Quote from: itsu on November 09, 2013, 05:21:19 AM
I calculated the input by using the "mean" voltage and current values as can be seen on the scope at around 2:40 minute in the video.  The mean values where 21.52V and 15.7A.

but that is very wrong because Mean Voltage * Mean Current <> Mean Power.

If your current and voltage were in-phase and proportional to each other, than the error would be tolerable, but your I & V waveforms are not proportional, e.g.: your current is increasing while your voltage is decreasing.

Take a look at the traces below. 
The current increases linearly from 0 to 100A giving a mean current of 50A.
At the same time, the voltage decreases linearly from 100A to 0A giving a mean voltage of 50V.

Naively 50A * 50V would evaluate to 2500W, but that's very wrong. 
Notice that when you multiply the current and voltage at each point in time (100V*0A, 99V*1A, 98V*2A, etc...) you get the blue curve, which represents the real instantaneous power flowing in a circuit (scaled down to fit the graph).

Now, when you average the blue curve, you get an average power of 1650W, which is correct.

1650W is only 66% of 2500W,  that's a huge difference or error!
And that's just the error with two inverted sawtooth I & V waveforms.  With other shapes the error can be much worse!

 
verpies,

Quote
Now if you average the blue curve, you get an average power of 1650W, which is correct.
1650W is only 66% of 2500W,  that's a huge difference or error!

Ok, i agree in your extreme example, but in my case, only the current fluctuates, the voltage is fairly stable (21.5V).
So when taken the average current being 15.7A, against the stable 21.52V, it must be within some bounderies we can work with.

Regards Itsu

Verpie
  I will remove the ceramic resistors, and replace them with the ones that I had on there originally.

Quote from: itsu on November 09, 2013, 11:28:55 AM
but in my case, only the current fluctuates, the voltage is fairly stable (21.5V).

The input voltage instability between the two horizontal blue lines looks like ~8V to me... and that's not counting the spikes.

If we were to trust that the scope readings are not falsified by stray inductances or EMI and those narrow spikes are really 65V and 74A high, we would get a whopping instantaneous input power of 4.8kW in those spikes, so it is worth investigating what causes such wild scope readings.

Also, I already mentioned some unexplained disproportionalities in the output waveform here.

Quote from: verpies on November 09, 2013, 07:17:38 AM
but that is very wrong because Mean Voltage * Mean Current <> Mean Power.

If your current and voltage were in-phase and proportional to each other, than the error would be tolerable, but your I & V waveforms are not proportional, e.g.: your current is increasing while your voltage is decreasing.

Take a look at the traces below. 
The current increases linearly from 0 to 100A giving a mean current of 50A.
At the same time, the voltage decreases linearly from 100A to 0A giving a mean voltage of 50V.

Naively 50A * 50V would evaluate to 2500W, but that's very wrong. 
Notice that when you multiply the current and voltage at each point in time (100V*0A, 99V*1A, 98V*2A, etc...) you get the blue curve, which represents the real instantaneous power flowing in a circuit (scaled down to fit the graph).

Now, when you average the blue curve, you get an average power of 1650W, which is correct.

1650W is only 66% of 2500W,  that's a huge difference or error!
And that's just the error with two inverted sawtooth I & V waveforms.  With other shapes the error can be much worse!

Nice lesson of electricity! Love it!