[woodturn]

Hi,

I have a LC4896 that I am trying to convert to Linux and I think the proper voltage for the servo motor is 48. The sticker on the servo reads:

MCG:

Part no 2234-M4130
H50R10-0243

MFG code 000769 0243
Torque Cont 50.00 oz
Current, Cont 1.99 amps

As you can see there is no voltage listed....does anyone know if 48 is the proper voltage for this servo? Just wanted to check before I blow it up....

Thank you....

[awerby]

You might ask these guys - they seem to handle MCG motors: https://www.automation.com/en-us/pro...-commutated-dc

[woodturn]

Cool....thanks...I'll give them a try

[woodturn]

I just sent them an email for information....fingers crossed....

[joeavaerage]

Hi,
there is no such thing as a 'proper voltage', but rather there is a voltage that will reliably drive 1.99A through the motor at some given speed.

A motor produces a 'Back EMF', or a back voltage the faster it goes. It is characteristic of a DC motor, often called the Voltage Constant, usually so many Volts per 1000 rpm or similar.

Lets say the voltage constant of your motor is 10V per 1000rpm. Then at 2000rpm the EMF is 20V. Thus you'd need a power supply of a little over 20V, say 25V to drive some current through the motor
and produce some torque to drive the mechanical load. If you want the motor to do 3000rpm, you'd have to have a power supply of 35V to do the same thing.

The servo amplifier is what produces the voltage, and it is matched to the motor to provide the required torque.

The correct question then is not what voltage the motor is but rather what is the max safe voltage for the servo amp. It might be that the servo amp says 48V, then anything up to 48V would work.
Lets say you only have a 24V supply, but you connect that to the servo amp, the servo would work fine but not really go beyond 2000 rpm, whereas if you have a 48V supply then it would
go all the way up to 4000rpm.

The bottom line is that it is current that produces torque but is is voltage that overcomes Back EMF and therefore sets the top speed. Current=Torque, Voltage=Speed.

Craig

[woodturn]

Wow….thanks for that explanation, I believe my servo runs max speed at 3000 rpm. So if I had a variable voltage power supply I guess I could increase the volts with the motor un coupled from the lead screw, put a tachometer on it and when I reach the 3000 rpm I know what the max voltage for that servo is?

Am I thinking right?

Thanks again

[joeavaerage]

Hi,
yes that is essentially correct.

It is, in effect, measuring the voltage constant, usually written in the specs as K_v, while the current constant is K_a, in Nm per Amp.

What you are doing is establishing the voltage required to get 3000rpm at no load. That is in effect a measure of K_v. The voltage at 3000rpm divided by 3000 is the rpm per volt aka K_v.

K_a you already have, 1.99A at stall torque of 50 oz.in or 0.42Nm ergo K_a=0.21Nm/A

Craig

[woodturn]

Awesome …… thanks

Really appreciate the time for the response

[woodturn]

I found the information on the servo....the only voltage constant I found was the V/K R.P.M as expressed as a value of 19.5. If that's 19.5 per rpm per volt for a 3,000 rpm motor that works out to about 154 volts which seems way too high. Maybe the motor is capable of running that fast but techno decided to detune the motor to run it slower? There was not tap in the electronic cabinet for anything close to 154 volts. the options I have were 48,24,12 and 5 volts. I hope the attachment went through....see attached....thanks in advance.

[joeavaerage]

Hi,
you are reading the voltage constant incorrectly.

K_v=19.5 V/krpm. Read that out........19.5Volts per thousand rpm.

Max speed is 3000 pm so max Back EMF = 3 x 19.5 =58.5V


This manufacturer has used so many volts per thousand, while other manufacturers use so many rpm per volt.
There is no particular standard, and each manufacturers spec need be looked at to determine what units have been used and
what 'sense' ie volts per rpm or rpm per volt is meant.

Craig

[woodturn]

Awesome…..that makes perfect sense. Thank you so much for taking the time to explain that to me!

[joeavaerage]

Hi,
you now have the back EMF at 3000rpm. But that is not the whole picture. If you connected a 58.5V power supply to the motor, it might indeed get to 3000rpm, but it would have no torque left over
to generate any torque to drive a load.

According to the specs K_a=26.33 oz.in per Amp. Thus if you want rated torque (50 oz.in) at 3000 rpm then the current needs be:

A= 50 / 26.33
=1.9Amp

The DC rotor resistance is 5.73Ohm. Thus the voltage over the rotor will be 1.9 x 5.73 = 10.89V. This is over and above the back EMF of 58.5 V or a total of 69.4V.

This would allow the motor to do max rpm AND produce rated torque while doing so. What happens if however you want the motor to be able to deliver its full overload torque
at 3000 rpm?

The peak overload torque is 300oz.in, thus the peak current is 300 /26.33 = 11.39A. So the voltage required is 11.39 x 5.73 + 58.5 =123.78V.

I rather suspect that while that calculation is correct the motor will probably object to that sort of treatment. Lets say you di have a servo amp and supply capable of 124V. If you were to hook
direct to the supply without the amp cutting the voltage down it would try to run at 124/19.5 (x1000)= 6.35k rpm. The motor is likely to destroy itself.

Craig

[woodturn]

That makes sense.....seeing as the machine only came with a 48v tap on the transformer, I suppose I'll just run that and see what kind of performance I get. Thanks again for your insights.

[joeavaerage]

Hi,
you can draw the speed/torque diagram now.

A 48V transformer tap will result in a 67VDC supply. 67VDC sets both the potential max speed and also the potential max current and therefore torque.

Craig

[joeavaerage]

Hi,
can work out the knee position now.

At 50oz.in, the current is 50/26.33 =1.9A. Thus the torque component of the voltage is 1.9 x 5.73= 10.9V. With 67V available then 67 -10.9 =56.1V is available to meet the back EMF
and therefore the speed at which that occurs is 56.1/19.5=2.87krpm.

The other knee is the speed at which you still have all the overload torque available. 300/26.33 =11.4A V=11.4 x 5.73 = 65.28V Therefore the available voltage to meet the back EMF is 67-65.28 =1.71V
Therefore the speed at 1.7V back EMF is 1.7/19.5=0.087krpm.

Craig

[woodturn]

Very nice explanation....I will have to digest that for a little bit....in the mean time I realized I was, once again, incorrect in my statement that I had a 48 volt tap.....as I learned more about transformers I realized that my DC voltage was more like 51.7 after the ac was rectified, so my ac tap is closer to 37 vac. Here is a link to a video I made to illustrate that point.

Thank you again

https://www.youtube.com/shorts/-X_hEtjj1Q4