Build two independent circuits with one switch

Question

I have 2 DC sources 12V and 5V, I want to switch them using one switch as shown on scheme. I used here 2 diodes to isolate circuits and it works fine on simulation, but in real scheme 5V (actually rather 4.7V from USB) becomes 5.4V. How can I make circuits run independent (ie. they do not affect each other) in this case?

Simulation with open/closed switch - works fine on simulation but not in reality

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Using a single-pole normally-open switch to select '5 V' or 12 V supply.

• SW1 open: 5 V is supplied to the load via D1. A silicon diode will drop 0.7 V. A Schottky diode will drop about 0.3 V.
• SW1 closed: 12 V is supplied to the load. D1 prevents backfeed into the 5 V power supply.

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^{simulate this circuit}

Figure 2. Switch to disconnect two loads.

• SW1 will disconnect the common return from both loads to the common ground at the power supplies.
• D1 prevents V1 back-feeding into V2 when SW1 is open. It will cause a voltage drop in normal operation as described above.

This isn't an elegant circuit. Switching negatives in a negative ground circuit can lead to all sorts of confusion. It is, however, common on automobile circuits.

Answer 2

Here's how, with the switch 'S1' being used to control the 12 V load and a 12 V electromagnetic relay to control the 5 V load.