Chief Delphi - CIM no-load current vs applied voltage

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CIM no-load current vs applied voltage

Does anyone have actual measured data for CIM no-load current (and no-load rpm) vs applied voltage?

I know that the motor curve spreadsheet here just scales it down linearly, but I've seen other sources that indicate no-load torque (and thus no-load current) of DC motors is a function of the square of the speed, or somewhere in-between square and linear.

And yes, I realize this is a very small current and it can safely be assumed to be a linear function of applied voltage for all practical applications for FRC, but I'm interested in the nuances.

Thanks.

Re: CIM no-load current vs applied voltage

Haven't found any no-load current vs voltage data for the CIM (AKA Chiaphua AKA Atwood AKA Chalupa) after a bit of searching. However, slide 5 of this 2006 presentation from a DEKA guy gives some measured data for the 2005 KOP Fisher-Price motor. The relationship appears to be a linear one.

This is reasonable, since:

Code:

EDIT: Shamefully bad math deleted! See Eric's post below...

(Some related threads)

Re: CIM no-load current vs applied voltage

Hi Nate. Thanks for your post.

Could you clarify a couple of your statements please?

Quote:

Pmech ~ Pelec (the mechanical (input) power of the motor is approximately proportional to the electrical (input) power of the motor)

What do you mean by the phrase "the mechanical (input) power of the motor"? The only input power to the motor is electrical.

Quote:

Pmech ~ RPM^2 (the mechanical power is proportional to the square of the rotational speed of the armature)

What mechanical power are you referring to here?

For a constant load torque, the mechanical output power of the motor is proportional to the rpm, not the square of the rpm.

For a constant applied voltage and a varying load torque, the mechanical output power first increases and then decreases as the load torque is varied from 0 to stall.

Quote:

RPM ~ V (the rotational speed of the armature is proportional to the input voltage)

The speed is proportional to the applied voltage only when the load torque is zero and the no-load current is small enough to be ignored. Since my question was about not ignoring the behavior of the no-load current, using this relation would be circular reasoning.

Re: CIM no-load current vs applied voltage

Quote:

Originally Posted by Nate Laverdure (Post 985201)

Code:

 Pmech ∝ Pelec (the mechanical (input) power of the motor is approximately

                proportional to the electrical (input) power of the motor)



 Pmech ∝ RPM^2 (the mechanical power is proportional to the square of the 

                rotational speed of the armature)



 Pelec ∝ V * I (the electrical power is proportional to voltage * current)



   RPM ∝ V     (the rotational speed of the armature is proportional to the 

                input voltage)



=> V^2 ∝ V * I



=>   V ∝ I     (therefore, voltage must be proportional to current)

(Some related threads)

Nate - Thanks for the link. I worked with Kurt when I was still in the frozen manchester wastelands, and can confidently say that he is always a good source of information.

Although I agree with your conclusion, I have to disagree with the way you got there. The statements made are all true under specific circumstances, but are inappropriate in this situation without very explicit statements about the assumptions made.

Pmech(output) ∝ Pelec(input) is extremely un-true: Refer to the efficiency curve on http://www.usfirst.org/uploadedFiles...or%20Curve.pdf . For this statement to be true, efficiency would have to be (roughly) constant.

V^2 ∝ V * I is true for resistors, but is not a generally applicable equation for motors. Back EMF effects dominate these interactions, and can not be ignored.

I'd go at it with this path instead:

RPM ∝ EMF = V - I*R = (Applied voltage - free current * resistance)
I ∝ Torque
Torque = f?(RPM).
RPM ∝ V - F(RPM)

We can plug in our assumptions of what F looks like and then solve for I as a function of V.

If we assume it is a linear function, we get:

1) RPM*A = V - B*RPM
Where RPM*A = EMF and RPM*B = I. Note that B combines resistance, the torque constant of the motor AND the torque per RPM all into one mega-constant.

2) RPM= V / (A+B)
3) RPM*A = V - I*R
4) V/(A+B)*A = V- I*R
5) I = V * (B/(R*(A+B))

Or basically, that I is directly proportional to input voltage.

If we assume it is an affine function instead, we get:
1) RPM*A = V - B*RPM-C
Where C is the affine component of the torque, in mega-constant form.
2) RPM= (V-C) / (A+B)
3) [(V-C) / (A+B)]*A = V - I*R
4) V/(A+B)*A = V- I*R
5) I = V * (B/(AR)) + C/R

Or basically, that I is affinely proportional to input voltage.

