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I have a 0.08A 12V fan that I'd like to run on a 24V PSU. I have a 1/4W resistor kit with several different resistor values and a max input 23V buck converter. Can I add a resistor in series before the buck converter to bring the voltage of the PSU safely down to 22V? I've calculated that I need a 25ohm resistor, is that correct? I've included an image of my proposed setup.

This is a datasheet for the buck converter I have https://matts-electronics.com/wp-content/uploads/2018/06/MINI-360.pdf.

24V to resistor then buck then 12V fan

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    \$\begingroup\$ DON’T! Get a proper DC/DC rated for at least 24 V + 50 %. If you absolutely must use it, put a 6.8 V-ish Zener in series. \$\endgroup\$
    – winny
    Jun 20 at 16:42

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This is not very reliable. The MP2307 is rated at 23 V which is an unusual value and indicates there might be a problem operating above this. Your 24V supply will have a tolerance and could be 25 V.

When unloaded, the series resistor won't drop the voltage you expect. The fan rating is probably the MAX loading.

Your calculation is incorrect anyway. The DCDC buck converter will consume approximately ½ the load current at the input - i.e. it will convert 24 V at 0.04 A to 12V at 0.08 A. Thus you need the R to drop 2V with 0.04 A ==> 50 Ω.

A better method would be to put a 4.7 V zener diode in series with the 24 V.

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  • \$\begingroup\$ As an alternative to the zener diode you can use 4 standard diodes in series. This will also drop 2 V, depends on the components you have \$\endgroup\$
    – Jens
    Jun 20 at 16:21
  • \$\begingroup\$ @Jens Does using 4 standard diodes have have the same drawback when the resistors are unloaded? \$\endgroup\$
    – Fallen
    Jun 20 at 16:44
  • \$\begingroup\$ @Fallen No, the minimum current drawn by the converter without load is high enough \$\endgroup\$
    – Jens
    Jun 20 at 16:49
  • \$\begingroup\$ So 4 50 Ω resistors in series? Wouldn't that drop the voltage by 8 V? \$\endgroup\$
    – Fallen
    Jun 20 at 17:03
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    \$\begingroup\$ For the benefit of others - it is only possible to carry this so far. If you try to add resistors in the input that drop more than ½ the supply, the system will become unstable and unpredictable. This is a consequence of the NEGATIVe input impedance of a DCDC converter \$\endgroup\$
    – jp314
    Jun 20 at 17:56

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