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There's a bunch of posts about this, but they're all for very specific scenarios. I'm a newbit, and some components, like op amps, require both a positive, and negative voltage.

But how do you get negative voltage?

I'm about to buy what I think are my last "essential" components, an assortment of voltage regulators, and an assortment of logic gates. Aside from a collection of those, I have most breadboard components.

  • I've read you could use transistors kind of like a charge pump,
  • I imagine you could use wind a transformer 1:1 and hook on the "wrong" sides of the secondary?
  • Could you use a NOT gate, or does that give 0 V?

Like from a day to day, really basic stuff, just giving -5 V to an op-amp, how do you get negative voltage?

I haven't played with the op-amp either, but they're everywhere, and I never see how they manage to give it the negative voltage the schematics seem to say it wants/needs...

I'm just putting my kit together, and preparing to embark on projects, so I don't have to order parts, and wait, and it's the wrong part, etc...

I just figure there's an easy way to to it, or there's a part, or collection of similar parts I need to be able to get negative voltage when I need it...

Some parts want +5V and -5V and ground, so I can't just flip +5V and GND, either... can you even do that?

Anyone can help me understand?

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    \$\begingroup\$ Basically, you get a negative voltage by calling the positive terminal of the power supply "Zero volts". \$\endgroup\$ Apr 6, 2021 at 0:08
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    \$\begingroup\$ If you desire a serious amperage, then two power supplies/batteries connected in series is a good choice. The middle point is Gnd. U may use one transformer with two equal windings or one winding transformer with middle point to save a space. \$\endgroup\$
    – user208862
    Apr 6, 2021 at 4:08
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    \$\begingroup\$ Not much to add to other opinions but for low currents (<10mA usually) a charge pump is good, otherwise go for an inductive converter: an coil inverter usually and a transformer one for serious loads \$\endgroup\$ Apr 6, 2021 at 6:39
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    \$\begingroup\$ @poorandunlucky Okay. That's a good-enough reason. I just wondered. My first design was a power supply. So I can't cast any stones. It's just that people are often interested in something else and the power supply is just a means to an end. No the end, itself. In your case, I think you are right to pick this project. But do NOT make it a switcher. I strongly recommend that go with a standard transformer with a center-tapped secondary. The center-tap will be the ground. The other two leads will provide the (+) and (-) outputs. \$\endgroup\$
    – jonk
    Apr 8, 2021 at 1:48
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    \$\begingroup\$ @jonk That is very useful, thanks 😊 Transformers are easy to understand, and characterize, and knowing this gives me a simple, go-to, robust solution I can easily understand and implement. Much appreciated 👍😊 \$\endgroup\$ Apr 8, 2021 at 1:53

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Op-amps don't really 'care' if they have a negative voltage, they just need enough voltage so that they can produce an adequate swing. Early op-amps like the uA741 needed a lot of voltage to give useful range, and so are designed to run on +/-15V or so.

Some newer op-amps can swing very close to the rails and thus can get by with much lower voltages (even logic-friendly ones like 5V or 3.3V) than the old '741.

Some op-amps even go as far as to say they are designed for single ended, like the LM324, but will happily run on split voltage too. What makes them 'single ended' is their ability to swing close to the (-) voltage, like the LM324 can.

That said, bipolar (positive and negative) supplies are convenient for realizing signals that swing above and below ground. Then each stage can use ground as the signal reference along the way. High-performance and high-power systems (like big audio amps) favor this approach.

In contrast, single-ended systems will use only a positive voltage rail and synthetic midpoint reference (sometimes called 'virtual ground') for the signal chain. These tend to be lower cost.

Creating a negative rail is nearly the same as creating a positive one. With some planning the cost impact is not too bad.

It's even possible to create split-rail supply (e.g., +5V, -5V) by connecting two positive supplies 'head-to-tail' and taking the connection between the two as 'ground'. This can be done with any power supply that has an isolated secondary, such as a wall-wart, USB charger or off-the-shelf DC supply.

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  • \$\begingroup\$ This was really informative as to the real-world operation of op-amps, theory usually uses ideal components, and this really furnishes my understanding of op-amps. One of the reasons I asked is because I think I'd like to practice with audio signals, and I know audio signals fluctuate between +/-. Your solution of just using two isolated DC supplies to make a sort of "center-tap" secondary is a good "off the shelf" solution, I have a bunch of those, and I think I could try using them to make just that. At least, this, and other comments, give me enough to be confident enoygh to try. Thanks! \$\endgroup\$ Apr 8, 2021 at 2:05
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Negative voltages are a pain in the ass.

