motor Ke and Kt Quiz

#1

*Here are the specs for a hypothetical ideal brushed motor, whose performance is strictly linear between the free and stall points:

The specs at 12 volts are:

FreeSpeed = 400 radians/sec

FreeCurrent = 0 A

StallTorque = 0.4 Nm

StallCurrent = 20 A

Calculate Ke:

Ke = 12/400 = 0.03

Calculate Kt:

Kt = 0.4/20 = 0.02

Question:

Why are Ke and Kt not equal?

#2

First I think some people need to hear that Ke and Kt are exactly equal theoretically.

#3

Pin = VI = Ke * w * I
Pout = T
w + heat = Kt * I * w + I^2 * R

Pin=Pout
Ke * w * I = Kt * I * w + I^2 * R

Therefore Ke must be larger than Kt, as some energy is lost to heat.

#4

Kt tells you the torque per amp, how hard the motor can push vs. the current it will draw, which assumes the motor is under load, but ke gives you 1/kv, which measures rad/sec per volt, or how fast the motor will spin based on the voltage it is supplied with. There will be more losses when a motor is at high current, high torque than when it is at it’s free speed, but isn’t this an ideal motor, so kt = 1/kv?

kt is measured in Nm/A which works out to a kgm^2/(As^2).

ke is measured in V/(rad/s), which again, works out to be the same thing.

#5

Well, it depends on how you define an ideal motor. I am having trouble grasping what it would mean if a DC motor had no internal resistance and still followed the traditional equations with Kv and Kt and the rest–does this mean that it would always spin at its free speed for a given voltage, no matter what the torque or current going through it?

#6

Can I get me one of them there motors please?

I think it will also help to know
Ke = Motor Voltage constant
Kt = Motor Torque constant

In theory, Kt = Ke * (some constant)
(The value of the constant depends upon the units used).

#7

Could you (or someone else) give an example of how Ke and Kt are used? Whenever I’ve wants to do motor calculations, I’ve just assumed free speed directly varied with voltage, that current drawn=free current + (free current-stall current)(torque load on motor/stall torque), and power=voltagecurrent=Speed*torque/efficiency. Those have basically gotten me where I’ve needed to go in terms of motor calcs. Where does Ke and Kt come into play?

#8

This is how I understand it (disclaimer I am just starting to learn this stuff).

Ke is the constant that relates the back emf generated by the motor to the angular velocity, Eb=Keomega. Kt is the constant that relates the torque output of the motor to the current drawn, Tau = KtI. In an ideal motor Ke and Kt are equal.

You can use these numbers to calculate the current drawn by the motor under a given load (tau) or speed (omega). Using the earlier definition of Kt, calculating the current under any torque is simple.

I = Tau/Kt

Solving for a given speed is slightly more complicated. Say you had the following circuit:

We are assuming the motor has been running long enough such that the inductive effects of the motor windings are ~0. (This is called steady state operation)

E is the applied voltage to the circuit, Ra is the resistance of the motor, Eb is the back emf generated by the motor and Ia is the current through the circuit (what we are trying to solve for).

Using Kirchhoff’s loop rule we know that ΔV of a loop is 0. The voltage drop across the resistor is IR (from ohm’s law) and Eb = Keomega; from here we can solve for Ia.

ΔV = 0
E-IaRa-Eb = 0
E-Eb = Ia
Ra
(E-Eb)/Ra = Ia
(E-Ke*omega)/Ra = Ia

If Kt = Ke the two currents should be the same

(sorry if this is poorly written, its late)

#9

I’m not really an electrical guy, but this is my understanding.

The biggest point of ambiguity is the definition of an ideal motor. I think a more accurate way to state the problem is that the motor is “perfectly efficient”. Its nice because efficiency actually has a definition that relates to the performance of the motor, as opposed to the word “ideal” which can mean a variety of things (inductance of motor is disregarded? heat effects are disregarded? I’ve usually seen “ideal” motors modeled as resistors).

If the motor is assumed to be perfectly efficient and linear, then only one datapoint is needed to characterize the entire motor (because the power transfer would be constant, meaning that any excess power that results from a loss in velocity must be made up with an increase in torque). You provided two datapoints, two datapoints that do not describe a perfectly efficient motor. The motor described by your datapoints operates most efficiently at halfway up the speed vs. torque graph (200rad/sec, 0.2 n-m).

If I understand correctly, the quiz is basically nonsensical because conflicting information is given. The statement is basically:
“this motor is perfectly efficient and the power input is constant, when it is going fast it outputs 60 watts, when it is going slow it outputs 40 watts. why is that?”

