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#75 Reply
Posted by
rstofer
on 08 May, 2017 17:59
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Since Torque = Kt*I, does it make sense that without friction and in the case of very small inductance similar to none, there would be a spike of large torque and then the motor produces zero torque while it keeps spinning at constant speed?
I also tried to display the torque curve by multiplying the current by Kt. As I recall, the shape looked like the spike current curve due to the constant Kt.
Absolutely! If current is zero, torque is zero (regardless of Kt) and, since we don't have any friction, keeping the armature rotating at a constant speed doesn't require any torque. Torque T, like force F, is only required if we want to accelerate something. F=ma, T=Iw' where w' is rotational acceleration and sometimes just called alpha and I is the moment of inertia.
Maintaining a constant RPM in the absence of friction doesn't require torque. It wouldn't! There is no need to accelerate the moment of inertia so w' is zero (no change in velocity).
This math thing works out pretty well!
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#76 Reply
Posted by
rstofer
on 08 May, 2017 20:16
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My DE-5000 came in so I made some tests.
First, I wanted Rs so I pushed LCR Auto until I got Rp and then pushed SER/PAL until I got Rs and measured 6.54 Ohms. This is very similar to what I got on my bench meters. Things are looking good!
Next, I wanted L at 100 Hz so I pushed LCR AUTO until I got Lp and FREQ until I got 100 Ohms. I got a reading of 6 mH and a Q of 0.58. Ls doesn't seem to be an option at this point.
Let's cross-check using Q=2*pi*f*Ls/Rs
Ls= (Q * Rs)/(2*pi*f) or 0.58*6.54/200*pi or 6x10-3 = 6 mH. It checks.
So, my little robotics motor has 6.5 Ohms series resistance and 6 mH of inductance. Remember, I measured 5 mH using the oscilloscope method. So, there is some basis for using the DE-5000 to measure the inductance of a motor.
Try Rs on your motor with the DE-5000. I still wonder about that 0.09 Ohm series resistance.
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My DE-5000 came in so I made some tests.
First, I wanted Rs so I pushed LCR Auto until I got Rp and then pushed SER/PAL until I got Rs and measured 6.54 Ohms. This is very similar to what I got on my bench meters. Things are looking good!
Next, I wanted L at 100 Hz so I pushed LCR AUTO until I got Lp and FREQ until I got 100 Ohms. I got a reading of 6 mH and a Q of 0.58. Ls doesn't seem to be an option at this point.
I followed your procedure but I had a different experience.
First, I tried to push LCR Auto until I got R
p. However, R
p never showed up. Only R
s. I have tried several times already.
R
s 1.035 Ohms. (The value is different from the one I got in my first trial.)
Then, I pushed SER/PAL. It changed to R
p. R
p 1.358 Ohms. Again, I got a different value compared with the previous trial.
Next, since I wanted L at 100 Hz as well, I pushed LCR AUTO. I got DCR 0.97 Ohms. Pushing LCR AUTO and pressed FREQ until 100 Hz was shown. This time, I got: Rs 0.97 Ohm. Pressed LCR AUTO twice got: 100 Hz, 3.6 degrees, 0.96 Ohms. Pressed LCR AUTO again, got: APO Q 0.065, Ls = 101uH.
I could not get to L
p by pressing LCR AUTO repeatedly. Each time I saw L
p, it quickly turned to L
s.
To cross check like you did, 2*pi*100*101e-6/1.035 = 0.0613 ~ 0.065 shown under Q on the display.
Ls = 0.065 * 1.035/(2*pi*100) = 1e-4 = 101uH. Checked.
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#78 Reply
Posted by
rstofer
on 08 May, 2017 22:32
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But your resistance reading is 10 times more reasonable. One Ohm is a lot more comforting than 0.09 Ohms. For one thing, the stall current would be just 12A, not 120A. That's a lot more believable.
Your motor has 1/6th the resistance but that's ok, the wire gauge seems much larger. But not 1/100th of the resistance...
Your motor has much less inductance (1/50) but, again, there is a lot less wire around the armature.
Looking at the constants presented in the two links, we really don't know what kind of motors they were testing. I can imagine that a 0.2H inductance might imply a fairly large motor.
