Let's Make Robots!

Electric braking

I'm working on an RC car and was thinking about the braking. The idea is: I'm driving at full speed and need to turn. Obviously, because my car is going to break the land speed limits ( ;) ), it cannot take the turn at full speed so I'll need to slow down. Now, just releasing the throttle won't be enough, so I'll have to brake extra somehow. I was thinking about giving slight reverse throttle, so that with PWM, I can handle the braking intensity. Now I don't know what this would mean current-wise. I know this will kind of short-circuit the motor (motor makes its own voltage when rotating, so when you apply a voltage in the different direction, you short-circuit it). I don't know however what currents I'd might me expecting. The motor will run at 8V, draws 1.5A cont without load, 15A when stalled.

Anyone with experience in this department?

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This is very doable and probably with what you already have. I had not done any braking until I got a hold of the Wild Thumper controller and now it seems, I can't live without this feature.

Check my math on this, but it seems that you simply need to run your PWM signal to both your driver pins at the same time instead of PWM on one pin and say, low on the other. This working or not working can depend on the layout of your motordriver (the L293 for example, is effected by the enable pin and how it is wired) but should be simple to figure out.

If you look at the schematic below -and I might be hugely mistaking here- it seems that if you use PWM on both pins, you just cut of power to the engine when they're both at 50% duty cycle's (both high and low on the same time). If they're not at he same duty cycle, this might work.

But then another problem arises: if the car is halted, I want it to go backwards (forward pin LOW and backward pin PWM) when I pull my throttle downwards. When the car is already running forward, I want the car to brake (both pins PWM, as you said) when I pull the throttle backwards. So how should the car know what to do?


Also, do you know what it would do to the current draw (because of the short-circuit I explained above) or is my assumption wrong?