I'm planning to do something you will probably advise me not to - diy a sub $50 lifepo4 charger.

So far (I think) I've got it figured out - I'll use boost converters in parallel, which should work if they have current limiting capabilities. The ones I got were advertised as 600W modules, but I see they have a 10A fuse on the input, so that holds true only when the input is 60V.. I think.

My battery is 36V nominal, so to get the maximum out of my boost converters I need to supply their inputs with a voltage as close as possible to the lowest charging voltage. I plan to set my BMS to cut off the load when a cell reaches 2.8V, so lowest possible battery voltage will be (12x2.8) 33.6V.

The charger will need to be able to provide 40A of charging current, or about 1500W. In order to do it with the "600W" boost modules, I figured I'd need 4. If each can take 10A with input voltage around 30V, then that'd be close enough.

The modules cost $10 each, so in order to meet my sub $50 criteria, I need an ultra cheap power supply solution. I can get used PC PSUs for pretty much next to nothing locally and after some digging around I believe they can be modified slightly and connected in series to produce higher dc voltage.

Finally, on to the question. If I connect 3 PSU's 12V rails in series I'll get 36V. The problem is that this is greater than the theoretical lowest battery voltage of 33.6V.

Would it be possible to use forward biased diodes in series to reduce the voltage?

I have a bunch of IRLZ44N mosfets laying around from old projects and what not, which the datasheet says have a diode voltage drop of 1.3V. Is it possible to use their internal diodes, as in would they be able to handle 20A, with heatsinks of course, and possibly active cooling.


simulate this circuit – Schematic created using CircuitLab

I could use the third PSU's 5V rail to make (12+12+5) 29V, but at 20A 3 volts are 60W, so I'll take that if I can. I'm not at all concerned about efficiency only about supplying the boost converter with as much power as possible.

Also, are there any better methods of reducing the voltage in this setup?

  • \$\begingroup\$ Sounds like a nice way to set your house on fire \$\endgroup\$
    – bobflux
    Dec 21, 2023 at 0:12
  • \$\begingroup\$ @bobflux I've thought about this as well, I'll charge outside! \$\endgroup\$
    – php_nub_qq
    Dec 21, 2023 at 0:17
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    \$\begingroup\$ If you want to use MOSFETs as diodes you need to tie gate to source. Otherwise gate is floating and drain-gate leakage will turn the FET on, which gives your suspiciously low threshold voltage, and it also conducts in both directions which is not desirable in this case. \$\endgroup\$
    – bobflux
    Dec 21, 2023 at 0:22
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    \$\begingroup\$ Can you share the datasheet for these 600 W boost converters with 10 A input fuses? \$\endgroup\$
    – The Photon
    Dec 21, 2023 at 0:40
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    \$\begingroup\$ Also, if these boost converters are so problematic, why do you think the solution is to add another power converter stage, which will reduce efficiency and add cost, instead of just choosing a more appropriate converter to begin with? \$\endgroup\$
    – The Photon
    Dec 21, 2023 at 0:42

1 Answer 1


You’re making a bad assumption, that dividing the output power by the fuse rating gets you the input voltage.

First, the fuse is going to be rated higher than the input current, as it’s meant to open on an overload condition. You don’t know if that fuse is rated 20% over expected current or 100% over.

Second, there’s the efficiency of the supply. The input power is going to be greater than the output power, at 80% for example a 600 W supply will need 750 W in.

So you’re going to have to rethink this a bit.


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