I'm planning to use a chip with two 5V output pins to provide power to two systems: a MCU (attiny85, operating voltage: ~2V - ~5V) and a USB device (input current: 3A). On top of that, I'll have an alternative 5.3V input source to power these systems.

I think the chip's 5V pins will need to be protected from current flow when the alternative source is active. I believe I could perhaps use two sets MOSFETs back-to-back and two Schottky diodes for the required circuitry. The diode will likely have a 300mA - 400mA voltage drop, which is why the alternative source is boosted to 5.3V to accommodate the voltage drops.

I'm very new to designing transistor circuits, so please bear with me. Is the following a good circuit for the task? I'm thinking of using Si2305CDS, which I just happened to find on Digikey and seems to have the right Vgs.

enter image description here

I added a 100r to each gate and a 1M to GND by following this answer. The idea is to block current flow in PIN 1 and PIN 2 when VDD is active. Is this the entire circuitry required? Are there things one should look out for?

Any advice will be much appreciated.


EDIT: Fixed MOSFETS. Thanks to @mkeith.

  • \$\begingroup\$ Is VDD the 5.3V source? And do I understand you right that you want to be able to turn on/off the supply for each system separately? Because then you would have to use your diodes in common anode config, not to mention the double drop of forward voltage from VDD to output2. \$\endgroup\$
    – christoph
    Sep 5, 2015 at 17:02
  • 2
    \$\begingroup\$ What is the "chip with two 5V output pins"? \$\endgroup\$ Sep 5, 2015 at 17:14
  • \$\begingroup\$ @christoph Sorry, yes it is. I've fixed it now. Is this the common anode config you're suggesting? Each system doesn't need to be switched on/off separately. The both can be powered up/off when VDD is/isn't in place. \$\endgroup\$
    – Kar
    Sep 5, 2015 at 17:14
  • \$\begingroup\$ @BruceAbbott It's the BQ24295. To be precise, the MCU (SYS) pin gives a max of 4.35V only. I thought I'd simplify issue by just taking it as 5V. Hopefully that won't affect the design. \$\endgroup\$
    – Kar
    Sep 5, 2015 at 17:20
  • \$\begingroup\$ The title mentions reverse currents, under which circumstances and where do you expect these? \$\endgroup\$
    – christoph
    Sep 5, 2015 at 18:10

2 Answers 2


If you are going to use a mechanical switch anyway,then this one is simple and cheap. Both your supplies will be mechanically isolated.


If you want electric circuit, and if you are sure that 5.3V supply will always be greater than (Pin1 - Vf(diode) - Vf(schottky)) and Pin1 & Pin2 supply won't be less than (5.3V - Vf(diode) - Vf(schottky)), then you can use this diode based circuit.You just need to used diode with high forward votlage drop. And schottky aren't that useful in that case.


simulate this circuit – Schematic created using CircuitLab

Else the @next-hack's Answer is the solution.


You need back to back MOSFETs connections. I suggest the attached diagram. You should use logic-level MOSFETs.

Let's discuss only the first output. The second one is similar.

Let's consider all the 4 cases.

  1. 5Vin_1 off (open), SW1 on: the gates of M2 and M1 are at 5.3V. The sources of M1 and M2 are at 4.2V (5.3V minus the forward drop of D1 and the forward drop of the body diode of M1, i.e. 5.3V - 0.4V - 0.7V). M1 and M2 are off. Out_1 is 4.9V. No back current to 5Vin_1.
  2. 5Vin_1 on, SW1 on: the gates of M2 and M1 are at 5.3V. The sources of M1 and M2 are at 4.3V (5V minus the forward of the body diode of M2). The body diode of M1 is off. Still, no back current to 5Vin_1.
  3. 5Vin_1 off (open), SW1 off: the whole circuit is unpowered.
  4. 5Vin_1 on, SW1 off: R1 brings the gates at 0. Initially, the body diode of M2 will bring the sources of M2 and M1 to 4.3V. Because now VGS = -4.3V (one should use logic level MOSFETs) then both M1 and M2 will turn on. (And now, since RDSon is small, Vs is brought to 5V, therefore Vgs becomes -5V). D1 is off.

Also the transitions between powered states (i.e. when you insert another power source, when the other is already active) must be checked.

SW1 ON-->OFF transition (With 5Vin_1 present) The gates of M1 and M2 are brought to 0. Then the MOSFETs start conducting following point 4 of the previous list.

SW1 OFF-->ON transition (With 5Vin_1 present) The gates of M1 and M2 are brought to 5.3. Then the MOSFETs will eventually turn off. Only the body diodes can conduct, but eventually we will go back to case 2.

SW1 ON. 5Vin_1 was open, then it is connected The gates of M1 and M2 are already at 5.3. The MOSFET remain OFF. The insertion of the 5Vin_1 will just bring the source (through the body diode of M2) to a slightly larger voltage.

SW1 ON. 5Vin_1 was connected, then it is removed The gates of M1 and M2 are already at 5.3. The MOSFET remain OFF. The removal of 5Vin_1, will eventually bring the source voltage to 4.2V (due to leakage currents). No problem.


simulate this circuit – Schematic created using CircuitLab


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