I just measured a 1 HP 24VDC motor:
Rs=0.35 Ohm
L=180 uH
Q=0.32
f=100 Hz
It all checks.
The 0.35 Ohm versus your 1 Ohm is also a lot more reasonable. This is a very large motor so it is reasonable that your resistance measurement would be larger. I'm a little surprised at the inductance but that's what I got. I wonder what kind of motor has 0.2 Henries?
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Thanks rstofer. I measured the diameter of the wire and it is about 1mm. I gathered that at a lower voltage, the dynamic resistance of the motor is 0.123 Ohms while the EMF/Torque constants are 3.55 V-s/rad and 3.55 Nm/Amp. So, even the resistance should be the same regardless of supplied voltage and current, my 0.09 Ohms seems OK? As far as I understand, the EMF/Torque constants should be the same regardless of the applied voltage. So, even I drive the motor at a higher voltage, the Back EMF should be 3.55 V-s/rad. However, using the method we have been discussing, I got only 0.0042 V-s/rad ?!
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#80 Reply
Posted by
rstofer
on 09 May, 2017 00:56
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I hope I get this right...
I wonder about 3.55 V-s/rad. That implies that the back EMF would be 3.55V if the motor was turning 1rad/sec or 1/6 RPS or 10 RPM whereas I would think the value would be the applied voltage divided by the steady state RPM. Less friction, of course.
It is the back EMF at some RPM that balances the applies voltage.
Edited to account for seconds and minutes. Duh...
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So, which inductance and at which frequency are correct? I got different inductance values each time I measured it. However, the values are always in uH (so small that it is like none) range.
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#82 Reply
Posted by
rstofer
on 09 May, 2017 13:29
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I would tend to go with the scope method simply because the work is being done in the time domain. We are interested in the step response, not the frequency response. The thing is, these turn out to be the same at low frequency. Or, they did for my motor, give or take 15%. I would probably use the 6 mH number I got from the meter knowing that I got a slightly higher number from the scope approach. I don't think I want to use a higher frequency value simply because nothing is happening at high frequency. I didn't spend any time comparing inductance versus frequency, I picked 100 Hz from the beginning.
I have two independent methods of determining the inductance and they both agree within a few percent. At least they're not different by orders of magnitude.
As to KE, going back to your reply #20, I get a value of 0.004 V-s/rad. This makes sense because we can calculate the back EMF from the applied voltage. If we ignore friction (or any load), the applied voltage and the back EMF are identical because there is no current flowing. So, from the applied voltage and RPM, we can convert to V-s/rad.
I wonder what you would get if you just spun the motor at some high speed and measured the terminal voltage.
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#83 Reply
Posted by
rstofer
on 09 May, 2017 14:46
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Thanks. The motor you have gave you 6mH but the one I have gave me Ls = 101uH. So, I use 101uH for Inductance?
As for the resistance, should I use the calculated 0.09 Ohms from the stall test or 0.97 Ohms from the LCR meter?
From previous discussion, we found that KE = 0.004 V-s/rad and KT = KE. How come the KE is much smaller than the one I was given (i.e. 3.55 V-s/rad at lower voltage)? Shouldn't the KE constant be the same regardless of the supplied voltage?
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#85 Reply
Posted by
rstofer
on 09 May, 2017 19:48
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Thanks. The motor you have gave you 6mH but the one I have gave me Ls = 101uH. So, I use 101uH for Inductance?
As for the resistance, should I use the calculated 0.09 Ohms from the stall test or 0.97 Ohms from the LCR meter?
I don't see any reason to not use the 110 uH value. My 1 HP motor had a similar value. We either accept the readings or we don't. The fact that I have two different approaches yielding similar results makes me a little more comfortable for both resistance and inductance.
I still have a problem with the 0.09 Ohm value. Yes, your calculations appear to justify it but I just can't get to 90 milli-ohms. That's a VERY low value. Were I you, I would use the 1 Ohm value from the LCR meter. My LCR meter agrees with my bench meters so I have no reason to doubt it.
From previous discussion, we found that KE = 0.004 V-s/rad and KT = KE. How come the KE is much smaller than the one I was given (i.e. 3.55 V-s/rad at lower voltage)? Shouldn't the KE constant be the same regardless of the supplied voltage?
