If i bypass BMS for discharging, does that mean my battery will get charged with regenerative braking but BMS won’t balance the cells? In this case, is it better if i just set regen to 0?
In that case I can only wonder what happens when you push for example 50V back through discharge port to the fully charged battery.
if you’re going faster than max speed of your board down the hill, yes you’ll be overcharging while braking. If your bms cuts off power you may be without brakes going down.
I noticed on VESC 6.6 that to have breaks I have to increase amperage in " Battery current max regan " otherwise break are not working , even if I got -60A at " Motor current max Break " So how can I set breaks without recharging battery ?
I had the exact though the other day. Say we have a 10S battery which is ~37 volts around the time of the ride Say we have a 100KV motor when we are riding it, the max rpm of the motor is 3700 RPM. Now lets just cost with no power to the motor. That 3700 RPM is going to produce a back emf of 37Volts, So when we braking under regen we “rectify”(because it’s sinusoidal) that bemf and feed it to the battery. But the thing is, that 37 volt won’t charge the battery. It has to be higher voltage. Unless VESC has a charge pump to increase that voltage to something higher. Does it?
So how does it exactly work?
Okay, so it turns out there is a charge pump because the motor winding is itself an inductor. When you short and unshort the motor it produces a voltage higher than the backemf(due to motor rotation), now this higher voltage is rectified and used to charge the battery. When this switching is going on 2 things happen:
- When you short the motor you are doing non regen braking,
- When you un-short the motor you are doing regen braking Of course the regen braking only works when the bemf + induction spike > battery voltage.
From wikipedia: "The key principle that drives the boost converter is the tendency of an inductor to resist changes in current by creating and destroying a magnetic field. In a boost converter, the output voltage is always higher than the input voltage. A schematic of a boost power stage is shown in Figure 1.
(a) When the switch is closed, current flows through the inductor in clockwise direction and the inductor stores some energy by generating a magnetic field. Polarity of the left side of the inductor is positive.
(b) When the switch is opened, current will be reduced as the impedance is higher. The magnetic field previously created will be destroyed to maintain the current towards the load. Thus the polarity will be reversed (means left side of inductor will be negative now). As a result, two sources will be in series causing a higher voltage to charge the capacitor through the diode D.
If the switch is cycled fast enough, the inductor will not discharge fully in between charging stages, and the load will always see a voltage greater than that of the input source alone when the switch is opened. Also while the switch is opened, the capacitor in parallel with the load is charged to this combined voltage. When the switch is then closed and the right hand side is shorted out from the left hand side, the capacitor is therefore able to provide the voltage and energy to the load. During this time, the blocking diode prevents the capacitor from discharging through the switch. The switch must of course be opened again fast enough to prevent the capacitor from discharging too much."
2 Likes
You break something, then it regenerates and then you break it again and it regenerates again. This would be the future.
@b264 I fixed the title for you.
Who am I kidding. I did it for me.
2 Likes
Does anyone knows an effective way to disable regenerative breaking??? I’m using FOCbox!
So u can brake without charging by shorting or is it just part of the regen process at high speed…it seems. u need to do regen to brake as the energy has to go somewhere. While there all kinds of bms typically people use the discharge type where if a cell goes over a set voltage,maybe 4.2, then that cell alone is discharged super slow with a resistor. Whether you’re braking or charging will be the same mechanism for balancing
Why the hell would you want to disable the brake?..and dont give me that crap that it puts your pack out of balance.
1 Like
I don’t recommend doing this, because energy is purely wasted as heat in the motor winding and the mosfets but basically you decrease the battery reverse current.
Yeah, you can brake without regen. Set your battery reverse current to 0 Amps. But that means increased heat in motors and mosfets.
3 Likes
come on body! I’m trying to see if my cells will still go unbalanced without that regenerative thing! This sh*t still happens on my brain new lipo Pack! I already can’t charge it all the way anymore 
sounds more like a bad bms/charger issue then anything to do with the brakes
U put each cell up to a multimeter? But I didn’t read the thread! That’s be my start
Not using the regen you wouldn’t be involved with the battery. No regeneration of energy back to battery
A video of u showing us would get to the bottom
What would produce highest regen?
Wouldn’t it be a huge amount of heat produced without regen and kill fets quickly? I heard that somewhere but though it was assumed we were always regaining as much as possible.
Would having a motor’s waveform more a certain shape better to suit the esc get better regen?
Braking at high speed, low battery voltage and having a large regen battery current.
But of course you know that setting too high of a regen battery current would damage the battery. That too high of a current depends on battery type and cells in parallel.
I’m forgetting the name but how does setting the motor brake current effect regen?
The motor braking current is directly proportional to the braking force as long as the temperature limits and the regen limits are not hit.
If you set that low, you’ll have weak braking.
1 Like