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Anonymous Poster

# VFD drive

11/29/2009 3:00 AM

In VFD input current and output current is different i.e.input is less than to output current.what is formula or in what ratio of both current.

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#1

### Re: VFD drive

11/29/2009 4:44 AM

IOUT = (IIN x VIN) ÷ (VOUT x E)

Where E is the efficiency.

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#3
In reply to #1

### Re: VFD drive

11/30/2009 1:00 AM

Its a funny thing VSD input and output loading but when you look at it you are correct, output voltage will vary as will the current depending on the load, I suppose the only time that the input and output current will be the simular is when the motor is at full load, and at the frequency that the motor was rated for and even then then they wont be the same because of the efficency. Power factor has been corrected so of coarse current will be different.

Also JohnDG, sorry have not got back to you on my bressbrake application, have had to go overseas a few times since we last communicated, and I havent had to much time to think about the application further. Next I am off to China for a project then Veitnam then the Philipinnes, when I am in the Philipinnes I will have more time to look at the pressbrake application.

Cheers

Joe

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#4
In reply to #3

### Re: VFD drive

12/02/2009 6:32 PM

No sweat, Joe.

Be safe,
John

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#2

### Re: VFD drive

11/30/2009 12:23 AM

There is no ratio between the input and output current of a vfd.

The reason why the input current is lower than the output current is because of the difference in the power factor to the vfd and to the motor.

The power factor in to the vfd is constant at approximately 0,97 regardless of the percentage load on the motor. The power factor however to the motor changes anything between 0,2 to 0,8, depending on the percentage load on the motor shaft.

If you use the formula P = √3 * Vl * Il * Cosø to calculate the line current Il you will find that the input current to the vfd is lower than the output current, and that the motor current change as the power factor changes. The input current remains constant with the change in load on the motor.

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