What does this op amp circuit do and why is Vout considered to be "virtual ground"?

Can someone identify this op-amp circuit?

What is it called, how does it work?

Vout is also called virtual ground.

simulate this circuit – Schematic created using CircuitLab

• I think it's meant to buffer a weak voltage reference. The extra capacitor to ground may be decoupling for whatever comes next in the circuit. Context would help though. Commented Oct 6, 2023 at 14:23
• "Called virtual ground" by what source? Commented Oct 6, 2023 at 22:37

The circuit is called a non-nverting amplifier.

It amplifies a constant reference voltage (Vref) to produce V0ut at a higher current capability. $$Vout=(1+\frac{R_2}{R_4})V_{ref}$$

$$\C_1\$$ compensates for input capacitance.

$$\R_3\$$ isolates the output from $$\C_2\$$ to prevent instability.

Many folks call Vout a virtual ground in the context of rail-splitting. I prefer the phrase mid-point bias voltage, although this circuit can provide a clean dc voltatage at any value determined by Vref and the gain. Capacitor $$\C_2\$$ provides an ac path to ground to provide a zero ac reference at higher frequencies than the op-amp can handle.

In op-amp circuits with split supplies, real ground (0 volts) is halfway between +VCC and -VEE. For single supply operation, a mid-point bias is used as if it was real ground.

• Why is it even virtual ground? How does virtual ground also mena mid-point bias voltage? Commented Oct 6, 2023 at 14:45
• See my edit @JoeyB Commented Oct 6, 2023 at 14:55
• Your equations are right, but I've never heard anyone calling Vout a virtual ground, because it simply isn't. The definition of a virtual ground (a concept really intended to ease newbies into op-amp circuits) is a point in the circuit which the op-amp will actively maintain at close to 0V. If it's not at 0V, it's not a virtual ground. Commented Oct 7, 2023 at 8:00
• There are two usages right or wrong @Graham. The TLE2426 rail splitter is advertised by Texas Instruments as a "Precision Virtual Ground." The OP has heard of the usage and is asking for clarification, which I have done. Stating that "There is no virtual ground in this circuit." is not helpful. The OP is looking for a reason why. I agree that the usage should not be used, but who am I to argue with TI. Commented Oct 7, 2023 at 16:09

Vout is not a virtual ground. This is a non-inverting op amp so there is no virtual ground in this circuit (normally that is the inverting input of an inverting op amp circuit).

Some component values would help, but this looks like the output driver stage of (possibly) an audio circuit or some such. R3 will often be about 47 ohms, R2 and R4 a few k each, C1 small, used for stability. C2 is odd. It is not normal to have capacitance from an op amp output to ground, but it might be very small (a few 10s of pF), perhaps to try to keep external HF noise out, if a long cable is attached.

As I said, component values and some context would help a lot.

There is no virtual ground in this circuit.

This type of circuit is usually used to buffer a reference voltage to be able to supply a significant current to the succeeding stage but also with significant capacitance or requiring decoupling for high frequencies. Normally a capacitance at the output of an amplifier will affect stability and can cause oscillation.

Resistors R2 and R4 provide DC feedback from Vout and set the gain between Vref and Vout.

Vout = Vref * (R2 + R4)/R4

R3 is to isolate the capacitance of the load from the amplifier by allowing the voltage at the output pin to change without being restricted by charging capacitor C2. If connected directly it would cause additional phase shift because of the output impedance of the amplifier and the capacitor. The resistor avoids this and provides a circuit node for AC feedback to the input with capacitor A1.

By suitable selection of values the circuit can be made stable although it is common to end up with a hump in the frequency response.

• How does R3 isolate the capacitance of the load from the amplifier? Commented Oct 6, 2023 at 14:35
• @JoeyB - I've added some text describing how R3 helps. Commented Oct 6, 2023 at 16:01
• Why is AC feedback important and how does it ensure stability Commented Oct 6, 2023 at 16:20
• @JoeyB - AC feedback doesn't ensure stability but it is a requirement for stability that the gain around the loop is less than unity at the frequency where the phase shift is zero. The feedback through R2 alone will usually not meet this requirement. This is a complex area of circuit design that can't be explained in a comment. Commented Oct 10, 2023 at 22:12

Possible context, if you want a symmetrical output but only have a single supply, you have to create a reference point, usually in the middle of your supply range. (f.i. 2.5v reference in a 5v supply)

Any signal that is symnetrical around this signal, can be output using the reference voltage as 'ground' (or more correctly 'return'). On an SMA connector you would connect the reference voltage to the barrel.

• This isn't an answer but a comment. Commented Oct 9, 2023 at 7:15

C1 is used mostly to avoid an oscillations - 47-100pF. V out is not a virtual ground - V out​ (the output) of the op-amp should not be perceived as a virtual ground. The output voltage V out is the result of the amplification and is not constrained to be close to 0V (or ground potential) like the virtual ground concept dictates for the inverting terminal in the inverting configuration.