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Topic: How To Control A 130vDC Motor With Arduino? (Read 5372 times) previous topic - next topic

SebastianZamarron

Jan 24, 2014, 10:29 pm Last Edit: Jan 25, 2014, 12:01 am by SebastianZamarron Reason: 1
Hello! I'm kinda of starter on using arduino, and i need a little advice of how can I control a Permanent Magnet Motor with arduino. The motor specifications are.

130vDC
18 Amp
6700 RPM

Thank You!

Zapro

Writing HELP PLEASE! in capitals will not get you any help.

You need to supply WAY more information if anyone will help you.

Describe Exactly what you want to do, and which parts you have at hand for the project.

// Per.

TomGeorge

Hi ,you can edit the Subject of this thread.
I suggest;
How to control 130Vdc 18A Motor with arduino.


That should improve your hits.

Hope to help.
Tom.... :) :)
Everything runs on smoke, let the smoke out, it stops running....

SebastianZamarron

Thank You for replying.

I have a little project and I'm planing to use that motor to move a conveyor belt, but I want to control it with arduino and a push button. I need to know if there's a chance of control it with arduino, but not at its entire power because maybe it will move to fast, I need that the conveyor goes kinda slow. I hope it's pretty clear. What do I need? or What do you suggest I should do?

Thank You!

Chagrin

The only practical way to do this is to find a motor controller that takes AC input and outputs ~130V DC. For an 18A controller you're going to be paying ~$150 USD. A few examples of DC motor drives. A 130V DC motor is a bit odd as most controllers on a 110V AC line output 90V DC; you should find it acceptable to use a 90V drive -- you just won't get full speed out of the motor.

Turning these on/off with an Arduino is pretty simple using a relay. Adjusting the speed is a bit trickier; the most straightforward method would be to use an RC servo to manipulate the potentiometer.

MarkT

What's the part number?   Please provide details like part numbers or the actual datasheet link, it really saves time and effort for those attempting to help.

This sounds like a seriously big servo motor - brushed or brushless though?  (Servo motor
in the industrial sense, not the hobby sense).  Just a motor? encoder as well?

Proper industrial controllers for that sort of motor are a lot of money, way beyond $150.
Alas 130V is a lot more than most budget motor drivers can handle in my experience - but
if you don't need the full 6700rpm then you can use a lower supply voltage.  For permanent
magnet DC motors (brushed or brushless) voltage is proportional to speed, current to torque.

I make that motor (from the available data) about 0.18 N-m/A, 50 rpm/V, 3 horsepower.
Assuming 18A is the continuous rated current, not the stall current.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

SebastianZamarron

The model number B4CPM-085T, I took it from a treadmill, and it had a transformer conected but it doesn't have the model or specifications, but I don't know if it still works or what it is for, maybe that can help, but I don't want to risk the motor.

jremington

Here it is for US$ 40 on ebay: http://www.ebay.com/itm/Used-Treadmill-Motor-Wind-Turbine-Permanent-Magnet-2-5HP-130VDC-PT-130730-/170989358286

Yep, the 18 amps will be the running current (that would be 3 Hp in for motor rated at 2.5 Hp out, which sounds about right). The stall current could easily be 200 amps.

This permanent magnet motor might make a nice wind turbine, as the ebay poster suggests by the title.

SebastianZamarron

Yeah that's it but I don't know if this can be useful to a conveyor belt. I'm planning to do a process of fried chicken and I don't really need a high velocity the conveyor will carry the chicken to a fryer.

Chagrin

130V / 30A sounds like you're well overboard on power. If you can move the conveyor belt by hand you're looking at around 1/3 HP.

SebastianZamarron

I think you ' re right  but what I really need it' s torque, I think Im gonna search for a 24vDc with the enough torque to move the conveyor. I think that might work better.

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