2
\$\begingroup\$

Background

I would like to make a simple battery-powered circuit with two alternatively selectable linear voltage regulators to set the output voltage.

Question

Given the circuit below should I worry about the two regulators "upsetting" each other in any way or is this going to work just fine?

Alternatively Selectable Voltage Output

\$\endgroup\$
  • \$\begingroup\$ The two regulators are quite safe in the circuit as shown, you see this arrangement on many MCU boards for both multiple voltage and USB verses Walwart powered boards. \$\endgroup\$ – Jack Creasey Mar 15 '17 at 19:13
  • 1
    \$\begingroup\$ You do realize when on the 5V you will waste 44% of the battery used and when on the 3.3v the loss will be 66%. This is not the job for a linear regulator. The LM317 is better than the two, but neither is a good choice. \$\endgroup\$ – Misunderstood Mar 16 '17 at 0:04
  • \$\begingroup\$ @Misunderstood Thanks for your note, good point. So this is basically just for testing out subcircuit designs -- not an actual implementation. I just quickly wanted to throw together a selectable voltage, battery-operated "power supply" to have to hand at workbench from the parts I had lying around. \$\endgroup\$ – pfabri Mar 16 '17 at 18:26
2
\$\begingroup\$

There is nothing wrong with your schematic the voltage regulators will not upset each other in any way.

However, it would seem if you change your LM7805 to an LM317 (or very many others) you could simply use a single regulator and change the output voltage. Look at the datasheet for any adjustable regulator such as the LM317.
This may also help you.

You'll see many arrangements such as this: enter image description here

Here they used jumpers to select the voltage, but you could just as easily use a switch.

Additional notes on "Reverse Current" regulator failures:
There is considerable mythology about this issue for this type of linear regulator and many times it's put down to these being older designs. Nothing could be further from the truth, this applies to even the most modern designs.
Looking at the datasheets for the LM7805 and the LD1117V33 you can actually map the current paths from output to input if the input is shorted. You'll notice for both there are multiple paths. In the 7805 there is a low current path that starts conduction at only about 1.6 V while the LD1117 has a full reverse Vbe of about 6 V before conduction.
enter image description here

For more modern LDO devices that use P-Channel FETs the reverse voltage capability is even lower, about 800 mV due to the intrinsic diode in the FET.
For example the LP38511 clearly shows the intrinsic diode across the power FET: enter image description here

Your design has to allow for this reverse current and this is a failure vector if you have short faults or a crowbar on the input side of the regulator. Controlling the output capacitance can help to reduce the transient currents to a safe level, but this is critical when multiple regulators feed an output as in the OP's circuit. This circuit is quite safe because the input capacitor for each regulator is not in any way clamped. So while there may be an initial current flow when changing from one regulator to the other, the respective input capacitor initially holds up the input voltage and will discharge due to load current, and then can be charged by the reverse current and still be within ratings.

To be safe against reverse current failures in the regulator you can add a Schottky diode from input to output. This provides a single well defined high current path when the output is above the input. However you have to carefully consider this in your overall circuit design as the output can now only be about 250 mV above the input under any condition.

\$\endgroup\$
  • \$\begingroup\$ Excellent, thanks for the schematic. The LM317 looks nice, but I it's these two separate voltage regs that I have to hand right now. Wow, nice link too! \$\endgroup\$ – pfabri Mar 15 '17 at 19:26
4
\$\begingroup\$

They will likely upset each other so why not consider using a single adjustable regulator and controlling the feedback resistors using the switch. Should be fairly painless. Consider the ubiquitous LM317 as a starting point.

\$\endgroup\$
  • \$\begingroup\$ Yep -- potentially, if the resistors required for the desired outputs line up nicely, you could even make it digitally adjustable by the system with a NFET to switch in a given resistor. \$\endgroup\$ – Krunal Desai Mar 15 '17 at 18:55
  • \$\begingroup\$ How might they interfere with each other? I was thinking that neither oscillation, nor higher voltage on the output than the input were impossible and only one of them is connected at any given time. I'm trying to understand the "evil" underlyings here. \$\endgroup\$ – pfabri Mar 15 '17 at 18:55
  • 1
    \$\begingroup\$ An 7805 can be destroyed by a 'reverse current'. I don't see how this could occur with only a 330nF at the input, but still. \$\endgroup\$ – Wouter van Ooijen Mar 15 '17 at 19:04
  • \$\begingroup\$ @WoutervanOoijen. Providing the output voltage does not rise above about 6 - 7 volts there will be no reverse breakdown within the 7805 or the LD1117. Even if the output voltage were taken above the breakdown voltage in the configuration used by the OP the input capacitor would simply be charged. It is only by shorting the input or using very large input and output capacitors that this failure mode occurs. In this case the circuit is quite safe. \$\endgroup\$ – Jack Creasey Mar 16 '17 at 3:57
3
\$\begingroup\$

If you want to use your current circuit, I suggest you put the switch at the outputs of the regulators, instead of at the inputs. That avoids all (potential, real or hypothetical) problems.

But a 317 would probably be a better idea.

Note: you mention 'battery powered' and your circuit shows a 9V source. Be aware that a 9V battery will very soon drop below the input voltage required by an 7805 (7V or 8V, depending on the brand).

\$\endgroup\$
  • \$\begingroup\$ Thanks for the note on 9V batteries. It's just for the schematic, though. In reality I'll use high-capacity 1.5V cells in series. \$\endgroup\$ – pfabri Mar 15 '17 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.