I'm a little confused about why these exist and what their distinction between regular positive voltage regulators are.

It seems to me like if I wanted to get -5V from -12V and 0V, that I would connect a normal 7V regulator, with -12V to its GND and 0V to its \$V_{in}\$.

Is the current flowing in the other direction? The positive regulator would fry or cause an open circuit?


3 Answers 3


Simple answer: You would get 5V in your case, but referenced to -12V instead of ground. In other words, you would have -7V, not -5V. In addition, the regulator would only source current onto the -7V rail, not sink it as would be expected for a negative voltage.

If you want to run some circuitry from 5V between -7V and -12V (the -12V will be the ground for this circuitry), then you can use a positive regulator as you described. If however you want to run some circuitry between ground or higher and -5V, then you need the negative regulator.

  • 2
    \$\begingroup\$ The OP mentioned using a 7V regulator, so he would have 7V referenced to -12V = -5V to GND. The important fact in this answer is the 'source' vs 'sink' aspect of LDOs. \$\endgroup\$
    – mxt3
    Jun 3, 2019 at 12:32

So let's take your 0 V and -12 V example. From the positive regulator's point of view, you don't have 0 V and - 12 V. Its GND, that is to say it's local 0 V, is whatever is connected to its GND pin. From the point of view of everything that's attached after the regulator, regulator's GND is its own GND and regulator's Vcc is its own Vcc.

What you've done here is basically replacing the rest of the circuit with Thévenin's generator. The load doesn't care what you have on the other side because it will simply see the positive voltage.

Let's take a look at this simplified circuit:

positive regulator circuit +7 V

Here we have the -12 V connected to the GND pin of the regulator, the outside GND connected to Vin of the regulator and we have a load connected to the regulator. The voltage on the load in this case will be +7 V, because the load's ground is same as regulator's ground. We've basically created a new virtual ground from which we can count voltage, however the main point is: There's no big difference from load's point of view between this and just calling the GND +12 V and the -12 V GND. You already need to have the negative voltage from somewhere and negative regulators are usually there to solve the somewhere problem. Also while the voltage at the load will be -5 V from the point of view of the main ground, it will be +7 V from the point of view of the load.

Here's another circuit which uses the same reasoning as the previous one:

positive regulator circuit +5 V

The main difference here is that the load is actually seeing the voltage of +5 V with respect to its own ground. That voltage is -7 V with respect to the main ground of the circuit, but the load itself won't actually see that, since the load's ground will be tired together with the regulator's ground.

Now let's take a look at a 7905 circuit:

negative regulator circuit

You have same ground between the load side of the regulator and the input side of the regulator! This is important! In almost all situations where the term negative regulator is important it is so because you already have the positive regulator somewhere. Devices which need to use both positive and negative voltage (such as operational amplifiers in some configurations) will need to have the positive voltage with respect to the common ground and negative voltage with respect to the common ground. While with your example, you will get -5 V with respect to the main ground, you'll already need to make the negative voltage somehow.

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    \$\begingroup\$ The basic problem is that you could do what Steven wants, and it would work, but then the load attached to the negative supply has to be floating with respect to the rest of the circuitry, assuming the device has positive and negative supplies. That will make interfacing other parts of the circuitry and external devices problematic, to say the least. Sometimes a floating output is exactly what you want, though, in which case regulating ground is a perfect solution (I've seen it done in some test equipment.) The circuit above makes a fine floating 7 volt supply. \$\endgroup\$
    – MattyZ
    Nov 20, 2011 at 11:41
  • \$\begingroup\$ @Bitrex Are you sure I'll get -7 V? From my thoughts, it should provide -5 V... If I'm wrong, I'll edit the part of the post out. \$\endgroup\$
    – AndrejaKo
    Nov 20, 2011 at 11:47
  • \$\begingroup\$ Whoops, you're right. 5 volts across the resistor. Sorry! Since it's floating, though, whether it's 5 volts or -5 volts depends on which way you connect the probes. \$\endgroup\$
    – MattyZ
    Nov 20, 2011 at 11:56
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    \$\begingroup\$ The '12V' in the last circuit really needs to be '-12V', to change from positive voltage to negative voltage would require a switch mode supply rather than a linear regulator. \$\endgroup\$
    – timrorr
    Nov 21, 2011 at 10:39
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    \$\begingroup\$ In your last picture (the 7905) the voltage at the Vout pin is what? -5V? No, the specsheets I see say Vin must be <-7V. So if Vin was +12V it would fry or not work. \$\endgroup\$
    – Steven Lu
    Jun 19, 2012 at 5:52

It seems to me like if I wanted to get -5V from -12V and 0V, that I would connect a normal 7V regulator, with -12V to its GND and 0V to its Vin.

There is IMO no 7V regulator commonly available, and your solution would behave unexpectedly in the case that the -12V is unregulated:

Lets say the voltage climbs to -10V because of higher current. Now the output would be only -3V in your case. A 7905 regulator would still output -5V.


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