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Hi people -

I'd like to bounce this off some fellow engineers/gearheads...

I've built a turntable system to add to my Shapeoko 3-Axis CNC machine. The idea is to turn the machine into a CNC lathe of sorts, turned on its side. The motor for my turntable is a 100 RPM (max) brushed DC gearmotor, rated at 60W, and having a 12 - 13.5 VDC input. In that I need the speed to be variable, I have also purchased a speed controller. I intend to drive the system with a 12 Amp Linear 270W power supply with a regulated 13.8 Volt DC output. The seller from whom I bought both the motor and speed controller (MakerMotor.com) advised me to add a 12V battery (lead-acid is fine) in parallel with the power supply in order to protect the speed controller from current surges coming from the motor when slowing the system down. The speed controller is a MD25HV "25Amp 7V-58V High Voltage DC Motor Driver Speed Controller", per his website.

I can see the physics behind using the battery in this way, but what about instead putting a big diode in the + line so that current can only flow FROM the power supply TO the motor and not the other way around? It seems if it tried to do so it would "bounce" back & forth between the diode and the motor until it dissipated in the lines between them. It seems like a way to avoid needing the battery.

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  • \$\begingroup\$ Hmm, is this motor intended as a 3rd or 4th axis, freely command-able from CNC? That will likely determine which method(s) are viable. If stopping slowly, with little mass, surges will not happen. The more mass, and faster deceleration, the higher the surges. \$\endgroup\$
    – rdtsc
    Jan 11, 2022 at 21:30
  • \$\begingroup\$ Hello rdtsc - thank you for your reply. No - the turntable is not intended as a 4th axis, per se. It's not a CNC device, but a steady state rotating deck that would be switched on before the CNC is started. I think your point about the slow deceleration with low mass is the key though. Because of the roughly 2:1 ratio between turntable diameter and motor pulley diameter the max speed of the turntable is only 50 RPM (~.83Hz) max and it weighs 5.6 lb. Its kinetic energy is only .6J (i.e. .6W-s) so it seems it won't try to backdrive the system much. \$\endgroup\$
    – Mike
    Jan 12, 2022 at 3:40

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If you need to fast decelerated, the motor will generate power and voltage will rise. A battery can absorb this energy and be used later, but when you switch off the main supply, the controller will remain powered, and some axiliary circuit is needed to avoid this. A diode in positive line will cause that voltage will increase and can trip the controller because overvoltage. Another solution with no battery is dynamic braking: voltage supply is compared with a limit higher than rated value, and if it is over will active a mosfet connected to a resistor that can disipate this energy.

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  • \$\begingroup\$ Bravale - many thanks for your input. To answer your question, no - I don't need to decelerate quickly. What I think will happen is that I will find an optimal speed (or narrow range) at which to run the motor, and I will leave it at that speed when the system is in use. I'm not sure I understand what happens when I shut it off. It seems you are saying that even with the system switched off the remaining kinetic energy in the rotating turntable will still backdrive the motor and create a voltage & current. Is that correct? I understand the concept of your last solution - an active shunt. \$\endgroup\$
    – Mike
    Jan 12, 2022 at 2:14
  • \$\begingroup\$ The system will be feeded by a linear power supply from the grid, and the battery will be also connected to these DC wires. When you shut down the linear power supply (AC input from the grid), the battery will keep feeding the controller until it is discharged. So, if time is long until the next start, the battery could be completely discharged and demand high current from the linear power supply. \$\endgroup\$
    – Bravale
    Jan 12, 2022 at 6:28
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I can see that the answers so far have got the wrong end of the stick, and the advice you were given by the seller was misleading at best to begin with.

The problem is that if you want to decelerate the motor quickly, the speed controller needs to apply reverse voltage to the motor (it's an H-bridge, so it can do that). But since the motor is still spinning, its back EMF is "pointing the wrong way" — instead of opposing the voltage from the power supply, it is adding to it. This creates a huge current surge, and the purpose of the battery is to protect the power supply from this overload, not the speed controller.

