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Dec 26, 2020 at 23:19 answer added whitegreg56 timeline score: 2
Dec 23, 2020 at 11:20 comment added gimpo @whitegreg56: final product will be powered from a 12V motorcycle battery. I was worry that the current pumped back could damage the driver board with some kind of voltage spike. So it seems not the case.
Dec 22, 2020 at 0:45 comment added whitegreg56 @gimpo: What is supplying the power to run the motor? A battery?, a 24V power supply?, or something else? If it is a battery, no need for a capacitor....the current kicked back from the motor (when decelerating) will charge the battery!
Dec 21, 2020 at 17:06 comment added gimpo @whitegreg56: Power supply? Do mean the driver, yes? At the moment I'm using the Big Easy Driver from Sparkfun. I will take a look at the schematic. P.S. I also looked at the stuff from Trinamic. Cool and professional stuff but I'm not at that level at the moment..
Dec 20, 2020 at 17:46 comment added whitegreg56 You don't need a non-polarized capacitor. An electrolytic (polarized) capacitor would be just fine. In fact, there may already be one. Look at the schematic for your power supply!
Dec 20, 2020 at 10:40 comment added gimpo @whitegreg56: there is no way to use a big non-polarized cap (ceramic?) to temporarily absorb the current kicked back from the motor? In that way the cap should act as an "equalizer" between the current pumped in from the driver and the (eventual) current pumped back from the motor even if the windings are energized. This is just my first thought without going deeply into the matter...
Dec 19, 2020 at 3:29 comment added whitegreg56 If the deceleration is small enough, there is no need for protection. The diodes will dissipate some of the energy developed during deceleration. But, mostly it is a matter of pumping current backwards into the power supply. If the deceleration is rapid enough, the current from the power supply will go negative and any capacitor on its output will be "pumped up" in voltage.
Dec 13, 2020 at 8:21 comment added gimpo @whitegreg56: Anyway, the diodes should work in this circumstances. i.e. the energy should be (slowly) dissipated into extra-heat by the diodes because of the voltage drop. Or I'm wrong on this?
Dec 13, 2020 at 8:13 comment added gimpo @whitegreg56: this would happens with any stepper (even the ones of my cnc-router machine) when the rpm are changed by the program while working. Nevertheless I never noticed any particular protection adopted in many project that I saw on Arduino/Sparkfun forums...
Dec 12, 2020 at 22:53 comment added whitegreg56 If the motor is decelerating the load, it is acting as a generator! The kinetic energy has to go somewhere!
Dec 12, 2020 at 17:25 comment added gimpo @BrianDrummond: so it seems that I have to wait for the delivery of the stepper... Then I can do some measurements of current and voltage when I quickly turn the knob. Anyway, I was expecting that if the windings are already energized then the stepper cannot turn itself into a generator. After your comment I started thinking that I'm deeply wrong...
Dec 12, 2020 at 14:13 comment added user16324 True, but you can easily measure voltage (DC easiest, after the diodes)
Dec 12, 2020 at 14:12 comment added gimpo @BrianDrummond it is difficult to evaluate the rpm-speed: if I turn manually rapidly the knob of 90° then the shaft of the motor will make 1,25 turns (because of the 1-to-5 spur gear). But... what is the speed of my hand???
Dec 12, 2020 at 12:20 comment added user16324 At higher speeds, you do have to watch what happens when you're pushing voltage back into the power supply. Especially if that voltage exceeds the supply rating. Consider whether you need to dump generated power into a load resistor or charge a battery with it.
Dec 12, 2020 at 10:48 comment added Kartman The diodes in the A4988 driver chip should be sufficient
Dec 12, 2020 at 10:37 history asked gimpo CC BY-SA 4.0