Peak power supply rating and regen braking - Page 1
Forum section: electronics / projects

author: Rutherford
author_rank: Newbie
timestamp: September 26, 2021, 06:47:56 pm
content: Hello everyone,I've got this project, but I'm unsure if my high level overview is even correct.I have a CNC machine that is moved with a BLDC motor. Currently, it has a braking resistor as its power supply can't handle all that back-emf. I would like to replace said braking resistor with a regeneration control device, ideally without having to change the motor controller. Basically now, whenever the backemf voltage is high enough, the power supply is disconnected and the braking resistor is connected.I would like however to create a simple regenerative braking board in order to capture some of that energy back for immediate later use. Furthermore, I was wondering if I could charge up those capacitors beforehand in order to supply some of the peak power demand in motor direction. This would allow me to replace the power supply (the one I'm using is a loan) with a much smaller and less expensive one.Without getting into the numbers, is this the right direction for what I'm trying to accomplish?

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 26, 2021, 08:04:56 pm
content: No. Wrong direction. (Without even looking at it).A power resistor to burn away surplus joules is simply the right way to go, and it is so for several reasons.1. It is simple and reliable.2. You want a DC bus voltage close to the maximum of your motor drivers for max performance.3. You have therefore little tolerance for overvoltage on the bus.4. The amount of energy you can recover is minimal.Just take an envelope and a pen, turn the envelope around and write down the power consumption of your whole CNC machine, and make an estimation of amount of energy you can recover from regeneration during braking. Also add the cost of electricity and the cost of "whatever-you-want-to-implement".Maybe, just maybe the equation is a little bit different if the CNC machine is standing in the middle of some bush somewhere and you have to run it from solar cells and a battery. But even then it is probably better to just set the acceleration and deceleration parameters a bit lower. This means no high current peaks during acceleration, and a deceleration that is about equal to friction, so there is nothing to recover in the first place.

author: Benta
author_rank: Super Contributor
timestamp: September 26, 2021, 08:16:18 pm
content: Quote from: Rutherford on September 26, 2021, 06:47:56 pmWhy wouldn't the power supply be able to handle the back EMF? BEMF will never be higher than the supply voltage, so what's the problem?The brake resistor is there to make the motor stop quickly. Whether it makes sense to regenerate is up to you, but the "return on investment" doing this will be extremely low, most probably negative.A mobile phone charger across the brake resistor could perhaps give you a few more minutes talking time

