Hello Everyone,
I have modelled a 3-phase motor drive, and now I would like to calculate its input and output power for the purpose of efficiency calculation. I am using a power sensor at the DC side in which I am averaging the power at every 0.001 sec, and using a three-phase power sensor at the inverter’s output. The issue which I am facing is that the inverter’s output power is somehow higher than input power, approximately 10 times higher. Could you please provide your insight on how do I rectify this issue, or what could the problem causing this? For three-phase power sensor, the settings are set at balanced load.
Moreover, I am also using a mechanical power sensor at the motor end to calculate mechanical power, and in order to calculate overall efficiency of the drive, i would normally divide the output power by input power. I have saved the scope data and transfered it into excel, where when I performed division between output and input, i get different efficiency values for different points which is expected, but at some points, the efficiency is more than 100% apparently. Could you kindly provide assistance on how should I approach this issue to find effciency? should I average the vlaues first?
The scope screenshot is hereby attached.

Note: A drivecycle is used as a reference that is why there is no output until 20Sec.Hello Everyone,
I have modelled a 3-phase motor drive, and now I would like to calculate its input and output power for the purpose of efficiency calculation. I am using a power sensor at the DC side in which I am averaging the power at every 0.001 sec, and using a three-phase power sensor at the inverter’s output. The issue which I am facing is that the inverter’s output power is somehow higher than input power, approximately 10 times higher. Could you please provide your insight on how do I rectify this issue, or what could the problem causing this? For three-phase power sensor, the settings are set at balanced load.
Moreover, I am also using a mechanical power sensor at the motor end to calculate mechanical power, and in order to calculate overall efficiency of the drive, i would normally divide the output power by input power. I have saved the scope data and transfered it into excel, where when I performed division between output and input, i get different efficiency values for different points which is expected, but at some points, the efficiency is more than 100% apparently. Could you kindly provide assistance on how should I approach this issue to find effciency? should I average the vlaues first?
The scope screenshot is hereby attached.

Note: A drivecycle is used as a reference that is why there is no output until 20Sec. Hello Everyone,
I have modelled a 3-phase motor drive, and now I would like to calculate its input and output power for the purpose of efficiency calculation. I am using a power sensor at the DC side in which I am averaging the power at every 0.001 sec, and using a three-phase power sensor at the inverter’s output. The issue which I am facing is that the inverter’s output power is somehow higher than input power, approximately 10 times higher. Could you please provide your insight on how do I rectify this issue, or what could the problem causing this? For three-phase power sensor, the settings are set at balanced load.
Moreover, I am also using a mechanical power sensor at the motor end to calculate mechanical power, and in order to calculate overall efficiency of the drive, i would normally divide the output power by input power. I have saved the scope data and transfered it into excel, where when I performed division between output and input, i get different efficiency values for different points which is expected, but at some points, the efficiency is more than 100% apparently. Could you kindly provide assistance on how should I approach this issue to find effciency? should I average the vlaues first?
The scope screenshot is hereby attached.

Note: A drivecycle is used as a reference that is why there is no output until 20Sec. motordrive, powersensor, efficiency MATLAB Answers — New Questions

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