Stalling the BaneBot 775 motors

[Leav]

Hi all,

Part of our design calls for holding the BB775 motors at stall at ~20%-40% of the motors maximum stall torque, by controlling the voltage applied to the motor with the Jaguars.

reading this thread (http://www.chiefdelphi.com/forums/showthread.php?t=84014&highlight=stalling+a+motor), I understand that if properly tested, this method may work with the fisher price motors.

I would appreciate your comments and insight from experience regarding attempting this with the BB 775.

Things we think we have going in our favor:

The motors are designed to work at 18v, and we are using them at 12v, which should increase the margin for error.
by redcuing the effective voltage to the gearbox with the PWM from the Jags, it is our understanding that the effective current will be decreased as well, helping us avoid a burnout due to thermal failure.


Again, your thoughts and insight are highly appreciated.

-Leav

[fox46]

No motor is designed to be stalled for any amount of time. Any power supplied to the motor when it is not turning turns directly to heat. All of the BaneBot motors and FP motors are fitted with a cooling impeller inside the motor. Unless the motor is running at a half decent speed it lacks sufficient cooling to allow operation. You may get away with it for a little while but you WILL eventually cook your motors. I would suggest you look into a wormgear drive or ratcheting system to prevent backdriving rather than relying on stalling the motor.

[fox46]

Also, the Jaguars have a "brake mode" which shorts the inputs to the motor when it is not running. If you spin a motor shaft, it acts like a generator. By shorting the leads you essentially load this generator and it resists rotation of the shaft. This may help prevent backdriving the motor if there is a good gear reduction on it.

[Chris is me]

As an add on to this thread - if the motors are running at 5-10% of stall torque does this problem still really exist?

[fox46]

Well at 20% of stall current (130A) and 12V, you have a power consumption of ~300W. To put it in perspective, thats about the same amount of heat generated by 10 soldering irons. This heat is localized in the armature and brushes of the motor which can melt the laquer that insulates the windings and overheat the brushes. If you insist on stalling them, I would make sure you have a couple muffin fans blasting air into/out of the motor case as much as possible. But again, stalling the motors is a poor engineering practice. I would strongly suggest you modify the design to include a ratchet, one-way bearing, worm or similar gearset to limit backdriving.

[Chris is me]

The RS 775 motors do not have a stall current upwards of 500 amps...

[fox46]

?
See above ^ 130A 130A*0.2*12V = 312W

[Ether]

Well at 20% of stall current (130A) and 12V, you have a power consumption of ~300W.

The RS775 as rated for 12V operation (which is how we are using it) has a stall current of 86.7 amps and a stall torque of 112.84 ounce-inches.

If stalled at 20% of this rated torque, the voltage will be 20%*12 and the current will be 20%*86.7, so the input power will be (0.2)(12)(0.2)(86.7)=41.5 watts, not 300 watts.

Even if the OP meant 20% of the 18V stall torque, the power would be (0.2)(18)(0.2)(130) = 93.6 watts.

[fox46]

Ah, quite right - I just took the spec off BaneBots and forgot it is listed at 18V. I'm not sure I agree with you on the calculation method. With your method you would be calculating power at 4%.

Max Power = Volts * Amps = 86A * 12V = 1032W
20% of 1032W = 1032 * 0.2 = 206.4W (312W for 130A)

By your method, (86 * 0.2) * (12 * 0.2) simplifies to 0.04 * (86 *12), the 0.04 representing 4% rather than 20%

My understanding is that the Jaguars control output by pulsing the power from the battery. They don't actually control the voltage or current. They are essentially a high speed switch which opens and closes at a specific frequency. The output of which is the percieved power to the motor and is essentially the RMS power of the frequency that it is switching at. At full power the switch is closed and supplying full power from the source. At 50% the switch is only closed 50% of the time and only 50% of the power is getting through per unit time. Likewise for other power factors. This is why the old Victor 883s had that annoying whine to them. Same with a cordless drill when you run it slowly. That is the mechanical manifestation of the switch power vibrating the motor windings and conductors just like the 60Hz humm you hear in electrical transformers and panels.

Then again, my background is Mechanical - I could be wrong but this is my understanding of how they work.

[Ether]

By your method, (86 * 0.2) * (12 * 0.2) simplifies to 0.04 * (86 *12), the 0.04 representing 4% rather than 20%

If you apply 12V to an RS775 that has a locked rotor, it will draw ~87 amps. This is straight ohms-law stuff. The DC resistance of the windings is 12/87 = 0.138 ohms

If you only want 20% of that current, you need to reduce the effective voltage to 20% of 12. So (0.2)(12)(0.2)(87) is the power draw at 20% of stall torque.

