I want to power 5 fans in parallel (12V 0.80A, each) using a adjustable power supply (DIY, made with an LTC 3780 connected to a 24V 15A PSU and mini Voltmetre). I read that a fan could destroy the LTC 3780 because of the induction effect. If I'm not mistaken, when it's ON, the fan store energy in a magnetic field (because of it's winding), which can create a reverse current when we turn it OFF.

So do you confirm that I should protect a circuit from the fans ? If yes, how can I modify the DIY setup to prevent burning my circuit ? The thread mention adding a Schottky Rectifier (the 1n5819). Is a Schottky diode (supporting the right amount of voltage and current) the best way to protect my circuit ? If yes, can I simply add it on the main wire that will power the 5 fans?

More generally, should I protect any circuit from any motors ? If no, what are the rule to follow ?


The selected fans have brushless motors with electronic control circuits built in. The motor coils are not powered directly by the input power and should not transmit any inductive kickback to the external supply. However there is no indication that the motors are designed for variable voltage operation.

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  • \$\begingroup\$ Thanks! Does it also means that the fan circuit convert DC to AC? To make it short I did not mention that I will use the DIY adjustable power supply as a constant power supply (12V). But I wanted to add a dimmer to change the speed of one of the 5 fans by adjusting the voltage, which according to what you wrote, seems risky. \$\endgroup\$ – MagTun May 6 '19 at 13:43
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    \$\begingroup\$ The internal fan circuit converts DC to pulsed DC that is similar to AC. Depending of the specifics of the design, reducing the input voltage may provide speed control with an unknown lower limit and an unknown relationship between voltage and speed. The risk of damage is also unknown. \$\endgroup\$ – Charles Cowie May 6 '19 at 13:58

Motors have a surge current that depends on acceleration and fans have a steady current that is linear like a resistor. V/I= constant at steady state RPM above start. So the BLDC motors convert DC to AC and the Vdc/motor DCR determines the starting current. But as the fan speeds up backEMF reduces the current.

So you have a Buck regulator rated for 80W and 5x 10W fans= 50W. So you will need an inrush current limiting ICL to reduce the start current to 6A. (*12V=72W)

e.g. https://www.digikey.ca/product-detail/en/ametherm/SL12-10006-A/570-1294-ND/5975280

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  • \$\begingroup\$ Thanks but I am not sure to understand the part about the ICL. Do you mean that I need a Negative temperature coefficient (NTC) thermistor that will first limit the current (high resistance when it's cold) and then slowly reduce the resistance to allow the 6A needed by my fans to pass through it (low resistance when it's warm)? \$\endgroup\$ – MagTun May 6 '19 at 13:44
  • \$\begingroup\$ Yes to avoid overcurrent protection OCP starting up fans \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 6 '19 at 13:59
  • \$\begingroup\$ I can't find the same NTC that you linked to in your answer but I was able to find 8 Ohms 6A NTC here. If my thinking is right, I only need a 2ohm resistance at start up (12V/6A) so this NTC should be enough. What do you think? \$\endgroup\$ – MagTun May 19 '19 at 16:15
  • \$\begingroup\$ No margin for warm effects. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 19 '19 at 18:54
  • \$\begingroup\$ How do you know that there is "no margin for warm effect"? Do you infer this from the 8 ohms or from another specification? And to be sure, does "no margin for warm effect" means that the NTC will heat too much ? Thanks for your patience! \$\endgroup\$ – MagTun May 20 '19 at 7:35

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