Vfd input output current relationship

[fornhamspark]

I have noticed that the input current 3ph to a vfd drive is lower that the output, can someone please explain why this is.
I know the input is converted for the dc bus.

 

[ScottyUK]

In simple terms, the drive power input is equal to the drive power output plus the losses within the drive. These losses are usually fairly small, 5% or so depending on the size of the drive and the drive technology used.

Neglecting the losses, the input power is delivered at a certain voltage and current. The output power is delivered at a different voltage and current. The output voltage is reduced linearly as frequency drops, so at 25Hz (30Hz if you are in the US) the output voltage is half of the value it would have at rated speed. If the power input and output are assumed to be approximately equal, the output current must be double that of the input current because the output voltage is halved for this case. Similar logic applies at 25% speed, 75%, and so on.

This explanation neglects losses, power factor, harmonics and a whole lot of other things, but you hopefully get the idea.




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[fornhamspark]

to add to this, i recently installed a 160kw ABB drive, via the vfd drive control panel the current read out was say 220Amps but when measuring the input current to the drive with a clamp meter(not rms type) the measured current was considerably smaller, i cannot undestand this as i would expect a higher current input due to the losses of the drive.

 

[aolalde]

What you should evaluate is the power consumption. The drive is changing the electric parameters ( frequency, voltage and wave form) but finally is converting energy, the drive efficiency and PF are high so the current is close to a minimum but finally you will have some losses on each transformation step.

Pr =1.732*I*Vll * PF

Pr = Effective or Real Power ( watts or kW) (consumption billed by the utility)
I = Average RMS Line current (Amps)
Vll = Line to line average voltage (Volts)
PF = Circuit Power Factor

You could do measurements of real power (watts or kW) for both the driver input and the motor input, then compare the results and you will get the driver’s efficiency.

Do not be confused with the apparent power (VA) which does not consider the PF.

VA = I * V11

 

[cbarn24050]

Hi, the current you measure at the motor has a reactive component as well as the real component whereas the current supplied to the drive is allmost all real power. You can see this effect better if the motor is unloaded your drive input current will be allmost zero. You need a true rms meter to measure the drive input current accurately.