There are a lot of factors: - Harmonics; - Power factor; - The voltage on the output will be of a other value then the input value
If you measure with an equipment withs is measuring effective sinewave on the output of a converter you measure a wrong probably higher value because it is a Square wave not a sine wave on the output. Always measure with an oscillosope and let it calculate the true values or calculate them by youre self.
You are almost right, but there are some points to be clarified:
> If you measure with an equipment withs is measuring effective sinewave on the output of a converter you measure a wrong probably higher value because it is a Square wave not a sine wave on the output.
The output voltage of a VSD is not square wave, but pulse width modulated. You also have to consider that the motor acts as a filter, therefore the current waveform is almost sinusoidal. So, the wrong *current* measurement will probably not arouse from the output, but from the input (if it is measured by a non true rms device) which is not sinusoidal due to harmonics.
Even with a true rms device, you can still measure higher output current than the input. If you ignore the losses on the VSD, kVA input must equal to kVA output (kW values may be and will be different). Since kVA is current x voltage, if the output voltage is lower than the mains voltage (which is normal if you operate the VSD to control the motor below the rated speed), than you get an output current higher than the input current.
Another possible, but exceptional case is the regenerative mode of the motor. If the load drives the motor, output current (in opposite direction) can be higher than the input (which will be almost zero), regardless of the voltages.
> What causes the output current to measure higher than the input current on a variable frequency drive?
Besides the other views, the first thing is often forgotten (below the fieldweakening or base speed):
If you think the issue from the transferred power point of view and remember the law P=U*I (putting it simple). In ideal cases, the transferred power through the input bridge and the output bridge is the same. Now if you bear in mind the formula and compare your constant AC input voltage to surely lower output voltage, you must get a higher output current than the input current, right ? E.g. a case where a supply of 575 VAC and 10 Amps with 10 Hz output. This gives the output voltage of (10 Hz/60 Hz)*575 VAC = app. 96 VAC and this results for the output current of app. 60 Amps !
I'am not sure, if this view was what you were looking for, but it's often neglected. BR, Sami
