BLDC Regenerative Braking

Discussion in 'The Projects Forum' started by tom niebel, Nov 13, 2015.

  1. tom niebel

    Thread Starter New Member

    Nov 13, 2015
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    Hi all

    First post here, i'm studying for my Motorsport engineering degree which is effectively just an applied mechanical course so my electronics knowledge is there but not very deep :eek:

    We have a Project to implement the regenerative aspect of KERS on a r6 motorbike. We have a small 2.2kw outrunner motor as a test motor. The aim to rectify and regualate this before charging capacitors.

    When you short out 2 or 3 of the leads from the motor, the motor becomes a brake and you can feel this, however when i built a bridge rectifier with 6 schottkey diodes the motor no long brakes when the two 'dc' terminals from the rectifier are shorted. I understand that the torque in the motor is proportional to current which matches up with the motor locking with a dead short. I was expecting the same behaviour by shorting the wires after the regulator.

    I have 60A 600v diodes which I suspect may be the problem as they will have some leakage and 1-5% of 600V is alot for a DC system.

    Hooking the output of the regulator up we get 5V which is expected from the 200kv motor hooked up to a cordless drill at 1500rpm - ish however the measly 1A is causing no noticeable braking on the motor

    Any help would be greatly appreciated

    Thanks

    Tom Niebel
     
  2. bwilliams60

    Active Member

    Nov 18, 2012
    722
    88
    Can you post any schematics, pictures of your motor etc? In most regenerative braking system the motor itself provides braking by turning into a generator (driven instead of driving). I'm not sure what you are shorting or why, but maybe if you can provide more information, we can help you out.
     
  3. tom niebel

    Thread Starter New Member

    Nov 13, 2015
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    Hi, thanks for the reply, here is the set up we have, the power meter measures voltage and current before leading onto a small 12v drill as the load.[​IMG]

    Im stumped as to why this isnt behaving as expected.

    Thanks in advance
     
  4. bwilliams60

    Active Member

    Nov 18, 2012
    722
    88
    I still don't see a wiring diagram or anything to show what you are trying to achieve. If you disconnect everything from the motor and spin it, do you get any voltage coming out of any combination of the three leads? Is it AC or DC voltage? Can you get past battery voltage by spinning it while disconnected?
    What model number is this motor. I am assuming it is a brushless motor?
     
    Last edited: Nov 14, 2015
  5. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
    744
    Sounds like its a 3 phase output, like an alternator, using 6 diodes for a full wave bridge, so there will be 1.4 v needed to keep the diodes on, its possible the diodes are absorbing the brake current.
     
  6. tom niebel

    Thread Starter New Member

    Nov 13, 2015
    4
    0
    [​IMG]Thanks Guys, this diagram isn't mine, its stolen from google but saved me doing an upload.
    [​IMG]
    This is the diagram we are using and the voltage we see direct from the motor is around 5V which correlates well with the 200kv motor at around 1000rpm (cordless drill)
    Ignore the power supply and battery in the photo, the only source of power is the rotational energy from the drill which is applied direct to the end of the motor.
    If you were to short the pins on the load ( + - ) then effectively its the same as shorting the three wires from the BLDC just through an array of diodes. However this doesnt seem to be the case and no noticeable braking occurs.
    We are only interested in harvesting, not delivering the power otherwise we would have chosen the mosfet option which suits both needs. If this option works it eliminates any need for the motor position sensors and control regarding the quadrants.
     
  7. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,509
    2,368
    If it is an outrunner it is a BLDC motor usually.
    I think the 200Kv in post #6 should be 2.2Kw?;)
    I take it the motor is going to be used as a generator and no motor action at all?
    There will be some considerable drop across the diodes at that low a voltage.
    How do you intend switching the braking in, could you not do it on the AC side?
    Max.
     
    Last edited: Nov 14, 2015
  8. tom niebel

    Thread Starter New Member

    Nov 13, 2015
    4
    0
    the Kv represents the relationship between rpm and voltage output(rpm per volt). Then the current is then related to the torque (negative in regenerative applications). This ties in nicely with:

    "P = Torque * Angular Velocity"
    and
    "P = Current * Voltage"

    The voltage we have is good and expected, with 1000rpm / 200rpm/v = 5 v
    The problem is the current, the next step i guess is to find some smaller diodes and see if this effects anything.

    As for modulating the braking in relation to the riders braking input, we were looking at a PWM controlled mosfet after the regulator, however now you mention it, with respect to FMEA it will be better to have it on the individual wires direct from the motor, this would also reduce current in the mosfets allowing smaller ones to be used.
     
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