Everyone needs a negative voltage rail eventually, but most only have a single rail supply. This project will show you how to build a negative voltage generator that runs off a single rail supply!

The Need for a Negative Rail

Many hobbyists will have come across an op-amp circuit that uses a dual rail supply. In order to provide the dual rails power, two power supplies can be used but many do not have access to such equipment or the design needs to be portable and thus two individual power supplies may not be ideal. One solution is to split the incoming power supply and create a virtual ground between two resistors in a potential divider.

However, this solution has some issues. Firstly, the current sinking / sourcing capability of the virtual ground is limited and, secondly, the supply voltage has been divided by 2 (for example, a 9V battery could become ±4.5V but 4.5V is the maximum achievable voltage).

But using an oscillator, some diodes, and a clever arrangement of capacitors, we can exploit an effect called capacitive coupling whereby a negative voltage near equal to the supply voltage can be generated... from a single rail supply!


Some Capacitor Theory

Capacitive coupling is an effect where a capacitor “attempts” to keep the voltage across its terminals constant. For example, if both sides of a capacitor are at 0V and plate 1 is suddenly brought to 10V, the voltage on the other plate will immediately begin to rise to 10V. The same goes for the opposite: If the first plate is brought from 10V to 0V, then the other plate will attempt to drop by 10V to keep the voltage across the capacitor constant.



Notice how the negative change gave a -10V when the left plate was charged to 10V and the other plate kept to 0V. This is the sequence that will be exploited to generate the negative voltage.

But how do we charge the left side to 10V without the other plate also charging to 10V? Remember that when you get the first plate to 10V the other side will also try to get to that voltage. This is where diodes come in!


Diodes - Removing Positive Voltage

Remember how diodes only conduct in one direction? Well, if we use a simple (but clever) arrangement with a capacitor and two diodes, we can generate a negative voltage on a capacitor plate!



As the input steps up from 0V to 10V, the negative plate on C1 will also try to match due to capacitive coupling. However, once the plate reaches 0.7V, D1 conducts and this prevents the plate voltage from getting higher than 0.7V. Now with the positive plate at 10V and the negative plate at 0.7V, the potential difference across the plate is 10 – 0.7 = 9.3V. When the positive plate drops rapidly from 10V to 0V, this 9.3V potential difference must be maintained due to coupling. Therefore, the negative plate drops to -9.3V. D1 cannot conduct as it is reversed biased, so this negative voltage remains on the capacitor.

D2, however, is now conducting because the anode (connected to C2 negative plate), is at a higher potential than the cathode (connected to the -9.3V). Therefore, the voltage at the negative plate of C2 will reduce until D2 stops conducting. D2 will stop conducting once the voltage drop across it becomes less than 0.7V. This will happen once the negative plate of C2 becomes -9.3 - -0.7 = -8.6V. 

But if an external device attempts to use the negative voltage, it would very quickly be used up! This is because of the relatively small sizes of C1 and C2 as well as the current limits of the circuit driver. The solution is to connect an oscillator to the input so that a negative voltage is constantly being generated at C1, which keeps C2 topped up.

C2 is used as a decoupler for external devices using the negative voltage to try and keep the value as constant as possible. For our project, C1 will be a 10uF capacitor and C2 will be a 100uF capacitor.


The Oscillator

The oscillator can be anything from a 555 timer to a microcontroller, but remember that the negative voltage will be equal to the oscillator's max voltage – min voltage. In this project, we will look at using two different oscillators: a 555 astable and a 74HC14 Schmitt trigger-based oscillator.


555 astable oscillator


74HC14 Schmitt trigger-based oscillator


Bill Of Materials

BOM - Diode Pump Circuit

Component Identifier Quantity
Diode (Generic Silicon) D1, D2 2
10uF Electrolytic Capacitor C1 1
100uF Electrolytic Capacitor C2 1


BOM - 555 Oscillator

Component Identifier Quantity
555 Timer IC 555 1
5.6kΩ Resistor R1 1
47kΩ Resistor R2 1
100nF Capacitor C4 1
10nF Capacitor C3 1
Assorted Wires - -
Breadboard - 1


BOM - 74HC14 Oscillator

Component Identifier Quantity
74HC14 IC (DIP) 74HC14 1
5.6kΩ Resistor R1 1
10nF Capacitor C3 1
Assorted Wires - -
Breadboard - 1


Creating the circuit is very easy and requires at most an hour. Start with the oscillator and then build the capacitor pump circuit. As the 74HC14 oscillator only has one capacitor and resistor, the values are those listed in the BOM. The 555 timer uses a 5.6kΩ resistor for R1, a 47kΩ resistor for R2, 10nF capacitor for C1, and 100nF capacitor for C2.

