# Programming An Inverter 1.5kw, 380v

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### #1 AB2005

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Posted 27 September 2007 - 03:02 AM

I am programming an inverter for an induction motor 1.5kw, 380V, 4pole for driving a roll in a wax application machine which is recently installed in our Corrugator machine. Application is simple. I have 0-10V signal and 0-50HZ output with no feed back. I haven’t a sufficient experience for programming the inverter. I have some questions which are as following;
(1) What is IR Compensation and how can we calculate this value (here I have 0-30% range)
(2) How can I calculate the value for Slip Compensation (here I have 0-10 range)
(3) In the “Control Mode” I have three options Linear V/F Control (Scalar), Quadratic V/F Control (Scalar) and Sensorless Vector. What are these three parameters and what should be the setting.

I have read about these parameters in the manual but still have some confusion. Please explain me in detail.

Thanks and regards

"Don't assume any thing, always check/ask and clear yourself".

### #2 jraef

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Posted 27 September 2007 - 04:51 AM

IR (drop) compensation is a rarely used feature that allows you to program the drive to compensate for any known voltage drop (IR losses) in the cable if there is a significantly long cable run from the VFD to the motor using marginally sized conductors. In other words, don't worry about it unless you discover you have a problem after installation.

Slip Compensation is also rarely used and is just a way to improve the motor performance if you discover that the motor has more than normal slip. Again, don't worry about it until you know you are having performance issues thjat don't seem to be resolvable by other means.

Scalar V/Hz is for simple constant torque applications, your wax applicator being a likely prime example. No need for precision control, not likely you will use it at speeds less than 6:1. Quadratic V/Hz is only for variable torque loads such as centrifugal pumps and fans where energy savings is more important that robustness. Not likely to apply to your application, don't bother. Sensorless Vector is a possibility, but maybe a bit overkill for a wax applicator unless you need 10:1 speed reduction. Still, it doesn't necessarily hurt anything if you want to try it out.
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### #3 AB2005

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Posted 28 September 2007 - 03:11 AM

Thank you Jraef.

I have programmed the inverter and run the motor but motor didn’t start. So I have increased the torque boost and found that 7% increment in torque is sufficient for turning the roll from steady state position to run. I have scaled the drive as 0-10v input and 0-50hz out put. Now I want that some one make clear my concepts;
(1) Increment in “Torque boost” means we are disturbing the V/F ratio of motor. The normal V/F ratio for 380v/50hz is 7.6v/1hz. Am I right?
(2) I have checked the speed reference verses out put frequency which was 1V=5hz and 9V=45hz, but at 10v speed reference, output frequency found only 46.5hz as I have input frequency 50hz and Max output Frequency set at 50Hz.
For finding out the problem I have scaled down the reference and set at 1V=2.5hz and found 9V=22.5hz but 10V=23hz only.
Is there any problem in inverter or I am making some mistake? (Inverter is WEG brand Brazil).

"Don't assume any thing, always check/ask and clear yourself".

### #4 marke

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Posted 28 September 2007 - 06:46 AM

Hello AB2005

When you run in a V/Hz mode, you will get a reduction in flux at the very low frequencies. This is due to the impedance of the motor being higher than ideal at the low frequencies. If the motor impedance was purely inductive, there would not be a problem, but the motor impedance is a combination of inductance and resistance.
At line frequency, the resistance is much less than the inductive reactance and so it is insignificant and does not affect the flux. A constant V/Hz results in a constant flux.
As the frequency is reduced, the inductive reactance also reduces, so the resistance becomes more significant and reduces the flux in the stator. The result is that the torque at very low frequencies is less than at the higher frequencies.
It is not uncommon to provide a voltage boost at low frequencies to increase the flux and increase the torque. This does not necesarily mean that the flux is higher than normal. If you could separate the inductive reactance and the resistance and ensure that the voltage across the inductive reactance followed the V/Hz rule, you would have constant torque right down to almost zero. - This is what a vector drive tries to achieve. If you boost the voltage excessively at low speed , and then operate continuously at low speed, you can overheat the motor due to excessive flux. Set the boost to the lowest figure that works and all should be OK.

Sounds like the input is saturating at less than 10 volts which is not good. The input would normally saturate above 10 volts.

Best regards,

### #5 jraef

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Posted 29 September 2007 - 08:36 PM

And just t finish that most excellent explanation (which cleared up a thing or two for me as well), most VFDs will automatically discontinue the torque boost at some level, usually around 10% of the programmed base motor speed. Others provide a user programmable level at which it is discontinued, but it isn't a good idea to go above 10%.
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### #6 AB2005

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Posted 01 October 2007 - 03:13 AM

QUOTE(marke @ Sep 28 2007, 11:46 AM) <{POST_SNAPBACK}>
Sounds like the input is saturating at less than 10 volts which is not good. The input would normally saturate above 10 volts.

Best regards,

Thank you Mark for your good explanation.
I have checked the speed reference and found constant 10V but out put frequency remains 46.5HZ.

"Don't assume any thing, always check/ask and clear yourself".

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