My understanding is that the Jaguars control output by pulsing the power from the battery.

Correct. It's called pulse-width modulation (PWM).

They don't actually control the voltage or current.

They control the effective voltage and current by turning it off and on at 15,000Hz and varying the "on" portion of the duty cycle.

They are essentially a high speed switch which opens and closes at a specific frequency.

Correct.

At full power the switch is closed and supplying full power from the source.

Correct.

At 50% the switch is only closed 50% of the time and only 50% of the power is getting through per unit time.

Not correct. If you reduce the voltage by 50%, the current will also be reduced by 50% (for a locked rotor which acts like a DC resistance with coil inductance), and the power will be 25%.

[fox46]

I dunnooo... I know and totally agree with you on Ohm's Law and how it affects current. My thinking however, is that when the rotor is stalled, for each pulse the applied voltage per pulse would still be 12V and the current drawn would still be 87A for each finite pulse. If those pulses occurred only 20% of the time, they would result in 20% of the electrical power consumption. I have no doubt the "percieved" amperage and voltage change but if you look at the microcasm (?) of each pulse, the resistance of the Jaguar does not vary so neither would the supplied voltage/current per unit pulse. I don't mean to argue, i'm just trying to understand it from an operational aspect. If the Jaguar controlled power output like a Variac or a transformer with a rectifier - changing the actual DC voltage then I would completely agree with you but it doesn't. The jaguar doesn't actually change the voltage of the applied power, only the duty cycle it is applied.

I'll suppose I have to read up on speed controllers a bit more. Do the Jaguars have any published data that shows output power vs duty cycle?

[fox46]

Hmm, seems like there's more to this than either of us are accounting for - seems inductance and chopping frequency play a part in this too :P
I hate Inductance... and capacitance...

I still think that a 20% output command sent to a Jaguar will result in a ~20% power output for that motor and, 20% of power consumption at stall.

In any case- Stalled motor = heat buildup = bad

[Leav]

If I understand correctly, the motor does draw full power during the on period, but since there is an off period as well the effective power would be the time-wise average of these two values.

Thanks everyone for the replies. we'll be treating this with as possible, but with extreme caution.

I'm reading the 2011 manual to try and figure out if cooling fans are allowed.

[Hawiian Cadder]

what about a system where the motor is being quickly oscillated between the tow extremes, thus driving the lift down and lifting it. then it would be moving, and it would move the lift or whatever you are using up and down a few inches.

[Leav]

what about a system where the motor is being quickly oscillated between the tow extremes, thus driving the lift down and lifting it. then it would be moving, and it would move the lift or whatever you are using up and down a few inches.

The problem I see with that, is that the banebots motors require high speeds for their fans to be effective, so the lift would bounce up and down quite a bit.

Additionally, i'm speculating that a fast switch from "drive down" to "drive up" would probably load the motor heavily, maybe even more than just stalling it (since you are trying to turn it one way, but the inertia is pushing it another).

-Leav

[AdamHeard]

The power draw would be less than what was calculated before, as he said 20-40% of stall torque, which would mean the voltage is lower as well.

Assuming 20%, that is 17 Amps @ 2.4V, which works our to 40W.

Not as bad a number, but far less than we plan on drawing; Getting that down to just 10% of stall torque bumps you to 10W, a much nicer number.

[Leav]

The power draw would be less than what was calculated before, as he said 20-40% of stall torque, which would mean the voltage is lower as well.

Assuming 20%, that is 17 Amps @ 2.4V, which works our to 40W.

Not as bad a number, but far less than we plan on drawing; Getting that down to just 10% of stall torque bumps you to 10W, a much nicer number.

Sounds like a typical engineering trade off, but one I could definitely live with.

[dtengineering]

Try strapping a cooling fan to the motor. A few zap ties and one of the little fans might make a big difference... you could even try making a little duct work to blow it all in one end of the motor... but that is probably overkill.

Our nerf ball shooter in Aim High was an FP that ran about 5-10 amps. It would be on for a fair bit, but not all, of the time during a match, it would be turning... and it would be... not hot, but definitely warm, by the end of a match.