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 26, 2021, 08:25:26 pm
content: So you've got a powersupply on loan, and have to replace it anyway.You're aiming for a smaller power supply, presumably to keep the cost down.So what sort of budget does that leave for your regenerative braking regenerator device?I don't know the size of your motors or the power supply you have or need.A good old chunk of iron with some copper wire is still a good choice as power supply for a CNC machine.An SMPS has to be overrated, because it collapses if you draw too much current out of it, while the chunk of iron / copper wire can give you some extra power in short bursts. It's not so sensitive for overloads. Also with a bridge rectifier and a big electrolytic capacitor bank, you already have your regenerative power storage device. Excess power will be pumped into the capacitors and the bridge rectifier stops any backwards flow into the transformer.Now you still have a power supply on loan, it is a good moment to do some measurements. How much current does the power supply actually deliver to the motors? How much does the bus voltage rise during deceleration? How hot does the braking resistor get? There is a good chance you don't need a braking resistor at all. Maybe this does set some small limit on the on the deceleration parameters for your motors, but that is just a software setting.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 26, 2021, 08:45:35 pm
content: Can the controller not be told to ramp slower, so as to not exceed the power supply's rating? Or is the interest that it could be started faster as a result (which also means the controller is currently running below rating / has spare capacity)?Depending on power supply rating, the energy storage can be as simple as a big ass capacitor on the power supply output. Preferably it should be rated for continuous current limiting (as opposed to hiccup mode current limiting), so it doesn't have trouble charging the cap initially; and able to handle a modest swell in output voltage, as that's where the energy will go, into excess output voltage. Idle current drawn by the supply and controller mean the voltage will bleed down between cycles, so you only get maximum benefit when stopping then immediately restarting; but you still get some advantage from starting from cold, as the power supply having charged the capacitor initially, means the motor can use that basically as a flywheel to restart faster (even if not overcharged by regen).Or a secondary controller could be built, effectively a capacitor multiplier, to be wired in parallel with the power supply. It would make use of the full 0-100% voltage range of the capacitor (or at least most of it, as opposed to the, say, 70-120% range of the above method), and might have a simple state input telling which way it should be charged, in preparation for whatever action is about to happen next (so that it can draw a little current from the supply to charge up, or apply a resistor to bleed down). It would respond by acting to stabilize the power supply, say if it drops below 95% of nominal voltage, it delivers current to stabilize it there; or if over 105%, draws instead.I'm not aware of such a module being generally available, or what a common name for it would be (some kind of booster maybe?). So this would be a fair bit of design work.A switching scheme seems unnecessary, and prone to failure: you need to switch it over exactly in time, without dropout; you need to worry about charging bypass capacitors (which can be quite substantial by themselves, making this a difficult task for semiconductor switches to handle), and you need a state machine to handle it, built on assumptions about how much energy is flowing where. It's brittle, poorly scalable, or full of parameters that need to be tweaked. It can be done, but how it should be done is non-obvious.A simple solution is communicating the state of load or charge by allowing the supply voltage to vary, giving a continuous analog indication of operation, to all parties involved. Hence the above proposal.Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 26, 2021, 08:48:43 pm
content: Quote from: Benta on September 26, 2021, 08:16:18 pmPresumably, by "regen", it's meant that the controller acts to boost raw motor bEMF up to the supply voltage, thus braking it faster, assuming the power has somewhere to go.Tim

author: Rutherford
author_rank: Newbie
timestamp: September 26, 2021, 09:07:44 pm
content: Quote from: Doctorandus_P on September 26, 2021, 08:04:56 pmThank you, you've given me much to think about. Some quick numbers I've come up with is that I theoretically can recover around 20 joules from the deceleration, while using around 60 joules from the acceleration + travel (low power use). Though I probably overestimated the recovery by quite a bit... (this is kind of a pet project of mine and someone wants to hear that this is going to save them money while it most likely won't...)Quote from: Doctorandus_P on September 26, 2021, 08:25:26 pmI believe the power supply delivers around 300 watts to the motors and the voltage rises around 20 volts from a nominal 40. This is however, without the breaking resistor. It's the other way around, I'm considering this because the braking resistor is engaging constantly (as the mass of the head is higher and the friction of the axis is lower from the original). While I think about what you said about the redneckpsu, even if I can't recover much with regen braking do you think I could get away with a smaller power supply if I get a big capacitor bank? From what I gather I only need around 60 Joules for a complete cycle, would it be possible to charge at a fixed rate some caps so they have the burden of the acceleration and acceleration peaks?Thank you for all your help, I clearly have a bias in my thinking and was looking for someone to set me straight.

author: Rutherford
author_rank: Newbie
timestamp: September 26, 2021, 09:11:25 pm
content: Quote from: Benta on September 26, 2021, 08:16:18 pmSorry I should have been clearer, I was refering to the back emf generated due to a sudden stop with high inertia thus becoming a generator with the possibility to generate bmef greater than the psu (as far as I know, I may be wrong though).

author: Rutherford
author_rank: Newbie
timestamp: September 26, 2021, 09:31:16 pm
content: Quote from: T3sl4co1l on September 26, 2021, 08:45:35 pmExactly, there's interest for it being driven harder while maintaining lower rated psu. (and avoiding buying higher rated ones)Quote from: T3sl4co1l on September 26, 2021, 08:45:35 pmThis is exactly what I had in mind. Some common cycles involve the head going up and down for calibration purposes, with a long travel time (2 sec) (with very low power consumption) and sudden stops (acceleration and deceleration). Even if I couldn't save much using regen my hope is that I could get away with a smaller power supply. Do you reckon this would also need complex switching strategies?Quote from: T3sl4co1l on September 26, 2021, 08:45:35 pmThank you for your answer. The second idea is perhaps out of my league, but you've given me much to think. Greatly appreciated!

