I think we can assume that for input power measurements we should put the current probe left of the junction with the diode (position 3) to measure the magnetization phase current, so we can collect some data to calculate this input power. Yes? / No?
I use this diagram with the probe positions green: current probe, yellow: voltage probe, purple: my normal input power probe points:
I toke several zoomed in screenshots showing the data we need to calculate input power.
Screenshot 1 shows the timing relations of the full cycle (1500KHz) and the on-time pulse:
screenshot 2 shows the measured average voltage (yellow), current (green) and power (red) of the ON-TIME pulse of 62.5us:
Data on screenshot 1 shows we have:
a full cycle of 667us = 1500Hz
an ON-time pulse of 62.5us
a duty cycle (DTC) of 9.37%
Data on screenshot 2 shows we have In between cursors A and B:
an average pulse voltage (Vpave) of 33V,
an average pulse current (Ipave) of 1.158A (roughly half of Ipeak 2.313A)
an average pulse Power(Ppave) of 38.52W in this 62.5us pulse ON-TIME.
From this we can calculate the average power over a full cycle as:
Pfave = Ppave x DTC = 3.61W.
As a double check we can use multiple cycles (6) for measuring the average input power confirming the 3.6W input power like:
And finally we can use the earlier proposed method (here:
https://www.overunityresearch.com/index.php?topic=4312.msg99687#msg99687 ) (method 4) to calculate the stored energy in a coil using this formula: E = 1/2 x L x I²
https://physicscalc.com/physics/inductor-energy-calculator/For L= 945uH (L1), I= 2.3A we get 0.0025278J (Ws), so for 667us we get 0.0025278/0.000667=3.79W
Which is close to the 3.6W but deviates somewhat due to the 945uH being measured outside the circuit at 1KHz while we work at 1.5Khz etc.
Measuring the input power, the way I normally did, can be seen in screenshot 3:
Where we have an average input voltage of 36.1V (yellow), an average input current of 107.4mA (blue) and a calculated average input power of 3.87W (red).
This 3.87W input is some 260mW more than the 3.61W calculated earlier and can be explained by some losses like:
# the heat loss in the 24K resistor (P=U²/R = 36²/24000 =) 54mW
# Vce(on) saturation voltage loss of the IGBT (typical 3V according to the IRG4PH30K datasheet)
So I think we can conclude that this 260mW loss is marginal and that the different shapes of the voltage and current signals caused by the filtering components has no influence on the energy supplied except for some losses and certainly will not cause the COP to go from <1 to >1.
The earlier calculated efficiencies of the used circuit of around 80% still stands IMO.
The next question will be if the by me used output probing points and procedure (measuring loads Lamp1 and Lamp2) are the correct ones for measuring output power.
Any comments are very welcome!
Itsu
Author
Topic: A Melnichenko effect replication (Read 8326 times)
