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If I have two different voltage regulators both powered from different sources is it safe if they are both outputting in parallel? If one is on and the other is off?

I made a basic schematic to show what I mean here .

Schematic:

enter image description here

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  • \$\begingroup\$ Most likley there is an body diode which will conduct backwards though U2 if say your RAW is 0 and 16.8V is 16.8 V. On top of that, the highest output voltage will take presense and you may need output resistors to get current sharing. The usual business. \$\endgroup\$
    – winny
    Commented May 23, 2018 at 10:55
  • \$\begingroup\$ 7805 - don't think those have body diodes. But, they do have problems when the output is higher than the input. \$\endgroup\$
    – JRE
    Commented May 23, 2018 at 11:01
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    \$\begingroup\$ I've seen it done with multiple switching regulators that operate with a phase offset so they never conduct at the same time, but for this you really need to know what you are doing. \$\endgroup\$ Commented May 23, 2018 at 12:16
  • \$\begingroup\$ Multiple voltage regulators don't seem to be even needed. What you need is to combine RAW and 16.8V into a single voltage source which you would then regulate down to 5V. \$\endgroup\$ Commented May 23, 2018 at 12:36
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    \$\begingroup\$ RAW and 16.8V are different entirely different one is powered by a battery and one is powered by rectified 15VAC \$\endgroup\$
    – user163268
    Commented May 23, 2018 at 13:50

3 Answers 3

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No, this is not a good idea.

Check the datasheet of the regulators, but most don't like reverse voltage.

Parallel voltage regulators can't be counted on to share current. One will always have a little higher setpoint than the other. This could also lead to instability, depending on the nature of the controllers in the regulators.

A better way to solve this is to combine the two voltage sources before a single regulator. At this low voltage, you can use Schottky diodes. Put one diode in series with each voltage source. Power will then automatically be taken from the higher of the two voltage sources. Make sure to put something like a ceramic cap after the diodes physically close to the input of the regulator.

You can still use multiple regulators to spread the dissipation and to reduce voltage drops to distant parts of the circuit. You bus around the higher voltage out of the diodes, then regulate that locally as needed. However, you don't tie the outputs of multiple regulators together. You have each power a different part of the circuit instead.

If you want to minimize local dissipation, you use a buck switcher after the diodes. This makes a little more than the minimum input voltage of the regulators. You filter that a little and bus that around. Then you make the regulated voltage from that as needed locally. For example, if using a 5 V LDO that requires 5.5 V in, you might bus around 6 V. Each local regulator would then be 83% efficient.

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  • \$\begingroup\$ So I should put a shottky diode after both of the regulators to prevent reverse current? \$\endgroup\$
    – user163268
    Commented May 23, 2018 at 13:54
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    \$\begingroup\$ @9291Sam: No, you are missing the point. The answer explicitly says "better way ... single regulator". And the diodes are before it. \$\endgroup\$
    – MSalters
    Commented May 23, 2018 at 13:56
  • \$\begingroup\$ Ok thanks so it should look like this. i.sstatic.net/M1RpU.png [Link updated by a moderator.] \$\endgroup\$
    – user163268
    Commented May 23, 2018 at 14:07
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Analog Devices has a great article that discusses current sharing in parallel voltage regulators here. The gist is that you need to add extra circuitry in order to have the regulators share current relatively equally. A quote from the article is shown below. They are discussing Low Dropout Regulators (LDOs), but the same applies standard regulators like the 7805.

I would also worry even more about current sharing when powered by two different voltage sources, since the regulators would be even less matched in that case. Also, in the case of one of the voltages being disabled, if the regulator does not have reverse current protection, current can flow from the output back to the voltage input that is off.

Current sharing with linear regulators is traditionally not as simple as connecting the parts in parallel. Two voltage reference-based linear regulators set to the same output voltage and with the outputs tied together will not share current equally. An LDO's output voltage is determined by the reference voltage multiplied by a gain factor based on the feedback resistors. Due to tolerance errors in the voltage reference and feedback resistors, the output voltages will be mismatched. With unmatched outputs, the LDOs will not share current; one LDO will provide the majority of the current until it hits current limit, thermal limiting or its output droops low enough for the other LDO to begin supplementing its current. These three situations present circuit operation challenges and can pose reliability concerns, leading to possible premature failure of the overstressed LDO.

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    \$\begingroup\$ I don't think the goal is to have current sharing. The combination of "battery and 15VAC" seems to have alternatives. In fact, it might very well be the intent to not share current, and rely on the external voltage when present to save battery power. \$\endgroup\$
    – MSalters
    Commented May 23, 2018 at 14:00
  • \$\begingroup\$ You are correct the Goal is to not share current. \$\endgroup\$
    – user163268
    Commented May 23, 2018 at 14:26
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This will depend on what the datasheet of the regulator says.

In general: No, you cannot operate them in parallel, because supply and load changes will have different effects on your regulators, and thus they will start regulating "against each other".

Even more basic: No two regulators are exactly the same, even under perfectly constant conditions, so one is bound to constantly "pull up" the voltage, while one is constantly pulling it down.

Whether you can just put them in parallel with one unpowered: In general, I'd say "no". Again, if the datasheet doesn't contain clear information that this is possible, it probably isn't. Read the datasheet.

LM7805 (and clones thereof) are relatively "dumb" and slow, so you might be in luck and operating them in parallel might under non-oscillating load conditions actually work relatively well. Do not power one but not the other: current will run through the unpowered one!!

So, no, don't do that, especially not with 7805s. Instead, consider just having one regulator, whose input you switch between the different sources.

Also, it's probably a very bad idea to regulate 16.8 V down to 5 V with a linear regulator: Your waste heat is (16.8 - 5) V · I_out = 12 V · I_out, and if your output current is let's say 250 mA, you're already burning 3 W, which heats up your device by 36 °C. If you use more than 250 mA, you get hotter.

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  • \$\begingroup\$ Linear regulators (at least positive voltage ones) don't generally try to 'pull down' the output at all, do they? Although as noted they may not cope with Vout > Vin. \$\endgroup\$
    – nekomatic
    Commented May 23, 2018 at 12:27
  • \$\begingroup\$ @nekomatic of course they do. Not much of a regulator if you can only regulate voltages that are too low! \$\endgroup\$ Commented May 23, 2018 at 12:29
  • \$\begingroup\$ But the intended use of a linear regulator is to supply a load that draws current, not one that tries to push current back into the regulator. I'm looking at the internal schematic in the 78xx datasheet and I don't see which path in that circuit can sink a significant current from Vout to ground. \$\endgroup\$
    – nekomatic
    Commented May 23, 2018 at 12:34
  • \$\begingroup\$ @nekomatic the intended use of a linear regulator is to regulate – even in the presence of a changing load, which would inherently lead to occasionally rising voltages, especially when you consider reactive loads! But you're of course right, the LM78XX is a dinosaur, and is actually pretty bad at down-correcting. More modern regulators (especially LDOs) have a much more versatile error feedback loop. \$\endgroup\$ Commented May 23, 2018 at 13:29
  • \$\begingroup\$ Good to know that newer devices do this better, thanks! \$\endgroup\$
    – nekomatic
    Commented May 24, 2018 at 10:24

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