit termination).  That is part of what we see that explains the reversing current.

If we look at it in terms of the MOSFET drain-source capacitance, then of course it gets charged to a very high potential because of the energy spike that hits it at one tenth the speed of light.  This capacitance at very high voltage then contributes to the reverse current pulse as it discharges.

We do not see a propagation delay between the high voltage spike and the reverse current that we observe through the shunt resistor.  There are a few reasons for this.  In contrast to the energy spike travelling downstream through the transmission line to smash into the MOSFET, the reverse current does not necessarily travel through the load resistor transmission line.  Note the voltage of the spike goes up about 400 volts in 20 nanoseconds, for a slew rate of 20 volts per nanosecond.  That sounds pretty fast to me.

This high slew rate is probably fast enough to completely bypass the load resistor transmission line in the reverse direction through capacitive coupling - effectively giving the high voltage spike a "short circuit" around the load resistor so that the reverse current can start instantly.

When you add up the capacitive bypass of the load resistor plus the reflection from the end of the open-circuit transmission line and the drain-source capacitive discharge you get a reverse current that is proportional to the spike voltage.  I am a bit shaky here and my modelling may not be perfect.  The bottom line is we know that we have a reverse-current spike.

Note that the various timing diagrams on the 100 nanoseconds per division time scale that clearly show the big positive spike are showing you the delayed voltage spike coming out of the load resistor where the delay is around 70 nanoseconds.   To repeat the simple analogy:  The MOSFET switches off but the big energy discharge from the inductor happens upstream.  70 nanoseconds later the spike has finally traveled downstream and hits the MOSFET.

These 100 nanosecond per division timing diagrams clearly show that the MOSFET drain pin is at a very high potential and current is flowing out of the pin.  (i.e.; current is flowing out of the MOSFET drain-source capacitor)  This explains why the DSO records the MOSFET power as being negative, which indicates that power is flowing out of the MOSFET.  The DSO data shows negative MOSFET power because that's what is really happening.  However - we now know that the source of that power is a delayed energy spike that travels along the roughly three meters worth of load resistor wire at about 1/10th "c."

Again, I am probably not completely correct in every statement that I am making, but I really think that this is on the right track.  Perhaps .99 and Hoppy can add their corrections/additions/deletions if I messed up somewhere.

Assuming this mini treatise is correct, it gives us insight into the energy flow vs. time.

Finally, for the COP > 17 or COP > 1 people, we are converging on getting a good handle of what's happening, and this will help us make accurate battery or power supply power output measurements.  We have already noted for a few trials that the 3.7% 2.4 KHz waveform is not showing any special heat production in the load resistor.  Honestly, it does not look too good but the jury is not completely out yet.

Double finally, I hark back to Aaron doing one of his first rat's nest "oscillation" demos and then good old Peter Lindemann chimed in with a victory jig.  This was about three months ago.  I posted right away in response saying, "Hold your horses, Aaron would have to do a full transient analysis of the circuit and account for all signal timings on a nanosecond by nanosecond basis and account for any potentail excess energy before you can make any definitive statements."  Well, we are seemingly here, getting a lot closer to the conclusion.  How about them apples!

MileHigh

[poynt99]

Hey MH.

I think you may have nailed it . Now, IF this is indeed what is happening, how do we compensate for this to obtain accurate measurements

And just to complicate things a wee bit more (not that we need it any more complicated), model the shunt resistor itself as an inductive resistor having an inductance somewhere between 0.2uH and 1uH.  I've modeled this in SPICE and it does exactly what we're seeing (it shifts the shunt voltage negative, in a square-like fashion during the pulse), with the exception of some added ringing.

btw, my shunt resistor measured to 1uH. I'll bet Glen's is the same or even higher as he did not use the non-inductive version according to his posted part number.

And just one more...  Now add in a strong electrostatic coupling from the Drain (including all the wire leading from it) to the surrounding earth ground paths    You touch the Drain with your finger and the wave forms change quite a bit, including, and most importantly, the shunt wave form. Same goes for the load resistor itself, as of course it is going through a large fast voltage swing as well.

Cheers,
.99

[poynt99]

We have assumed that the shunt resistor is ideal, i.e. no inductance.

Well, the ones I and Glen are using, DO have significant inductance, relative to the load resistor inductance.

This could be the "correction" needed that Hoppy mentioned. I'd rather get rid of the shunt's inductance all together by buying one that is guaranteed not to have any inductance (or as little as practical). Also, I'm going to try the 0.25 Ohm shunt installed just before the load resistor on the high side. The Source shunt will be removed and replaced with a "short".

For that matter, the shunt could be inserted between the load resistor and the Drain too. If it works, I'm not going to complain.

.99

[Hoppy]

We have assumed that the shunt resistor is ideal, i.e. no inductance.

Well, the ones I and Glen are using, DO have significant inductance, relative to the load resistor inductance.

This could be the "correction" needed that Hoppy mentioned. I'd rather get rid of the shunt's inductance all together by buying one that is guaranteed not to have any inductance (or as little as practical). Also, I'm going to try the 0.25 Ohm shunt installed just before the load resistor on the high side. The Source shunt will be removed and replaced with a "short".

For that matter, the shunt could be inserted between the load resistor and the Drain too. If it works, I'm not going to complain.

.99

I too think that MH has nailed it. I favour the shunt (yes definately non inductive!)being placed between the drain and load resistor with all interconnections kept as short as possible. This would physically be the mosfet drain pin connected directly to one side of the shunt resistor, with the other side connected directly to the load resistor.

Hoppy