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    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    My Sonorbo is consistently dissipating LESS electrical power with the rotor up at speed than it does with an identical drive signal and the rotor at standstill (TDC). For instance, I am using a uniform duty cycle of 280 microseconds and a uniform active clamp time of 2.78 milliseconds. This is the same regardless of whether or not the rotor magnets are triggering this signal or the microcontroller is. Absolutely the same active signal. With a 125 HZ signal and the rotor at standstill, the Sonorbo's electrical circuit dissipates 463 milliwatts. With the rotor triggering the same signal, the electrical circuit dissipates 432 milliwatts at 130 hz. So, this thing uses less power when encountering windage and bearing losses while also cycling at a higher speed.
  1.  
    So how much does it dissipate with rotor spinning at 260 hz (doubletime)?
    •  
      CommentAuthorAngus
    • CommentTimeJan 19th 2010
     
    What position is the rotor in when it is standing still?
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    The rotor is at max magnet attraction. It also fires actively at max attraction (TDC). I have tried this also at 140 hz with the same type results.
    • CommentAuthorjoshs
    • CommentTimeJan 19th 2010
     
    First you want to be certain that you are making your power measurements correctly. If you are then you can start working out what is different between the situations. Energy from each of your fixed duration pulses can go one of two places:

    Towards charging up a magnetic field, some of which can propel your motor OR
    I^2R dissipation OR
    Iron loss in the coils.

    Iron loss should be very low, leaving mostly energy stored in the magnetic field or dissipated in the circuit resistances.

    When the thing is rotating the inductor charge time will vary from the value at stall. Where the rise-time takes longer, then the input power will be lower. The miracle of BEMF would make that so for rotating motors. What you report for your motor is that it weakly exhibits higher stall current and power than running. Off-hand, that seems expected to me.
  2.  
    At a guess I would say increased momentum, which means the magnets pass the coils faster which means less time for the coils to cause drag on the magnets.

    The more important question, are you getting more power out than power being consumed.
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    Kind of the only thing I can think of really. The energy is converted from what would be an ohmic loss. The circuit must be cooler when the motor is running.
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    No, not more power out than in I'm sure. I do not know what the rotor is dissipating however.
    • CommentAuthorjoshs
    • CommentTimeJan 19th 2010
     
    Posted By: sonoboyKind of the only thing I can think of really. The energy is converted from what would be an ohmic loss. The circuit must be cooler when the motor is running.
    Do you have scope captures of voltage and current under both conditions?
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    No But I can run it all again I guess. Don't have time tonite.
  3.  
    Sounds like a normal situation to me. Put a load on a normal motor the cost will increase. To get a motor to speed it has a cost to do so.. Input to a normal rpm state. To get it over that speed, energy has to be added.
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      CommentAuthorAngus
    • CommentTimeJan 19th 2010 edited
     
    If you slow it you are increasing friction and windage. Increased power would not seem unexpected? If you do that until it stops, why would increased power then be unexpected? I expect it goes to Joule heat and extra hysteresis.
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    From what I'm seeing in the signal, I'm going to be able to re-use some of the energy in the toroid, and it looks like it will be the same amount regardless of whether or not the thing is rotating.
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    If I slow it down with my finger it uses less and less power.
    • CommentAuthorjoshs
    • CommentTimeJan 19th 2010
     
    Posted By: AngusWhen you stall it you are increasing friction and windage. Increased power would not seem unexpected? If you do that until it stops, why would increased power then be unexpected? I expect it goes to Joule heat and extra hysteresis.
    In an ordinary DC brush motor input and output power are both comparatively low at any speed where the BEMF approaches the input supply voltage. Input power is high and output power goes to zero at stall.
    • CommentAuthorjoshs
    • CommentTimeJan 19th 2010
     
    Posted By: sonoboyIf I slow it down with my finger it uses less and less power.
    Why does this surprise you? Or does it surprise you?
    • CommentAuthorsonoboy
    • CommentTimeJan 19th 2010
     
    It doesn't surprise me.
    •  
      CommentAuthorAngus
    • CommentTimeJan 19th 2010
     
    Posted By: joshs
    Posted By: AngusWhen you stall it you are increasing friction and windage. Increased power would not seem unexpected? If you do that until it stops, why would increased power then be unexpected? I expect it goes to Joule heat and extra hysteresis.
    In an ordinary DC brush motor input and output power are both comparatively low at any speed where the BEMF approaches the input supply voltage. Input power is high and output power goes to zero at stall.


    I think we're all agreeing.
    • CommentAuthorsonoboy
    • CommentTimeJan 21st 2010
     
    Bump for Al
    •  
      CommentAuthoralsetalokin
    • CommentTimeJan 21st 2010 edited
     
    Uh-huh.


    "If I slow it down with my finger it uses less and less power. "

    It may not surprise joshs but it may surprise me, I'm not sure yet. Kind of like good squirrel chili.
    I think it should use the same power. But I can't remember right now: is your microprocessor controlling pulse width as well as frequency? That is, do you keep the pulse duty cycle the same or do you keep the pulse duration the same, as frequency is varied? The average power will be different in the two cases.

    I can't quite figure out why this might be happening in your case; could I be permitted to examine some raw data?
    Not that I don't accept its validity; I just want to see what the data look like going in to the power calculations.
    Are you using a scope with onboard realtime trace math including integration?