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  1. #1
    Hi, I run an eShapeoko, set up for cutting balsa sheets for model boat kits. All I needed for a spindle motor was a small Johnson 12v can motor and a brass 3.2mm collet set - which worked fine.

    Then I needed to cut some aluminium, and wanted to raise the capability a bit - so I bought one of the cheap 775 motors with an E11 collet that you can find on eBay. This would cut 1/8" aluminium fine once I found the right alloy for CNC machining. But it's a bit powerful for standard light balsa use, and I thought about using dynamic braking to cut the speed rapidly when I turned off.

    At this point I have a question. Can I just short the motor leads? How much energy will this dump into the windings if I am running at full tilt (16,300rpm)? Do I need to connect through a braking resistor? And if so, what value should it have?

    If anyone out there has addressed this question before, I would be glad to hear of their experiences....

  2. #2
    I have no idea of what a 775 motor is but I wouldn't just short the leads. Anyway, do you have to stop it so often that it is a problem? In any case, the motor spinning at that rpm generates a lot of electric energy.

  3. #3
    Quote Originally Posted by A_Camera View Post
    I have no idea of what a 775 motor is but I wouldn't just short the leads. Anyway, do you have to stop it so often that it is a problem? In any case, the motor spinning at that rpm generates a lot of electric energy.
    It's not for general use, but for the emergency stop. I was not impressed by the time this larger motor took to spool down, compared to the little toy I was using earlier.

    A775 is still in the 'toy' range for real spindle use, but it's a bit larger - see https://www.ebay.co.uk/itm/393219444...SABEgIbnPD_BwE

    And so the question of what value for a braking resistor is still a valid one...

  4. #4
    Shouldn't be too difficult to calculate.... First you'll need to know the rotational inertia of the spindle; with that and the rpm you calculate the rotational energy in Joules. Then you decide how quickly you want to stop it in seconds which will give you a figure in Joules/sec i.e. watts. Then size your resistor using watts = V2/R where V is the back emf of the motor. If you're slowing it down quickly you probably don't need to have a resistor that's continuously rated at the power, more like the peak power rating.

  5. #5
    If this is a 12V motor, then the maximum voltage you'll get out of it will be.......12V or whatever voltage you were driving it with. This would allow you to calculate the peak dissipation you'd ever see, although this would rapidly fall off as the speed fell, so the nominal power rating could be a fraction of that, depending how often this would be happening. Bottom line is, does it even get warm in normal use?

    You can work out the internal temperature rise if you dumped all the kinetic energy into the thermal mass of the resistor winding - this would be the very worst case condition and the skill is not to approach the melting point of the wire. But instead of doing fancy sums, it is simpler just to select the right kind of resistor. That would be a wirewound or foil resistor (they have a decent mass of metal in them), rather than a thin film resistor.

    The motor will stop more quickly into a resistor than a short circuit. You'd want to make sure you didn't switch the short across the motor when the drive was working.

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  7. #6
    Shouldn't be too difficult to calculate.... First you'll need to know the rotational inertia of the spindle.....

    Not only do I not know that, but I'm not even sure what units it should be measured in!

    The only things I know about the motor are:

    Voltage: 12v-36V (I'm running at 36v with a PWM controller)

    12V Load Current: 1.7A (starting current 2.2A)

    Stall Current: 4.7A

    Load Power: 20.4W

    Speed: max is16,300 rev/min at 36V with 100%PWM

    Looking around, sites seem to suggest anything between about 0.5-10 ohms, and 10-50 watts. I'm not sure how critical such values are. The resistor would be wired into the crash stop switch, so the motor would either be connected to power, or the shorting resistor, with no possibility of any other linkages...

  8. #7
    Doing a quick re-arranging of some equations, I come up with the rotational energy E = 1/4 * m * r2 * w2 where m is the mass of the spindle armature, r is the radius of the spindle armature and w is the angular velocity which is about the rpm * 0.104. To work out the mass of the spindle armature it's the density (say about 5,000kg/m3) * pi * r2 * length. Alternatively I suspect there ought to be an empirical way of doing it....Measure the run-up time at full power (36V), and the current it takes at start up: now I guess slowing down should be pretty well the reverse of this. So say it takes 3 sec to get to full speed at 36V and is drawing 4 amps, a resistor of 36/4 = 9 ohms ought to slow it down in roughly 3 sec. The DC resistance of the motor will influence this, so you might want to measure that as well, but if it's considerably less than the value you come up with then it obviously won't have a huge effect.

  9. #8
    That's probably the most useful approach so far! It gives a starter value for experimentation. I suspect a bit less than 4A, so perhaps 12 ohm 100W resistor would be worth trying. Given it's very intermittent, I could probably get away with 50W...

  10. #9
    Just checked the data sheet for 50W aluminium cased WH series resistors from Welwyn and they're rated at x10 overload for < 1 sec and x5 overload for < 5 sec

  11. #10
    Dodgy Geezer,
    You mentioned that the point of wanting this braking device is for use in an emergency stop. I would also suggest you look at some kind of interlock to prevent an unwanted start of the spindle. In the most serious accident scenarios, including an unexpected start during a manual tool change, the spindle would have already caused serious injury before you ever managed to hit the ES button. Just thought I'd throw one more worm into the can!

    Kit
    An optimist says the glass is half full, a pessimist says the glass is half empty, an engineer says you're using the wrong sized glass.

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