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I am planning on using a BLDC motor to drive a small go-kart like vehicle. I would like to implement regenerative braking, but am having trouble finding certain details.

My main concern is being able to quickly stop the vehicle. Is there a way to control the braking speed? I assumed that when powering a load with a motor (when being used as a generator), the motor would become harder to turn, which would act like a brake. Is this the case?

If so, I figured I could use a PWM-like circuit to essentially change the duty cycle of when energy is being drawn from the motor. A higher duty cycle would mean faster braking. Would this be an appropriate solution?

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4 Answers 4

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You control the de-acceleration the same way by sensing current according to demand using PWM but boosting the voltage as speed reduces with a DC converter.

Friction braking typically is about 5-10x faster than acceleration even with all wheel drive so regenerative braking can do no better.

If cycling from full e-brake to full acceleration , current is typically 5x current at full power so extra cooling is needed. Adding a power dump resistor reduces current and e-braking effect.

Regenerative charging must boost voltage as speed drops and is more complicated than simply reversing the direction of bridge current because V is proportional to RPM with no load. Thus as speed reduces, so too does e-braking effect reduce . The braking reduces, just as acceleration reduces towards full speed.

Of course in theory you could use degenerative braking by reverse current but that is not practical.

Also if the vehicle kinetic energy is recovered it is only to assist the primary friction brakes and a dead time must be designed to prevent shoot-thru.

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  • \$\begingroup\$ Okay, that's what I figured. However, I fear that the output voltage from the motor will not be very much and therefore I won't be able to transfer energy quickly enough. Unless, of course I find a very small value resistor as P = V^2/R. Is this something to be concerned about? \$\endgroup\$
    – John Leuenhagen
    Commented Aug 7, 2018 at 1:40
  • \$\begingroup\$ The voltage will be no different to the last applied at constant speed. A Short cct would dissipate the heat in the winding resistance just like very low PWM switch accelerating from start at 100% so no difference in g level \$\endgroup\$
    – Tony Stewart EE75
    Commented Aug 7, 2018 at 1:47
  • \$\begingroup\$ Ahh, that makes a lot of sense. So if I short the motor leads then it should effectively brake? If so would this be the fastest way to brake? And would opening the circuit result in very low mechanical resistance? \$\endgroup\$
    – John Leuenhagen
    Commented Aug 7, 2018 at 2:06
  • \$\begingroup\$ Yes for full brake and of course after disabling driver otherwise poof. Coast when open \$\endgroup\$
    – Tony Stewart EE75
    Commented Aug 7, 2018 at 2:09
  • \$\begingroup\$ Alright, great! I assume this would heat up the motor quite a bit when braking a lot, so I assume it would be wise to do something with this energy? Some suggested dumping it onto an resistor. Personally I plan on implementing regenerative braking. Should this reduce the heat on the motor by a noticeable amount? \$\endgroup\$
    – John Leuenhagen
    Commented Aug 7, 2018 at 2:15
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The effective drive voltage is VBATT * D, where D is the PWM duty cycle.

The effective motor voltage is based on Kv and RPM. If you arrange for VBATT * D to be less than Kv * RPM, you will be in the regeneration mode. This will charge the battery and cause the vehicle to decelerate.

Be careful you don't over-charge your battery. Generally batteries have max charge rates much lower than their max discharge rates. This can limit how much regen you should apply, especially at high speeds.

Basically, if you lower the duty cycle enough, the car will slow down.

