I followed your REF solution, but find a problem with the CH1 and CH2 values.
Your CH1 (yellow) signal (1 Ohm csr screenshot above) shows less amplitude (20.42V) compared to your CH2 (blue)
signal (20.56V), while mine show the other way around (CH1 yellow = 16.64V, CH2 blue = 16.47V), see my 1 Ohm csr screenshot
The result CH1 - CH2 = negative (-35.39mV mean).
When continueing (using CH1 x REF), the final result is shown in the screenshot and shows a input power of 279mW (mVV).
(the green trace is the current probe)
Why is your CH1 amplitude smaller then CH2?
Are you sure the first math should also be in mean? (instead of rms).
Good luck with the shopping, perhaps your local ALDI still has some.
Thanks, itsu
Itsu, That is a good question and I was going to test for this but went shopping. We went to Aldi's and about a half dozen other stores and found no Cree lamps but I see they are available on Amazon. Anyway, the difference is in the channel gain variations. I've attached scope pix taken one after the other to show what I mean. The first scope screen is the same as previous where CH1 is the SG side of the csr and CH2 is the output side of the csr. The second shot is with the probe connections reversed along with the Math. The reason CH2 is larger in voltage than CH1 in my previous post could be due to one or both of the following reasons. First, the gain of CH2 is slightly greater than CH1 and second, the input truly is negative raising the output side of the csr above the input side. In this case, it is a combination of both. Viewing the first scope shot of the CH1-CH2 input, we see the p-p of CH1 at 20.45v and the p-p of CH2 at 20.64 for a difference of .19v p-p. Viewing the second scope shot with the probes reversed or CH2-CH1, we see the p-p voltages are equal. From this we have determined that CH2 has a slightly larger gain than CH1. To determine what the voltage differential is, we can apply dv = (d(CH1-CH2)-d(CH2-CH1))/2 = (.19-0)/2 = .095v p-p. Now, we can use dv to recal our CH2 while assuming CH1 is accurate. So, with CH1-CH2 we now have CH1 = 20.45 and CH2 = 20.64-.095 = 20.545 all p-p. Likewise with CH2-CH1 we have 20.26-.095 = 20.165 and CH1 = 20.26 again all p-p. Now we can compare the rms differentials in these voltages to see how they would be used in the final power calcs. Therefore, CH1rms-CH2rms = (20.45*0.5*0.707)-(20.545*0.5*0.707) = 7.23-7.263 = -0.033v rms and CH2rms-CH1rms = (20.165*0.5*0.707)-(20.26*0.5*0.707) = 7.128-7.162 = -0.034 rms. The slight difference is from rounding but is sufficiently accurate. So, we see that now we have a higher output rms voltage on the csr than input which means we truly are receiving energy somehow from the LED array and the input energy is negative. I think it would also be correct to say that we could average the negative input powers from the scope measurements to arrive at a corrected input power. So Pin' = (0.4046+0.05055)/2 = -0.228w corrected mean input power. Anyone is welcome to correct my figures or logic above so we all can learn. Regards, Pm
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