I would suggest that you try running the motor at your predicted loads on a test bench in the shop before you even mount it on the robot. Just lock the shaft and run the motor at your expected current. Actually, since this is a test... start at half your expected current. Hold it there for two minutes. It would help if you had an IR thermometer to track case temperature, but remember that the real heat will be forming in the coils and will take a while to get out to the case. This will be fairly simple to do if you have a clamp-on ammeter, but if you don't, I believe the Jag can be convinced to give you feedback on your current consumption... you may even be able to just send a "give me 5 amps" command to the Jag. The real heating situtation during a match will be quite different, of course... the motor will draw a lot of power as you lift the arm, and then, perhaps before that heat has a chance to be "blown away" by the motor's fan, you'll be stalled... so it may not be just the stall heat that you have to deal with. Of course, then you'll lower the arm and the motor may have a chance to rest and cool, but perhaps you can run it through a few cycles like that over a two minute stretch.

Then keep in mind that you may or may not have time to cool your motor between matches, particularly during practice matches. We've been fortunate to clear tech fairly quickly from time to time and get practice match after practice match after practice match. One year we had to pull out from the practice field because the drive CIMs were starting to overheat! Just think what would have happened to one of the little motors!

Jason

P.S. Another time your motors will be subjected to extended heating and use will be in the post-season while doing demos.

[AdamHeard]

Try strapping a cooling fan to the motor. A few zap ties and one of the little fans might make a big difference... you could even try making a little duct work to blow it all in one end of the motor... but that is probably overkill.

Our nerf ball shooter in Aim High was an FP that ran about 5-10 amps. It would be on for a fair bit, but not all, of the time during a match, it would be turning... and it would be... not hot, but definitely warm, by the end of a match.

I would suggest that you try running the motor at your predicted loads on a test bench in the shop before you even mount it on the robot. Just lock the shaft and run the motor at your expected current. Actually, since this is a test... start at half your expected current. Hold it there for two minutes. It would help if you had an IR thermometer to track case temperature, but remember that the real heat will be forming in the coils and will take a while to get out to the case. This will be fairly simple to do if you have a clamp-on ammeter, but if you don't, I believe the Jag can be convinced to give you feedback on your current consumption... you may even be able to just send a "give me 5 amps" command to the Jag. The real heating situtation during a match will be quite different, of course... the motor will draw a lot of power as you lift the arm, and then, perhaps before that heat has a chance to be "blown away" by the motor's fan, you'll be stalled... so it may not be just the stall heat that you have to deal with. Of course, then you'll lower the arm and the motor may have a chance to rest and cool, but perhaps you can run it through a few cycles like that over a two minute stretch.

Then keep in mind that you may or may not have time to cool your motor between matches, particularly during practice matches. We've been fortunate to clear tech fairly quickly from time to time and get practice match after practice match after practice match. One year we had to pull out from the practice field because the drive CIMs were starting to overheat! Just think what would have happened to one of the little motors!

Jason

P.S. Another time your motors will be subjected to extended heating and use will be in the post-season while doing demos.

You could do this test by locking the shaft to force stall, and setting the voltage to the proper value. To get 10% of stall, set 10% of 12V.

[Gdeaver]

We plan to provide assist to our arm. Gas shock or cylinder to take the load off the arm motor. Constant high temps on the motor can weaken the motors magnets cutting power output and increasing the amp draw.

[Al Skierkiewicz]

Fox and Leav,
I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor, full power is being applied during the pulse. i.e. 12 volts * 86 amps=1032 watts. That power is being dissipated only 20% of the time so the average power is 213.6 watts over the full period of the PWM. This does not account for other losses. That amounts to a lot of heat over a two minute match. Please remember that the current is also heating the circuit breaker. Since the breaker is a heat sensitive device, it may trip well before the motor hits the fail point. As the breaker resets fairly quickly, that adds more to equation than simply the motor and the more a breaker resets, the hotter it becomes. Also if you are using a Jaguar, you may trigger a fault condition there due to high temperature or sustained high current which shuts the Jaguar down for 3.5 seconds.
Leav, to answer your other question, fans are allowed if they follow all the other electrical rules for wire and breakers.

[Jon Stratis]

FWIW, we've done something similar in the past...

In 2008, we used a FP motor to lift the trackball on an elevator. Once at the top, we had to stall the motor in order to keep the elevator from drifting down. While it worked and we never blew that motor (after all, we only needed to stall it up there for a few seconds at a time), it did get very hot - hot enough we put a shroud around it to prevent people from getting burned by accidentally brushing it.

