For a given op amp, in my case a OPA197 by TI, I'm trying to calculate the output impedance for a near DC signal.

The datasheet gives an "open loop" output impedance of 375 Ohm, to me that seems high, unless I have lulled myself into believing op amps are more ideal then I though. I expected something in the 10s of Ohms. Also Wouldn't an output impedance be gain dependent?

Can anyone give me the proper way to find/calculate output impedance? Note I'm not looking for a method to measure it.

This question may have the answer and I am just not understanding...

Output impedance of Op Amp with Voltage Divider?

enter image description here


When the amplifier is operated with the "loop closed" i.e. it is run at (say) a gain of unity or ten or other normal sorts of gains, then the effective output impedance drops massively from 3 kohm to sub 1 ohm at DC.

This is due to the effect of negative feedback. So, you look at the open loop gain (about 120 dB or 1 million at dc) and factor this and the circuits target closed loop gain to calculate closed loop output impedance.

If your target closed loop gain were ten then the DC output impedance would be about 3 kohm / 100,000 = 30 milli ohm

At 100 kHz the open loop output impedance is about 400 ohms but the open loop gain is now reduced to 40 dB (100) so now the closed loop output impedance is about 40 ohms (assuming the closed loop gain is ten).

If the closed loop gain were unity then the closed loop output impedance at 100 kHz would be 4 ohms.

One more thing to remember is that the open loop output impedance is a real measure of the output transistor's ability to deliver current to a given output load so consider this: If the DC output impedance is 3 kohm and the load is 1 kohm, the maximum peak dc voltage that can be produced at the output is severely limited by this potential divider irrespective of the chosen closed loop gain.

Personally, I think that TI are potentially pulling the wool over people's eyes given that they state in the DS (page 8) that the maximum volt drop with a 2 kohm load (to mid rail) is only 500 mV. They refer to this as: -

Voltage output swing from rail

And, it can only be for high frequencies else it contradicts the open-loop output impedance graph. Be aware of this.

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  • \$\begingroup\$ Good explanation, I'd just want to point out that the output impedance is a small signal parameter. The voltage divider equivalent only holds for AC signals. The maximum DC voltage depends on the driving capabilities of the output stage (max. current and swing). The swing on page 8 is given in terms of headroom required for the output stage, so for a 2k load the output can swing as close as 500mV to either of the supply rails. For no load it's typically just 5mV which is considered a rail-to-rail output. \$\endgroup\$ – Mario Jul 8 '17 at 14:26

Consider example of gain , Av=100 or 40dB

  • if Aol=120dB ( which has wide tolerance) from spec.
  • Aol-Av=80dB feedback (4 decades) @ DC

  • if Zdc (o.l.) =350 Ω (spec)

    • then for Av=40dB, Zdc (c.l.) = 35 mΩ

However it is current limited to less than the short circuit current.

Isc, Short-circuit current ±65 mA @Vs = ±2.25 V from spec.

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  • \$\begingroup\$ So for a unity gain system, the impedance would be calculated as ~350Ohm/120db=0.35mOhm? So essentially it can be ignored then for my use case. I'm following the amp with a precision voltage divider and wanted to ensure the output impedance gave less then 0.1% added error. \$\endgroup\$ – MadHatter Jul 8 '17 at 5:12
  • \$\begingroup\$ correct................ not counting R tolerances for gain error and total Input offsets \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 8 '17 at 5:35

The datasheet gives an "open loop" output impedance of 375 Ohm, to me that seems high, unless I have lulled myself into believing op amps are more ideal then I though. I expected something in the 10s of Ohms. Also Wouldn't an output impedance be gain dependent?

On a non rail to rail opamp, you can expect an output stage like the one on the left:

enter image description here

This emitter follower topology has rather low open loop output impedance. However, it is not rail to rail. The output can't go higher than Vcc minus one Vbe, minus the top current source's minimum voltage.

The schematic on the right is rail to rail, but then we have a common emitter output stage, and the output is taken from the collectors.

This means the opamps' output is essentially a high impedance current source. Its high impedance is kept in check by an internal local feedback/compensation circuit, of which the only part shown here is the capacitor. Due to the capacitive coupling, this scheme does not work at low frequencies, which explains why the OL output impedance of your opamp rises at low frequencies.

So, yeah, without special circuitry, a rail to rail opamp will have higher open loop impedance due to the topology of its output stage. It counts on feedback to lower it.

Low OL impedance in a rail to rail opamp indicates the designers employed some clever local feedback around the output stage to improve performance. It's a compromise wrt cost, idle current, etc, as usual, can't have everything.

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