# Why should the input terminal be shorted to ground when analyzing the output impedance of an op-amp?

I read this question about analyzing the input impedance of an op-amp.
It says to short the input to ground when analyzing the input impedance.

I've also read about using a two-port network, for which to analyze an op-amp model, I should adopt Z-parameters because the op-amp model is a voltage controlled voltage source.
According Z-parameters, the current of input should be 0, which means the input should be open.

Why should the input terminal be shorted to ground when analyzing the output impedance of an op-amp?

The z-parameter theory is not especially useful for opamps. They are highly non-linear devices. But when they are used in a well designed feedback circuit, the circuit as a whole can operate linearly or actually so closely linearly that we can accept the result.

The simplistic model of an opamp (a voltage controlled voltage source which may have an internal series resistance in the output) is handled properly in your linked example and gives the closed loop output resistance of the shown circuit. The simplistic model isolates the input from the output, the input is an open circuit which has no current, no matter what's connected to the output. The Z-parameter equations are useless because zero input current cannot cause any output. The circuit could be said to be a singular case for the whole Z-parameter idea. Thus also Z-parameter rules are useless just in this case.

The simplistic model doesn't at all present the real internal circuitry of an opamp which for example doesn't at all work if the input DC currents are prevented by leaving the inputs open.

Of course, you can use a real opamp in a well designed stable circuit. There in the idle state every part finds its operating point and can be linearized for using the linear circuit theory. Z-parameters may well work when the parameters are calculated for the operating point and the signals are small. Making such linearization is a hefty task and it's needed for a dense enough collection of operating points. But surely it's done by the designers of the opamp IC. The rest of us use some simplified model of the opamp - maybe something as simple as used in your linked example. If a circuit doesn't work with the simplistic model, it very likely doesn't anything useful with practical opamps.

Instead of the simplistic model one should use something which predicts better such effects as gain reduction vs frequency, phase lag vs frequency, voltage and current clipping and some other important non-linearities in the internal circuitry, noise, input current and power consumption. Opamp manufacturers offer for us circuit models to be used in circuit analysis programs. Often the given model predicts pretty well how a built circuit would behave although the model isn't the schematic of the real internal circuit, it's something much simpler, only an approximation of the behaviour of the real IC.

According Z-parameters, current of input should be 0, it means input should be open.

Yes - that is correct. However, it concerns the SIGNAL voltages only. On the other hand, the device should work under normal DC bias conditions.

Therefore: Input terminals of the amplifier (with suitable feedback) should be short to ground. Without DC feedback, offset compensation is necessary for a real opamp.

• And a real op-amp’s offset may well not be constant even if the temperature and supply voltages are constant. It shifts with the common mode voltage - usually only slightly, but more dramatically when the input stage transitions between N and P devices (whether BJT, FET or MOSFET). A very high precision (high Aol, excellent input stage) op-amp’s input offset may move say 100nV over 1V of the common mode range, as an example. That’s a 0.1V change on the output. Offset compensation then needs to be a DC servo rather than constant, and we’re back to DC feedback. No way around it, usually. Commented Nov 9, 2023 at 14:03