I have a generator that outputs upto 20v dc, to power a 12v led light I tried a buck+boost converter (xl6009 based), light requires 1Amp on 12v.

Problem is when engine is not accelerated, generator outputs around 4v but enough amps, buck boost tries to boost them and burns diode.

What would be minimal circuit to cutoff supply to buck boost incase of voltages below 6v? Can I use n-channel mosfet so it doesn't let enough amps through when not fully turned on?

  • \$\begingroup\$ You need an Under Voltage Lock Out circuit with an hysteresis of at least 2 Volt. Hysteresis is very important in order not enter hic-cup mode. I designed and tested one circuit like this in the past. If I find it, I'll post it here. \$\endgroup\$ Nov 19 '21 at 11:04
  • \$\begingroup\$ Like this: electronics.stackexchange.com/questions/4967/…? \$\endgroup\$
    – devnull
    Nov 19 '21 at 11:32

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