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Many datasheets recommend this, for 'stability.' I've never understood this. Googling the subject I've seen much talk about "adding additional poles" but, not being an electrical engineer I don't know what they mean by this or why it helps.

Can anyone provide some physical intuition? It seems that the op amp will 'see' the capacitor whether or not you put a resistor on the end of it.

Another question: how does one choose the value of this resistor?

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schematic

simulate this circuit – Schematic created using CircuitLab

The effect of the decoupling resistor is as follows:

It does not attenuate oscillations - it prevents the start of oscillations. In this case, the keyword again is "loop gain". Together with the finite output resistance r,out of the opamp the load capacitor C forms a first-order lowpass. In many cases this causes an additional phase shift for the loop gain that can bring the opamp (under closed-loop conditions) to its stability limits.

Now - what is the effect of the additional resistor R? It prevents that the loop gain for very large frequencies is approaching zero; instead the loop gain reaches a finite value. This is identical with a phase enhancement which is necessary to avoid instability. In terms of system theory: We have created an additional "zero".

Example: Without such a resistor R the lowpass has the function:

$$ H(s) = \frac{\frac {1}{sC}}{r_{out}+\frac{1}{sC}} $$ $$ H(s) = \frac {1}{1+s(r_{out}C)}$$

With resistor R we have:

$$ H(s) = \frac{R+\frac{1}{sC}}{R+r_{out}+\frac{1}{sC}} $$ $$ H(s) = \frac{1+sRC}{1+s(r_{out}+R)C} $$

$$ \lim_{s \to \infty} H(s) = \frac {R}{r_{out}+R} $$

As you can see, as frequency increases, the lowpass function now approaches \$R/(r_{out}+R) \$

In other words: The lowpass has changed its properties - the phase does not continuously go to -90 deg but increases again towars zero (real voltage division between r,out and R). As a consequence, the phase of the loop gain (including the phase shift of the opamp) does not reach the critical value of -180deg at the cross-over frequency (loop gain 0 dB). This stabilizes the whole circuit.

The value of this additional resistor R should be chosen to produce a zero app at 25%---50% of the cross over frequency (for unity gain feedback identical to the opamps transit frequency).

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  • \$\begingroup\$ Thank you. one question: you say the phase doesn't go to -90 anymore but increases "towards zero"-- the way I understand this, H(s) (the gain) no longer approaches zero at high frequencies, thus the phase shift never gets to -90? \$\endgroup\$ – Paul L Feb 23 '15 at 3:16
  • \$\begingroup\$ @PaulL Not exactly. Note in the first expression for \$H(S)\$, the \$r_{out}C\$ term dominates the denominator. When you take the limit as \$s \rightarrow \infty\$, you end up with \$\frac{1}{s(r_{out}C)}\$. Because there is an \$s\$ in the denominator, the expression has a -90° phase shift. In the stabilized version, the \$s\$ in the numerator and denominator cancel, leaving a completely real function (i.e. 0° phase shift) \$\endgroup\$ – Shamtam Feb 23 '15 at 6:14
  • \$\begingroup\$ Paul, I spoke about the phase of the RC circuit only. Of course, the phase of the opamp´s open-loop gain still approaches -90 deg (for gain values above 0 dB). Without the decoupling resistor, the total phase shift would go through -180 deg, which could be very critical because this is identical to -360 deg (0 deg) due to additional phase inversion (180deg) of the feedback inv. input. \$\endgroup\$ – LvW Feb 23 '15 at 9:10
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The resistor limits the amount of current that the op-amp has to / can sink or source, ensuring that the output transistor/s are never driven into saturation even when the op-amp is charging a completely empty capacitor (which briefly resembles a dead short to ground, which will almost certainly cause saturation).

A 'pole' is the frequency at which a filter starts have a noticeable effect; for example and very broadly, a low-pass filter with a pole at 100Hz would cause signals with a frequency of more than 100Hz to be attenuated, while frequencies of less than 100Hz would be unaffected.

Pretty much any electrical circuit exhibits some filter-like behaviour, and has poles (and their complement, 'zeroes'). A circuit made up of a resistor in series with a capacitor to ground is the archetype of a low-pass filter; it creates a pole at a frequency 1 / (2 * pi * R * C).

If an op-amp starts oscillating (usually seen as the output madly swinging between minimum and maximum as fast as it can go, at very high frequency), typically due to too much positive (self-reinforcing) feedback making it unstable, then a low-pass filter on the output will attenuate the high-frequency oscillations, which makes the op-amp more stable - less likely to start oscillating and more likely to stop oscillating if it does start.

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  • \$\begingroup\$ I think, it´s not quite correct to say that the lowpass at the output "will attenuate the high-frequency oscillations". Instead, the phase of the loop gain is enhanced, thereby preventing oscillations from the beginning. \$\endgroup\$ – LvW Feb 22 '15 at 9:22
  • \$\begingroup\$ @LvW I was trying to make the explanation more approachable. \$\endgroup\$ – markt Feb 22 '15 at 9:52
  • \$\begingroup\$ The low pass filter on the output (Ro and load capacitance) is what makes the op-amp potentially unstable (by reducing the phase margin). The key thing is that the feedback is taken before the (undesirable) lag of the added resistor and load capacitance. \$\endgroup\$ – Spehro Pefhany Feb 22 '15 at 12:39
  • \$\begingroup\$ Quote markt: "An obvious follow-up question may be "why don't all op-amps just have a low-pass filter on the output to make them more stable?" This would not cure the situation - just the oppopsite is true. Both elements R and C must be located OUTSIDE the output terminal of the opamp (as shown in the circuit diagram). Otherwise the circuit does not work as desired. \$\endgroup\$ – LvW Feb 22 '15 at 17:59

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