I want to build some circuits to learn about op-amps, and so I would like to set up a negative rail.

Are there any problems in creating a negative rail by using an 18 V bench supply and then simply dividing the voltage using, say, two 10 MΩ resistors and setting ground to the mid-point so I have +9 V/GND/-9 V?

Please keep in mind that I am a newbie when it comes to electronics!

EDIT: In further reading guided by all the kind and helpful responses I received here, I just want to signpost to other newbies that there is a great overview of the implications of using a voltage divider as a power source in The Art of Electronics, Horowitz and Hall 3rd Ed Section 1.2 on the Thevenin equivalent circuit

  • \$\begingroup\$ Thanks justme kd9pdp aaron arnisz - so the learning point for me is the impedance on the ground. using an op amp to buffer is something i understand and can experiment/learn with, thanks all \$\endgroup\$
    – JohnnieL
    Commented May 18, 2021 at 18:31
  • \$\begingroup\$ 10M is extremely high for most circuits. As suggested use the OpAmp. use the two 10K resistors and connect to the + input, the - input goes to the output which is now your ground referenced. If that does not support enough current you can always add some transistors in an emitter follower configuration. \$\endgroup\$
    – Gil
    Commented May 18, 2021 at 19:35
  • \$\begingroup\$ This can work (ignoring the issue of the resistor values for now), but there are many places where things can go wrong. Many, many years ago, a fellow engineer was designing an analog circuit with typical symmetric supplies, +/-15V IIRC. He had it all working, when the program decided, in the interests of costs, to eliminate the -15V supply. He created a virtual ground using some of the techniques described here, but had problems with oscillations because of the high gain involved. This was for a satellite application. \$\endgroup\$
    – SteveSh
    Commented May 18, 2021 at 23:43
  • 4
    \$\begingroup\$ This article at Tangentsoft is a great introduction to various virtual ground techniques and their pros and cons. \$\endgroup\$ Commented May 19, 2021 at 3:01
  • \$\begingroup\$ Almost. What you are looking for is a circuit called a "virtual ground". There are several ways to do it, and it's well covered here : electronics.stackexchange.com/questions/560103/… \$\endgroup\$
    – danmcb
    Commented May 19, 2021 at 12:22

5 Answers 5


Good question that is great for new people to electronics to think about!

Let's look at that circuit:


simulate this circuit – Schematic created using CircuitLab

You can think of that as two supplies, a positive and negative supply, and the Thevenin equivalent of each independently looks like:


simulate this circuit

(In reality you can't consider them independent, but just for illustration here, you can see the effect. In reality, both the voltage source and output impedances are functions of what the other polarity's load is - which is complicated.)

So you can see, that's not that great of a voltage source - the output impedance is way too high. As others have suggested, you can use an op amp buffer. Just make sure that the resistors are chosen such that the current through them is 10 or 100x more than the input bias current of the op amp you choose.


simulate this circuit

EDIT: See comments, I added R3 so that capacitive loads are less likely to cause oscillation. You can also add decoupling capacitors between V+ and virtual ground, V- and virtual ground like you would in a standard power supply.

  • 8
    \$\begingroup\$ Using an opamp to create the virtual ground must be done carefully...very often there will be DC supply bypass capacitors to "ground". Or other capacitors to "ground". The ground-generating opamp can see a substantial capacitive load, which it finds disturbing, so that it oscillates. You may need to add some series resistance to tame it. \$\endgroup\$
    – glen_geek
    Commented May 18, 2021 at 18:51
  • \$\begingroup\$ @glen_geek yeah I just +1ed that. Especially the oscillation part. I'll add that as an edit to the answer. \$\endgroup\$
    – KD9PDP
    Commented May 18, 2021 at 19:25
  • \$\begingroup\$ So just to understand this point as a noob: if there is a capacitative load, this will cause voltage at the op amp output to oscillate, therefore causing the feedback to oscillate and so the virtual ground will also oscillate? So by adding the resistor it will dampen that voltage fluctuation caused by the capacitor? Many thanks \$\endgroup\$
    – JohnnieL
    Commented May 18, 2021 at 20:26
  • 1
    \$\begingroup\$ @johnieL If you have a non-zero output impedance from the op amp, and you have a capacitive load, then higher frequency noise voltage at the non-inverting input will actually be inverted at the op amp output (or at least phase shifted). That is then fed in to the inverting input, so a -1 times -1 becomes a +1, and you get amplification of those frequencies. Adding R3 just "adds a zero" to the freq. response, effectively reducing that phase shift at higher frequencies so you don't get amplification. \$\endgroup\$
    – KD9PDP
    Commented May 18, 2021 at 21:04
  • \$\begingroup\$ This is better done using an open loop buffer without feedback, instead of an opamp, for the reasons that glen-geek mentions. The BUF634 is a traditional choice \$\endgroup\$
    – tobalt
    Commented May 19, 2021 at 4:12

Yes, some problems.

First problem is that creating a new midpoint voltage (the would-be 0V ground in your circuit) with 10Mohm resistors has very high impedance, so it would be quite unusable as a ground return. It would have to be buffered with an op-amp to have any reasonable sinking or sourcing ability for any input or output terminal that uses the ground as current return path. Compare that to two separate 9V batteries creating the full 18 volts, there would be a relatively low impedance 0V ground if you used two 9V batteries to generate +9V and -9V supplies.

Another problem is that the power supply output would have to be floating. It may not reference the output 18V to any other equipment, including mains earth/ground. When it is floating, this is possible, then the divided half-voltage can be the 0V ground for any external circuits. However, if it happens that for some reason the power supply output is not floating, but it for example connects the negative output terminal to mains earth ground, then unfortunately it means your device would use 9V referenced to mains earth as ground and it can't be connected to a say computer that expects 0V to be same potential as mains earth ground.

  • 2
    \$\begingroup\$ It'd also be a noisy ground, 10 MΩ would give a fair bit of Johnson-Nyquist noise. Even buffered with an op amp you could have ground noise problems depending on what you're doing. \$\endgroup\$
    – Hearth
    Commented May 19, 2021 at 14:38
  • \$\begingroup\$ @Hearth right. 10 MΩ is definitely excessive, though Johnson-Nyquist is actually not that problematic in this case because you can use big capacitors to filter it from the reference before buffering. \$\endgroup\$ Commented May 19, 2021 at 18:57

You can do this, but you need to check your impedances. If the "-9V" is going to supply power to an opamp, how much current can it get through a 10MΩ resistor? Answer: Not much!

Look at your opamp current consumption spec, and design your voltage divider to allow that amount of current to flow without a significant voltage drop. And then also check the power in those resistors, you'll probably need to use some 1W or there abouts.

There are better/more efficient ways of doing this, but this is a simple dirty way of making a split power rail.


yes of course but the impedace of your new virtual ground will be fairly high. Your can either amplify it to lower the impedace

aplyfied ground

or use e.g. a so called "rail split transistor" (TLE2426) witch does essentially the same.


In addition to the other answers (put a buffer after divider to create virtual ground), a common tactic that is somewhat simpler is this:

For ac signals such as audio, instead of splitting the supply, you can move the signal up to the dc midpoint voltage. To do this feed the signal into a large value resistor divider via a capacitor.


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