Graham Gunderson's Energy conference presentation Most impressive and mysterious

[TinselKoala]

TK,

I have no issue with Gunderson measuring his Pin after the H-bridge. I think we can cut him some slack here, as obviously the transformer gadget is the DUT. It just requires a particular input wave form.

If the transformer itself is measured to have a COP>1 that is all that counts, right?

If you had a DUT that required a square wave input, you would measure the output power of your FG, not its AC input, agreed?

Look at it this way. The measurement point is measuring the input to the transformer. But this same point is also the _output_ of the H-bridge. If the H-bridge is putting out "0.000 Watts" but is still able to power something downstream, like a 10 watt brake light bulb.... wouldn't you want to know what the _input power_ to this magical H-bridge is? And where its input power is going?

Maybe it is the _H-bridge_ that is the special DUT and the transformer is just another Red Herring.  (insert tongue in cheek emoticon here.)

[TinselKoala]

So why was the information that MH posted on GGs past removed from this thread?-->is this thread now being censored ?.


Brad

What's the matter? If "Open Source" now costs two million dollars (USD, that is 2.68 million AUD), and some incomplete information about it is being sold as DVDs, videos, downloads, etc., then what's surprising about information being censored?

[tinman]

edited

[poynt99]

It depends. If the FG was only providing a clock input then it might be acceptable to measure only its power output. But in this case the H-bridge is providing the power to the transformer, and as such is a critical part of the device. OK, I'll accept that you can ignore the timing input to the H-bridge, but you cannot ignore the input of power that is being switched by the bridge.

I've obviously encountered this issue before, and I've built oscillator-driver circuits that are powered by the overall power supply to the DUT, or even from the bank of batteries that the DUT is supposed to be charging (hence "output")  so that there is no question about where the power is coming from.

I say again, if a component or subsystem is necessary for the device to operate, the power supplied to those components or subsystems must be included in the "input power" of the device. For a simple square wave clock signal, obviously what is generating this signal could be a mains powered FG, or it could be replaced with an oscillator-driver that is powered from the DUT's own power supply or even its output power. But in this case the H-bridge is supplying power to the transformer over the link where a measurement of "0.000 Watts" has been quoted, yet the device keeps running, and this link cannot be broken if the device is to keep on running. Hence the H-bridge is a critical component of the DUT and its input power must be included in the COP calculation.

OK, I will allow the power to the _clock_ signal generator for the H-bridge to be ignored, but not the power that is being switched by the H-bridge, this must be included.

As far as I can tell, the special transformer (DUT) itself requires no auxiliary power source or switching input of its own. (yes I know about the active diode network on the output, but I'm not referring to that) Would you agree?

Therefore you can think of the h-bridge as the input power supply, just like the battery is in a variety of other dc-powered devices such as the Joule Thief. As such, the Pin measurement should be captured at the output of the h-bridge, as it is the power source. In other words, Gunderson is measuring at the proper point for input power.

[TinselKoala]

Obviously I disagree.

The input to the transformer is indeed switched or cycled, that is, the H-bridge is being clocked at a definite mix of frequencies. It is not simply providing a +/- voltage input to the transformer. As Smudge has determined, the H-bridge waveform is a sinusoid at 75 kHz but is being chopped or switched or gated at 50 kHz so that the "input" to the transformer is one full cycle of a near-perfect sinusoid (from an H-bridge!) followed by a dead time equal to one half-period of the sinusoid. See the attached scopeshot below.

Furthermore, what do you make of this statement:

Also, Gram said that by adding a 100 pf Silver-Mica capacitor between the Source and Drain of one of the Back End FETS would completely destroy the OU effect.

If by "back end FETs" he means the H-bridge, then it becomes very clear that the exact specifics of the H-bridge and its drive parameters are critical to the performance of the entire unit. Therefore the power supplied to the H-bridge is part of the power necessary to produce the effect and must be included in the "input power" for the COP calculation. Ditto for the FETs in the "synchronous diode".