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I want to supply a µC with 4x AA alkaline batteries or via USB, with the USB source being dominant. Meaning when USB is on, the batteries should be disconnected. For the USB path, I intend to use a Schottky Diode.

For the Battery path, I initially thought about using a "oring" IC or one of those "powerpath" controllers. The problem with most of the ones I've seen so far is, that they only have one integrated FET, so there could be a reverse current flowing into the batterie, when the batteries drop below the usb voltage and the forward voltage of the FET body diode.

I want to use two back to back FETs with common source, but I'm not sure what driver would be suitable.

One way I thought of would be to use a mosfet driver like MAX1614 and let it drive two Fets. But the Max1614 only goes down to 5V Vin which would be too high. Something like 4V would be better to get the most out of the batteries. Vin is between 4V and at least >7V, if possible >12V.

But most of the drivers I find are for Lowside configurations and some barely go above 5.5V.

Then there would be ones like the MAX5048 or the MAX15070 that go down to 4V.

Or I could use a oring controller with external Fets, something like a LTC4412 or LTC4359 (looks like the 4359 is hard to get in germany) and then use the enable pin to shut them down when the USB voltage is present.

I tried to design a discrete circuit with Mosfets and Transistors but that didn't work very well, I lack the experience.

Does someone have tips or ideas on a good approach? Or alternative mosfet driver ICs?

Edit: This was my general idea for a discrete circuit, but I didn't choose proper parts yet. When I simulate this in LTSpice, I get a huge spike for Vout when it switches from USB to Batt. (from ~4.7V down to 0V and back up to 4V) I think an output capacitor is needed to prevent that spike.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I think I've dealt with this exact problem before. It is kind of a pain because VBATT can be higher than VBUS. If you just use PMOS high side switches, VBUS may not be high enough to turn them off, and the battery may drain. Still, that might be your best bet. \$\endgroup\$ – mkeith Aug 12 '17 at 16:49
  • \$\begingroup\$ @mkeith I edited my original post with the discrete circuit I came up with. The parts are so far randomly picked, it appears to somewhat work, but I am sure there is a lot to improve there. \$\endgroup\$ – Winter Aug 12 '17 at 19:18
  • \$\begingroup\$ The way you did it, with two single transistor inverters, can definitely work. I probably would have just used two NMOS transistors (like BSS138) instead of one NPN and one PMOS. The problem with adding those transistors is that you will need to consume some battery current to keep the two back-to-back PMOS fet's on. So you will have a constant battery drain. You can get rid of both Q1 and M3 and just directly connect the gates to V2 (USB). That is basically what I did in my answer, below. \$\endgroup\$ – mkeith Aug 12 '17 at 23:57
  • \$\begingroup\$ Your powering the circuit with USB using diode and using USB you are enabling/disabling that path of Batteries to Load. with quick check, the circuit is OK. But is it OK for you to power your circuit with diode drop and power loss in diode with USB. \$\endgroup\$ – user19579 Nov 22 '18 at 6:04
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I think this will work. The only drawback is that maybe M1 and M2 will turn on just a little bit when VBATT is high, even though VBUS is present. This could lead to the batteries discharging just a bit. They will stop discharging when the voltage drops a little. This can be partially controlled by selecting M1 and M2 with kind of medium range Vgs(th).

Another alternative is a more complicated arrangement for switching the gates of M1 and M2. Maybe a buffer powered directly from VBAT, with VBUS as the input. This could insure that the gates get fully turned off when VBUS is present.

schematic

simulate this circuit – Schematic created using CircuitLab

R1 is just there to limit battery charge current in the event M1 or M2 should fail. You don't want to charge alkaline batteries. You could bump it up to 100k if you like. Or even 1M.

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With quick simulation, this kind of circuit should work. The FETs take a little time to open after the USB voltage goes away, so you need a bulk capacitor to offer current for that time.

Make sure that the P-FET conducts with 5 V V_GS

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You might check out the TI tips 22860. It's an integrated, low leakage, high side switch. Tie the input to the USB +5VDC Though a resistor. It's also pretty cheap.

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I routinely do this with use an isolated power source/regulator powering a low-side gate drive (any low-side gate drive method will work...nothing fancy here) with an optocoupler to float the control signal to the now floating gate driver.

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