Good job Brad triple checking the Pin measurements! It's good to make your measurements in more than one way to ensure one is not fooling one's self.

Along that train of thought, I went back through the thread and I couldn't find any Pin measurement you've done using the scope and the tried and true method of AVG[v(t) x i(t)]. I think you are familiar with this method, and I would encourage you to do this on one of your DUTs in question. I have confidence you will see quite different results.

The problem with the technique you have been using for Pin, is that assumptions are being made about the voltage and current used in the calculation. If the load driven by the signal generator was purely resistive, then the technique is valid. However, you are driving the primary of a transformer, and all bets are off in terms of peaks and duty cycles. Picking a Vp from the input voltage and ignoring everything else present on that wave form is not valid here. Using the tried and true AVG[v(t) x i(t)] method takes care of all the issues present with this type of scenario.

The Pin measurement for Rosemary's circuit is different in that the power supply "Input" is an assumed steady DC voltage. With that scenario, we can measure the average current with the DMM and multiply the result by the DC input voltage. So I led you astray somewhat in my last post, apologies. With your circuit one can see that the "Input Voltage" is anything but a steady DC state, and there is a reactance involved.

Have you yet purchased a decent meter that can do RMS measurements to say 100kHz? One cool thing that can be done with a true RMS meter is to determine the exact power being transferred to the secondary of your transformer. If you use a very low inductance CVR you can then determine the RMS current in the primary circuit. From this, you can now calculate the power dissipated^{1} in the primary of your transformer (in your case a primary resistance of 1.6R). Subtract that from the AVG[v(t) x i(t)] measurement, and the result is the remaining power (VA) going to the secondary. Of course you have the fancy scope that can do RMS of a trace, so buying a separate fancy true RMS meter is not necessary.

You have the scope and I would suggest you use it for these measurements, even as a double-check for those that seem like no-brainers.

^{1} This is the power dissipated (W) in the transformer primary, not the total power (W + VA) in the primary. P(W)(prim)= I(rms)^{2} x R(prim).

Hi Darren

Yes,i am a little confused here regarding this AVG[v(t) x i(t)] method you are talking about.

I thought that is what i had done all through the thread here,where the voltage during the 20% on time was averaged,and the current during the on time was averaged,and then the two multiplied to give our average P/in.

Should not the average P/in be calculated by only using the values during the 20% on time,as that is the only portion of the complete cycle the DUT is receiving energy from the SG. The rest of the waveforms through the remaining 80% of the cycle is recycled through the primary via the circuit as a whole.

If we look at the scope shot below,we see the on time where power is delivered to the transformer from the SG(between the red lines).

Between the green line's shows the inductive kickback cycle,where the voltage is inverted,but the current continues to flow in the same direction,and the last small portion before on time is C2 discharging

Now,the average voltage in that scope shot is the average voltage across both the CVR and primary of the transformer,not just the transformer it self,and that is why it cannot be used to calculate average power dissipated by the transformers primary only--the joys of having the scope and SG sharing a common ground

So,when making my calculations during the 20% on time,i subtract the voltage across the CVR from the total voltage to get my voltage across the transformer. I confirmed this value by using the circuit below,where i could measure the voltage across the transformers primary only during the 20% on time portion of the cycle.

This showed me that my calculated voltage across the transformer in previous tests was correct.

Im not quite sure what you mean by this AVG[v(t) x i(t)] ,as i thought that was what i was doing throughout the whole thread here,but you say you dont see where i have done it ?.

If you mean average V over time X average current over time,then that is what i have done throughout the thread here,where i averaged the input voltage across the coil during the 20% on time to the full cycle time,and the same for the current.

Now,if you mean using the average voltage X average current from the scope's measurements,then lets use the average values shown in the supplied scope shot.

Average voltage is 150mV,and average current is 316mV across 10 ohms for an average current of 31.6mA

Now our average P/in would be 4.74mW

Our P/out is 20v across 15kohm=26.66mW

We now have a COP of 562%

I am happy to test the DUT anyway you see fit,as all my testing so far gives COPs that should not be.

Brad

Never let your schooling get in the way of your education.