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Re: Thyristor Conroller

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  • Kerim F
    ... Hi Jack, In theory your idea should work. For instance, when I was 25 years, after my graduation, I thought charging my 12V acid battery (60 AH) by using a
    Message 1 of 7 , Feb 18, 2013
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      --- In Electronics_101@yahoogroups.com, "oakite2000" <stackerjack@...> wrote:
      >
      > Hi,
      > I want to run a 12Volt D.C. motor from the mains power.
      > Has anyone got any idea what would happen if I connected the motor to my half wave thyristor power controller,( which is designed for 240VAC operation), and turned the output down to minimum?
      > If I connect my 240 VAC electric drill to the controller, it doesn't start turning until I rotate the control pot. by about 30 degrees.
      > I don't want to try it without checking first, in case it burns out either the controller, the motor, or both.
      > Thanks Guys
      > Jack
      >

      Hi Jack,

      In theory your idea should work. For instance, when I was 25 years, after my graduation, I thought charging my 12V acid battery (60 AH) by using a power triac, not thyristor, and a full diode bridge to let the charging current of the two half-cycles equal. I measured on my DC ammeter about 15A. But soon later my battery died. The reason is that the triac has to conduct during a relatively very short phase (about 1ms, if I remember well but I will likely simulate the idea in the near future on LTspice). During this period of time, a very high triangular pulse (about 150A peak) stroke the battery cells... that couldn't cry ;)

      In your case, the pulsing will be much lower for two reasons:
      (1) The inductance of the motor windings
      (2) The DC resistance of the motor is much higher than of a big battery.

      I assume you will use a diode bridge to provide a varying DC voltage to supply the motor and its thyristor controlling it. In general, it is not a good idea to let the AC current unbalanced; like passing it during one half only. Anyway the exact shape of the relatively high pulsed current depends on the motor power, its speed and applied torque. But for the same DC current, I_dc (if 12V DC is used), the heat generated inside the motor will be much higher. Assuming the current pulse is square with amplitude I_max with a duty cycle K_dc and the internal resistance is R_motor:

      In case of DC:
      P_dc = I_dc * I_dc * R_motor

      In case of pulsing:
      I_max = I_dc / K_dc
      P_pls = I_max * I_max * R_motor * K_dc
      P_pls = I_dc * I_dc * R_motor / K_dc

      P_pls = P_dc / K_dc

      So if the on phase is 1ms:
      Full wave:
      K_dc = 1ms/10ms = 1/10
      P_pls = P_dc * 10

      Half wave
      K_dc = 1ms/20ms = 1/20 (if half wave)
      P_pls = P_dc * 20

      Hope this little helps.

      Kerim
    • lists
      In article , ... Smoke, I expect. -- Stuart Winsor Only plain text for emails http://www.asciiribbon.org
      Message 2 of 7 , Feb 18, 2013
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        In article <kftotb+bb5q@...>,
        oakite2000 <stackerjack@...> wrote:
        > Hi, I want to run a 12Volt D.C. motor from the mains power. Has anyone
        > got any idea what would happen if I connected the motor to my half wave
        > thyristor power controller,( which is designed for 240VAC operation),
        > and turned the output down to minimum?

        Smoke, I expect.

        --
        Stuart Winsor

        Only plain text for emails
        http://www.asciiribbon.org
      • lists
        In article , ... If you re anything like me you have accumalateda large collection of the things over the years! Saw a
        Message 3 of 7 , Feb 18, 2013
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          In article <006901ce0e02$5debeff0$19c3cfd0$@com>,
          Rick Sparber <rgsparber@...> wrote:
          > If your 12V DC motor doesn't draw too much power, I suggest you look at
          > buying a used "wall wart". I paid $3 for mine and use it to power the
          > motor which moves the table on my milling machine.

          If you're anything like me you have accumalateda large collection of the
          things over the years!

          Saw a lovely cartoon a while back.

          --
          Stuart Winsor

          Only plain text for emails
          http://www.asciiribbon.org
        • Jan Kok
          First of all it s a real shock hazard. The motor insulation probably isn t designed to withstand 240VAC. There could be leakage to the case. If you DO try it,
          Message 4 of 7 , Feb 18, 2013
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            First of all it's a real shock hazard. The motor insulation probably isn't
            designed to withstand 240VAC. There could be leakage to the case. If you DO
            try it, please connect a safety ground wire to the motor case.

            Second, not all thyristors are suitable for controlling motors. In
            particular, triac-based incandescent "light dimmers" may not always turn
            off between cycles of the AC line with inductive loads such as motors. The
            result is that you could end up applying full AC line voltage to your motor.

            Third, if you run the motor at the full rated torque you'll likely burn up
            the motor due to I^2R heating. Let's say the motor delivers rated torque
            when the shaft is spinning at 2/3 of the unloaded RPMs. That tells us that
            the "back EMF" produced by the motor is 2/3 of the rated voltage, i.e. back
            EMF = 8V = 2/3 of 12V. And let's say the current is 1A at full load.

            Now, assuming that 12V _DC_ is supplied: The voltage drop through the
            resistance of the windings and brushes is 12V - 8V = 4V. Thus, we can
            determine the motor resistance is R=V/I=4V/1A=4 ohms. And the power
            dissipated by the motor as heat is I^2R=1A^2*4 ohms=4W.

            But what happens when the current is supplied in narrow pulses? Torque is
            proportional to current. To get the same _average_ torque, you need a much
            higher _peak_ torque - and corresponding peak current.

            The pulses are approximately triangular (the edge of the sine wave). For
            50Hz, I calculate the triangle width as .87 ms wide and 23 amps at its
            peak. (You can verify that that gives an average current of 1A.)

            The peak power is I^2R=(23A)^2*4ohms=2.13KW. The average power _during the
            pulse_ (integrate the power) is 1/3 of that, 711W. But the duty cycle is
            .87 ms/ 10ms so the overall average power dissipated in the motor is about
            62W. Considering our hypothetical motor was designed to dissipate 4W, smoke
            seems likely.

            Bottom line, running a 12V motor with a thyristor controller from a 240V
            line does not seem like a good idea. Likely to be harmful to the motor, the
            controller - and you! :-)



            On Mon, Feb 18, 2013 at 10:37 AM, oakite2000 <stackerjack@...>wrote:

            > Hi,
            > I want to run a 12Volt D.C. motor from the mains power.
            > Has anyone got any idea what would happen if I connected the motor to my
            > half wave thyristor power controller,( which is designed for 240VAC
            > operation), and turned the output down to minimum?
            > If I connect my 240 VAC electric drill to the controller, it doesn't start
            > turning until I rotate the control pot. by about 30 degrees.
            > I don't want to try it without checking first, in case it burns out either
            > the controller, the motor, or both.
            > Thanks Guys
            > Jack
            >


            [Non-text portions of this message have been removed]
          • Dave C
            Pray tell please describe the cartoon if you can t find a link to it... Dave -=-=-=- ... [Non-text portions of this message have been removed]
            Message 5 of 7 , Feb 18, 2013
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              Pray tell please describe the cartoon if you can't find a link to it...

              Dave

              -=-=-=-

              On 18 February 2013, at 2:24 PM, lists wrote:

              > Saw a lovely cartoon a while back.
              >
              > Stuart Winsor



              [Non-text portions of this message have been removed]
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