How does an OPA work?

I am a bit confused about how OPAs work. My questions are as follows:

Let's say I use an OPA with dual, simmetric supply. I do not put anything to its inputs. If there is no voltage difference between the two input pins, how will output work? Will it be 0V or something else?

What if I use the OPA with single supply? The same case ( no voltage difference between inputs ). What voltage will I measure on the output? Near 0V? ( I know, the voltage on the output cannot reach

0, except the OPA is rail-to-rail. Let's say it is a non rail-to-rail kind. ) Half of the supply voltage?

What if I use the OPA with +10V for positive and -3V for negativ supply. Would it work at all?

Thanks for your replies, Zsolt ( from Hungary )

Ideally, the output is at zero volts when the difference between the two inputs is zero. In real life, this isn't what happens.

For many op-amps the output really follows a rule like:

Vout = Vneg + K1 * (IN1 - IN2 + Voff)

where: Vneg is the minus supply of the op-amp K1 is a huge number like a million IN1 is the non-invering input IN2 is the inverting input Voff is the offset voltage of this op-amp

Because K1 is so huge, even a tiny voltage difference between IN1 and IN2 matters more than it does, so we can drop it from the equation when we don't need a perfectly accurate description. This makes the equation:

Vout = K1 * (IN1 - IN2 + Voff)

Voff is usually such a small number that in most cases, we don't need to worry about it so we can drop it too.

Vout = K1 * (IN1 - IN2)

Now lets use this to figure the gain of a simple inverting op-amp.

First we need to say that the inputs of an op-amp draw no current. This isn't really true but the current they do draw is usually so small that it can be ignored.

ASCII ART

R1 R2 Vin-----/\\/\\------+-------\\/\\/-------+-------Vout ! ! ! ! --!-\\ ! ! >------------ GND-------!+/

Lets make R1 and R2 equal and 1K and K1 a million.

We will work from the output back to the input. Assume Vout = 1V

Remember: Vout = K1 * (IN1 - IN2)

Rearrange:

(IN1 - IN2) = Vout/K1

IN1 = 0 so IN2 = -1V/K1 = -0.000001V

R2 has a voltage of 1V - (-0.000001V) = 1.000001V on it

R2 has 1.000001V and is 1K so I= 1.000001mA

R1 has 1.000001mA and is 1K so it drops 1.000001V

IN2 is at -0.000001V so the Vin must be at:

(-0.000001) - 1.000001 = 1.000002V

The gain of this circuit is Vout/Vin so it is

1V / 1.000002V = 0.999998

This is so close to R2/R1 that there is no reason not to just say it is R2/R1.