Op - Amp and Unbalanced Input Differential Signal

Question

Let's consider for instance an Open Loop Op-Amp (for instance with double power supply: VDD and -VDD), which is a good comparator circuit.

The sign of its input differential signal (Va - Vb) determines if its output will be VDD or - VDD.

Surely it works if Va and Vb are voltage sources connected between, respectively + and - of Op-Amp, and GND.

But what does it happen if we put a source of value Va - Vb between + and -? The difference is that it is not balanced with respect to GND. It is something like putting a battery between + and -.

I'd say that the circuit will not work, since the base - emitter (or gate - source) part of each transistor of the differential pair at the input stage wants a signal referred to GND.

The circuit may work if at the input of the op-amp is present a sort of BALUN, is there it?

Answer 1

The difference is that it is not balanced with respect to GND.

You mean: not referenced to ground. And that is actually the issue, for example if you would do this:

^{simulate this circuit – Schematic created using CircuitLab}

then the Vfloat battery only sets the voltage difference between the opamp's inputs. The voltage (or potential) at the opamp's inputs relative to GND is NOT defined. The inputs are actually floating relative to GND.

An ideal opamp (or a model of an ideal opamp used in a simulator) would only consider the voltage difference so the floating inputs are no issue.

However, in the real world floating inputs is an issue. It means that the input voltage relative to GND will depend on leakage currents, for example from the ESD diodes that are present in the opamp. This could mean that the voltage relative to ground "floats" to a value that the opamp cannot work with, for example a voltage that is very close to GND or the supply voltage. Not all opamps can deal with that.

Besides the question if the opamp can deal with it or not, it is simply bad practice (to leave inputs floating) so it should be avoided. There is a simple solution, just define the voltage with high value resistors like so:

^{simulate this circuit}

This sets the input voltage relative to GND of the - input to Vsupply/2. You can also set the + input instead of the - input, you can also choose different resistor values and resistos value ratio to set a different voltage than Vsupply/2 as long as the voltage is in a range that the opamp can use.

So yes, the opamp's inputs need voltages that are referenced or have a reference to GND. If you don't apply that then leakage currents etc. will take over and define it for you. That's something to avoid as it goes out of your (the designer's) control. You can never know what will happen. So always define the voltage.

Answer 2

If the opamp inputs are bipolar transistors the operation is impossible because one transistor cannot feed the mandatory few nanoampere bias current for the other. Both base current directions cannot be right at the same time.

If the opamp has fet inputs the gate voltages must be defined in relative with the fet substrate. That means some connetion is mandatory between the gates and the rest of the fets. As said already by other's, leakages would take the place of the missing connection and what voltages happen to be in series with them would render the result unpredictable.