Problem with Motor & protection circuit

Discussion in 'The Projects Forum' started by simo_x, May 7, 2011.

  1. simo_x

    Thread Starter Member

    Dec 23, 2010
    200
    6
    Hi all,

    I need a little help about solving a problem..
    I have this circuit

    [​IMG]

    R35 is 1 Ohm and the Voltage reference is connected to a comparator, which is an overcharge detector that will be active when motor exceeds power.

    Well, the problem is the peak voltage created by the motor when power is supplied, because at start up for a very short period of time I measured a maximum of 230mA with my tester, than the current decrease rapidly to 70 ~ 80mA with 0.8V output from R35..
    I think the current should be higher, because the voltage reference for overcharge detection is about 260mV and when the power is supplied, I have a continuous overcharge detection which make impossible to run the motor correctly.

    However I am just testing the circuit with a motor which consumes 70 ~ 80mA, because our professor told us that we will try it with a 12V 25W motor :eek: , so, if I have a current peak 3.3 (or more) greater when I turn on power with a small motor like I have now, how much of current peak will I have with a motor like that?

    Maybe is obvious that I have to use an RC snubber circuit. I putted one about 1K and 100n but it doesn't work because off course is not well calculated.

    Which snubber can I use to solve the problem and to use both motors?
    Thank you for your help..
     
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    Last edited: May 7, 2011
  2. simo_x

    Thread Starter Member

    Dec 23, 2010
    200
    6
    Someone knows how to implement a snubber circuit to limit the peak current on turning on a motor?

    Thank you.
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,789
    945
    Current draw in a motor is a function of back EMF. At start up there is no generator action occuring(since the rotor is not moving) and therefor only the resistance of the windings is responsible for the current being drawn. AFTER the rotor gets moving, counter EMF will reduce the current flow.

    You need to have some extra resistance in the line at start up if you wish to reduce the current draw.
     
  4. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    Something like this concept may be helpful. When power is first applied Q will be on (shunting R35) until C charges. You could use the same concept with a FET.
     
    Last edited: May 9, 2011
  5. simo_x

    Thread Starter Member

    Dec 23, 2010
    200
    6
    Hi Kermit2 & CDRIVE, I have just seen the post now..
    I solved with a 10K resistor & a 100nF capacitor after R35, and I increased the threshold voltage of the first comparator to 2,14V substituing R6 with a 2K2 resistor, since the motor I will try will consume 2A.

    Since I have used the resistor & capacitor which works fine for the 70mA motor, I hope the 10K will be larger enought for the starting peak current draw of that motor (12V 25W), however I will increase that too if it will be required :rolleyes:.

    The solution posted by CDRIVE is interesting too but, if I am not wrong, its purpose is to create a short delay to running the motor, making the starting high peak current draw negligible, right? Very interesting, maybe I will use it. :D

    Regards,
    Simon
     
    Last edited: May 8, 2011
  6. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    I would naturally try your method first because it does not use any active components however, the answer by CDRIVE is probably more valid. Just to be clear, It does not create a delay in running the motor. It shunts the error signal from R35 to ground for a moment when the power is first turned on. The advantage is that it will not delay a fault signal that happens during run time, only at start-up. Your method delays fault signals at all times.
     
  7. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    Actually, I like simo_x's concept better for two reasons. His circuit will behave much like a 'Slo-Blow' fuse, in that it won't pop during brief surges anytime. Secondly, I don't know what I was thinking when I submitted that circuit. simo_x said that R35 will develop about 260mV drop at startup. It's doubtful that Q will saturate enough for Vce to be much lower than that, if at all.

    As it stands, I didn't give good advice.
     
  8. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    Good point on the saturation voltage. I suppose it would require a MOSFET or a mechanical relay to have a lower impedance than R35.

    as for whether to make a start-up stupifier or a slo-blow function, I leave that choice to the O.P.
     
  9. simo_x

    Thread Starter Member

    Dec 23, 2010
    200
    6
    Well, finally the motor used for test the circuit in front of professor was the 70mA, the same I used for the test in my home.. It worked, obviously, perfectly :D
    To test the overcharge & protection circuit, my professor shorted the circuit connecting the two motor terminals between them, all finished well. Project approved.. :D
     
  10. CDRIVE

    Senior Member

    Jul 1, 2008
    2,223
    99
    You did that on you're own. Congratulation!
     
  11. simo_x

    Thread Starter Member

    Dec 23, 2010
    200
    6
    Thank You CDRIVE, however, in a first time the suggestion about putting a resistor and capacitor after R35 was given by another person (it was about 1K), but I increased it to be larger because in the first moment did not work etc.
    It has been the first time I deal with a motor and high peak current at start up..
    I discovered measuring with tester and it was a new problem for me which I never dealt.
    Again, I lernt something new! :D
    Thank you for your support, you are very gentle people!
    Regards,
    Simo
     
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