7-1/2HP on single phase update

I bought a 10hp Hitachi inverter, $662 from driveswarehouse.com, to power my

7-1/2 lathe spindle motor from my single phase 240V 30A circuit. I dug out my Amp clamp to get Amp readings on the input side and am using the Amp display on the drive to get the output amp readings. Just running the motor at idle 60hz showed approximately 14amps going into the motor, the clamp on Amp meter showed approx. 5.3 Amps input. Running the motor at 30hz still showed 13-14 Amps out but the input amps dropped to around 2.7A. These findings go along with what Jim Pentagrid (Cox) said about the power factor problems showing up on the output side being handled by the drive and the input drawing closer to the true power from the source. I cut metal with this setup but I didn't have my Amp clamp attached at the time. I had red hot stainless steel chips flying off at a decent rate and nothing kicked out or got warm on the drive or motor side. So far my conclusion is that VFD's are a great way to run 3 phase motors from single phase plus you get the benefit of variable speed.

Thought this info might be useful to others.

Roger N

Attaching your amp clamp on the motor leads between the drive and the motor will not be at close to the actual current. Amp clamps are designed for

60Hz sine wave and the modified PWM sine wave gives them fits. The only way to measure the actual current is to connect the the meter to the input leads of the drive. Many drives have a builtin meter that will display the motor current. I am not sure that all the low cost drives have this feature.

Yours is on a CNC machine, IIRC. I have a very similar drive and used the voltage input feature to set spindle RPM. If I reduce spindle RPM quickly, I get a spindle kick out. Instead of writing: M3 S3000 ... S500

I have to write: M3 S3000 ... M5 G04 L1 M3 S500

I'm told a large braking resistor would solve this problem.

Karl

They want several hundred for a 17 ohm 800 watt resistor that my VFD calls for. Will something like this really work??????

Karl

Consider a cheap electric cooktop element designed to give 1 kW at

120v.

E = V^2/R

R = V^2/E

R = 120*120/1000 = 14.4 ohm.

That's close enough to your ohm rating. If you find a slightly smaller power element, like 800 watt, you would have a 18 ohm element.

i

You should consider putting it in some sort of protective enclosure or getting an electric hot plate for this purpose. Dump some kinetic energy into it and it will get HOT. ;-)

I would consider buying a cheap electric cooking stove and just wiring it into the VFD. Make sure to mount it in a safe manner so that nothing is ignited by a hot element.

For my own VFD, I have not done it, but my specs call for a 80 watt,

200 ohm resistor. I have some ceramic insulated 50W, 100 ohm resistors that are gold in color and mount to heatsinks, and I will simply use two of them in series and will mount them to a small heatsink. i

When our new Haas TL-1 showed up, there was a nifty slotted metal enclosure with a stove enlement inside. Looked like exactly the right place to put a coffe pot. :)

I just measured the >>The braking resistor's I've seen are nothing more than a cooktop burner

One of each in parallel should be about right. I'll take them. Your SO won't notice, will she?

Karl

I just MIGHT have some spares stashed away in the attic storage space................... It's currently dark, dusty, and VERY hot up there. Might require suitable refreshment for me to get up the ambition to check. :)

Actually, if you could get >>I just measured the ones in my stove: the 6 inch/4 turn ones were 45 ohms,

Karl -

If you are looking for some resistors still - I might have what you want.

Let me know - and I'll check. Hate to get rid of them - but to an person in need - ok.

These are new, never stressed or burnt.

I have to verify the wattage. They are like two loves of bread on a bracket.

Mart>> The braking resistor's I've seen are nothing more than a cooktop burner

If you go stove top - they are insulated. Also - buy new at Home Depot and get a socket there. Wire socket - plug in the resistor.

Martin

Mart> >>> The braking resistor's I've seen are nothing more than a cooktop burner

I have a pile of unused heating elements from an applience called "baconizer" (a really stupid idea, which is why I have this pile of new elements from 20 years ago) - if these strike your fancy, give me a yell - get my email from my web page,

I don't have my drive connected to the CNC controls yet but that is on the TO DO list. My drive came with default accel and decel times of 10 seconds. I lowered the accel to 2 or 3 seconds, if I set the decel less than 3 seconds, it kicks out if I'm in the higher gears. I looked at the list of dynamic braking resistors available and they varied in ohms and watts, I don't think the ohms are super critical but there should be a minimum or else you could just use a shorting jumper! If you go with too high resistance, the drive will kick out but probably not as easily as it does now, if you use too low of a resistance it might limit the current to the breaking resistor not allowing it to dissipate as many Watts as it would with the correct resistance. McMaster Carr used to sell heating elements for different voltages and Wattages, might be something to look at. Or you could plug in a toaster oven, program a bunch of accels and decels to cook a pizza :-)

RogerN

Very good demonstration of the way VFDs handle non unity PF.

A bit more explanation may be interesting. The non unity PF "wattless" current that flows in and out of the inductive component of the motor load is basically the magnetising current required to maintain the design chosen magnetic flux density in the rotor to stator airgap.

In the VFD, the ratio between output voltage and frequency (the V/F law) is tailored to always operate the motor close to it's design flux density. Because this needs constant magnetising ampere turns the magnetising (wattless) current taken by the motor is little affected by motor speed. This is why Roger's change from 60Hz to 30Hz makes little change to the motor current.

However, with roughly the same torque load at half the motor speed the lower delivered HP shows up directly as reduced VFD input current.

Jim

snip

Attaching your amp clamp on the motor leads between the drive and the motor will not be at close to the actual current. Amp clamps are designed for 60Hz sine wave and the modified PWM sine wave gives them fits. The only way to measure the actual current is to connect the the meter to the input leads of the drive. Many drives have a builtin meter that will display the motor current. I am not sure that all the low cost drives have this feature.

snip

VFDs output a modified PWM sine wave VOLTAGE waveform and the peculiar shape of this waveform makes conventional meter voltage measurements unreliable.

This NOT true of the VFD output CURRENT waveform. The VFD output current is filtered by the comparatively large motor winding inductance. The result is a pretty respectable approximation to true power frequency sinusoid. This can be accurately measured with clampon ammeters so Rogers results are valid.

VFD manufacturers are pretty coy about the waveforms that their VFDs actually deliver. I have yet to find a single manufacturer that publishes realtime pictures of their output waveforms.

Single shot pictures of both VFD voltage and current waveforms are includes in the new 2nd Edition "Electic Motors". The voltage waveform picture is little short of frightening but, apart from a few wriggles, the current waveform is pretty good. It HAS to be good because the motor efficiency would suffer if there were any major deviation from the wanted power frequency sinusoidal shape.

Jim