Electric-Light
Senior Member
I am still having a hard time to understand this. This is quite opposite to what I thought before. As all others pointed out, output power is always less than input power. If the voltage is the same for both input and output, I would think, the output current should be less than input.
Of course, the power factor gets into the mix so "I think", it might not be always the case but...I also have seen other VFD manufacturers data sheets show input current is "WAY" higher than output current. I am totally lost here..
Black = V
blue = I
The front-end topology decides the input waveform.
For each of those waveforms you can computer the RMS values separately and the simple product of the RMS values is the VA
The shape and magnitude of current waveform changes but electronic loads like VFDs tend to not shift phase wise.
To get the wattage over the same time period integrate in multiples of line cycles, or you beak it into many pieces, multiple instantaneous V*I for each segment and average at the end. The voltage waveform stays close to the same at least for the input end. The shape o the current waveform affects the Irms value even if the power is the same.
The latter value is always less than or equal to Vrms * Irms. When you have simple coils and capacitors such as across the line motors, current is higher than what power usage in watts divided by V gives you because the current lags behind. The waveform was close to sine wave so you just had to determine the phase shift to get watts. The actual output current and voltage waveforms from ASD is not a clean current waveform that stretches and shrinks like a bench function generator.
So this has much to do with the current mystery you're seeing.
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