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I have read a lot about capacitive loads on the output of an op-amp and the possibility of unwanted oscillations and instability. my knowledge on phase margin and frequency domain analysis is a little weak and I can't figure out if my circuit bellow is gonna be unstable or not:enter image description here

The Op-amp supply is connected to 5vdc and GND. and here is it's datasheet:

I read in some application note that placing R19 would help the stability issue by increasing phase margin or something like that and fortunately I already had that in my circuit. but as you can see C10 capacitor also has a very high capacitance value. is there an analysis or rule of thumb which would determine if this circuit is safe? should I change my op-amp IC? or should I use other methods as well to insure my circuit's safe operation?

I can provide additional information if you need it. Thanks.

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4 Answers 4

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Most OAs have limited capacitive load capabilities. The origin is quite intuitive. Consider the simple circuit below. On the left there is a real OA. On the right, the OA has the same characteristics (poles, etc.) but the output resistance is externally shown.

This will have the effect of creating a pole, with time constant \$R_{out}\cdot C\$.

Such pole will reduce your phase margin of your system.

Limiting case: C is so large that it can be considered a short. Then you're actually removing the feedback, and you get a comparator (assuming non zero feedback Z)!

Instead, if there is a "large enough" resistor in series to C (R19 in your circuit), the series R-C will add a pole (with time constant \$(R_{out}+R_{19})\cdot C\$ ) and a zero (with time constant \$R_{19}\cdot C\$ ). If, as said, R19 is large enough, the pole and zero time constants are close enough and the overall effect is negligible (limiting case: R19 is infinity).

Intuitively, after the zero frequency, the series R19-C will act as a load, and it will not open the feedback (it will change the feedback attenuation factor, but if \$R_{19}>>R_{out}\$ then this variation will be negligible as well).

EDIT: In your case, the circuit will be stable.

EDIT2: the following consideration is no more valid with the new edits to the schematics made by the OP:

I assume that "Input Signal" in your schematics is a current signal (i.e. not a connected voltage source), otherwise the circuit won't work.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Why would a voltage source not work? \$\endgroup\$ Commented Sep 18, 2017 at 12:00
  • \$\begingroup\$ The OP has edited the schematics (See question edit revision history), and now R18 is not going to the output, but to the GND. Now it's stable. If you leave the original R18 connection to the output, you notice that if you apply a voltage (e.g. 5V), then V+ will be stably larger than V-. \$\endgroup\$
    – next-hack
    Commented Sep 18, 2017 at 12:16
  • \$\begingroup\$ @VladimirCravero and therefore it will saturate. With a current source, this could not happen. In fact you have \$V_o-I_i R_{18} = V_o D + 2.5V (1-D)\$, where \$D=R_{15}/(R_{15}+R_{16})\$. Then, solving for \$V_o\$, you have \$V_o(1-D) = I_i R_{18} + (1-D) 2.5V\$ hence: \$V_o = I_i R_{18}/(1-D) + 2.5V\$ (I hope there are no errors on these quick inline calculations) \$\endgroup\$
    – next-hack
    Commented Sep 18, 2017 at 12:33
  • \$\begingroup\$ Thanks for the clarification, I didn't see the original schematic. \$\endgroup\$ Commented Sep 18, 2017 at 12:49
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Lets cut through the hand waving, and provide a specific answer to Rseries, so as to dampen the ringing.

OpAmps have excellent Rout at low frequencies, with that Rout increasing continually above the open loop gain rolloff frequency, often 10Hz or 100Hz. The increase in Rout also shows a 90 degree phase shift. Result is the OpAmp has inductive Zout, above that 10Hz or 100Hz corner frequency.

That inductance, given a capacitance on Vout, will ring unless adequately dampened. Can we predict Lout (output inductance)? Some opamps provide a chart of Rout versus freq; if so, grab the Rout at UGBW (you may have to extrapolate), and compute the Rout = 2 * pi * Freq * L, solving for L.

An opamp with 100 ohm Rout at 1MHz (many ordinary opamps), has inductance of

L = 100 / ( 2 * pi * 1e+6) = 16 microHenry.

How to dampen? Use the formula Rdampen = sqrt(L / C), derived from considering the various equations for Q=1 for R and L, and R and C, and combining to eliminate Frequency as a variable.

Given 1uF Cload, and 16uH output inductance, the proper Rdamp is sqrt(16/1) = 4 ohms.

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Some opamps can have an issue when driving a capacitor to ground or supply directly meaning, without any additional series resistance.

You do not have this issue as long as R19 is present and larger than 500 ohms or something in that order.

When an opamp has a large capacitor to ground or supply directly at the output and the output is also used in a feedback configuration then instability can occur or the circuit might even work as an oscillator.

Both opamps do have feedback loops but the large capacitor C10 is not directly on any opamp output so I do not see any potential stability issues with this circuit. Here R19 saves your day as it separates the opamp circuit from the large capacitor C10.

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Tim Green wrote a quite nice and hands-on tutorial on stability of opamps. There are very little formulas and he tries to give an intuition by using lots of graphs and drawings.

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    \$\begingroup\$ This answer will be useless when TI happens to rearrange their website, something that companies are prone to do. It would work if you incorporate the solutions into the answer, and link to the file as an additional source. \$\endgroup\$
    – pipe
    Commented Sep 18, 2017 at 12:47
  • \$\begingroup\$ Yes the link will be dead. But you can still google for it. Or go to archive.org. Yes, I could incorporate the solution into my answer. But do you really want to have a 100 page essay here? Opamp stability is not a trivial issue. You cannot explain it in a paragraph or two. There is a reason why any textbook that even mentions this problem requires you to have at least good understanding of signal theory and of control systems. Tim Green's tutorial is the most simple explanation I know of. \$\endgroup\$ Commented Sep 18, 2017 at 12:57
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    \$\begingroup\$ Stack Exchange are for complete answers to a specific question - not complete essays on a given topic. As per Stack Exchange policy, your answer should answer the question and only use links for further reading/reference for material. If you think a question is too broad, please flag it as such so it can be dealt with using the moderation tools available to high-rep users. \$\endgroup\$
    – pipe
    Commented Sep 18, 2017 at 13:58
  • \$\begingroup\$ Well, the further reading/reference material was my intention. next-hack already answered the immediate question, in my opinion. Though not very well, as the knowledge that you need to understand it is quite broad. Ok, maybe I should have done this as a comment to his answer. \$\endgroup\$ Commented Sep 18, 2017 at 17:15

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