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Anonymous Poster

Does higher rated VFDs have more losses

05/30/2006 8:13 AM

Chethan writes:
It is advised in induction motors that right sizing of motor is important for lesser losses. That is due to the no load losses of the motor. But is it the same case with the VFDs (variable frequency drives)? If I have a motor of 15 HP will I be having more losses by using a VFD of 50 HP. Is there any concept of no load losses in VFD?

Need some help.

regards,
Chethan

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#1

I don't think so

05/30/2006 10:36 AM

This is an interesting question. Power losses in a VFD have two components: switching losses and conduction losses. Switching losses are incurred each time the power devices switch on or off. The switching rate in the drive is primarily a function of the voltage output, and hence the motor speed, not the power rating. At low speed the motor back-emf is smaller so the inverter needs to switch more to lower the effective output voltage and hence regulate the motor current.

Conduction losses are a function of the forward conduction drop of the semiconductor switches, which is a function of the switch technology (e.g., MOSFET or IGBT) and to some extent the switch rating - higher rated switches tend to have lower conduction losses.

So, while the exact answer depends on the specific drive being used, I would not expect the VFD losses to be worse for a lightly loaded drive than for a fully loaded drive at the same power level. In fact the losses for the larger drive may be lower, depending on the details of the switches being used.

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#2
In reply to #1

Re:I don't think so

05/31/2006 12:15 AM

I agree but a piont for consideration.
What ever the type of drive, higher power component draws more residual-power [off-load].
It may be a DC VSD, Phase-controlled drive, AC-DC-converter-VFVV-Inverter, Brush-less DC Motors or conventional motors [AC|DC] it may be

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#3
In reply to #1

Re:May be!

05/31/2006 12:28 AM

As for as my exp. goes I have always believed that with a VFD will always result in optimising the frame size of an Induction motor. Thus under the circumstances of using a larger frame size motor (50HP?) for a duty corresponding to (15HP), I feel the core loss would be higher, even with constant V/F control, by the basic fact that one needs a larger magnetising current for the larger frame size, I feel the noload losses would be higher. This is more pronounced when the voltage applied to the motor is higher and the waveform of the voltage applied to the motor is distorted (Harmonics).

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#4

does higher rated VFDs have more losses

05/31/2006 2:02 AM

The electric VF motor itself with a higher or lower load is having nearly the same efficiency ( net work done / energy absorbed ). If we look at the system: much is due to the efficiency of the frequency generator. The risk is to go loosing in the inverter what is gaigned by the motor. Generally variable frequency generators can have a better efficiency than a traditional induction motor, but they can't be low cost (& low perfomance) type.

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#5

Does Higher Rated VFD's Have More Losses

05/31/2006 8:17 AM

Losses aside, VFD's have self limiting outputs. Most have fault conditions and error codes for such, to protect the motor the VFD is running. With a 50HP VFD on a 15HP motor, you can way over drive the motor. I recently found we had a bad gearbox (new from the factory) on a mixer we had installed on a tank, built at the company I work for, that overloaded the motor driving it. I disciovered this quickly because the correctly sized VFD faulted within ten seconds every time the motor was started. I got my clamp on current meter and discovered an over load of about 3 times rated current for the motor. Thanks to the faulting of the VFD, no damage to motor or VFD, and the manufacturer replaced the gearbox. Happy ending to the story.

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#6
In reply to #5

Re:Does Higher Rated VFD's Have More Losses

05/31/2006 4:35 PM

Phys makes an excellent point. Assuming that the (oversized) VFD has an adjustable trip point, make sure to set it to trip out at the rated motor current, rather than the maximum drive current. Otherwise you'll end up with an unhappy motor.

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