Switching between two power sources

Question

I am designing a circuit, which can use either 5V from USB, or some higher voltage (7-12V) from a battery. If both power sources are connected at the same time, I want some kind of electrical switch, that would disconnect the battery and power the circuit just from the 5V USB. But the circuit still has to work, when only one of the sources is present. I tried to design a switch using two mosfets, but I was not able to figure out a working circuit.

Can such a circuit be build by only using two mosfets? Also, do not worry about the voltage regulation, that is taken care of.

Answer 1

There are dedicated chips that you can get which will do this but doing it with discrete parts would look something like this:

^{simulate this circuit – Schematic created using CircuitLab}

All part numbers are the defaults rather than recommended parts.

Without the USB disconnected R1 ensures M3 is off. R5 turns M2 off and R2 turns M1 on. The end result is that the output is connected to the battery.

With USB powered M3 turns on. This pulls the gate of M2 down turning it on which in turn pulls the gate of M1 high and turns it off. Power then flows through D1 (which should be a schottky diode) and to the output.

D1 also protects the USB from over voltage while the voltage on C1 drops to USB levels. Note, if the battery is under the USB voltage then the body diode of M1 will feed power into the battery. This is outside of your stated operating range but if it is a possibility add a diode to prevent it.

C1 should be sufficiently large to prevent the output voltage dipping too far during switch over.

And now I wait for everyone else to point out the problems with this circuit (or point out how to do it with half the parts) since I'm sure I've overlooked something...

Answer 2

The simplest way that I can think of uses 3 parts. A CPC1117N, a resistor and a diode (eg. 1N5819).

^{simulate this circuit – Schematic created using CircuitLab}

Operation should be self-evident- the presence of the +5V USB source turns off the 9V battery source and D2 prevents back-feeding the USB +5.

This circuit has no provisions against brown-out in the USB+5 input (for example, a 3V input could switch off the battery and leave only 2.5V at the output). If you need that, add a power supervisor chip to switch the SSR. An LM431 and two resistors would work too.

Answer 3

That's the job of power monitors.

You can either buy a ready-made circuit that is pretty clever about that, for example the LTC4412, which will ensure low switchover transients etc.

Or you can actually build this yourself, as you said, from MOSFETs. In principle, yes, one or two might suffice, if you got a few diodes to spare to ensure a few voltage drops and avoid current flowing into the USB port.

^{simulate this circuit – Schematic created using CircuitLab}

Downside of this is clearly that operation from battery wastes energy by letting voltage drop over D2 – but that's necessary to ensure that M3's V_GS is always positive when V_G = V_Bat, even when there's a high voltage drop across the load.

You can build a more elegant version of this circuit essentially by employing the CMOS ideology – but you'd be, logically, building nothing different than two logic circuits: one that conducts power from the battery if (not USB voltage ), and one that conducts power from USB if (USB voltage).