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, 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.
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.
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.
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.
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.
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.
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%.
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?
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
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.
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.
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).
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.
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.
