[saint gillis]

Hello, here's a rail splitter circuit :


If I'm not wrong, the 1K resistor protects the op amp's output from big capacitive load, and the reference is taken on the inverting pin to preserve low output impedance, correct?

Can we double this circuit this way :

[JohnRoberts]

That looks like one op amp too many... If you need more drive current you can add discrete transistors to output side.

I had to combine a relatively high current 5V digital supply with split supplies for audio path from a single transformer winding without a center tap so needed to balance out the +5V draw on the negative side.  My version included a power transistor turned on by the op amp output in the direction needed to balance the two supplies.

JR

[squarewave]

The 1K just limits the amount of current that can be sourced / sinked from the op amp. In small signal circuits this would be called a "build out resistor" or "zero impedance output". To understand it better, imagine an errant circuit pulls VGND down or up to a rail. For example, let's say down to -15V in which case the OA output will go high to say +13V before it clips and now you have 28V / 1K = 28mA. It will be a crappy 28mA because the OA will be maxing out but below that it should be clean and 20mA is pretty easy going for an OA.

So this circuit will be limited to pretty much biasing things like the + terminal or feedback network of an OA but it should do a pretty good job of it incidentally. However, you would probably not want to use this VGND to source or sink current to / from a line driver output for example (unless the line driver is differential in which case you're not sourcing / sinking to VGND).

You could double the current handling in the way you show. But there would still be a significant current limit of 50mA or so whereas a real ground is only limited by the physical characteristics of the ground connection (trace thickness, wire gauge and such).

A dual RC4580 like you've drawn with 680 ohm resistors could probably do 80mA cleanly.

[PRR]

With NFB taken at load end of the build-out resistor, it is still unstable with capacitance on VGND.

1K seems a high value. This would be fine if the load on VREF were several 100K bias resistors.

But you seem to want double-up. What is VREF expected to drive?? If the return from a headphone, 1K is larger than it needs to be. One 19-cent opamp can dump over 20mA. Batteries are usually low-voltage. 9V batt with 3V swing, 3V/20mA= 150 Ohms. Actually 100 Ohms may be fine.

Yes, you can use a quad opamp, even 47r for each, and get over 100mA VREF current.

However all this VREF current just heats the opamp and Rs. On battery, waste heat is bad battery life.  When you reach this level of power, it may (and may not) make more sense to design the audio circuit as a bridge, double your available output swing. (The specific case of 3-wire stereo headphones is difficult.)

[saint gillis]

I'm not thinking of a specific use for this, just trying to understand, and sometimes a dual op amp is as cheap as a single one.
So yes we have to consider if there is a battery and if we need to spare current.

Is there a specific interest to take VREF at the inverting input rather than just making a regular buffer and putting  the limiting resistor at the output?

And also with a 100 Ohm (for instance) limiting resistor on a tl071 or 4558, have much capacitive load can it handle? like few 100s µF? or up to 1000s µF? Is there an interest in filtering VREF or should we rather filter the input voltage between V+ and V- ?

[JohnRoberts]

It is better to start with an objective in mind..

This too is mature technology, they even made dedicated ICs http://www.ti.com/lit/ds/symlink/tle2426.pdf  While I never bought or used one.

JR

[saint gillis]

Sure...
But at least I have 2 specific questions...

[squarewave]

Is there a specific interest to take VREF at the inverting input rather than just making a regular buffer and putting  the limiting resistor at the output?

No! You need the feedback or the voltage at VREF will not be equal to what is on the + input. Current drawn though the current limiting resistor will cause a voltage drop.

Also, looking at your dual circuit again, I'm not sure if it will work correctly. Normally when you use multiple op amps to increase power, you mix the outputs through small resistors. Meaning the feedback is inside the resistor and not outside like you currently have it. But like I just said, you can't do that because you need feekback. There's probably a clever way to combine the two but I just can't think of it at the moment (and like I said before, it's not really going to make much of a difference if the circuit  will always have a current limit).

PS: The RC4558 is a horrible op amp. Maybe you mean RC4556 which is another OA that can source / sink more power.

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