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I want to experiment with amplifiers but I need a negative power voltage source.

I looked for some circuits but as far as I found they either are not simple & reliable or require an IC.

Is there any simple way to get negative power supply without using a railsplitter IC or AC power source?

Also I read about these 'virtual grounds' but I couldn't figure it out. Is it like ; we choose a positive voltage point as ground so technically the real ground is more negative relative to positive point we chose?

ac to dc dual power supply

This one works well but needs an AC power supply and high capacity and voltage capacitors and they take a lot of space but so far best I could find.

Virtual ground resistor circuit

This one looks simpler and doesn't need an AC source or big capacitors but either I can't understand how it works or it doesn't work that well.

Edit: If what I want is not possible, can you recommend some ICs for it? I can order them from abroad because they're not common in my country and this why I asked about circuits that don't involve them in the first place.

Edit I think I was not clear about what I want:

I specifically don't want a circuit which requires an AC power supply because yes I know it's so easy to make a circuit like that.

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    \$\begingroup\$ there is no real ground ... a ground is the point in the circuit that you choose to be your reference point ... you place the negative terminal of your voltmeter on that point \$\endgroup\$
    – jsotola
    Aug 25, 2019 at 22:02
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    \$\begingroup\$ How much current does the negative supply needs to source? \$\endgroup\$
    – User323693
    Aug 25, 2019 at 22:25
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    \$\begingroup\$ I don't know yet as I said I want to experiment it but between 0.1A to 1A . I never use more than 3 A because test probes & cables starting to melt.Also my power supply is 5-12v 20A (computer power supply) \$\endgroup\$
    – Mordecai
    Aug 25, 2019 at 22:28
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    \$\begingroup\$ @Mordecai You could look at something like this. Note that you are asking for what amounts to a power rail-splitter. And those are not commonly found in ICs. \$\endgroup\$
    – jonk
    Aug 25, 2019 at 22:58
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    \$\begingroup\$ @jonk thanks for the circuits but as I said , I'll only use ICs if what I want is considered impossible otherwise I don't want them because delivery will take minimum 60 days \$\endgroup\$
    – Mordecai
    Aug 25, 2019 at 23:12

5 Answers 5

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enter image description here

I believe that this is by far the simplest way to get negative voltage, if you can simply use two batteries in series and treat the mid point as GND. then the + side of the 1st cell will be your positive rail and the - side of the 2nd cell will be your negative rail.

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    \$\begingroup\$ Downside of this is that if your loads aren't drawing exactly the same current, your batteries don't drain evenly. \$\endgroup\$
    – Hearth
    Oct 1, 2020 at 19:38
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The little rail splitter circuit with two resistors and 2 capacitors is a poor design:

schematic

simulate this circuit – Schematic created using CircuitLab

With R1 and R2 equal, you have a resistor divider which gets you a potential half-way between its top and bottom potentials. C1 and C2 simply try to keep that voltage steady as the power supply wobbles up and down slightly, but asside from that, they aren't much help.

The problem is that any current being drawn via R1 or R2, due to things connected to that mid-point (for example, Rload in the circuit on the right), will cause the center potential to to deviate from the required half-way value, so you have to be super careful not to "load" it in any way. Even worse, as the load varies (and it probably will), the mid-point potential varies with it.

This is a reasonably good starting point though, because at least you get a nice 6V reference potential, that you can use to create a stronger, more robust 6V supply point. You can do this with a single opamp, connected as a voltage follower:

schematic

simulate this circuit

The opamp input draws no current from the divider, so you needn't worry about it ever loading and upsetting the +6V reference. It uses negative feedback to counter any influence that a load presents at its output, to maintain a steady 6V there even under modest load.

You can power this circuit from any DC source over 4V or so, reassign your ground point, and re-label the supply rails to more meaningful names. For example:

schematic

simulate this circuit

This will work quite well as long as you don't ask for more than about 10mA from the opamp (sinking or sourcing). More than that and you'll need a beefier solution.

Also, beware the dreaded ground loop. If your source of DC power is not isolated from mains Earth, you must not connect your new ground point to that same Earth, because it has a different potential.

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  • \$\begingroup\$ With that solution, if I want to connect, let’s say, a comparator and a microcontroller, which ground should I use? \$\endgroup\$
    – ludicrous
    Apr 6, 2023 at 13:07
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You could try using a MAX1044/ICL7660.

