Electronic – Switching Between Two Voltage Supplies

diodesfetpowerprotectiontransistors

I have been trying to work out a circuit something like the following:

schematic

simulate this circuit – Schematic created using CircuitLab

I want VOUT to be 3V when the switch is open and 5V when the switch is closed. The simulation seems to give me what I want when the switch is open, but not when it's closed. What am I doing wrong? Is there a better way to do this? I was avoiding a common cathode diode pair to avoid the forward voltage drop.

Best Answer

I came up with the following circuit that does what you ask:

Rail Switcher

5 volts from the "SWITCH" source will turn on M1 and M2, connecting the 3 volt rail to the load, and pulling current out of the right hand transistor of the differential pair Q3 and Q4. This turns off Q2 and shuts down the 5 volt rail. When Q1 turns off by applying 0 volts to "SWITCH", Q4 is biased on by the voltage divider comprised of R7 and R8, which turns on Q2 and the 5 volt rail. R2 then pulls up the gates of M2 and M1, shutting the 3 volt rail off. The arrangement of M1 and M2 prevents conduction from the 5 volt rail into the 3 volt rail when the 5 volt rail is connected.

In a real circuit there should probably be stopper resistors on the MOSFET gates.

Another way, in response to the "too many components" criticism:

comparator switch

If you don't want to use discretes, you can do it with a cross-coupled dual comparator. One IC and 4 resistors, plus the switches.