The output impedance of a close loop OpAmp(voltage amplifier) is:

$$ Z_{out(CL)} =\frac{Z_{out} } {1+A_{VOL} B} $$

How is this equation derived?

The Z_out(CL) decreasing by a factor of (1+AB) is like implying that the current increases by the same factor.

I don't get why current will increase since in my eyes, the current is just a simple:

$$i_{out} = \frac{V_{diff}*A_{VOL}} {Z_{out}} =\frac{V_{out} } {Z_{out}} $$

Thus Zout(CL) is just equal to Zout.

What am I missing?

  • \$\begingroup\$ For calculation of the output impedance you must connect a voltage (or current) source to the OUTPUT (and short the input node) and measure/calculate the corresponding current (or voltage) at the output node. \$\endgroup\$
    – LvW
    Jul 12, 2020 at 8:50
  • \$\begingroup\$ It's a good practice to use lower case letters to denote small signal voltage and currents to differentiate them from large signal or DC values. The capital V in Vdiff struck me as strange at the moment I looked at it. \$\endgroup\$ Jul 12, 2020 at 14:24
  • 1
    \$\begingroup\$ The output impedance was derived directly from a negative feedback theory. If we "sample" the voltage the output impedance will decrease by the factor of (\$1 + A_{OL}\beta\$). Look here: electronics.stackexchange.com/questions/297409/… or here site.iugaza.edu.ps/mabuwarda/files/2017/09/… \$\endgroup\$
    – G36
    Jul 12, 2020 at 15:32

3 Answers 3


Draw the small signal equivalent of the circuit as shown below:


simulate this circuit – Schematic created using CircuitLab

Write KVL as follows: $$-ABv_x + i_xZ_o = v_x$$ $$Z_{oCL} = \frac{v_x}{i_o} = \frac{Z_o}{1+AB}$$


Just go back to the definition of the output impedance: if the output is excited with a given small signal current, how will the small signal output voltage look like? Than take the ratio. It is not important in theory whether you use a voltage or current source.

So "just" derive the equation of the small signal output voltage with a small signal load as variable, when the input is grounded. Then just express vout/iout.


You have a model of the opamp, including the gain_bandwidth plot that has high gain at DC, and then perhaps at 10Hz? begins to roll off, with a 90 degree phaseshift.

Now insert a Rout, perhaps 100 ohms or 20 ohms, in series with the voltage_controlled voltage source that is the opamp high_gain model. Ohhh I recall a opamp with 10,000Hz UGBW and the resistor Rout of 80,000 ohms, using only 1microAmp.

With the Rout, you now have an adequate model to derive the Zout with the phase shifts and frequency variation.

Now --- alter the model to become a CLOSED LOOP, gain of +1.

Now --- drive the Closed Loop model with a Current Source.

Now --- work thru the equations, and find Vout/Iout as the frequency varies. And you must bring along all the phase information.


Now for the surprise.

The falling gain of the opamp, and the 90 degree phase shift, and the Rout lumped component in the model --------- result in an INDUCTOR result.


Let us use the excellent answer (almost a 1_line derivation) of sarthak. And make the open_loop output impedance be just a resistor Ro. Now we can see:

  • Zo = Ro / (1 + A * B ) where B is ratio of 1, and A has -90 degree phase shift and A becomes very small as frequency approaches UGBW.

which becomes INDUCTIVE.

As frequency increases, the Zo increases just like an inductive reactance increases with frequency, while keeping +90 phaseshift.


For fun, evaluate the maths, or run a .AC simulation, with a external Capacitor shunting to Ground.

Notice for frequencies well below the opamp's UGBW, the Zo is very close to a pure inductor, thus there is NO DAMPENING for the ringing generated by the resonance of an external capacitor.

Its is YOUR RESPONSIBLITY to pick a DISCRETE RESISTOR, external to the silicon opamp, that will dampen.

I suggest you consider Rdampen = sqrt(L / C).

Can you now compute L?


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