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 26, 2021, 10:08:26 pm
content: Quote from: Doctorandus_P on September 26, 2021, 08:25:26 pmSwitching supplies can also be designed to handle short periods of higher peak current. That said, adding regenerative capability to a power supply based on a mains frequency transformer is likely easier than you think by implementing synchronous rectification. Could possibly work using a LT4320 with the positive output sense connected to a circuit that fools it into allowing reverse current at the peaks. (Have not tested it, I should since it may very well be the world's simplest grid tie inverter that works!)Maybe consider running some other kind of load for regen braking? One that comes to mind is that since CNC generally makes a lot of dust, a vacuum cleaner or air compressor would be very useful.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 26, 2021, 11:28:59 pm
content: If a bulk capacitor will do, its value is determined as follows:The energy in a capacitor is E = (1/2) C V^2. We are moving from one voltage to another, say 40 to 50V (so, just reducing the overshoot modestly from what it is now, so it would seem), so we have a change in energy that is the subtraction of two cases of this relation.E_regen = (1/2) C (Vmax^2 - Vmin^2)60J = (1/2) C ((50V)^2 - (40V)^2)C = 120 / 900 = 0.133 F = 133 mF = 133,000 uFSo, that's pretty substantial. I would guess on the order of 10mF is in the circuit; which seriously brings into question the 60V figure (actual overshoot would be ~100V, if everything works as expected). Perhaps the controller won't backfeed above that point and you're not getting full performance, perhaps something else is absorbing that power (destructively?). Or, if that's with the braking resistor in circuit, which I'm guessing just turns on once in a while depending on applied voltage or something?The figure for a capacitor fully discharging (0 to 40V, what would be used in the complicated capacitor-multiplier booster) is about half that: 75mF, still pretty substantial.Note that going from 0-40V takes such-and-such amount of energy, but going to sqrt(2) times higher (+41% or 56.6V) takes exactly the same amount again -- so it shouldn't be surprising that we don't gain much (i.e. the value is only about half) by going to a higher voltage swing.In any case, these figures are in the range where a typical power supply will have some warnings (perhaps it won't start up smoothly, or at all?). So, a bulk cap might work, might not, maybe it'll just be fiddly.And of course, this assumes 60J is the worst-case maximum. If it can spin faster, that's more energy -- again, quadratically so, as energy goes as velocity squared. You'll exceed the predicted overshoot, and be back with the same problem, except more electrical energy to dump into the controller in the event it fails (satisfying sparks, but do wear eye protection).The energy can also be calculated from the mass and velocity of everything, including the motor's rotor, gears, shafts, etc. which, as rotating loads, need to be accounted by inertia rather than mass (E = (1/2) I \$\omega^2\$). Easy enough if they're solid cylindrical parts, more or less, but you need to sum up everything.Tim

author: Rutherford
author_rank: Newbie
timestamp: September 28, 2021, 07:05:42 am
content: Quote from: T3sl4co1l on September 26, 2021, 11:28:59 pmYour capacitance figure of the circuit is dead on, however I don't see how you get at the 100V overshoot value, what am I missing? The 60V figure is based on the 15J of energy from the deacceleration. As far as I know this is without the resistor engaging (manually disabled), which indeed is only active when the voltage rises above a certain value.Quote from: T3sl4co1l on September 26, 2021, 11:28:59 pmAs Doctorandus_P guessed this is indeed a switching power supply (SMPS) and it is continuous current limiting. I might be able to test some rudimentary capacitor controller / bank with it. If I ware to is size the capacitor bank for a 70%/120% voltage range there wouldn't be as much as inrush current right? Any program suggestions for simulation or just cold hard math (and dust the books for those transients modeling equations?)Quote from: T3sl4co1l on September 26, 2021, 11:28:59 pmI'm pretty certain the energy value (60J) for a calibration cycle that includes the CNC head going from one end of the table to the other is a pretty good theoretical approximation that has taken into account most of what you've said. But as you've correctly pointed out, if this cycle becomes shorter the duty cycle changes and that would change the calculations.