This is a DC-DC converter I use for bipolar PSUs in audio circuits sometimes. You can use a 9V wall wart or battery to get a -9V output for example. Just make sure to filter it enough and connect the "Boost" if you're working with audio frequencies. The boost will bring the switching frequency above the audio band so you don't hear that noise.

There are some example schematics in the datasheet you might find useful. I often use the one from Figure 6 on page 8.

Hope this helps!

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One solution is to use two "wall wart" (or other) DC power supplies, and connect the positive terminal of one supply to the negative terminal of the other, and call that "Ground/Zero volts". The free positive and negative terminals are your bipolar supply.

This will work with the typical "wall wart" supply, as their outputs are "floating" - neither side is connected to any "Ground". Caution may be required with other sorts of supply, as their negative terminal may be connected to a metal case, and thence to "Safety Ground/Earth Ground".

Another solution would be to use an isolated DC-DC converter - connect its positive output to the negative terminal of your main supply, then its negative terminal is your negative supply. Some DC-DC converters are not isolated (their input and output ground terminals are connected) so cannot be used here.

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    \$\begingroup\$ The OP said they are specifically looking for a solution that does not require ac power. \$\endgroup\$ Aug 25, 2019 at 23:00
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    \$\begingroup\$ @ElliotAlderson: I believe he said he didn't want something that needed AC because he couldn't get a transformer - I wanted to point out that easily-available DC supplies can be connected to provide a bipolar supply. He's really limiting his options if he won't consider a couple of wall-wart supplies. \$\endgroup\$ Aug 25, 2019 at 23:09
  • \$\begingroup\$ I think I didn't explain myself enough , I have a lot of ac or dc wall adapters or transformers , this not a problem. Problem is I don't want to limit myself with these options , like everyone knows how to get negative supply with ac but If you designing a portable device with batteries( for exp : a mp3 player) then how can you make the amplifier without ac supply ? I just want to know if there is a another ways to do it . \$\endgroup\$
    – Mordecai
    Aug 25, 2019 at 23:24
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    \$\begingroup\$ @Mordecai: if you are using a battery supply, you simply call the mid-point of the battery "Ground" - For example,you could connect two 9 volt batteries in series, and call the connection between the batteries "Ground/Zero Volts", or you could use a single battery to provide the positive supply, and an isolated DC-DC converter for the negative. \$\endgroup\$ Aug 25, 2019 at 23:33
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    \$\begingroup\$ @Mordecai Your second image (the one you said you didn't understand) uses a single battery and a voltage divider to create a "virtual ground" (VGND). Its the same solution using a single battery. Disregard the capacitors and I think you can understand it better. It "doesn't work well" because of the resistor divider and whatever you connect is a varying load, but the concept is there... \$\endgroup\$
    – Ron Beyer
    Aug 25, 2019 at 23:43
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You could get ahold of an old ATX power supply, which has -12V on pin 14.

That's probably your cheapest and easiest route if you don't care about the noise from a switching supply. You'll need to add a dummy load to get it to start up most likely (google for details, there are a number of folks using these for a poor man's bench supply ). I really can't condone getting inside where the 400V mains-derived voltages live, but some do it, and many, I assume, live to tell the tale.

Once you have +/-12V you can regulate it down to lower voltages, but many op-amps are just fine with +/-12V.

Note: Such a supply does not really have the protection and current limiting of an expensive bench supply so you should at least fuse the outputs so that they can't burn up any wires you attach.


If you want to try a "rail-splitter", chip the TLE2426 is designed for this task, but note that you'll end up with limited ground sink/source capability (read the datasheet carefully), an oddball chip, and your split 'ground' will not be at the same potential as your power supply ground, possibly causing problems if you are using things like USB-to serial cables, oscilloscopes and USB-powered microcontroller in-circuit emulators that are earthed.

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    \$\begingroup\$ I believe ATX power supplies typically can't handle much current on the -12 rail, though, so be aware of that. \$\endgroup\$
    – Hearth
    Aug 25, 2019 at 23:01
  • \$\begingroup\$ The OP said they are specifically looking for a solution that does not require ac power. I don't think an ATX power supply is a reasonable answer. \$\endgroup\$ Aug 25, 2019 at 23:01
  • \$\begingroup\$ @ElliotAlderson He specifically said he's using a PC power supply, so a suggestion to use a different one is appropriate. \$\endgroup\$ Aug 26, 2019 at 0:13

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