Here are some methods to obtain a negative supply

  1. Use a bench supply with positive and negative outputs to power your circuit
  2. Clip a isolated bench supply's positive output to the ground node of your circuit. This is pretty much like #1 but with a caveats that are designed to be implicitly handled in #1:

WARNING: This bench supply MUST have ISOLATED outputs if your circuit is anything else also plugged into the wall. This includes another bench-supply, and any oscilloscopes or instrumentation that is wall-powered that you connect to your circuit including USB connections to your computer.

  1. Clip an AC-DC wall wart's positive output to the ground node. This is identical to method #2, except AC-DC wall-warts are all isolated and are a lot cheaper than bench supplies.

  2. Throw on batteries onto the other side of the ground node (batteries are inherently isolated).

  3. Capacitive charge pump - mainly a solution you put onto a board with everything else. Small size, most often in the form of a premade IC and very low power so normally only used to provide just enough current for the negative biasing required on some analog circuits, such as op-amps. Sometimes only -0.3V to -1V just so that the circuit itself can output a true 0V since a transistor isn't a perfect conductor so cannot connect to GND to produce a true 0V; The voltage is always a tiny little bit higher

  4. Inverting/negative switching converter - another board level solution capable of higher currents than the capacitive charge pump, but complex. Lately, might be sometimes available as a single component you can just drop into your system.

  5. Isolated switching converter - another board level solution similar to the isolated bench supply method listed above. Like #6 is also more complex and capable of higher current. Lately, might be sometimes available as a single component you can just drop into your system. More flexible than #6 since the isolated output allow you to connect it however you want to produce a positive or negative voltage and it doesn't even have to be referenced to GND; It can be relative to whatever node voltage in the circuit you want.

If you understood what I wrote, you'll see that a negative supply isn't a board component you just have lying around in your stock. If it's lying around it is usually in the form of a a battery, bench supply, or wall-wart.

I recommend getting some AC-DC wall-warts with compatible DC barrel plug adapters to convert it to alligator clips or screw terminal blocks to make the output more accessible. Sparkfun, Solarbotics, and Adafruit carry such adapters.

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    \$\begingroup\$ Not just might--it's very easy and not too expensive to get isolated switching converter modules off the shelf. \$\endgroup\$
    – Hearth
    Apr 6, 2021 at 1:57
  • \$\begingroup\$ So another easy solution is to have two separate V+/G and V-/G rails, where your V- rail is really G, and your G is a positive voltage, just that if you were to connect your V- "ground" to the V+ ground, then your V+ ground would become another V+ rail, and you'd have no more ground for your positive rail? Am I understanding this correctly? \$\endgroup\$ Apr 8, 2021 at 1:58
  • \$\begingroup\$ Just a question, though: You warn about the need for the outputs to be "ISOLATED", like what do you mean by that? How do you know your supply is solated, how do you isolate it if it's not, and is there a way to protect against accidens, like putting a diode somewhere, or something? I just want to be sure what "isolated" mean, I understand you don't want your ground to become positive so V+/G become V+/V+ (that there's no voltage on the ground the rest of your circuit uses), but could you expand a bit on isolation, please? If you have time, of course... \$\endgroup\$ Apr 8, 2021 at 2:11
  • \$\begingroup\$ @poorandunlucky Isolated means that DC current cannot pass directly from the input to the output. The consequence of this is that the two output terminals produce floating voltage only relative to each other and nothing else which means you can connect either terminal to anything and the other terminal will be ensured to maintain a voltage relative to it (similar to how you can connect one end of a 9V battery to any voltage and you know the other end of the battery is 9V above or below whatever voltage you connected the other end to, depending on which end). \$\endgroup\$
    – DKNguyen
    Apr 8, 2021 at 3:00
  • \$\begingroup\$ If you connect power supplies without this, you get short-circuits since the output negative/GND on both supplies will be connected to the GND on the input side. And when you connect the positive output of one to the negative output of the other, you end up with a short-circuit (draw it out with boxes and lines) \$\endgroup\$
    – DKNguyen
    Apr 8, 2021 at 3:02

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