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I'm working on a project that operates at 5v - and can draw UP to 10 amps. Under normal conditions, it won't be doing that - probably closer to 3 amps, and (if configured for low-power) only 450mA or so.

It's designed to be powered from USB when in low-power mode, but I also want to have 2 barrel jack to add external power supplies (5V, 8A from aliexpress). Obviously I don't want the power supply backfeeding the USB, or the USB backfeeding the power supply. I want the option to plug in two of these power supplies because most of the time, I won't need to activate the 10A mode - so just one power supply would be sufficient.

Considering these power supplies are not designed to be paralleled (no master/slave setup, etc), how effective would an ideal diode be in enabling these various sources to all operate in parallel (to increase max current)?

Note that I don't need the load sharing to be perfect, but I'd like to avoid serious issues. I originally thought I could just use a P-Channel MOSFET, but after some research have learned that when they are "on" they allow current to flow both ways, which makes them non-ideal non-diodes (less than helpful).

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  • \$\begingroup\$ "Ideal diode" is usually a mosfet of some kind any way, so unless you can't switch with infinite speed you will have the same problem with backfeed \$\endgroup\$ – Anton Ingemarson Mar 20 at 15:30
  • \$\begingroup\$ "Unless you can't switch with infinite speed" - I'm not sure I'm following? Really I just need to, when I plug in one of the external power supplies, have it add it's available current to the mix. If I unplug something (and doing so should turn "off" the diode) I don't care about backfeeding any longer. \$\endgroup\$ – Helpful Mar 20 at 15:31
  • \$\begingroup\$ to get as much balance as possible between the supplies I would go with a diode with common cathode, to have temperature balance and let the current over Vf balance out. \$\endgroup\$ – Anton Ingemarson Mar 20 at 15:32
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    \$\begingroup\$ physically the mosfet barrier takes some time to close, so what I mean to say is that it will always be some backfeed unless it's perfectly balanced. \$\endgroup\$ – Anton Ingemarson Mar 20 at 15:37
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    \$\begingroup\$ Not necessarily silly, since there are circuits designed for these situations, though the design isn't great. I've read a couple of Linear Tech appnotes describing ideal current-sharing circuits etc. Anything with such a low voltage shouldn't really draw that much current because the voltage drops will be large relative to the nominal voltage. \$\endgroup\$ – pipe Mar 20 at 15:57
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You need two back-to-back P-channel MOSFETs for each power input. Here is an example from an Atmel eval board, though you'll want to scale up the MOSFETs considerably, and maybe reduce the resistor values to something like 10K:

enter image description here

The VCC_USB input is dominant if both are present (in other words, if only one supply is present, that one will be used, if both are present, then VCC_USB is used).

The back-to-back MOSFETs are "ON" when the common gate connection is at ground (assuming the left source is high) and are off with the gate connection equal in voltage to the source.

Because of the intrinsic body diodes you need two MOSFETs to be able to block in either direction. This is the common configuration you'll see in hot-swap controllers and similar devices.

The connection to R613 turns Q109A/B on if VCC_USB_EDBG is low. If both supplies are present Q110A/B are off (because their gates are high) and Q109A/B are on.

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  • \$\begingroup\$ Wow, thanks! I'm not sure what you mean by "dominant" in this context - as in, it will be the primary source of current (up to its max)? Additionally, I'm not entirely sure I understand the principles on which this is operating. Do you have a minute to explain? \$\endgroup\$ – Helpful Mar 20 at 15:36
  • \$\begingroup\$ Dominant means that if both supplies are present the power is to the output is supplied by VCC_USB, and the other one is disconnected. \$\endgroup\$ – Spehro Pefhany Mar 20 at 15:41
  • \$\begingroup\$ Ah. Hmm. The main problem with that idea is that I have ANOTHER external power supply, and I would want to be able to connect both of them at once (with out disabling one or the other). I don't suppose this circuit would enable me to do that (if I added another stage)? \$\endgroup\$ – Helpful Mar 20 at 15:43
  • \$\begingroup\$ If you want to share the current form the two external supplies you could combine them with Schottky diodes but unless they're specifically designed to share current gracefully the results may not be ideal (or usable). For example, one may hog the current and go into foldback shutdown. That's really a separate question, I think. \$\endgroup\$ – Spehro Pefhany Mar 20 at 15:52
  • \$\begingroup\$ Thank you for adding the explanation! Upvoted, but waiting to hear a few more answers before I select one. \$\endgroup\$ – Helpful Mar 20 at 15:52

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