In analog synthesis, to generate almost any signal with op amps, it is necessary to have positive and negative voltages. This allows the op amp to generate a signal that spans positive and negative voltage values.

this process is pretty simple when using 2 different power supplies, but to create it with 1 supply it is necessary to create a voltage divider, then to reverse the polarity of supply.

This instructable will inform you how to do this.

## Step 1: Schematic

Here is the schematic of the voltage divider.

Note: The ground symbol is the virtual ground

Breadboards are used to create solderless joints on circuits. Here is the breadboard layout for create positive and negative voltages.

Note: To create a positive voltage, use the red column and virtual ground. To create a negative voltage, use the blue column and virtual ground.

## Step 3: ...and You're Away

With this, you can use this layout as a basis for all dual polarity voltage needs!
<p>I'm sorry but this only works for an &quot;ideal circuit&quot;. In the real world it's not this simple unfortunatly.</p>
<p>You need an amplifier at the middle-point to buffer the voltage division, then you can use the psu-original ground as -V/2. Voltage is relative to potentials.</p><p>To the other guys here in the comments complaining about the bad circuit with no stability: you are all CORRECT. ;-)</p>
<p>Thanks for sharing, good sir. I used your schematic to make positive and negative voltages using a UA741CN and two 10k resistors and it worked like a charm! </p>
<p>Here I made it with a UA741CN and 2x 12K .1% resistors and it works really fine.</p>
<p>IT WILL NOT WORK as when load is applied . resistor get solved in parallel with load as a result voltage will not be same on the two resistors R1 &amp; R2</p>
<p>Of course it works. Voltage dividers are common circuit elements.</p>
<p>No... It does not work as the load you apply will change the voltage divider resistance values and the voltages will not be symmetrical.</p><p>In addition to that, how much current can this supply?? Less than 0.5mA if I'm not mistaken as the circuit always have a 10K resistor in the way...</p>
<p>Then you admit that it works? Good! Because like I said voltage dividers are common circuit elements. Of course you cannot expect very much current out of them. But they do work as well as they need to.</p>
<p>As with anything in the realm of engineering, we want to make a simplistic and practical design. This fits simplicity, but it is not even close to practical. If a load is added to the system, the INPUT IMPEDANCE will change. That is insanity! As soon as ANY load is added, the rail voltages WILL change, thus it will not work to its specifications. If a power supply cannot sustain a stable voltage across a reasonable load then it does not work, simple as that.</p>
<p>Voltage dividers remain very common in many circuits. So they must work well enough. You do not use them to drive large loads either. There's one in the top left of this schematic</p><p><a href="https://www.instructables.com/files/orig/FVC/VEDE/GT44GZWP/FVCVEDEGT44GZWP.jpg">https://www.instructables.com/files/orig/FVC/VEDE/G...</a></p><p>It works fine.</p>
<p> You're both correct it's just a matter of semantics. Brett's point was clear none the less, you shouldn't use this to drive a load of any kind. Voltage dividers are used to provide a reference node for a high impedance circuit.</p>
<p>You can use it to drive a load. You just need to make sure it is a very small load. That's electronics.</p>
<p><a href="https://www.instructables.com/member/nunorvoliveira" rel="nofollow">nunorvoliveira</a> is right:<br>as soon as you apply asymmetrical load your virtual mass point will shift massivly. Of course a resistor divider still works as you. The only problem is as soon as the circuit you power from this divider is becoming more complex, your divider no longer is a simple divider, but a bigger more complex circuit.<br><br>So using a resistor divider might be okay for something very simple (say powering a few LEDs), but if you wanna power something meant for audio (opamps) or even mikrocontrollers, this is definitly a very bad idea.<br><br>A really good solution for low power applications is a opamp-buffered virtual ground point, this is however limited to the current the opamp can supply. If you wanna have more power (let's say 1.5A for each rail), one could use a combination of the LM317 and LM337 voltage regulator, which will even deliver a more stable power supply (see this circuit http://www.goldpt.com/virtual_ground_circuit.html...<br><br>If you are good with higher voltages and don't need to operate from battery you can also use a Transfromer with &plusmn;15V, which you rectify and stabilize again with LM317 and LM337 to create what is called a &quot;Linear Power Supply&quot;.<br><br>The differences between these different supplies is not only hearable and measureable, but they will also have a big impact on the stability of the circuit.</p>
<p>The voltage divider will work, but the voltage across each resistor will drop as the load increases, so if you are going to use light loads make sure the voltage is high enough so it won't go to low for your application. The total voltage will be the same, but how much is positive and negative is not predictable, but it works in a pinch. A much better, but slightly more complicated circuit for beginners is a LM317, LM337 dual power supply. The output voltage will not vary for any load, but current maxes out at 1 amp without adding transistor boosts. Check out this website where you can calculate the voltage drop across a voltage divider with a resistor in parallel with one load (though in this case both resistors have some load)</p><p>http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html</p>
<p>Here is my problem. I am starting with a couple of power supplies I can use in the same project. (laser projector) +12v and a +/- 24 supply. All grounds are common to avoid any float or stray voltages. This would in effect tie the -5 to the common system ground?</p><p>what I need is +12 in (or +/- 24 in) with +/- out not to exceed 15 volt to drive an opp amp. </p><p>Keep in mind all power supplies are commong negative ground in the system (not subject to change) </p><p>Thinking about just using one each of a + and - 5 Volt regulator on the +/- 24 supply so the grounds are common ( that would actually make the + side of the -5 common.</p><p>Thoughts?</p>
<p>I tried that before when I was working on an op-amp circuit. Everything was working fine at first. However, after adding few parts to the circuit, the voltage wouldn't split equally between both side and I ended up having a -22 v and +2 v instead of +/- 12.. I suppose the circuit loads should be equal on both sides for that method to equally split the rail voltage?</p>
<p>You have to load both &quot;shoulders&quot; equally if you want both outputs to produce symmetrical voltage.</p><p>In general, as others already said, it's a non-practical scheme because of its very-very high output impedance. You can get less 0.1mA (9/10k) at most with a catastrophic voltage drop. Now, most opamps has quiescent current MUCH higher than this (0.5-3mA per opamp in average). </p><p>Reducing the value of resistors will help, but will also increase the battery drain. I know the scheme is very attractive due to the very low component count. But it will not work.</p><p>For an opamp there are much better solutions: http://www.ti.com/lit/ml/sloa076/sloa076.pdf . While they also use voltage dividers, they are used the ways the schemes will work.</p>
Here's how to create +/- voltage with two power supplies:<br> <br> <a href="https://www.instructables.com/id/Dual-POS-NEG-Power-Supply/" rel="nofollow">https://www.instructables.com/id/Dual-POS-NEG-Power-Supply/</a><br> <br> I'm sorry, but I just couldn't resist :)
<p>i see why you had to, if you didn't have any resistors on hand than it would be obvious that you couldnt resist </p>
There are virtual ground ICs in transistor shaped packages that have closely balanced resistors inside so that the positive and negative rails have almost the same magnitude thus a less noisy signal from the amp/op-amp output
Hi. its a good idea but u cant supply any device by that. couse the current is too low!! <br>i suggest buck-boost convertor for this usage.
Hi, yeah, you can't supply anything of course,just wanted to show how I power my 9v projects :)
Neat. I usually do this with two diodes.