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.
"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."