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I want to power a symmetric amp op with a raw and cheap 12V wall power source.

Can I use a schema like this to create a symmetric output ? Or it will fail ?

schematic

simulate this circuit – Schematic created using CircuitLab

ps: I know I can use non symmetric amp ops, but I want to use symetric ones.

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    \$\begingroup\$ Zeners are the wrong way around, otherwise, as long as your virtual ground current never goes much above 5mA on average it will probably work okay. Given the 5mA limit your caps are unnecessarily HUGE. \$\endgroup\$ – Asmyldof Feb 28 '16 at 14:20
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    \$\begingroup\$ Note that the author edited the question to fix the orientation of the zener's. (commenting to prevent confusion since so many answers pointed out the incorrect orientation, which has now been fixed) \$\endgroup\$ – Johnny Feb 29 '16 at 1:09
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    \$\begingroup\$ An alternative approach is to use a chopper circuit and "pump" capacitors to the negative voltage. This may be a better approach if the incoming voltage source has one side effectively grounded. \$\endgroup\$ – Hot Licks Feb 29 '16 at 3:21
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You would do much better by using an opamp to make the ground point.

I sugget using a circuit like this:

enter image description here

Just use your 12 V instead of the 9V battery. It depends on the opamp how much current can flow through the ground connection.

This circuit is also much more efficient than your proposal.

To make it suitable for higher (ground) currents, use a power-opamp (some audio power amps are also suitable) or see WhatRoughBeats's answer for a discrete solution. Note that that solution has no over current / over temperature protection.

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    \$\begingroup\$ Also be aware that some op amps get very unhappy if asked to drive capacitive loads. \$\endgroup\$ – WhatRoughBeast Feb 28 '16 at 17:36
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Well, if you reverse the zeners it will work in principle. In practice, not so well. The problem is that zeners come with a tolerance, with about 5% as the norm. So your nominal 5.6 zeners, which drop a total of 11.2 volts, could produce a "real" voltage of anywhere from 11 to 12.2 volts. The larger voltage, obviously, will not provide good regulation, and the lower voltage will draw more current from the supply and dissipate more power in the resistor.

Rather worse is the effect of that 100 ohm dropping resistor. Since all your op amp current flows through it, any current at all will affect the supply voltages, and this is a recipe for your op amps oscillating. A total of 10 mA, for instance, will drop 1 volt in the resistor and the zeners will be completely ineffective.

A better virtual ground circuit would be something like

schematic

simulate this circuit – Schematic created using CircuitLab For the transistors shown, a virtual ground current of about +/- 50 mA seems reasonable, which is about 300 mW in the affected transistor.

As opposed to using a regulator (which is also an option), this will keep the virtual ground centered between the power supply + and -, which you may find preferable.

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Regular 9 V PSU wall-wart. Figure 2. Modified for +9/0/-9 V.

You can easily modify a standard wall wart supply as shown in Figure 2. Ripple voltage will be worse and max current on each supply will be half of original specification so I'd recommend some large caps or voltage regulators to eliminate hum.

If you want to keep a standard jack on the PSU then convert it to an AC PSU by removing the rectifier and capacitors and put the diodes and caps into your project case.

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  • \$\begingroup\$ That's useful, but in a product I will need to provide the power supply, and I wan't to use a Off-The-Shelf 9/12/15V DC adapter so I don't need to tap around with the transformer. \$\endgroup\$ – Cristiano Araujo Feb 28 '16 at 15:04
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    \$\begingroup\$ No problem. Use an AC wall-wart. Do the rectification in your product. \$\endgroup\$ – Transistor Feb 28 '16 at 15:15
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Your basic concept can work, other than the zeners are backwards. C1 and C2 also don't need to be anywhere that large.

It might be useful to use two LDOs to make clean and better regulated ±5 V from the ±5.6 V.

This concept will only work for low current levels. You are starting with 12 V, and the two zeners will drop 11.2 V, leaving 800 mV maximum across R1. The maximum current before the voltages start dropping significantly is therefore (800 mV)/(100 Ω) = 8 mA. If that's enough, then you're OK. If not, you need to decrease R1 or use a different topology altogether.

The limit on how low R1 can go is the maximum dissipation of the zeners. For example, let's say they are good for 200 mW (just made that up). (200 mW)/(5.6 V) = 36 mA, which is the maximum the zeners can pass without blowing up. (800 mV)/(36 mA) = 22 Ω, which is the theoretical minimum R1 can be. However, there are various souces of slop, so I'd probably not make it less than 47 Ω.

If you need more current than represented by the above, then use a switcher to make the -5 V supply from the +12 V supply, and a linear regulator like a 7805 to make the +5 V supply. The 12 V supply negative lead is then also the ground for everything.

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Your zener diodes has the right orientation. Nothing wrong there. But your circuit have some other problems. You will get in trouble when you only have an nominal voltage drop of 0.8V over your resistor. The current through the zener diodes will be much too sensitive to both input voltage and the voltage across the zenerdiodes. If the current through the zenerdiodes gets too low then you loose your virtual ground. It all depends on the usage. if you only need a virtual ground ref, then just use a resistive voltage splitter (e.g. 2x1k). I you need a bit of current then you can buffer the splitter with an opamp as allredy suggested. Best regards.

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  • \$\begingroup\$ I've fixed the zeners orientation so others don't get confused reading the question \$\endgroup\$ – Cristiano Araujo Feb 29 '16 at 1:02

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