How does an op-amp know where ground is?

How does anything know where ground is? Ground is just a symbol we stick on the schematic to make it easier to read. None of the components in a normal circuit read the schematic, so none of them know where ground is.

In the case of op-amp B, the output voltage will be either the maximum voltage the op-amp can output (limited by the supply rails), or the minimum, depending on the polarity of the voltage source on the input.

And building such a circuit in practice, you'd have a problem: there's no path from the voltage source on the input to anything else. As such, the actual values there will be defined by the input bias current of the op-amp and other non-ideal behaviors, so what you'll get is something strange that's mostly a function of the details of that particular op-amp.

You'll probably find it easier to think of op-amps not as amplifying the difference between their terminals. In practice, op-amps are usually operated with negative feedback: when they aren't, they tend to be called comparators. So, the op-amp tries to adjust the output voltage such that the two inputs are equal, and for the ideal op-amp with infinite gain, this is exactly the case: the inputs will always be at the same potential.

The input bias currents behave as below, where I1 and I2 are the respective input bias currents and I2-I1 is the input offset current.

^{simulate this circuit – Schematic created using CircuitLab}

The op-amp will only operate properly if the inputs are within a given common mode range (with respect to Vcc and Vee). That might be very close to the supplies or it might be a volt or two away from either or both supply.

As you can see in your right-hand example, there is no path for I1+I2, so the inputs will rapidly approach the supply rail (at which point the current sources stop being more-or-less ideal).

It's possible that some op-amps under some conditions might happen to sort-of work but it's not something you should rely upon. Always provide a DC path for both inverting and non-inverting inputs. The above example provides a path for only the offset current (I2-I1). The total bias current (I1 + I2) has no path.

As to what exactly the output would be- you can think of it as Avv_d(Vcc+Vee)/2, though the offset voltage of the op-amp times the gain is usually enough to saturate the output at either rail, so the mid-rail adder (Vcc+Vee)/2 is sort of arbitrary. Hopefully that makes sense to you.

This becomes clearer with some information on the internals (wikipedia).

That's the old style of bipolar-transistor input. A moderate input bias current (some microamps) flows through the transistors to the negative/positive supply rails.

FET and JFET inputs have much smaller input currents, but there's still a reference against the supply - across the insulating gate of the FET.

There may also be input protection diodes.

See the following elementary drawing of the internals of the opamp:

The input transistors need their base current - both of them! The current is usually less than 1uA. The opamp determines himself how much he takes, but it must be available and for both input it must be directed to inside the transistor. If you connect "something" only between the + and - inputs, then the currents can't be simultaneously towards the transistors because that "something" should create new electric charge. It's Kirchoff's law.

In practical opamp circuits the way for the input base current (=bias current )is some conductive part between the input and the supply voltage rail or the GND. In this case (see the arrows in the input transistor emitters) the -VE supply rail is impossible as the right direction input current supplier, but +VE rail is ok and also GND if its lifted above the -VE potential by either adding a battery betveen -VE and GND or by a resitor that is connected to +VE.

Fet inputs are not better. Without a galvanic connection to elsewhere than the "something" between the inputs, they soon drift to undetermined state due the accumulated leakage charge into the gates of the fets.

An op-amp has no idea where ground is.

Op-amps are differential amplifiers. They amplify the difference between the two inputs and (ideally) ignore any common-mode voltage. There's no difference between the two circuits in your diagram. Neither op-amp's output is referenced to ground. The output bias point is probably close to mid-way between the two supplies. You could measure try to measure it by shorting the inputs together, but you'll have to deal with the input bias voltage and current too. It's probably not worth the trouble.

Fortunately, you don't have to worry about the output bias point or the "real" reference voltage, because they don't matter for either of an op-amp's common uses. If you're using the op-amp as a comparator, you want the output to be either as positive as possible or as negative as possible, even for a tiny differential voltage. If you're using the op-amp in a linear circuit, you use negative feedback, which which causes the output to be referenced to the positive input.

Real physical op-amps aren't perfect differential amplifiers, so in real life common-mode voltage has a small effect on the output. As Phil's answer says, the construction of the op-amp matters too. But I don't think that's important for what you're asking about. Op-amps aren't built to do what your circuit is trying to make them do.

It doesn't, and if you don't ground one of the terminals the output voltage cannot be determined. Why? because of input bias current. Opamps are not perfect, they require a small amount of current. The input bias current can vary from device to device on each terminal.

If the input bias current is small enough and the input impedance high enough other currents can determine what the voltage on the terminals are.

If your doing any kind of sensing you need to ground one side of the terminals.

A thermocouple is just like a voltage source but if you don't reference it to ground, it could be floating anywhere. In the (a) example the voltage between the terminals is the voltage of the thermocouple (and of the voltage source) but the voltage common to both terminals could be 0V, 1V, -2.3V virtually anywhere .

As there is no negative feedback, the output will be:

Vo = +Vcc (if V+ > V-, like in the diagrams you provided) Vo = -Vee (if V+ < V-, like if you inverted the Vd input polarity )

Considering them as ideal op-amps (and regardless of which kind of supply it would use - if single or dual), this is independent of Vcc and Vee. But the thing is: the system doesn't need a "ground" to work because it just does its work with the voltage difference between both of them.

Some months ago I had to build a light-sensing "robot flower" which pointed to the strongest light source. It used four LDRs - one pair for looking up/downwards, and one pair for turning horizontally. Each LDR was connected to a current source and gave its potential difference to a summing amplifier.

One of the problems I had to face, was that the op-amp was one of the dual supply kind (TL084). I needed +/- 9V as source, and I could only have one battery. So I used an ICL7660 inverting switching source (they turn +9V into -9V); but the problem was that the input current was such that the output voltage fell (or rose) to -6V. And while feeding the summing amplifier with 9V and -6V, the circuit couldn't find its ground correctly and had to create itself an offset. See: in that case, ground should have been "(9V + (-6V))/2 = 1.5 V"... not zero (in fact, said offset was around 1.5 V)

But that's because this circuit needed a common ground to compare its output with the inputs, such being the objective of the process of negative feedback itself... and that common ground should be the midpoint between both power supply nodes. In the case of your circuit, it solely acts as a comparator so the output is just 9V or -6V depending on the polarity of the source Vd.

Sorry if my answer was too long! It's just that it's cool to share different experiences that maybe can help others... In fact; this is my first post! Hope it helped!