The hypothetical test situation I gave would indeed indicate OU. Now, a slight adjustment to the scenario; what if the DUT input power indicates 15W, and the control power is still 10W. What would be the conclusion from these results?
.99
As no one has yet taken a stab at this, I'll explain where I was attempting to go with it.
ION, you mentioned this:
The method and schematic supplied is a general purpose design for most devices where it is desired to have an estimate of input / output power into a resistor load. DUT losses can be easily calculated separately.
e.g 10 Watts input power and 8 Watts output power = 2 Watts lost in the DUT circuitry.
Be creative, the method I supplied is a rudimentary starting point for power measurement on a budget. It may not be an exact fit for the Tseung device, but can be adapted to it.
The case presented by POYNT of 10 Watts input and 10 Watts output is unique and unrealistic. In this case the device can be assumed to have some overunity and the output power plus DUT losses / input power will equal the COP. We have not come very close to needing to test for this unique case yet. It is, however easily accomplished as I stated in an earlier post.
I understand your points, however imho there are some assumptions being made there that could be throwing the baby out with the bath water.
First and foremost, when we are vetting any "black box", we should not make any assumptions regarding whether or not the DUT is OU, nor what degree of OU it may exhibit, no matter what prior tests on similar devices may have shown. We are simply making a purely objective measurement and assessment on the device based on the outcome. From the purely scientific standpoint, we can not make any assumptions or pre-judgments on the device we are about to test with the black box method.
My example of 10W input and 10W output should be considered a possibility. Judging it as "unique" and "unrealistic" is irrelevant to the test, if we are using the scientific method and we are being objective and unbiased.
e.g 10 Watts input power and 8 Watts output power = 2 Watts lost in the DUT circuitry.
This is also an assumption, which by its nature, eliminates another possibility. In the similar example I gave at the top of this post, I asked what the conclusion of these results would be. Did most conclude that the DUT was under-unity in this case? If so, that would be an erroneous conclusion. We did not measure the DUT as a whole, therefore we can not assume that we obtained the "whole" picture regarding what the total output power is for the DUT.
Black box testing assumes
all possibilities, therefore we can not exclude the possibility of an "OU device" before we commence testing any DUT. In my example where 15W was measured as a DUT input power and 10W at the DUT resistor, we are tempted to conclude that 5W is lost in the DUT circuitry, but we have not yet captured the whole picture. It is entirely possible that there is significantly more than 5W of power being dissipated in the DUT circuitry, such as 20W, 50W, or even 100W for example. This of course would represent poor power transfer to the resistor (especially if the goal was to heat that resistor as much as possible), however that is irrelevant.
The point being this: In order to obtain a true representation of the DUT's output power and in keeping with the scientific method and objective black box testing, the
entire DUT as a whole must be measured, not just the output element itself. In the case of the device in this topic, and with reference to ION's test setup, this would involve either encasing the DUT circuitry and placing a second series DUT thermal couple on the enclosure, or placing all the DUT components inside a single enclosure, including the output resistor, and measuring this enclosure with a single thermal-couple.
And that, in essence, was the reason for my original post suggesting the requirement of a second thermal-couple on the DUT switch.
Respectfully,
.99
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