These derivations look like huge wastes of time, but they are true within the confines of the assumptions made, and there is only one assumption made that is not generally considered to be a given - what is the relationship between speed and free torque.

If that relationship is known, plug in a new F and go through the same steps.

Or just measure the darn thing.

Re: CIM no-load current vs applied voltage

Aha. I see that I (very unscientifically) made up the assumptions to fit the conclusion. Time to re-take that physics course.

Re: CIM no-load current vs applied voltage

Piggybacking on Eric's last post, consider the following:

Code:



1) tau = F(NLS)  



2) emf = Ke*NLS



3) V = NLC*R + emf



4) tau = Kt*NLC

where:

tau: motor internal torque due to bearing and brush friction, windage, core losses, etc. Note: This is not the load torque (external torque)

NLS: no-load speed

Ke: constant (assumed). It converts NLS into back emf.

Kt: constant (assumed). It converts current into torque

NLC: no-load current through the motor

R: motor coil resistance

V: voltage applied across the motor terminals

F(): a to-be-determined function which converts NLS into motor internal torque

emf: the back emf generated by the motor's NLS

Taking equation (4) and substituting from equation (1),

we can get NLC as a function of NLS:

Code:



(5)  NLC = tau/Kt = F(NLS)/Kt

Taking equation (3) and substituting from equations (2) and (5),

we can get V as a function of NLS:

Code:



(6)  V = NLC*R + emf = (F(NLS)/Kt)*R + Ke*NLS

Equations (5) and (6) give the relation between motor no-load speed and no-load current and applied voltage.

The question is, what is F()? Is it a proportional function? A linear function? Or a non-linear function?

Remember that F() is the motor's internal torque (due to bearing and brush friction, windage, and core losses). It certainly isn't going to be proportional (since friction doesn't disappear at low NLS), and it will not be perfectly linear (since windage for example is not linear).

For the motors that are used in FRC, the no-load current is small enough, and the manufacturing tolerances which affect the nominal value of Kt (and Ke, which is numerically equal to Kt if SI units are used) are large enough, that the non-linear behavior of F() can safely be ignored for most purposes.

Globe has published an interesting primer on DC motors which addresses this topic:

Quote:

The no-load-torque value shown in this catalog for each
motor series includes all no load losses and can be considered
a nominal value over the speed ranges where it is anticipated
that the unit will be used. While brush and bearing friction are
relatively independent of speed, other factors such as grease
viscosity, windage, hysteresis and electrical losses will change
as exponential functions of speed. The most noticeable variation
from unit-to-unit or test-to-test will be caused by temperature
effects on grease viscosity. When more exact calculations are
required, you may assume that one-half of the no load losses
occurs at zero rpm and that these losses will follow a linear
curve from this point to the listed catalog value

(emphasis mine)

Re: CIM no-load current vs applied voltage

Quote:

Originally Posted by Ether (Post 985505)

[Globe has published an interesting primer on DC motors which addresses this topic

Check out page 5 of this presentation.

FRC 2005 KoP FP motor free current vs volts. Linear with y-intersept (similar to Globe's doc).

Re: CIM no-load current vs applied voltage

Wow, that's the first real data on an FRC motor I've seen.
However, there's a lot of detail missing:
Are the different data points different motors, or is it all the same motor?
What was the power supply? What was used to control the voltage?
Is this with the gearbox or without?
How fast did it turn?

Re: CIM no-load current vs applied voltage

Quote:

Originally Posted by kamocat (Post 986352)

Are the different data points different motors, or is it all the same motor?
At first I thought it was one motor with a lot of noise in the data, but now you have me wondering.

What was the power supply? What was used to control the voltage?
It was probably a variable DC supply.

Is this with the gearbox or without?
Did the FP come assembled with a gearbox in the 2005 KoP? If yes, then with; else without... would be my guess

How fast did it turn?
Yes, I would have liked to see that as well, to do a sanity check with the equations.

Re: CIM no-load current vs applied voltage

Do you know if a variable-voltage DC supply is usually made with a movable-core transformer, or with PWM?
If it's done correctly, it shouldn't make a difference. I was mainly worried that they were using a Victor 884 or 883. I know those have very poor control at low duty cycles.