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Assume, I need to use an opamp with +9 V and -9 V supplies. While prototyping, etc., I can and probably would use a DC power supply. Since it has multiple channels, it is easy to have +9 and -9 volts.

However, when it comes to finalize the project, making it a standalone box, which will get its power, let's say by batteries, how are +9 and -9 volts obtained?

I only can think of using two 9 V batteries with reverse polarities connected to a common ground. However, my insights tell me that there may be another way of doing that without using two batteries. Because I don't think in every module which consists +/- power rails, there exists two separate power supplies.

Peter Mortensen
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muyustan
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  • What type of circuit is this? Because most of an op-amp based circuit can work from a single supply. – G36 Dec 16 '19 at 15:12
  • @G36 push-pull npn-pnp complementary pairs to drive a speaker. So I will need +/- 7 volts which can also provide up to 1 A for speaker also, besides opamps' +/- 9 volts. – muyustan Dec 16 '19 at 17:21
  • @MarcusMüller I know ground is just a convention and what matters is relative voltage. The problem is to have both +vcc and -vcc. – muyustan Dec 16 '19 at 17:23
  • @muyustan "what matters is relative voltage": exactly! so, you just put ground in the middle of your battery pack. Bam, solved. – Marcus Müller Dec 16 '19 at 17:28
  • @MarcusMüller battery pack? What does it mean for example in case of a single 9v battery? – muyustan Dec 16 '19 at 17:53
  • @muyustan there's no chemistry that has 9V, and you want +-9V, so 18V in total, so two 9V blocks in series, and that makes it easy. – Marcus Müller Dec 16 '19 at 18:18
  • @MarcusMüller yes that is the obvious answer. However, for the supply of push pull structre, a battery solution will probably require a double 7.1V lipo then. For cases of pc speakers, for instance, how do they obtain that much current from the pc directly? – muyustan Dec 16 '19 at 18:24
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    They simply don't use an amplifier that requires a dual supply. – Marcus Müller Dec 16 '19 at 18:36

2 Answers2

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When using an AC input to the device a transformer with two seperate windings (or one winding with a tap in the middle) can be used to create a postive and a negative rail.

enter image description here
Source: https://www.quora.com/What-is-a-dual-polarity-power-supply-using-a-transformer-a-full-wave-bridge-and-a-Zener-diode

With only one DC input you typically use a DC/DC converter, like a switching inverter to create the negative rail (see e.g. APPLICATION NOTE 3844 by Maxim Integrated).

If the needed current is quite small it is a viable and very easy solution to just use a voltage divider with a buffer to create a virtual ground at half the DC voltage. The TLE2426 is actually a ready to use "rail-splitter" to do exactly this.

schematic

simulate this circuit – Schematic created using CircuitLab

When building the above circuit you should choose a Rail-to-Rail OpAmp, otherwise you are limited to quite low output voltages. This circuit only makes sense when the current is not significantly higher than 10 or maybe 20mA.

jusaca
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  • Thanks, can you please a little bit ellaborate the last paragraph? – muyustan Dec 16 '19 at 12:24
  • But, where will the opamp's power rails be connected to? – muyustan Dec 16 '19 at 13:03
  • The power rails are also connected to the 5V supply (or whatever is your DC input). As I wrote, use a rail-to-rail OpAmp for this and you will get pretty close to the +/- maximum values. – jusaca Dec 16 '19 at 13:23
  • should it be also a single supply type opAmp or no need for that? – muyustan Dec 16 '19 at 13:24
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    Single Supply capability basically just means, that the OpAmp can pull the output almost down to the negative rail - so every rail-to-rail OpAmp will fullfil this criterion. – jusaca Dec 16 '19 at 13:39
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If you have a single 9 volt battery and you need a regulated negative voltage (for op-amps etc.) then consider looking for a type of switching regulator called "inverting" i.e. an inverting buck regulator. This type of circuit takes a positive battery voltage and recreates a negative fixed voltage such as -5 volts: -

enter image description here

I think the keywork you are looking for is "inverting" so try the search engines at ADI (Linear tech), Texas, Maxim etc..

But, there is another option. You could boost the 9 volt battery to +18 volts then you have 0 volts, +9 volts and +18 volts to hand. Voltages are just relative so you can use the +9 volt from your battery as the "new" 0 volts (Mid rail): -

enter image description here

The circuit shown above has a +24 volt output but by changing the feedback resistors you can achieve 18 volts.

Andy aka
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  • Does it have to be a *buck* converter? What about boosting +9 V battery to +/- 15 V rails? – muyustan Dec 16 '19 at 12:41
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    Usually yes it does but I'm not ruling out other options. Try the search engines I mentioned. You can also use a flyback controller because it has an output transformer and can provide isolation therefore, you can connect its positive output to ground and get a negative voltage. There is also the SEPIC converter that can be used to generate negative voltages. – Andy aka Dec 16 '19 at 12:42
  • Regarding your last edit, '+' terminal of the 9V battery will be connected to Vin of the converter and '-' terminal will be connected to the GND pin. But after, I will use + terminal of the battery as common ground of the rest circuit, and will use GND pin as -9V and Vout as +9V. Correct? – muyustan Dec 16 '19 at 13:17
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    Correct, 0 volts maps to -9 volts, +9 volts maps to 0 volts and +18 volts maps to +9 volts. – Andy aka Dec 16 '19 at 13:23
  • You can also use any old isolated dc-dc converter to do this. Take two of them and tie the + of one to the - of the other. – Hearth Dec 16 '19 at 15:25
  • @Hearth the possible problem with DC/DC converters is that (usually) efficiency is lower than a buck or boost regulator and battery life will be reduced. – Andy aka Dec 16 '19 at 15:28
  • @Andyaka I'm not sure what you mean here, as buck and boost converters are both types of DC/DC converters? – Hearth Dec 16 '19 at 15:41
  • If it's an isolating DC/DC converter then it's likely a flyback type and this (usually) means it has less power efficiency than a non-isolating buck or boost. – Andy aka Dec 16 '19 at 15:45