Contrast that with 2009, where we used a Banebots motor (I don't recall the exact one) to power a conveyor belt. In our second competition, we burned a few of them out from balls getting stuck and stalling the motor. Not one of our finest moments!

[thefro526]

I'm reading the 2011 manual to try and figure out if cooling fans are allowed.

Leav, we're planning on using a fan to cool each of our 775's as well. There are 6 Small Fans, and 2 Large fans included in the KOP This year, and since they were included in the KOP, according to <R45 - A> they are legal.

<R45> Motors specifically permitted on 2011 FRC ROBOTS include:
A. all motors, actuators, and servos listed in the 2011 KOP Checklist,


The Fans I'm talking about can be found on page 12 of the KOP Checklist Rev B.

[fox46]

I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor, full power is being applied during the pulse. i.e. 12 volts * 86 amps=1032 watts. That power is being dissipated only 20% of the time so the average power is 213.6 watts over the full period of the PWM. This does not account for other losses. That amounts to a lot of heat over a two minute match. Please remember that the current is also heating the circuit breaker. Since the breaker is a heat sensitive device, it may trip well before the motor hits the fail point. As the breaker resets fairly quickly, that adds more to equation than simply the motor and the more a breaker resets, the hotter it becomes. Also if you are using a Jaguar, you may trigger a fault condition there due to high temperature or sustained high current which shuts the Jaguar down for 3.5 seconds.
Leav, to answer your other question, fans are allowed if they follow all the other electrical rules for wire and breakers.

Thanks Al. Good call on the circuit breakers. When I was reading up on chopper style speed controls last night the inductance/resistance of the motor was where things started to get messy, differential equations began popping up and my desire to keep reading was all of a sudden non-existent lol. I don't believe the rise/fall time would have a significant effect on current consumption but its definitely a factor, especially at low power settings when the chopping frequency is high.

[Ether]

Fox and Leav,
I believe your analysis is correct. Assuming that the pulse width is sufficiently long enough to over come rise time caused by the inductance/resistance of the motor,

I think that assumption is false (for the Jag/RS775 combination we are discussing here) and therefore the cited analysis is not correct. But I have been unable to locate an inductance spec for the RS775 so I can't state that for certain.

Al, what is your basis for assuming this is true for the Jaguar/RS775 combination? Have you measured the RS775 locked-rotor inductance? If so, would you please share that information with the CD community. If not, make an educated ballpark guess what you think the inductance is, and I will run a simulation model and post the results here.

[Al Skierkiewicz]

Ether,
I am basing the statement on the experience of rise time vs inductance with any motor. Since the Jaguar is switching at 15kHz, current through any motor may not be able to rise to full stall current during a short pulse. I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

[Ether]

I am simply suggesting that an estimate of power dissipated can be made if one ignores rise time for this discussion.

If you "ignore rise time" and assume that "full power is being applied during the pulse" you will get a very wrong answer if your assumption is wrong.

I do not know what the inductance of the RS775 is, but even if it is as small as 20 micro Henries, then power consumption at 20% duty cycle (with a Jag and locked rotor) will be roughly 40 watts, not 213.6 watts.

If anyone reading this has access to the necessary test equipment and would be willing to measure locked-rotor inductance of the RS775 and post the result here, I will run a simulation and post a graphic of the current waveform (and power calculation).

[fox46]

For the purpose of this discussion I think we are getting too far into the theory behind the component. It would be a very difficult piece of hardware to work with if it only output 4% power when you tell it to give 20%. For the purpose of this discussion and our use, I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions. In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

If the jaguar input controlled the switching frequency- linearly, then I think Ether is right. After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output wheras Ether's thinking is that the input commands control the switching frequency. I'm going to stick with the former so tell Regis my final answer is ~200W!

If you know what the frequency is (someone with an oscilliscope?), we know what the percieved output voltage, current and resistance is at full power, I can take a look at the guts of the Jag and see what the capacitance is. Can we not deduce the inductance?

[Ether]

For the purpose of this discussion I think we are getting too far into the theory behind the component.

This is pretty basic AC circuit theory, and it has a profound effect (40 watts versus 200 watts) on the answer you get, so it's worth understanding.


I am quite sure that the Jaguar compensates for the exponential effects of Inductance/switching frequency/rise time and changes its pulse frequency so that the output power varies linearly with the input instructions.

Sorry, it doesn't work this way. First, the output frequency does not change at all. It is a constant 15000 Hz. Only the duty cycle changes. The duty cycle varies linearly with input command (unless, of course, you are running closed loop, but that's another story entirely).


