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VFD operated from battery 3

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xj25

Electrical
May 7, 2011
110
Hi,

I saw recently a peculiar system, this is a 3ph custom VFD + 4 induction motors in parallel. The motors are labeled Vn=600Vrms

The curious thing is that the VFD can work in an "emergency mode" supplied from a (really big) 48V battery.
They use that to allow moving the thing quite slowly to a "controlled" position even if power fails.

It puzzles me so big difference of supply voltage (600 to 48), but even so it makes the work quite well.
The VFD has no boost circuit or DC/DC, just typical PWM VSI.

I understand that a high current can be supplied by the battery to the DC bus (it uses a 200A breaker), but voltage is very low, and as far I understood, motor flux depends mainly on peak voltage (and freq.). In a vector controlled drive I understand that torque is proportional to flux and "load" current so torque should be quite low comparing the nominal voltage (48/600?).

Just want to know your opinion on the topic about how affects so big voltage change to the motor performance. Maybe there is some control strategy to overcome that?.

Regards
 
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600V/60Hz = 48V/fHz
f=48x60/600
f=4.8
(Assuming 60Hz in the 600V mode.)

So with the battery V=48V, the "creep home" frequency is V/Hz limited to 4.8Hz.

No control strategy needed. This is the same thing the drive would provide if it was fed 600V. The only difference is that the drive doesn't need to PWM at an extremely low duty cycle to provide the equivalent 48V from the 600V.

The drive will change its modulation scheme because it would no longer be needing to synthesize 48V out of 600.


Keith Cress
kcress -
 
An active front end can very likely boost the input voltage to a more useable level. They apply a small boost inherently in their operation, but I'm not sure how much boost they can achieve.
 
And as to torque available, it should be identical to nameplate rated torque of that motor.

A Non-vector vfd will keep basic v/hz ratio at this 48 vdc input, so assuming 2v drop in drive, it will make a (48-2)/1.414 or 32Vac rms output, so hz out can be 60hz*32/600 or 3.2hz. so if assume a 1775 rpm motor then it will run 1800*3.2/60 or 96 rpm if very lightly loaded. Since this is above its slip speed of 25 rpm, it will run down to as slow as 71 rpm if putting out full nameplate rated load.
 
I'd agree with Mike. It would be 3.4Hz assuming perfect devices in the VFD so somewhere around 3.2Hz is likely the practical frequency.

Your torque will be the same as long as the VFD outputs 32V RMS and 3.2Hz.

Basically, the PWM has to be adjusted to account for the lower buss voltage. Also, the drive has to switch it's control power source from the ~850VDC bus when running on line power to the 48VDC bus from the battery. I doubt there is a power supply that can handle that big a range so it's likely switched some how when the main power is lost.


Keith, you forgot to divide by sqrt(2) to convert the 48V peak from the battery to the AC RMS output voltage you would get.

 
But other than that, he hit the concept dead on. [morning]

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
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"custom VFD" That is most probaby the explanation. Most COTS VFDs cannot do that.
I checked a VFD of similar size, a Siemens Micro Master 420, and it won't work at all at low DC link voltages.

Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
 
Just for reference, most vfd designs could probably do this if they had external connection for logic control power (some like Hitachi, refu, do) and gave access to dc bus min variable so it could be set down low enough. We have run most of our 120vac, 230vac, and 460vac servo drives with as low as 12vdc battery for bus input. Drives all operated identical with no changes other than permission with bus min setting not artificially limited to some higher value.
 

The drive is vector controlled, including speed sensor. It uses a separate power supply for the control, coming from the battery.

Then following your explanation, the motor would be working about the nominal flux (I now remember that one the objectives of v/f was to approx. get a constant flux), and the torque current will be vector-controlled by the command.

Regards
 
I ran on another doubt regarding power factor and efficiency.

From thread237-193052, I understand than reactive power is normally drawn from the DC bus cap, and due to this reason PF should be quite high even with low load.

In a regular AC installation you can get some cap-banks that must be sized to supply the needed reactive power, but with the DC bus cap it just doesn´t seem to need this "fixed" sizing. I guess that the motor gets what it needs from the cap, and it shall be sized for maximum power drawn.

The doubt is, it makes any difference supplying from the low voltage battery directly to the DC bus? (the cap will store much less energy than with nominal voltage)

Regards
 
You are worrying too much about the MOTOR power factor. The VFD does not actually "correct" it at the motor side, no need to. It's just that whatever the motor presents to the VFD is supplied by the caps, so the LINE side does not see it. Bottom line, you don't need to concern yourself with the motor PF, regardless of how the VFD is powered. In your battery system, the only difference it will make is in the Amp-Hours you will get from the battery, because that must be the entire source for all of the real power (watts) demanded by the load and motor/drive inefficiencies.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)

For the best use of Eng-Tips, please click here -> faq731-376
 
The rectifier keeps the reactive power from being sourced by the AC line. As for the DC side, it depends on the connection between the batteries and the drive. If it's diodes, then the same would apply as the AC lines. If it's a direct connection via a DC contactor, then the reactive power is likely mostly sourced by the battery bank. It doesn't really matter though, sourcing reactive power won't be discharging the batteries at least not very much. A bit of power would go into the resistive battery and connection losses due to the reactive current flowing.
 
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