If you have an oscilloscope, you can confirm that the oscillator works by probing the output. If you only have a multimeter, then you can measure the output using a DC voltage reading. There is a good chance that the circuit is oscillating if the reading is approximately half of the supply. For example, if you power your circuit from a 9V supply and the circuit is oscillating, then the output will read around 4.5V. This happens because the voltmeter takes multiple readings quickly and produces an average value. As the output is on for half the time and off for the other, half the average value is half of the max value. 


The Complete Circuits

555 driven negative voltage generator



74HC14 driven negative voltage generator



The Final Circuit

Here is the negative voltage generator using the 555 oscillator where the output is the blue wire on the right.



Here is the output from the negative voltage generator. Note how the voltage is approximately -7.2V because of the double diode drop (1.4V), and the supply was not a perfect 9V but closer to 8.5V.



Here is the construction using the 74HC14 oscillator:



The output here is -7.95V and is higher than the 555 timer circuit because the 74HC14 uses the excess gates to create a driver. This driver can source and sink more current than the 555's output and thus produce a larger negative voltage. Remember to connect all the unused inputs to ground on CMOS chips!



NVG Capabilities

This negative voltage generator can only source a small amount of current. If a regulator is used on the output (such as a 7905), don't expect to get more than 4mA ~ 5mA. But if left unregulated (such as in opamp projects), you can source up to 20mA before the output voltage begins to drop significantly. When using this generator in projects, try to keep the current draw from devices as small as possible so that the negative rail retains its voltage value.

The circuit can be made more efficient as well as increasing the output voltage by using Schottky diodes instead of generic silicon. This is because Schottky diodes have a smaller forward voltage drop (down to 0.1V), which results in a larger negative voltage.



Now that you have a negative voltage generator circuit what can you do?

  • Build a single rail dual supply : Using regulators and filtering components, you could make a dual rail supply that takes in a battery voltage or a wall wart adapter and converts it into a 5V and a -5V. If you use a large enough input voltage, you could also make a 12V and -12V line.
  • Op-Amp Projects : Many op-amp projects (such as synthesizers) require the use of a negative rail and now you got one!
  • Audio Projects : Audio components such as speakers and piezoelectric discs work better when they use both positive and negative voltages. This is because the negative voltage deforms the mechanism the opposite way the positive voltage deforms the mechanism. Therefore, the produced sound is much louder and clearer.
  • Portable Dual Rail : If you have a system where weight needs to be reduced (such as an RC plan) that requires the use of a negative voltage, this circuit can be used instead of having two separate supplies, which would be heavy

Note : This project will be used in the up-coming series on DIY synthesizers so make sure you get one built soon!


Give this project a try for yourself! Get the BOM.




  • Paul Rautenbach 2016-06-17

    Nice simple circuit. Thanks.

  • to3metalcan 2016-06-17

    I was going to comment that it’s odd that commercial power supplies don’t ever seem to be made this way, but then I realized that most switching power supplies ARE doing this, they’re just also doing it for the positive rail, too!

  • mjaa 2016-06-21

    Great article. SImple, educational and practical for hobbyists. Thank you!

  • ci139 2016-07-11

    is there any good reason to use pulsed varying voltage sources with extremely sensitive input devices such as Op Amp.-s - in adition to making their bias stable to temperatre and component aging effects - you should also try “a varying negative rail” which will fail - so a “virtual ground”—all of which does not let you gain precision / noise immunity that’d be available with a balanced stable supplies

  • buteman 2016-07-26

    I like this especially the 74HC14 one.
    I have used 2 USB wall warts, cutting the USB plugs off and wiring them so one gives the positive and the other the negative 5 volt supply. You could do it with 2 x 9 volt ones too.

  • PIZZAPLANET 2017-02-10

    How could one increase the current output?

    • Jeff-Russ 2017-08-11

      It seems the 555 output is the bottleneck. I’m experimenting with an op-amp based relaxation oscillator and getting much better results.

      • David Ward 2018-01-01

        I was wondering the same thing myself, would like to build something that can supply up to 1A. I thought about using the 555 to control a high side switch, as used in switching power supplies. I wonder how this would effect the choice of capacitors used?

  • Harsh Chittora 1 2017-06-03

    The circuit works fine when the output is open. However if I connect anything the negative voltage drops (in magnitude). I was trying to use the negative supply for my LM358 op-amp. the voltage drops to 6 volts as soon as I connect the op-amp. Is there a way I can maintain the negative voltage even after connecting op-amps?

    • Jeff-Russ 2017-08-11

      What I did was find a reasonable voltage I know it can put out even when it’s struggling a lot and put a zener diode clipper to hold it there.  I would post a schematic here but not sure how. It just involved a zener (of three volts) put in parallel with ground just like the final capacitor, and right after it. Normally you would reverse bias a zener but since the current is negative you just put the zener’s negative end at the connection to ground.

  • aroberts 2017-06-12

    Straightforward guide, thank you. I was wondering if C3 must be a polarized capacitor in the 555 driven generator? I am having a hard time finding a polarized 10 nF. I see in the 74HC14 C3 is nonpolarized.

