How can I set a wire to one of 3 voltages from a simple digital controller (e.g. Raspberry Pi)?

Question

Here's my problem. I have a device to control that needs to always have one of three voltages applied to it: 5V, or 6V, or 7V. And I need to be able to programmatically switch between them.

What I have available is a general mirocontroller with GPIO pins that I can set at some different voltage (e.g. 3.3V). I have a feeling relay circuits could do this but I can't figure out how to switch between my 3 different voltages.

Question summary:

What circuit would implement this with simple components? How should I create the 3 voltages I need? Assume I can buy whatever simple parts from Amazon etc.

Answer 1

Two SPDT relays can be used to select the voltages like this:

^{simulate this circuit – Schematic created using CircuitLab}

You will normally require transistors to drive the relays, as most microcontroller outputs cannot handle a relay's coil current.

With GPIO2 Low, "7 V" will be selected. GPIO1 will have no effect.

With GPIO2 High and GPIO1 Low, "6 V" will be selected

With GPIO2 High and GPIO1 High, "5 V" will be selected

Answer 2

I have shown 2 3 5 7 volts rather than 5 6 7 V.
lter to suit.

I showed MOSFETS for M 5 6 7 and then realised that when Vot is more than about 0.5V above other supply voltages the MOSFET body diodes of any lower voltage supply will conduct. This can be avoided with back-to-back MOSFET pairs (annoying) or by using bipolar transistors for M 5 6 7 plus an extra base drive resistor in each case from M 7 8 9.

Left hand circuit:

Each of 3 existing voltages can be switched to Vout.
Dissipation in M4 ...6 is low
= Rdson x Iout^2

Right hand circuit:

LM317 regulator canm be set to any desired output Voltage > 1.25V and < Vmax_LM317.
Vout = 1.25V x (Rx + R1)/R1
Rx = any of R2 R3 R4.
Vout will rise to ~= Vin if all inputs are off. This can be avoided by replacing one output switch with a resistor set to provide the highest voltage (here 7V) and the lowering this with the other switched resistors.

Dissipation in LM317 is significant at high currents
= (Vin - Vout) x Iout.
LM317 has about 2V "dropout voltage" at higher currents.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Here is a method that's a minor variation on Russell's method.

^{simulate this circuit – Schematic created using CircuitLab}

With A & B both set to be inputs (no pullup or pulldown, use ST mode), output voltage is about 7V

with A set as output 0V, B set as input as above, output voltage is about 6V

With B set as output 0V, A set as input as above, output voltage is about 5V

If you intend to draw substantial current from the output, pay attention to the power dissipation. At 5V out, 100mA will result in about 735mW of dissipation, about where you will want a heatsink on a TO-220.

The resistor values are chosen so that no more than 3.3V appears at A or B. D1 is optional protection in case V1 can be shorted or subjected to a heavy load on turn-off. C1 and C2 are input and output capacitors. R2 sets the current through the resistor chain at a bit over 5mA.