In this case, an instruction of 20% output would result in an intuitive 20% power (W) output and thus ~200W.

It doesn't work that way. A 20% command gives a 20% duty cycle.


If the jaguar input controlled the switching frequency- linearly, then I think Ether is right.

The Jag does not control switching frequency. The frequency is held constant at 15000 Hz, which gives a pulse width of 66.7 microseconds. What the Jag changes is the "on" portion of that 66.7 usec.


After one of our team members hooked up a jaguar backwards today, I opened it up to take a peek inside. The output has a capacitor on it- I believe this would effectively "smooth" the output power and generate a physically lower voltage/amperage. With the capacitance/inductance/resistance and switching frequency you effectively have an RLC circuit. Relative to the switching frequency, output would indeed change exponentially just like Ether's calculation.

Not sure what you mean by "exponentially" here. The effective voltage changes linearly with input command, and the effective current also changes linearly. The effective power thus changes as a square function.


I believe we are just thinking of two different parameters of which we are commanding 20% thereof- My thinking is that the input commands control the power output

They do, but not linearly.

wheras Ether's thinking is that the input commands control the switching frequency.


No, it controls the duty cycle, and thus the effective voltage and current.


I'm going to stick with the former so tell Regis my final answer is ~200W!

Are you a betting man?


If you know what the frequency is

It's 15000 Hz.


Can we not deduce the inductance?

If you knew how many windings, and the core material magnetic properties, you could make a rough calculation.

[AustinSchuh]

... The output has a capacitor on it ...

http://www.luminarymicro.com/index.php?option=com_remository&func=download&id=1352&chk=04b8b655f69ec0de40623005be63d748&Itemid=591

This user manual for the black Jaguar has the schematic. I'm not sure which capacitor you were seeing, but there is no capacitor on the output of the Jaguar. There is a bootstrap capacitor in the circuit, but will not generate the effect you are describing.

[fox46]

Whatever the capacitor is in the middle of the ring of transistors is.

For the purpose of this discussion I think we are getting too far into the theory behind the component.

This is pretty basic AC circuit theory, and it has a profound effect (40 watts versus 200 watts) on the answer you get, so it's worth understanding

I agree but seems like this is much more than a basic AC circuit when we are apparently not dealing with changes in frequency but rather changes in duty cycle. The original purpose of the OP's question was to determe if it was acceptable to stall the RS775 at 20% power. My OPINION is that it is not acceptable, but by all means, go for it- you are only burning up 40W right? Last I checked, a 20% position of a joystick did not result in a 4% robot speed.

The duty cycle varies linearly with input command
Okay but we are not concerned with the duty cycle but rather the output power which according to you is not linear to the input command. I disagree with this.

Not sure what you mean by "exponentially" here. The effective voltage changes linearly with input command, and the effective current also changes linearly. The effective power thus changes as a square function.

Sorry, this is what I meant- Power is not linear to input using your method, as I stated earlier. "Exponentially" was the wrong term to describe this.

In any case, I still think the output is closer to 200W than 40W. I'll bet you the OP's RS775 :P . Let me know what your simulation shows. Until we come up with some actual Jaguar output wattages relative to control inputs lets just agree to disagree. When we get our software up and running and if I get around to it. I'll measure the percieved voltage and current going to our RS775s and plot it relative to input value in the program. Then we can see if the output power is linear or nonlinear to input command.

[Ether]

Whatever the capacitor is in the middle of the ring of transistors is.

Here's (http://www.chiefdelphi.com/forums/attachment.php?attachmentid=9802&d=1295820836) the FET bridge portion of the Black Jag. As you can see, there is no filter capacitor on the Jag's output.


I agree but seems like this is much more than a basic AC circuit

With the rotor stalled (which is what we're discussing) it's just a switched LR circuit.


The original purpose of the OP's question was to determe if it was acceptable to stall the RS775 at 20% power.

Correct.


My OPINION is that it is not acceptable, but by all means, go for it- you are only burning up 40W right?

I do not know how long the RS775 can tolerate 40W without overheating. I suggested to Leav that he run some low-power tests to determine at what point the motor starts to become uncomfortably warm to the touch.


Last I checked, a 20% position of a joystick did not result in a 4% robot speed.

I don't know where you got that from. I hope you do not think I am claiming that.


Let me know what your simulation shows.

I already ran the simulation with 20 uH and it shows 40 watts. I do not know what the actual inductance is but I suspect it is higher than that.