Dear Professor
Measuring power out (into a load resistor or LED or combination thereof) vs. power in misses a part of the picture. How do you account for energy that may arise or be absorbed in the form of heat in the ferrite core or other device that is part of the system. You will miss it.
You can carefully measure the temperature rise of each component and try to sum these heat gains or losses.
I chose the dual chamber thermal balance bridge method in order to account for all gains and losses using a null balance technique against a known power input and heat source in a second test chamber.
This allows real time steady state monitoring and logging of real power in and out of the DUT.
As we have said before, other types of energy loss may be difficult to account for, such as radiated RF, but the test setup can be modified with absorbers to turn the radiated RF into heat energy, such that it can be measured as part of the black box I/O.
ION -- I would like to understand your method better, but found your other thread (which was closed IIRC) to be hard to understand. Since you are here, could you explain the method more thoroughly? Is it basically a calorimetric method as it appears?
I have suggested above in this thread a straightforward calorimetric test along these lines:
1. The initial energy is provided by a capacitor charged to a measured voltage, Vcap-start.
Ecap-start = 1/2 *C* Vin**2
2. The ENTIRE device is placed inside a precision calorimeter, with just one or two thin wires to permit "turning on" the device.
I have not worked out how this is done in detail; it could simply be that a switch is turned to on with a single thin wire.
3. From runs down previously, it is know how long it takes for the cap to go from Vin to a lower value (above the cut-off point for the DUT). For example, one could go from Vin = 2.5 V to Vstop = 2 V in 10 seconds, turning off the switch when the device is inside the calorimeter at 10 seconds. (The ending voltage Vstop is measured soon after the run.)
4. Ein is simply Ecap-start minus Ecap-stop.
5. Eout from the entire system is determined by the calorimeter.
6. The efficiency is simply n = Eout / Ein.
There is a concern that the operation of the device may be affected by the shielding of the calorimeter, but I think this approach is worth a try.
Do you see any problems with this method? And -- is it similar to your method?
@exnihiloest -- I somehow missed the detail on how you measured Eout or Pout. Pin is relatively straightforward -- but could you delineate just how you determine Pout or Eout? Thanks in advance.
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Topic: Circuit sj1. Terse and Technical only. (Read 167607 times)
