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motor Ke and Kt Quiz
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 Calculate Ke: Ke = 12/400 = 0.03 Calculate Kt: Kt = 0.4/20 = 0.02 Question: Why are Ke and Kt not equal? |
Re: motor Ke and Kt Quiz
First I think some people need to hear that Ke and Kt are exactly equal theoretically.
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Re: motor Ke and Kt Quiz
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Pin = V*I = 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. |
Re: motor Ke and Kt Quiz
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 kg*m^2/(A*s^2). ke is measured in V/(rad/s), which again, works out to be the same thing. |
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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). |
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Ke is the constant that relates the back emf generated by the motor to the angular velocity, Eb=Ke*omega. Kt is the constant that relates the torque output of the motor to the current drawn, Tau = Kt*I. 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 I*R (from ohm's law) and Eb = Ke*omega; from here we can solve for Ia. ΔV = 0 E-Ia*Ra-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) |
Re: motor Ke and Kt Quiz
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?" |
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In this context ideal means linear torque-speed tradeoff, NOT a lack of resistance. |
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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. |
Re: motor Ke and Kt Quiz
Let's try a different question. Here are the specs for the 2014 BAG motor: Spec Voltage: Vspec = 12 Calculate Ke and Kt from the above data, and discuss possible reasons why these values are not equal. |
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Adam Heard and Richard Wallace, please also feel free to post. I know you both have been holding back. |
Re: motor Ke and Kt Quiz
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. |
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If you designed it for FRC, I'd love to hear more about the process. |
Re: motor Ke and Kt Quiz
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 |
Re: motor Ke and Kt Quiz
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Re: motor Ke and Kt Quiz
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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. |
Re: motor Ke and Kt Quiz
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Re: motor Ke and Kt Quiz
So, for most small, high Flux DC motors there is this phenomenon called magnetic field saturation. It happens when the permanent magnet doesn't have enough field strength to drive the high amount of Flux through the moving parts of the motor. I am simplifying here to try to keep it short.
The bottom line is that at high current (aka, near stall), the magnetic field saturates producing less actual torque than ideal and this is one of the reasons for the difference between Kt and Ke. However, it is not the only difference. This can really only be found with motor testing. |
Re: motor Ke and Kt Quiz
From Ether's 3rd attachment, Ke = Kt + (other losses /Iw).
If there are "other losses" these will account for the difference. Of course, there's a lifetime of study in "other losses"... |
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"Kt equals Ke" is a theoretical result which is only approximately true for real-world motors. Firstly, the "linearity" (straight lines) assumption (notice the asterisks) used to compute Kt would not be valid when the magnetic field saturates. Second, "other_losses" such as bearing friction, windage & viscous damping, eddy currents, and hysteresis in the power equation account for the difference between Ke and Kt. |
Re: motor Ke and Kt Quiz
If Ke = Kt (which would be a perfect world scenario)
Then the max torque the motor could output would be: Tstall = Ke*Istall Which for a bag motor comes out to: Tstall =~ .32 Nm This is less than the .4 Nm in the motor specs, and given we have come to the conclusion that the real world Kt is always less than Ke, wouldn't the true stall torque of the bag motor be even less than .32 Nm? (I also don't understand why our calculated Kt is larger than Ke or is that a result of the curve actually being nonlinear?) |
Re: motor Ke and Kt Quiz
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