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#1
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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? |
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#2
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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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#3
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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. Last edited by Rauhul Varma : 19-05-2014 at 16:11. Reason: edited for clarity |
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#4
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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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#5
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Re: motor Ke and Kt Quiz
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?
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#6
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Re: motor Ke and Kt Quiz
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). |
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#7
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Re: motor Ke and Kt Quiz
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=voltage*current=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?
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#8
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Re: motor Ke and Kt Quiz
Quote:
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) |
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#9
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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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#10
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Re: motor Ke and Kt Quiz
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In this context ideal means linear torque-speed tradeoff, NOT a lack of resistance. |
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#11
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Re: motor Ke and Kt Quiz
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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. |
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#12
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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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#13
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Re: motor Ke and Kt Quiz
Quote:
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#14
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Re: motor Ke and Kt Quiz
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. |
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#15
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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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