Think about 3.55 V-s/rad at, say, 20,000 RPM. That's about 2100 rad/s so the motor would be generating:
3.55V/(rad/sec) * 2100 (rad/sec) = 7,400 Volts - that simply isn't possible!
ETA: The back EMF and applied voltage should be about the same if the motor is running at a constant speed and unloaded. If you put 12V into the motor, the absolute maximum back EMF is 12V (with 0 friction). There's no way you could put 12V into the motor and get a back EMF of 7400V. Since the voltages need to be similar (just call them equal), you can see where 12V couldn't even get that 3.55V motor up to 4 rads/sec.
You can get somewhat close to K
E by just applying a voltage to the unloaded motor and measuring the RPM. You could just use these values. Or you could simultaneously measure current through the 1 Ohm resistance. Subtract that value from the applied voltage to get the back EMF.
Take several readings and plot a graph. It should be a flat line but it probably won't be. Particularly at very low RPMs. If you are into the math, you could apply regression analysis to get an equation from the sampled points. The more points the better. You could try linear regression if the line is nearly flat or polynomial regression if necessary. I haven't tried it but it looks like Matlab can do this.
https://www.mathworks.com/help/matlab/data_analysis/programmatic-fitting.html
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Thanks. The simulink model seems to behave similar to the real one but there is a problem when I considered adding a gearbox. In the UMich model, I added a gain factor (of 1/G where G is the gear ratio) at the speed output so that the virtual scope displays a speed similar to the real one. At the torque output, I also added a gain of G so that the torque is amplified by a factor of G. The virtual scope also displays a torque similar to the real one. However, the actual turning does not reduce by a factor of G. I am sure that the current output also does not take the gear ratio and the actual load at the motor shaft into consideration.
For example, assuming that I have a test square input voltage signal of 1Hz. Before half a cycle (positive voltage) is completed, the motor have already completed way too many turnings already. This is probably due to the fact that although I amplified the output of the torque, the speed of the shaft is not geared down. What can we do to modify the model so that the gear ratio is taken into account?
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#87 Reply
Posted by
rstofer
on 21 May, 2017 16:24
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#88 Reply
Posted by
jmelson
on 21 May, 2017 16:51
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Yes. On Motor B, all you want to know is RPM and voltage. You can then compute Volts per thousand RPM, which is one of the important motor characteristics.
The lower voltage on motor A indicates your power supply was sagging under load.
The reason motor A had a higher voltage is it was under load (driving both itself and motor B.) So, the voltage on motor A was back EMF PLUS IR drop in the windings.
Jon
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#89 Reply
Posted by
jmelson
on 21 May, 2017 16:53
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You are simply measuring the voltage that motor 2 is producing.
this has nothing to do with back EMF.
Back EMF is defined EXACTLY as the voltage produced when a motor is spun with no load. (It gets way more complicated on series-wound motors, but I assume we are talking about permanent magnet field motors, here.)
Jon
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#90 Reply
Posted by
rstofer
on 21 May, 2017 18:02
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You are simply measuring the voltage that motor 2 is producing.
this has nothing to do with back EMF.
Back EMF is defined EXACTLY as the voltage produced when a motor is spun with no load. (It gets way more complicated on series-wound motors, but I assume we are talking about permanent magnet field motors, here.)
Jon
But, with no load. If you know the 'R' and measure the running current 'I' along with the terminal voltage, you can get back EMF under the operating condition.
Nevertheless, this part of the problem has been solved and an analog model created. Now the issue is adding a gearbox to the analog model. This should involve multiplying the torque output and dividing the RPM. The question is: How to model this in an analog model like Simulink.
A side issue: Less losses in the gearbox, which we can model as some scaling factor (related to torque?) and viscous damping (which we already know how to model) the horsepower should remain constant. The product of torque and RPM (divided by 5252) gives horsepower. So, torque goes up, RPM goes down and vice versa.
We have a pretty decent analog model:
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Thanks. I just added the scales and scopes. For your convenience, I highlight the parts that I have added.
https://flic.kr/p/TMWmoTThe scope outputs look correct but it is kind of cheating because even the speed is decreased as shown in the scopes, the actual speed output is the one without gearbox. Similarly, although the torque have been amplified by the Gear Ratio G, when driving by a periodic signal, it does not oscillate with the signal but oscillate at much higher speed. I guess the output of the current also does not match the real one that has both gearbox and load mounted to the unloaded motor that we simulate.
