I am trying to design a board in M.2 key B form factor to split the USB and PCI-E signals to serve two devices.

However, M.2 key B only provides 2.5 A current (5 power pins, and 0.5 A each), but the total power requirement for the devices is 4 A. The motherboard I am working on has a standby 5 V pin rated at 2 A. I wonder whether there is a way to satisfy the power requirement by sharing current between the two sources.

Some of my "ideas":

  1. Motherboard 5 V->buck converter to 3.3 V->ideal diode->load

    M.2 3.3 V->ideal diode->load

    The main problem I see in this idea is the step down of 5 V may not really match the 3.3 V generated by the motherboard, and the current sharing does not happen until the load voltage drops below the minimum of the two. I am not sure whether the over-current protection could be triggered before current sharing even starts.

  2. Active current sharing like UCC39002. May work, but seems quite complex and I am not familiar with the relevant field.

Would really appreciate evaluations on my "ideas" and/or alternative solutions to the problem.

  • \$\begingroup\$ Where is this standby pin located? how will you connect it to your M.2 module? \$\endgroup\$
    – Anas Malas
    Oct 19, 2022 at 5:53
  • \$\begingroup\$ The front panel header of the motherboard provides 5V standby. Was considering simply connecting a jumper to the module. \$\endgroup\$
    – Shuhao Tan
    Oct 20, 2022 at 6:31

1 Answer 1


Although the M.2 Key B has no input for external power other than 3.3V, if you are to connect somehow an external power to your board then you should consider 12V instead of 5V stby. Because:

  • Every motherboard has (or is supposed to have) 12V supply
  • 12V coming to the motherboard generally has (or is supposed to have) higher output power
  • Conversion from 12V to 3.3V is easier compared to 5V to 3.3V conversion e.g. the stress will be low due to the lower duty cycle.

Active current sharing like UCC39002. May work, but seems quite complex and I am not familiar with the relevant field.

I'm familiar with it. I kindly recommend you to stay away from UCC39002 as it'll bring you headache, hair loss and sleep disorder. I'm not dishing on neither the product nor TI, it's just the complexity of the sub-system and the potential problems it causes, and it'll be overkill for your application.

  • \$\begingroup\$ That's a good idea. If I am stepping down from 12V and higher, can I get away with simply using the 3.3V as ENABLE signal for the regulator and supply the whole module with it to avoid voltage difference? \$\endgroup\$
    – Shuhao Tan
    Oct 20, 2022 at 6:35

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