Not a lifetime, but more than 10 years now
I haven't started thread here. When people learn that energy from such device be in range of milliwats, nobody want investigate it anymore...

Harold Aspden and John Strachan put together a piezo-capacitor based device that put out respectable power levels from small temperature differences. Not quite the same, but Harold Aspden did some research along the lines you are investigating. Cyril Smith also wrote an interesting paper.

But.... in Gunderson's demonstration, isn't the "output power" measured at the bulb, rather than at the output of the transformer? The transformer output, as I understand it, is being fed through the synchronous rectifier and into a large capacitor bank before it reaches the bulb.

So how's this: If you want to only include the power at the transformer input as "input"... then let's just only include the power directly at the transformer output as "output". Fair enough?  Just disconnect everything downstream of the transformer output itself and measure the output power directly at that point. This will eliminate any possible contributions to the output measurement that may come from the synchronous diode arrangement and its power supply, pre-charged capacitors, etc.

As far as i know (or can work out),is the synchronous rectifier is self triggered from the output side of the transformer--as per Itsu's schematic at OUR. Not to sure on that one though,as i lost interest
when you discovered the current inputs being bypassed on the Clarke-Hess power analyzer.

We have both built devices that show an !apparent! 0 watt input TK,but on all occasions it is only measurement error,or incorrect measurement point's.

The only way to gain an accurate yes or no to GGs OU transformer,is to build it,achieve the same result's,and then work out where he went wrong with his power measurements.


Brad

I think he is measuring "input power" as input to the transformer, that is, between the H-bridge output and the input to the transformer.
There has been some discussion on this matter. It is my opinion that the input power should be measured as input to the power supply to the H-bridge and should also include the inputs to the other necessary power supplies, but obviously not everyone agrees with me.

I do and I think it is important as well to define the actual input (start point).

If we were to take even a device that is usually powered by grid connection to a place with no grid power we would quickly realise that to run the grid powered device we would need to generate the AC input ourselves which incurs losses.

In my opinion any device should have the input considered to be from a source of stored energy such as a battery. Doing that assures us that we can get the same result from a battery bank charged by solar or wind or whatever.

However if we consider the grid a source of stored energy then the input would be from the wall socket. Much easier to determine input at 60 Hz.

Using grid power is even cheating because it costs money, the input AC is bought with money. The grid is part of the system.

Using a H bridge then the power consumed by the H bridge itself must be considered as the H bridge is part of the system. Measuring the input to the H bridge should be easier than measuring the output from it shouldn't it ?

A transformer and load itself is not a system, it requires switching or an AC input.

..

What "bench top power supply" delivers a power signal to its load like this, at 50-75 kHz?

So how far back do you go?
When we are doing power measurements on the bench,do we go all the way back to the source,and take into account heat losses at the power generation plant?.

When doing power measurements where we are using a DC input from a bench top power supply,we measure from the bench top power supplies DC output-not from the AC input to the power supply. So in other words,we circumvent all the dissipated power within that power supply,and do not take that into account.

If the power delivered from the H bridge go's only into the transformer,then we would only measure the P/in after the H bridge,as this is our bench top power supply.
If !!if!! there is power being delivered to the drive mechanism of the synchronous rectifier that bi passes the transformer,where it may add to the P/out,then this power has to be taken into account as part of the input power.

Only once a full and correct schematic is presented,will you know what has to be measured where.


Brad

Good question, The way I see it with a setup that plugs into the grid we can do a couple of things, we can consider the grid as a source of stored energy which it kinda is.

1, We can use the grid and measure the device input at the wall socket or where the non standard equipment begins such as one socket powers the function generator/whatever changes the input frequency or wave shape and the actual device input. In other words whatever is needed to run the device from the wall socket.

2, Or we can use a DC supply and measure the device input right at the DC supply as with the grid.

In other words everything that is required to run the device from a standard AC wall outlet or a standard DC supply outlet.

When I look at the efficiency of stuff I think of it that way. I don't consider the losses in a DC supply as long as I can recreate it with batteries. eg. if it's a HV DC supply then I consider the losses in the DC supply.

eg. I don't consider the losses in the solar or wind system only the output of the batteries, however if a setup is 50 Hz for
here"Aus" I don't think we need to consider the losses in the DC to AC inverter either as we would most likely have one anyway
even if living off grid. But if the device is using a 60 000 Hz input to the transformer then whatever changes the 50 Hz AC to
60 000 Hz is part of the device as it isn't for anything else and it is necessary.

We have to start somewhere.

Still it is fine and dandy to measure the efficiency of the transformer itself. But what good is that if we cannot just plug it in.

As TK  said. Where do you buy a supply with an output like that. One that can power a decent load as well. 

What "bench top power supply" delivers a power signal to its load like this, at 50-75 kHz?

Why would it matter what type of wave form is delivered by the power supply?-it's still providing power,and so it is a power supply--and we always measure our P/in at the power supply output.

Just put all the components of the Hbridge inside a black box,with only the two output terminals pocking out,and forget about everything inside the black box--just like we do with any other power supply,be it a variable DC power supply,or some fancy wave forms from your signal generator.

1 question regarding your provided wave forms.
How is there current flowing without voltage for such a long time?.


Brad

Why would it matter what type of wave form is delivered by the power supply?-it's still providing power,and so it is a power supply--and we always measure our P/in at the power supply output.

Just put all the components of the Hbridge inside a black box,with only the two output terminals pocking out,and forget about everything inside the black box--just like we do with any other power supply,be it a variable DC power supply,or some fancy wave forms from your signal generator.

Fine. Let's just substitute some other power supply then. What happens to the claim of "OU" then?

1 question regarding your provided wave forms.
How is there current flowing without voltage for such a long time?.


Brad

Exactly.  You tell me. What kind of power supply will do that?