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OT: reducing current draw of fan

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  • OT: reducing current draw of fan

    Hello all. I installed a new cooling fan in this amp as the original died. The original was a 12v 0.07A and the new one is a 12v 0.11A draw and is quite a bit louder than the original. I want to reduce the current draw by adding a resistor in line with the positive lead.

    Am I correct by thinking the 12v divided by the .07A draw I want would equal roughly a 170 ohm resistor to be used? If I’m doing this math correctly what size resistor would I want? 1/4w, 1/2w, 1w?

    Thank you,
    Dan

  • #2
    That resistance is too high. If you want to get back near 0.07, use a 60 to 70 ohm resistor (e.g. 62 or 68 Ohms are common). With a 62 Ohm resistor, your back at 0.07 amps and the resistor will need to dissipate 0.3 W so a 1 W resistor should do fine. The fan will now see roughly 8 V.

    If you have any resistors laying around in the that range, a quick speed test with a 1/4 W won't harm anything.

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    • #3
      Thank you very much. Would you be able to tell me how you came up with that figure so I can figure this out for myself in the future? Thank you so much.

      Dan

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      • #4
        Or you can use diodes to drop the voltage
        Each diode will drop .7 volts

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        • #5
          Or you can restrict/ reduce the air flow across the fan to reduce the amp draw.
          John H

          Synergy Horn, SLS-85, BMR-3L, Mini-TL, BR-2, Titan OB, B452, Udique, Vultus, Latus1, Seriatim, Aperivox,Pencil Tower

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          • #6
            Originally posted by saabracer23 View Post
            Thank you very much. Would you be able to tell me how you came up with that figure so I can figure this out for myself in the future? Thank you so much.

            Dan
            Basic electrical equations:

            (1) V= I * R. Hence,
            (2) R = V / I.
            (3) P = V * I, substituting I * R for V (equ 1) yields I * I * R, hence
            (4) V = I2 * R.

            The original fan: V = 12 and I =0.07, using equation 2, R must equal 12 / .07 = 171.4 Ohms.
            The new fan: V = 12 and I = 0.11, equation 2 again R = 12 / 0.11 = 109.1 Ohms.

            With 12 volts to get 0.07 amps you need R = 171.4. If the new fan is 109.1, then you need to add 62.3 Ohms in series to get 171.4 Ohms. Use equ 1 to get the voltage drop across the fan and the resistor: So 0.07 a * 109.9 ohms = 7.64 V for the fan and 0.07 a * 62.3 ohms = 4.36 for the resistor

            Using equ 4, the power across the resistor will I2 * R = 0.072 * 62.3 = 0.305 W (a little less than a third of a W).

            Gee, that was fun. I could teach Middle School shop. Now if someone could only teach me how to write literature ....


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            • #7

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              • #8
                https://www.maximintegrated.com/en/a...ex.mvp/id/1784

                Originally posted by saabracer23 View Post
                ... What diodes would I use? ...
                Silicon diodes have a forward voltage of approximately 0.7 volts.
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                • #9
                  You can also consider this simple circuit to control fan noise. A couple of my DIY amps have them.

                  http://www.heatsink-guide.com/tempcontrol.htm
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                  • #10
                    Originally posted by jhollander View Post
                    Or you can restrict/ reduce the air flow across the fan to reduce the amp draw.
                    John, I think that will cause the amps to increase due to a heavier load - more air resistance. A DC motor increases draw with increasing load.

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                    • #11
                      Originally posted by Millstonemike View Post
                      John, I think that will cause the amps to increase due to a heavier load - more air resistance. A DC motor increases draw with increasing load.
                      You wouldn't believe how many bets I've won from this... less work less amp draw
                      John H

                      Synergy Horn, SLS-85, BMR-3L, Mini-TL, BR-2, Titan OB, B452, Udique, Vultus, Latus1, Seriatim, Aperivox,Pencil Tower

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                      • #12
                        So for my second fan a 47 ohm resistor would be good?

                        Original .09A and new .14A

                        Dan

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                        • #13
                          Originally posted by jhollander View Post

                          You wouldn't believe how many bets I've won from this... less work less amp draw
                          John is correct. Works best if you throttle the suction side of the fan, then it is working in a partial vacuum. Google "inlet guide vanes". It was a popular method of controlling system flow and static pressure before VFD's became affordable and reliable.
                          Craig

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                          • #14
                            Originally posted by jhollander View Post

                            You wouldn't believe how many bets I've won from this... less work less amp draw
                            Well is you meant changing the vane pitch as mentioned by PWR RYD then I totally agree. But you mentioned restricting the air flow. That's where I disagree. There is less air moving so there's less "work" being done on the air. But the motor will be using more amps and dissipating the excess as heat. That's why vacumm cleaner motors burn out if the nozzle is plugged - lots of work and no air flow to cool the motor itself.

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                            • #15
                              Inlet guide vanes do restrict the air flow and therefore they reduce the motor load up to a point (~25% if I remember correctly). They are not the same as variable pitch fan blades. I was an Application Engineer for a large European VFD manufacturer and in 90's I replaced an aweful lot of inlet guide vanes with AC drives.

                              A vaccum cleaner doesn't burn out because of an increased motor load. It burns out because that open frame motor relies on the air flow the vacuum creates to remove the heat it produces.
                              Craig

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