another motor question

[jackorocko]

I managed to grab an industrial 1.5 HP baldor DC motor for the right price out of luck today. I am trying to understand one important thing. How voltage and current relate to the motor.

From what research I have done to my limited understanding. Voltage controls the speed of the motor. The slower the RPM the lower the voltage. Current is proportional to torque. As the motor undertakes a load at a given RPM the motor will draw more current to maintain the RPM required by the voltage. Now this is where I start losing it. Current draw can not be endless, so does the typical motor controller limit the max current to the motor to save it from drawing an excessive amount of current (IE a stall state)?

The motor rating is 90V @14.5A. Most motor controllers I see for industrial use (IE KB electronics) state a maximum 12A @ 90V output for a 1.5 motor. This fits my above theory, but I just want to make sure.

 

[Harald Kapp]

A littel simplistic explanation:
A DC motor's current is determined by the ohmic resistance of the motor windings plus the counter emf (inductance, too plays a role).
The first part is easy to understand: The windings are made from (typically) copper and the length of wire used has an ohmic resistance. You measure that with an Ohmmeter while the motor is not running.
A running motor, however, will develop a counter emf. That is, the (rotor) coils turning in the magnetic field (of the stator) act as a generator producing a voltage that acts counter the externally aplied voltage. Therefore the current is reduced because only the difference between external voltage and counter emf can force current through the winding's resistance.
When the speed of the motor is reduced (by e.g. an applied load), the counter emf is reduced, too. Therefore the difference between applied voltage an counter emf is increased which in turn leads to a higher current.
If you stall the motor, there is no counter emf and the current is limited by the winding resistance of the motor,
The current will not be infinite - while it can be excessive leading to destruction of the motor. This is where motor protection relays come into play.

 

[jackorocko]

What does that mean for the ratings on the nameplate then? Under what condition is that measured?

 

[duke37]

The Bitish standard horse, given sufficient oats, can produce 746W. Other horses, ch, PS or CV differ.

Input 90V 14.5A = 1309W
Output = 1.5HP = 1119W
Efficiency = 85%

These measurements will be made at full power and should not be exceeded except for very short transients such as at start up. Overloading the motor can damage the commutator and burn out the windings. The commutator is very susceptible if the motor is stalled.

 

[jackorocko]

Ok, so when I picked up this motor it came with a SCR speed control circuitry. I managed to wrap my head around it and wired it up for the most part and got the motor working with the controller. But, I have one left over part! I believe it is an inductor and I am just wondering on what side of the controller it goes. The AC input side or the DC output side (my guess is the DC output). If it does go on the DC output side what side of the DC does it go on, the neg or pos?

I say it is an inductor because it looks like a transformer with only two wires. What else could it be? If you need picks, let me know.

 

[duke37]

My guess is that if it has an iron core, it should be placed in series with the motor. Either lead should be OK. It could be to smooth the current to ease the stress on the commutator.

If it is air cored or ferrite cored it should be in the input leads.

 

[jackorocko]

My guess is that if it has an iron core, it should be placed in series with the motor. Either lead should be OK. It could be to smooth the current to ease the stress on the commutator.

If it is air cored or ferrite cored it should be in the input leads.

https://www.electronicspoint.com/treadmill-repair-and-pleasing-wife-t47284.html

Well well, the answer was here all the time. LOL