    • kmarcel 2017-10-25

      C3 can be a non-polarized type.
      Polarized caps cannot be fully reversed biased, as they go out in flames of glory and a big bang (most times together with the board).
      The design requirements usually gives you a minimum capacity and voltage rating of a cap. You can always use a non-polarized version in any bias, if you have it. For the situation you only have a polarized cap, all schematics that include voltages beyound the ground rail show the positive terminal of caps, to let you know how not to reverse bias a polarized cap.

  • vanderghast 2018-06-14

    I have a problem with the “Capacitor Theory” section. Isn’t it that a cap can hold an internal voltage for months, if left unconnected, outside a loop? So, basically, the examples given are for a cap connected to a loop. But that would be dependent of the circuit in that loop. If there is a 10V battery, then the cap plate next to the positive terminal of the battery will be at 10V and the cap next to the “negative” terminal of the battery will be at 0V ( well, as long as we use a convention such that the DIFFERENCE is 10V). Now, what becomes my problem is with your statement: “If the first plate is brought from 10V to 0V, then the other plate will attempt to drop by 10V “... that has to be in the context of a closed loop. So, I assume that we somehow remove the initial battery which charged the cap, next… ?  We close the circuit again, somehow, but then, the cap become like a battery source, isn’t? But so, the voltages are still 10V at the left, and 0V at the right of the cap, no? Unless you have some special circuit in mind, that I can’t figure out. Or I am totally wrong?

    • RK37 2018-06-27

      I find your post a bit confusing, so I’m not sure how much this answer will help. The bottom line is that the voltage across a capacitor cannot change instantaneously. Let’s say you have a fully charged capacitor connected to a 10 V battery. One terminal (call it terminal A) is at 10 V, and terminal B is at 0 V. If you use switches to rapidly disconnect both terminals and then connect terminal A to ground (=0 V) and terminal B to some sort of load, terminal B will initially remain at a voltage that is 10 V lower than the terminal A voltage. Terminal A is now at 0 V, so terminal B must be at -10 V.

      • vanderghast 2018-07-20

        Keep the fully charged cap connected to the 10V battery. If you rename, label, the +terminal as being 0V and the -terminal as being -10V, you changed nothing short of taking a new “origine-reference”. That is what you end with in your second case, isn’t ? To “produce” a negative voltage of, say, -10V, WHILE KEEPING an access to the original +10V, (to power an OP AMP with a +10 and a -10V with a single battery, forgetting voltage regulation for simplicity) and doing this with a single battery of 10V (difference of electrical tension between its terminal of 10V) seems more likely what we are getting to obtain, isn’t it? A little bit like the ICL7660, but with minimal set of components.

        Furthermore, charging a cap at 10V and THEN changing the voltage at the terminals of that cap with a voltage source (of inverted polarity, by comparison with the first battery which charged the cap) is another way to say that we have an initial clock signal varying from +10V to -10V… to start with, already. So we don’t really create a -10V from a single 10V battery (to which we added a clock-like oscillation), doing so, isn’t?

        • vanderghast 2018-07-20

          Oh, unless we have, say, a closed circuit, a rectangular one, with summits A, B, C, D, and the cap between A and B, and the battery between C and D. We can charge the cap with the loop A-B-C-D-A then disconnect B-C and D-A to reconnect A-B-D-C-A. That new loop is now giving a total of twice the battery voltage… but that is a voltage doubler, not a negative voltage, no?

        • RK37 2018-07-20

          The comments section of an article isn’t a convenient place to have a discussion of this nature. I recommend that you post your question in the AAC forum.


        • vanderghast 2018-07-21

          RK37: you are right, sorry for the inconveniences. Can you delete this thread that I started here?  I will see if is still appropriate to start it in the forum that you suggested instead. Thanks for your time.

  • jptf59 2019-05-16

    I am at a loss how to understand how such a circuit could output as much as 20 mA.  Neither my test circuit could nor the various LTSpice simulations I did. The negative voltage on the output depends essentially on the load. It seems that the output current is rather in µA than in mA for me !

    • jptf59 2019-05-16

      Sorry, I have just understood : it was the particular oscillator I used that was limited in current !

  • qwertyqwq 2019-07-11

    hi folks!

    im going to lose my mind how can you applly 9V to sn74hc14 ic ?? in its datasheet ,absolute max rating is 7V !!!!
    am i looking wrong , please someone explain me   :x
    here is the datasheet link : http://www.ti.com/lit/ds/symlink/sn54ls14-sp.pdf
    another link : https://www.diodes.com/assets/Datasheets/74HC14.pdf

  • Privatier 2019-08-23

    Since the output voltage is about the negative of input voltage minus two diode drops, I suggest using Schottky diodes instead of silicon diodes, to gain about half a volt. This is particularly useful if the input voltage is 5V or less.

  • Basiru Ado 2019-08-24

    Where can I get the full report of this project?