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I want to ask a similar question to this, but a little bit more in detail in terms of execution and what I can expect from it.

* disclaimer, I'm quite new to the electrical side of things, so please bear with me;;;

I'm using a Meanwell 24V 125A power supply and Cytron SmartDriveDuo-60 to drive a couple of 24V, 750W motors. From the manual (and a forum post by Cytron), they suggest hooking up a battery in parallel with the power supply.

I'm a little lost on how exactly the battery absorbs the back emf; is it splitting it with the power supply, or is it receiving all on its own? Would the regen be immediately used when running the motors again?

Also, when choosing the battery, what factors do I need to consider (the previous post mentioned handling the required regen current; how would I calculate/measure this?).

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  • \$\begingroup\$ Regen depends on what you do with your motor. Basically when you use the motor as a brake it will either recover or burn as heat both the potential energy and the kinetic energy of whatever object is now forcing it to turn. So if you're lifting an elevator, the motor (used as a brake) will attempt to release a lot of energy on the way down. What are you doing with the motor? \$\endgroup\$
    – bobflux
    Commented Jul 12, 2022 at 17:47
  • \$\begingroup\$ @bobflux The motors are attached to worm gearboxes that turn a series of lead scres connected via gears. Since the motors don't have much attached directly to it in terms of mass, it shouldn't have that big of angular momentum. But I do want to decelerate quite quickly; does that factor in as well? \$\endgroup\$
    – lascooter
    Commented Jul 14, 2022 at 1:30
  • \$\begingroup\$ If your worm screws can't be turned backwards by whatever they're moving, at least you won't have the "elevator" problem of getting rid of potential energy. But yes if you want to decelerate quickly you have to get rid of the kinetic energy of all the rotating parts. If you set the PWM to zero and short the motor, it will be dissipated in the motor winding (and you don't need a battery). If you set it to reverse to stop even faster, then it will be dumped into your power supply, increasing its voltage. Which one are you going to do? \$\endgroup\$
    – bobflux
    Commented Jul 14, 2022 at 5:39
  • \$\begingroup\$ The gearbox ratio is high enough ratio that the screw can't (shouldn't) be backdriven. Currently I'm setting the PWM to zero on the arduino connected to the driver, but it's still tripping the over voltage protection on the PS, so I think it's getting dumped regardless (maybe the motor driver is reversing it briefly?). I think I'll add a battery to compensate for it. \$\endgroup\$
    – lascooter
    Commented Jul 15, 2022 at 0:54

1 Answer 1

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Adding a battery is like adding 10kF with unknown charge and low ESR to handle the 750W x 2 x10 for full voltage and acceleration.

So you must define your load acceleration and choose to adjust the voltage if poss. to get ideal float V. It must be precharged before connecting with suitable cables and maintained as per user manual for battery of choice.

Since the above is 15kW, you can understand why these factors are important as the battery only supplies and absorbs the large transient overloads and reverse currents.

To derive the maximum current addup all ESRs plus each motor DCR/2 into the applied voltage at rest.

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  • \$\begingroup\$ I'm a little bit lost on how the battey will function in the setup. My assumption was that the battery shouldn't be fully charged, since it needs to be charged due to the b emf (which will then be quickly used up during normal operation). Also, the surge current should be around 120A, but I don't know how this factors into determining the right battery for the system. \$\endgroup\$
    – lascooter
    Commented Jul 14, 2022 at 2:18
  • \$\begingroup\$ The battery just acts as a capacitor to supplement the supply which cannot handle the worst case torque currents. If the voltage must stay regulated and not trip the over-current threshold of the PS that implies usually < 1% of the voltage or 0.24V. If 120A then the net ESR of the battery and supply must be < 0.24V/120A = 2 mohms which is extremely low. If the battery recovers quickly from over current then a higher ESR can be tolerated as the voltage drops e.g. 10% drop means 20 mohms. Lead acid CCA of 500A is rated for a 5V drop or 10 mOhms. \$\endgroup\$ Commented Jul 14, 2022 at 11:52

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