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    \$\begingroup\$ Can you elaborate your answer a little more, please? Taking the extreme duty cycle of 0% how does power get back to the motor in the system you are thinking of? \$\endgroup\$
    – Transistor
    Commented Aug 6, 2018 at 6:43
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    \$\begingroup\$ Zero duty cycle is a dead short for the back EMF. I am not sure you get regen in that case, but in any event you would not want to do that at speed. In six-step commutation, one of the phase wires will be pinned to GND (low-side FET on). The other phase will be PWM'd to apply an effective voltage of VBATT * D. If the effective voltage is less than the back EMF, it will operate as a boost converter for the back EMF, and the energy accumulated in the winding when low will be delivered to the battery when the low-side FET turns off. There are papers about this. \$\endgroup\$
    – user57037
    Commented Aug 6, 2018 at 7:18
  • \$\begingroup\$ st.com/content/ccc/resource/technical/document/application_note/… \$\endgroup\$
    – user57037
    Commented Aug 6, 2018 at 7:31
  • \$\begingroup\$ So from what I've been reading, getting regen breaking on a BLDC seems to be done with one of two methods... The brute force method of PWMing to get the right voltage back into the battery from the Back EMF, and the 'elegant' method of using additional FETs to switch in an actual proper boost regulator IC to boost the Back EMF voltage up to the battery level, PWMing that to control breaking power. I'm planning to build my own BLDC driver, and I'm planning to go with the former method, hopefully it works. \$\endgroup\$
    – Pecacheu
    Commented Apr 5, 2021 at 20:30
  • \$\begingroup\$ Nobody uses extra FET's. You can get perfectly good regen using the same FET's that apply forward torque. The "L" of the motor windings is all the inductor you need. The "C" in parallel with the battery is all the output capacitor you need. The 6-FET's in the standard controller are all you need to turn the motor into a boost converter to charge the battery. \$\endgroup\$
    – user57037
    Commented Apr 6, 2021 at 3:39
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The more power you draw from a motor acting as a generator, the more power you take from the thing driving it.

There are two ways to regeneratively brake a BLDC.

If you have a sufficiently sophisticated controller, one with '4 quadrant' control, then it can already do it, and it's as simple as finding out how it can be told to so.

If your controller can only drive it as a motor, then the simplest way is to disconnect the controller during braking. Full wave rectify the 3 outputs to get a DC supply, which powers a seperate battery charger to charge the batteries. Obviously the charger has to be sufficiently clever to take the varying input voltage from the motor as the speed varies. The faster you charge the batteries, the heavier the braking.

As HenryCrun points out in comments, the energy you get back is fairly minimal, the big win is getting electric braking. For that, just dump the power into a beefy resistor, with PWM to control braking level, without the complexity of a battery charger.

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  • \$\begingroup\$ It can be much simpler to just dump the energy into a resistor via a pwm, and forget about charging the batteries. You get far less energy back than you would like to imagine, and simply having electrical braking is often the big benefit. \$\endgroup\$
    – Henry Crun
    Commented Aug 6, 2018 at 4:45
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    \$\begingroup\$ I often wonder if the car manufactures added regenerative braking because it was easier than having their marketing department explain why it wasn't worth the trouble. \$\endgroup\$
    – Transistor
    Commented Aug 6, 2018 at 6:31
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    \$\begingroup\$ I think having light regeneration is a good call in a vehicle, even if it has mechanical brakes. Good for recapturing energy in stop-and-go traffic. Also, the bigger the battery pack, the more likely the batteries can accept current from aggressive regen. \$\endgroup\$
    – user57037
    Commented Aug 6, 2018 at 7:04
  • \$\begingroup\$ Thanks for the detailed answer. My one concern is how fast I will be able to brake. To increase the power transferred from the motor, might it be a good idea to use a boost converter to convert to a high voltage? Thanks! \$\endgroup\$
    – John Leuenhagen
    Commented Aug 7, 2018 at 0:23
  • \$\begingroup\$ @JohnLeuenhagen No, it sounds like you don't understand something fundamental. If the object is fast braking, then you need lots of reverse torque, this will be limited by motor heating and mechanics. The highest passive torque is got from shorting the motor, though all the power becomes heat in the motor windings, so you normally settle for less braking and drive external resistors, or a battery charger (aka a boost converter to convert to the battery's higher voltage). You could also actively use battery power to generate reverse torque for even more (too much) braking. \$\endgroup\$
    – Neil_UK
    Commented Aug 7, 2018 at 4:33
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A simple silly answer is in the SSD safe stopping distance calculation done in highway design. Here you can get the amount of energy you need to dissipate and the braking time available. If The required braking power exceeds the capacity of your motor, it siMply will not take the load. Instead after absorbing the max power (capacity of motor/ generator) it will keep rotating so long as the total amount of mechanical power is converted to electrical and absorbed by the battery. By then you may have overshot your stopping point. A 10 kW generator will be effective braking tool for a bike. The tricks of higher voltage, dynamic voltage transforming , battery capacity will all come later. A civil engineer says !

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  • \$\begingroup\$ My e-bike, which does not have regenerative braking, has a 250 W motor. 10 kW would be scary. \$\endgroup\$
    – Transistor
    Commented Dec 26, 2018 at 10:53

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