I know about the matlab gearbox block. I think the effect is similar to what I did but adding scaling factors at the speed and torque output. Am I correct?
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#92 Reply
Posted by
rstofer
on 21 May, 2017 19:33
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The unloaded motor ('b'=0) generates no torque so there is nothing to multiply. It also draws no current because the applied voltage and back EMF are equal.
I put b=1e-3 and things get a little more realistic but we still have no load on the gearbox. I might add a bit of viscous damping to the torque output of the gearbox and feed it back to the same adder than handles 'b'. I don't know if I would leave the existing damping or just damp the output.
As it is, 'b' is the only load once we get up to speed. Of course, the inertia thing comes into play at startup but it doesn't last long.
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The unloaded motor ('b'=0) generates no torque so there is nothing to multiply. It also draws no current because the applied voltage and back EMF are equal.
I put b=1e-3 and things get a little more realistic but we still have no load on the gearbox. I might add a bit of viscous damping to the torque output of the gearbox and feed it back to the same adder than handles 'b'. I don't know if I would leave the existing damping or just damp the output.
As it is, 'b' is the only load once we get up to speed. Of course, the inertia thing comes into play at startup but it doesn't last long.
Do you mean if I connect the DC motor circuit so that the torque from the motor is used to drive an object, some kind of feedback loop is created which affect the current, speed and torque of the motor?
I recall testing it by using the torque output of the DC motor circuit to make an object to turn. The actual turning of the object happened much faster. For example, the object is supposed to make a full rotation in one second. However, in simulink, it completed the rotation in 0.004 seconds. Anybody knows what is wrong?
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#94 Reply
Posted by
rstofer
on 21 May, 2017 22:01
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Do you mean if I connect the DC motor circuit so that the torque from the motor is used to drive an object, some kind of feedback loop is created which affect the current, speed and torque of the motor?
I would think so... If I load the motor, it slows down. This reduces back EMF and increases current. The increase in current also operates on the internal resistance causing heating (about which we aren't concerned at the moment, it has to do with efficiency). About the only parameter that isn't affected, long term, is the inductance and its di/dt. This makes sense because we are talking about a constant speed.
Here is some more math:
https://www.precisionmicrodrives.com/tech-blog/2015/08/03/dc-motor-speed-voltage-and-torque-relationshipsOr a graph where RPM drops linearly with torque and current increases linearly with torque. Notice the equation a little farther down:
M
0=M
f+M
lTotal torque = Motor friction torque ('b') + load torque.
http://www.micromo.com/technical-library/dc-motor-tutorials/motor-calculations#numericalcalculationI can see another input to the adder where 'b' is applied.
I like your idea of doing units conversion. That's the one thing I don't have a good feel for in the simulation, units.
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Do you have an idea why an object driven by the torque of the DC motor model completed the rotation in 0.004 seconds in Simulink rather than 1 second in the real world?
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#96 Reply
Posted by
rstofer
on 22 May, 2017 00:47
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I have always had a problem with the value for J (we're calling it IL), the rotational inertia. The motor accelerates in near zero time because there is no inertia to overcome.
You calculated the inertia much earlier in the thread and the numbers seemed right but somehow they don't present enough of a load to slow the acceleration.
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Is there another good way to measure the moment of inertia of the rotor?
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#98 Reply
Posted by
rstofer
on 22 May, 2017 04:24
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I found a recommendation to just measure RPM versus time for a step input. The problem is, how to measure the RPM as it is changing so fast.
Maybe it can be done by connecting another motor as a generator but that will add even more inertia. So will any other physical method I can think of. But if the motor and generator have similar moments of inertia, maybe the math is pretty easy.
You still need a way to capture the voltage waveform from the generator. A scope is the way to do this. I know you're considering one so that may turn out to be good timing.
Are you confident in your measurements, units and calculation of 'J'?
BTW, your gearbox and ultimate load also contribute to rotational inertia. Not just viscous damping (drag) but another source of inertia. Probably modify the existing constant being used for IL
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Thanks. I increased the inertia in steps of 100 and tested. Even it is amplified to a 7 digit number, it is still spinning very quickly.
I have a tachometer to measure the rpm of the real motor.