The main power is USB charger 5V. The backup power is battery 12V. Output should be 3.6V Current goes to DC-DC LM2596 that gives 3.6V output.

So, I cannot use simple schematics with zenner diodes or MOSFET cause to fact: voltage on backup battery is higher than that on main line.

The ideal switch is relay. But it is so slow. enter image description here

Is one more scheme based on 2 DC-DC LM2596 with inverted OE pin. enter image description here

The next solution is LTC4412, but its hard to find.

Advise me solution, please. I know how to solder SMD and have a lot of different fets, transistors, zenner, etc.

What about https://electronics.stackexchange.com/a/131048/97927 ?


How about the following design?

(The 12V is constant, the 5V is switching on and off every second.)

Currents through diodes show clean switching between supplies.

When 5V is on, NMOS3 opens, pulling down gate of PMOS2, which thus also opens, pulling up gate of PMOS1, closing off 12V.

When 5V is off, NMOS3's gate is pulled low closing it, which leaves gate of PMOS2 pulled up closing it, which leaves gate of PMOS1 pulled down, opening it.

Note, that when 5V is on, there is only insignificant leakage from the battery (proportional to the leakage current of the MOS gates).

  • \$\begingroup\$ Sorry, mine not work.. easyeda.com/editor#id=CYR9rlDVd \$\endgroup\$ – Boris N Feb 12 '16 at 18:26
  • \$\begingroup\$ Please comment part numbers. What is your max load current? \$\endgroup\$ – Szidor Feb 12 '16 at 18:56
  • \$\begingroup\$ I simulated it with "common" mosfets. Not concrete parts. \$\endgroup\$ – Boris N Feb 12 '16 at 19:33
  • \$\begingroup\$ Oh yes, default values are crap. Choose mosfets from list with small Ron values and above 12V Vds. Thats what I did. \$\endgroup\$ – Szidor Feb 12 '16 at 19:35
  • \$\begingroup\$ What you need is small Ron, small gate capacitance and fast switching. Also schottky diodes would waste less power. \$\endgroup\$ – Szidor Feb 12 '16 at 19:38

First, let's look at couple of critical requirements from what you have described. The USB has a tolerance of 4.75V to 5.25V. The LM2596 you have chosen has a minimum operating voltage requirement of 4.5V. Therefore somewhere between 4.75V and 4.5V from the USB input, the switch over has to occur. And, whatever that goes between the USB input and the regulator input cannot drop more than 0.25V.


simulate this circuit – Schematic created using CircuitLab

An 1% or so voltage reference is needed to set the switch over voltage to the middle of 4.5V-4.75V range. A TL431 is a convenient circuit that takes the place of a voltage reference and a comparator.

The PMOS needs to be such that the voltage drop across is less than 0.25V at maximum operating current. It is connected backward (compare to normal practice) to have the body diode oriented in the desired direction.

Personally, I would look for something other than the LM2596. There are inexpensive switching regulators that operate up to 1.5MHz and therefore requires a much smaller inductor. Also, if the input voltage requirement is lower, then M1 can be replaced with a diode.

  • \$\begingroup\$ What would be the 12V leakage through the TL431? \$\endgroup\$ – Szidor Feb 13 '16 at 13:36

I have seen this solution a couple of times. Simple ORing schematic for this purpose:

ORing switch

  • 1
    \$\begingroup\$ This does not work, because BAT voltage is higher than that on Main. So, MOSFET is always open by high-voltage BAT. This scheme has collected all disadvantages! It works like simple zenners, but has excess FET. And BAT voltage drops on zenner, there no "lowPower" feature. FET+zenner should be changed to oppositely FETs. \$\endgroup\$ – Boris N Feb 12 '16 at 10:45

There is a lot of circuit configurations you can try, but most of them have current leakage in the 12V supply. If you want something simple you may try something like this circuit

As you can see the parts are simple enough. - BC846 transistor - plain smd resistors - 5V coil relay like this for ex.

  • When the 5V power is present, the relay coil is energized and goes to NO, so the 12V is disconnected from the DC-DC converter.

  • When the 5V power is not present the coil is not energized and so the Relay goes back to NC, so the 12V power is connected to DC-DC converter

  • \$\begingroup\$ your scheme is a copy from my question (1st idea with relay). \$\endgroup\$ – Boris N Feb 12 '16 at 12:37
  • \$\begingroup\$ Aren't there true break before make relays out there? We used to use them when switching controls and loads in embedded work. \$\endgroup\$ – user65586 Feb 12 '16 at 13:00
  • \$\begingroup\$ @Boris N Sorry, I didn't understand from the image that you suggested a relay based solution. This is circuit is very simple, thats why i posted it. If you want more ideas probably you have to be more specific. For ex, what "fets, transistors, zenner, etc" you have? Can you buy other parts? Do you have MCU? etc... \$\endgroup\$ – hoo2 Feb 12 '16 at 13:14
  • \$\begingroup\$ @jdv I can not understand your English :( \$\endgroup\$ – hoo2 Feb 12 '16 at 13:15
  • \$\begingroup\$ @jdv I am afraid that the relay is too slow. My handmade gadget consumes 0.3A 3.6V, so I need to install too big capacitor to compensate slow relay switch. And big capacitor looks like short circuit at first time, when power source charges it. \$\endgroup\$ – Boris N Feb 12 '16 at 13:16

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