Operational amplifiers

o takes three nodes, in this order:

o1,n+,n−,nout

No power rails, no supply, no gain figure. It is an ideal op amp: a nullor, which is the statement that the two inputs are at the same potential and draw no current, and that the output supplies whatever the rest of the circuit needs.

e,1,0,2
r1,1,m,10'k
r2,m,o,47'k
o1,0,m,o

The inverting amplifier. Node m is the summing junction, o the output.

So the output sits at −9.4 V: a gain of −4.7, which is −47k/10k, arrived at from the nullor conditions rather than from the formula.

Note the node order — 0 is the non-inverting input and m the inverting one. Swap them and the feedback becomes positive, which is a different circuit and will usually refuse to solve.

What the op amp reports

An o reports a current and a power, and no voltage drop — its input terminals are at the same potential by definition, so a drop across it would not mean anything. The current is the one it sources or sinks at its output.

When ideal is not enough

There is no finite-gain op amp element. Model one as a dependent source, which is what a finite-gain op amp is:

e,1,0,2
r1,1,m,10'k
r2,m,o,47'k
ea,o,0,100000*(0-vm)

ea is a VCVS of gain 10⁵ driving the output node from the differential input — here 0 - vm, the non-inverting input minus the inverting one. Solve it and the answer differs from the ideal one in the fifth figure, which is the honest way to find out whether the ideal model was good enough for your problem.

The same shape gives you finite input impedance (a resistor across the inputs) and non-zero output impedance (a resistor in series with ea). Once you are modelling it, the op amp is just circuit elements.