I have a policy on DIY eboard building:
Buy extra spare parts every time, that’ll save you months of waiting time for overseas shipments to repairs, replacement parts and more time shredding.
Buy better quality parts when it’s cost if slightly higher, you will buy them down the road sometime
This goes for motors, ESCs (specially VESC 4.12), extra battery cells and antispark switches
I believe people generally have -12 at most for 12s. I’m not 100% sure though
Why does regen current have anything to do with voltage? Not saying you’re wrong, I genuinely don’t know. I would have thought you determine regen current entirely by what capacity your battery is (for a 1-2c charge rate).
V=iR Voltage = Current x Resistance
So the resistance is for the most part constant. The current and voltage are the main variables here. This basically makes voltage and current directly proportional to each other.
So if I use my power supply to run a VESC. I can spin a motor up and apply the brakes. In doing this I can get a reading of about 43V from my power supply that only outputs a max of 30V. This is with regen current set to somewhere around -40.
I’m still not sure how that works. This is my understanding:
The motor generates voltage proportional to speed. The faster it goes, the higher the voltage. When you’re accelerating, at some point the voltage generated by the motor will be almost the same as the battery voltage and it won’t be able to go faster (because the current no longer wants to flow to the motor when there’s no voltage differential). This is why we have to use higher kv motors to reach higher speeds on the same gearing. I’m pretty sure that’s correct.
This is where I’m confused. Why would the voltage ever increase with increased load? As I understand it, braking is the VESC allowing current to flow back to the battery. It regulates that current (with a duty cycle on the fets? not sure) so you don’t blow up your battery and so you have controllable braking force. I’m not sure what resistance you would be talking about but the way I see it the main resistance is the fets. We’ll assume for simplicity’s sake that they’re either off or on (lots of resistance or very little). When they’re turned on, the resistance is low but the current is high. That means the voltage wouldn’t change. Same thing when they’re off, the resistance is high but current is low.
That might be oversimplifying it but I’m really not sure why voltage would increase with higher regen current.
What happens if you do the same test on your bench supply but set the regen current to -10?
it’s why people get over voltage errors. sometimes, when braking hard on a full charge the voltage can spike high enough to throw a fault code, or fry components depending on severity
I get the same general result, just maybe 6V less or so. Honestly I’m just getting into this topic in my classes, but it looks like V=L(di/dt) which is the voltage equation for an inductor. Since a brushless motor is many inductors placed in a circular pattern, this equation is relevant here. So the inductance is pretty much a constant here, which leaves us with the derivative of current with respect to time. I also found a formula I believe is for 3 phase motors, but I may be wrong V=I*X where X=angular frequency * inductance (L)
To simplify all of that, voltage produced by an inductor will in fact increase as the current produced increases.
To take an example, the inductance of one of LHBs motors is 13.7uH according to the VESC tool’s FOC detection. Lets say you have 40 amps of regen current, which realistically probably won’t happen, but it’s just an example. So V = (40)(5000)(13.710^-6) Basically 5000 RPM is how fast your motor would be spinning in this example. However, I’m only getting about 2.74V when I calculate that. I can only assume it’s the wrong formula, or you need to account for the number of inductors in the motor and multiply by that. If I do that, we have 14 in this motor, so now that’s 38.36V
I know the motors aren’t wired in series in a dual setup, but all I can figure is when you have two motors producing just shy of 40V a piece, bad things happen
If there are any Engineers on here who can correct me on that math or those formulas, I’d appreciate it. I’m only a second year, so I have a long way to go haha
That explanation makes me a lot more convinced haha. I have no formal EE training so that’s now outside my area of expertise.
Why does 40V from the motor break a VESC though? I’m not sure which bit dies so I don’t know which voltage rating to check. Is it the DRV?
If it’s XML at least give us text so we can view it in a GUI. An image of text won’t do…
I can only assume somehow this when being fed by the capacitor bank can create more power or something somehow. I truthfully don’t know how it happens, but the bottom line is something is exceeding its designed power limit and letting its magic smoke out.
Here’s a Dropbox link to the xml’s