Re: Stalling the BaneBot 775 motors
Whatever the capacitor is in the middle of the ring of transistors is.
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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
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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.
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The duty cycle varies linearly with input command
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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.
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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.
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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. |
