Hi Itsu,
This approach is the best to use but it needs to have a correct, two channel digital oscilloscope.
There is another approach for those who have analog or other scope types with no Math function.
If a home experimenter, using his scope can measure the peak current his coil draws at the switch off moment and knows the inductance of the coil, then he can calculate the instantaneous energy (and power) the coil just possesses.
Here is the well known formula to calculate how much energy a coil receives from an input current:
E = 1/2 * L * I²
Let's take as an example your 3 Amper peak current measured for your old L1 coil shown in your Reply #1 https://www.overunityresearch.com/index.php?topic=4312.msg99509#msg99509
Your L1 coil inductance was 726 uH (without the L2+core), switching frequency was 1.4 kHz with about 9 % duty cycle as you wrote.
Using this online calculator https://physicscalc.com/physics/inductor-energy-calculator/ for the stored energy in coils, we get E = 0.003267 Joule.
Your ON time may have been 64.2 us (based on the 9 % duty cycle, and one full cycle time was T = 1 / f = 714.2 us). So the L1 coil received 3 A current pulses at each 714.2 usec. Multiplying the stored energy, E = 0.003267 J by the 1400 Hz input frequency (or dividing it by T=714.2 us), we get 4.57 Watt, this is very close to your scope's Math calculated 4.63 W result for the input power by multiplying the instantaneous DC input supply voltage and the input current (2nd scope shot in your Reply #1 above).
So the available magnetic energy (at the switch-off moment of the input current) in your L1 coil was 0,003267 Joule. This amount of energy calculated by the formula does not include the switching and the coil (and core) losses, these should be separately considered to get the full input energy (and power) taken from a power supply. These losses are relatively low or can be kept low.
Gyula
Gyula,
thanks so much, so like we can calculate the energy in a capacitor (different formula of course), we can with your formula (E = 1/2 * L * I²) also calculate the amount of energy in a coil.
So we can add a 4th method to measure / calculate the input power:
1) using a voltage and current probe at the PS entry of the circuit.
This shows i have 36V DC @ 129.7mA rms AC for which the scope math function calculated the mean power to be 4.42W
2) using a method presented by Jagau "Urms_pulse=Vp√D" where for:
voltage my t1 is 72us and T is 770us thus D = 0.093, Vp is average 32.9V, so giving Vrms to be 10.06V rms
current my t1 is 72us and T is 770us thus D = 0.093, Ip is 2.8A, but triangled, so halved for square so is 1.4A, so giving I rms to be 427mA rms
Power then is V rms x I rms = 10.06 x 427mA = 4.29W
3) A 2th way to calculate I rms in the above 2) method is to use the triangle formula "Urms_triangle_ = Vp√t1/3T" which gives 2.8A x √72/3*770 = 2.8A x 0.1765 = 494mA rms
This multiplied by Vrms 10.06V gives 4.97W
4) Gyula's presented formula E = 1/2 * L * I² which for the data presented in method 2) gives 1/2 * 0.000951 * 2.8² = 0.00372792 Joule (using 951uH for L1 and running at 1.3Khz 10% duty cycle).
Divided by the 770us period, it shows 0.00,372792 / 770 = 4.84W.
So we got input powers: 4.42W, 4.29W, 4.97W and 4.84W by different measurement / calculations.
Each method has its own parameters, and although i tried to measure them as best as i can, none will be exactly right due to several fluctuations thus these outcomes.
But overall, if i would have to pick the most correct one i will vote for method 1.
Regards Itsu
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