#10

The prompt of the quiz is fine.

In this context ideal means linear torque-speed tradeoff, NOT a lack of resistance.

#11

I thought it meant something different because the linear properties of the motor were specified later, didn’t expect redundancy.

So in the situation presented Kt and Ke aren’t supposed to be equal because of motor’s internal resistance/losses?

Are there other ways that theoretical motors differ from ideal motors?

Sorry, I haven’t done a lot of work with electrical theory.

#12

*Let’s try a different question.

Here are the specs for the 2014 BAG motor:

**Spec Voltage: Vspec = 12

Spec Free Speed: Wfree = 1466 radians/sec

Spec Free Current: Ifree = 1.8 A

Spec Stall Torque: Tstall = 0.403 Nm

Spec Stall Current: Istall = 41 A**

Calculate Ke and Kt from the above data, and discuss possible reasons why these values are not equal.

#13

Do I get to answer or is that cheating?

#14

By all means Paul, please do jump in. I think there may be at least a small audience of interested students following the thread.

Adam Heard and Richard Wallace, please also feel free to post. I know you both have been holding back.

#15

Russ – I will join the discussion when I have data to share.

Paul – this motor is your design, and I want to go on record saying you did a fine job. It is a worthy successor to the venerable Globe, and about three orders of magnitude more flexible for a range of applications powering FRC mechanisms. You may have noticed I ordered a few of them yesterday – will share test data (and THEN my own answer to Russ’s question) when I can.

#16

Paul, did you design this motor for FRC? or was it a motor already in production (or not in production but available as a configurable item) that you picked out?

If you designed it for FRC, I’d love to hear more about the process.

#17

I’ll give it a try:

So from the given data we can write an equation relating current I to speed W:
I = (Istall-Ifree)(1-W/Wfree)+Ifree

We can also write an equation that relates T to W:
T = (Ts)(1-W/Wfree)

Kt = T/I, however if you divide these two equations you get Kt as a function of W, and this is because the equations have different roots. Theoretically they have the same roots as at T=0->I=0, but they do not because the equation for T only takes into account external loads on the motor and ignores the torques due to friction forces in the motor.

So the equation for T really is:
T = (Ts)(1-W/Wfree)-Tloss

Solving I=0 for W gets:
W(I=0) = Wfree(1+Ifree/(Istall-Ifree))

Then plugging that W into Tloss = (Ts)(1-W/Wfree) to solve Tloss gives:
Tloss = -(Tstall*Ifree/(Istall-Tfree))

Plugging Tloss back into T = (Ts)(1-W/Wfree)-Tloss gives us:
T = Ts(1-W/Wfree+Ifree/(Istall-Ifree))

Now since these two equations have the same root, we can divide them and get a constant:
Kt = T(W)/I(W) = -Tstall(Istall/(Ifree-Istall))/Istall

Plugging in for a bag motor yields:
Kt = 0.0102

——————

Ke is much simpler to solve for:
Ke = V/W

However you need to take into account the voltage drop through the motor’s windings at W, so we first need to solve for internal motor resistance which is:
R = Vspec/Istall

We now can rewrite Ke to include this drop:
Ke = (Vspec-I(W)*R)/W

Since we are given Wfree and Ifree we can plug those in along with R and get:
Ke = (Vspec-Ifree*Vspec/Istall)/Wfree

Plugging in for a bag motor yields:
Ke = 0.0078

As to why they are different… I would guess bad measurement, but its more likely I did the calculation wrong

#18

There is.

#19

A little bit of both. CCL, the manufacturer of the CIM motor, did all the detail design work. They did not have any motor like the BAG. In our VEXpro product discussions, before we even made one product, we all agreed that losing the Globe motor from the KoP was a big loss to FRC teams. We then finalized what we wanted from a power, torque at max power, and packaging perspective. I ran some quick numbers based on DC motor design formulas (I will get the exact book details that I prefer at another time b/c I loaned it out to someone recently) to get a rough idea of what was possible.

I then took the rough design to CCL and said, “construct this like the CIM motor, use all ball bearings, and hit these power and torque numbers”. I was fairly certain they could do it, but haven’t had to design a DC motor in a long time so left it to them.

But this definitely is a custom motor specific to the FRC application.

The MiniCIM was a similar process except we literally took a picture of a dismantled CIM and said make it only “this” long. I gave them torque and power numbers so it would match up 1:1 with a CIM and contribute the most power possible in the meat of the CIM speed profile.

#20

OK, I’ll toss this out for discussion

*