So no, a diode will do nothing to help this situation. But instead of a battery, you might consider using an ultracap. It needs to be big enough to supply this braking energy without allowing the terminal voltage to drop too low for the speed controller. But you'll still need some current limiting between the power supply and the ultracap if it isn't already built into the power supply.

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  • \$\begingroup\$ Thanks, Dave. I don't need to decelerate to decelerate quickly. The idea is that the motor will generally run at a constant RPM for 30 min or so at a time, and then I'll shut it off to load my next part. I think the seller's concern is what happens at that point. In his defense he's a mech engr by training, not an EE. The stored kinetic energy in the rotating system is low (<1J) so maybe this is all moot. UPDATE: Just found the mfr's datasheet for the motor driver. See screenshot. It says for a switching PS use a battery in parallel, but I like the sound of the ultracap. Buy @ Digikey? \$\endgroup\$
    – Mike
    Jan 13, 2022 at 18:50
  • \$\begingroup\$ manualshelf.com/manual/cytron/md25hv/data-sheet-english/… Couldn't get a screenshot in here, so I have added the link for the datasheet. The part covering backdriving/over-voltage surge is on pg 7 of this datasheet. I intend to use a linear power supply, by the way, not a switching type. \$\endgroup\$
    – Mike
    Jan 13, 2022 at 19:07
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The main thing you want to protect the motor driver and power source from is the inductance of the motor. Inductors don't allow current to jump instantaneously, so any attempt to do that would cause a massive spike in voltage, potentially damaging things nearby. If you've ever rapidly pulled the plug out from a running vacuum cleaner you might have noticed an arc form between the outlet and the plug as a result of this voltage spike.

Your attempt to put a large diode on one side of the motor doesn't do anything to alleviate this. Instead, what you want to do is provide a return path for the motor current to safely follow when the motor driver tries to stop providing power to the motor. Most likely the motor driver has part of this built-in already, however as part of the return path it needs somewhere to dump the current.

Here's a rough schematic of what the motor driver and protection system looks like (source):

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The diodes and MOSFET's are likely built into the motor driver because they are relatively compact and cheap to integrate into the driver. The energy dump is the capacitor C1. It needs to have enough capacity that you can't overload it, and also needs to be safe to dump current into (this is not safe to do with all battery types!). You are effectively adding the lead-acid battery to replace C1. You might be able to get away with capacitors instead of a lead-acid battery, however you would have to figure out how large these would need to be. Unfortunately, in order to figure that out you need to know the motor inductance and the amount of dissipation in the reverse current path, or experiment with a lead-acid battery initially and measure how much energy is dumped into the battery and choose a replacement capacitor accordingly.

As far as why can't you just use the power supply itself to absorb the reverse current, most DC power supplies are designed such that this isn't really possible without damaging the supply itself. Even just using batteries is not necessarily safe; for example trying to reverse the current into an alkaline battery is not a good idea. Luckily, lead-acid batteries for the most part work just fine with reverse currents (up to a point, of course).

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  • \$\begingroup\$ Helloworld922 - Wow, that was fantastic. I posted that question less than an hour ago and already have useful input from you and two others (Bravale and rdtsc). I figured it might take weeks! In any case, yes, I follow your analogy about unplugging the vacuum cleaner. The same phenomenon was used in old car ignition systems with points and condenser, eh? I have an oscilloscope, and I already have a lead-acid battery on order, so when I get the system up and running I'll see if I can capture what happens when I shut it off - or when I quickly rotate the potentiometer knob to slow it down. \$\endgroup\$
    – Mike
    Jan 11, 2022 at 22:35
  • \$\begingroup\$ long-term I think you really do want to replace the lead-acid battery with caps, however I just don't have any sense of scale for how big these should be personally. \$\endgroup\$ Jan 11, 2022 at 23:16

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