Electronic – Non-inverting UA741 opamp simulation on QUCS

Good question!

There is no ground leg of an opamp: there's a positive and a negative supply rail, and the current through each of them is roughly as follows:

$$\begin{aligned} I_{s+} &= I_q + \max(I_{out}, 0) = \max(I_q, I_q+I_{out})\\ I_{s-} &= I_q + \max(-I_{out}, 0) = \max(I_q, I_q-I_{out}) \end{aligned}$$

where Is+ is the current into the + supply terminal, Is- is the current out of the - supply terminal, Iq is the quiescent current specified in the datasheet, and Iout is the current flowing out of the output.

So, for example, if Iq = 1mA and Iout = 3mA, then Is+ = 4mA and Is- = 1mA; if Iq = 1mA and Iout = -3mA then Is+ = 1mA and Is- = 4mA.

This is because opamps are linear and use pass transistors in their output stages: positive output current = sourced current has to come from the + supply terminal, and negative output current = sink current has to go into the - supply terminal.

I say "roughly" because there are two other factors:

1. Input currents are never exactly zero -- if they're picoamps or nanoamps you can probably ignore them, but if the opamp is a bipolar opamp and the output is saturated (inputs are unequal) then the input current could potentially be larger than the specified input bias current, and could be significant enough that you can't neglect it. You'd have to know which polarity the input currents are: with PNP input stages, input current would be negative (current coming out of the inputs) and therefore add to the current from the positive power supply terminal; with NPN input stages, input current would be positive (current going into the inputs) and therefore add to the current exiting the negative power supply terminal.

2. Quiescent current isn't exactly constant, and can vary somewhat with operating point (especially if the opamp is in saturation, or if the output load is large).

What happens if there are resistors in those legs?

Then your supply voltage will vary somewhat due to the IR drops, depending on load current. Don't do this without two things:

1. You must understand completely what the load current is, and make sure it doesn't cause enough IR drops to make a significant difference in the circuit capability (e.g. op-amp saturation range contracting, or worse yet, opamp failing to work at all because supply voltage range drops below that for which it's guaranteed to work).

2. For goodness sake, put bypass capacitors between the supply voltage pins to your signal ground, so that AC supply current variation won't cause AC supply voltage variation that couples into your circuit.

When we talk about the input resistance of a circuit, we're describing how it affect the other circuit that is providing the input signal.

Specifically, we want to know, in order to change the input current by i amps, how much do we need to change the input voltage? That's why the input resistance is, by definition, \$ \dfrac{\mathrm{d}v_i}{\mathrm{d}i_i}\$.

So what's the input resistance of this circuit?

The key point is that in this configuration, as long as we avoid saturating the op-amp output, the inverting input of the op-amp is a virtual ground. The feedback in the circuit operates to keep that node at 0 V. So whatever input current we want, by Ohm's Law, the required input voltage is \$\mathrm{R_{in}}\times{}i_{i}\$. Therefore the input resistance is R_in.

Then there are answers like this that mention that input impedance is infinity.

That answer was talking about the input resistance of the op-amp, which was assumed to be ideal, not the input resistance of the whole circuit.