If I decide to have two motors in a drive train, but only apply current to one, will this ruin the Victor? I do not know if applying reverse current through a victor will damage it.
it shouldnt make a difference. You always wire the voltage into the victor the same way. Then through the RC you can manipulate whether to oputput current, how much, and which polarity
Hope that was some help
No, it won’t damage the Victor. It will however back feed a voltage to your battery. That shouldn’t cause any problems either.
EDIT: It should actually back feed a voltage to your Robot controller through the PWM connection. I think you should be okay though, as long as it’s a low voltage.
When I apply current through one motor, the second motor will now spin, acting like a generator. Is this OK?
oh yeah thats fine… You will get a fault light on the Victor though. To try this first, just go push your robot around when theres no power on. Stuff should light up.
It will be ok, I’ve done this before with both Victor 883’s and 884’s and have never had any problems. The victors are pretty sturdy and will take pretty much anything you throw at them…if their fans are running haha.
Don’t be mislead by the lights…
Victors will light if there is no power applied and you push the robot thereby spinning the motors and and generating reverse currents to a device that is not turned on(this is only when the brake jumper is in coast, I think). If the Victor is powered and is receiving a 127 from the RC then the brake jumper will determine what the output connection of the Victor will do. If the jumper is in coast, the output is essentially open circuit (all FETs are OFF)) if the jumper is in brake then the the output is shorted (all FETs in the negative side of the bridge will be ON). While powered, current may only return to the battery through the Victor during the time the FETs are switched on and then only if the generated voltage from the motor exceeds the battery voltage plus the voltage drops in the wiring and the FETs.
In the case of two motors on a common output shaft, the undriven motor should be in “coast” else the load on the drivetrain will be extreme. The “brake” mode will act as an additional mechanical/electrical load on the drivetrain and speed will decrease and current consumption will increase.
I have pusehd a robot several times before (while the main breaker is turned off) and EVERYTHING lights up. But I was just wondering, where does the electricity go besides lighting up stuff and spinning fans? It can’t go back to the battery because the breaker is off. Is it burned off as heat? Just curious.
Yeah, I would imagine that most of the electricity you generate while pushing the robot is largely lost as heat, and lighting up the LED’s and such.
Since the motors acting as generators, I assume the electrons are just “shoved” in a circle around the wiring, just like the robot is being shoved across the floor. The “friction” in this wiring loop is what creates the waste heat, I believe. As far as I understand it, electricity is a lot like water flowing through a pipe. The water/electricity doesn’t really go anywhere, it just gets moved around.
The current generated by pushing can only go where there is a complete circuit. The current flows backward through the resistances of the FETs, this is what lights the LEDs and turns the fans but the current will flow through anything that is connected together through the fuse panel which is connected to the power input of the victors. There is a direct relation to the amount of current passing through circuitry and the force it takes to push the robot. Push harder and more current flows. Since heat is work and work is measured in watts, and watts equals volts times amps…then yes as long as you push and make a voltage that generates current it will develop heat. Although the heat is small, it is generated in any part of the circuit that has resistance. Namely, LEDs, wire, other motors (fans), the victors and the RC will be where the current flows through resistance and heat is generated.
It is possible to connect two motors together and while turning just one of them, the current generated will turn the other one. Since the conversion process of turning the motor to generate electricity and using the electricity to make another motor turn, is a lossy one, the first motor must turn faster than the second one. These efficiencies are rarely above 96% and if you compare the efficiency curves of the motors you can get an idea of what speeds will produce the optimum results but it can be done. What work doesn’t convert to mechanical energy at the other end is given up as heat in the two motor’s windings. If you try this at home, test how hard it is to turn a motor that is not connected to anything and then connect the two and try the same motor to see if there is a difference. Report back, inquiring minds want to know.
Lego Mindstorms/Techic motors work well to try this out
As a side
It was mentioned previously that the current generated in the victors from the motors being backdriven would also ‘leak’ through the pwm wires and back to the controler.
I was under the impression that the Victors (and Spikes) had an optical isolation between the controller and its power circutry, such that there is no posiblty of a victor melting down and somehow sending 12v+ back through the pwm wire and toasting the RC by some freak accident.
Or perhaps I’m thinking of some other RC system?
It has been a while since I opened a Victor but I have recently opened a spike and those indeed are optically coupled. I would expect no less for the victors.
Yeah, it doesn’t make technical sense that the Victor could give power through the PWM cable. However, I have observed the an RC being powered by pushing power, and if my memory is correct the 120 amp breaker was in the open position…
Alright, I just realized the flaw in my thought process. It doesn’t matter if the switch was open, the power is being distributed through the ATC panel. So yeah, you’ll push a voltage back through the distrobution panel. From there, my classes tell me it should be equally distributed to all the components connected to the ATC.