1
\$\begingroup\$

It has been a long-lived tradition of mine to put an RC low-pass between the power supply and each power-in pin of each op-amp, where R is typically 1 to 10 ohms, and C is either 0.1uF ceramic, or that in parallel with a, say, 10uF tantalum:

schematic

simulate this circuit – Schematic created using CircuitLab

My thinking is: the decoupling capacitor is supposed to be a low-impedance path to GND for switching/high-speed signals that originate inside the op-amp, so that they do not propagate to the rest of the system; by placing a resistor in series I not only make the path to GND low-impedance, but the path to the rest of the system now has substantially higher impedance compared to just connecting directly.

Does this make sense in general? Or am I wasting my time, money, and PCB space in placing next-to-useless resistors?

\$\endgroup\$
1
  • 2
    \$\begingroup\$ If there is a long trace in the power supply line, the 1 Ohm resistor is good at providing damping by lowering the Q of the resonance between the inductance of the trace and the decoupling capacitor. The ESR of a tantalum capacitor can do the same thing, if it is high enough. \$\endgroup\$ Dec 19 '16 at 5:04
4
\$\begingroup\$

I found an article that discussed using resistors in the power lines of op amps (top of page 47). They say that in troublesome cases, you can add inductors to the power lines along with the decoupling capacitors. Or for a cheaper solution you can use a series resistor in the 10‎Ω to 100‎Ω range to form a low-pass filter. The drawback is it will reduce the rail-to-rail voltage range (since there is voltage drop across the resistors).

My interpretation is that adding resistors can be done if necessary, but adding them by habit is probably more harmful than helpful.

Edit: article is "The PCB is a component of op amp design", Bruce Carter, in Analog Applications Journal - Texas Instruments, Aug 2000, p 42-47.

\$\endgroup\$
2
  • \$\begingroup\$ Please add the relevant parts of the article to your answer (imagine the article gets removed in the future, your post would be less valuable). \$\endgroup\$ Dec 19 '16 at 7:52
  • 2
    \$\begingroup\$ He did --- everything in the first paragraph after "They say that ... " is essentially what the article says. Maybe it could be a good idea in general to include the exact title and author(s) of the article, so that if it is moved, someone determined to find it could actually find it. \$\endgroup\$
    – Cal-linux
    Dec 19 '16 at 13:35
1
\$\begingroup\$

I have seen this approach used but never used it myself. The case it seems to make sense is when you have long power leads whose inductance can form a LC resoantor with your bypass caps. This is a parallel LC resonance, so on resonance the impedance is very high. Adding a bit of damping will lower the Q, and therefore lower the supply impedance seen by the amplifier at the resonance frequency -- so yes, adding a resistor can lower impedance, at least at some frequency. It of also raises the impedance at low frequency, but if done properly, not enough to be a problem, and opamps have really great PSRR at low frequency.

I wouldn't do this unless i had some measurement or simulation that showed it was necessary and helpful for a given circuit.

What is more common, and which I have done is to use ferrite chips instead of the resistors. These are technically inductors but their Q is so low that at high frequency they just look like resistors (typically 10s or 100s of ohms). They still have a low DC resistance, so they don't have much voltage drop from the DC supply current.

Neither of these should be needed for typical audio frequency opamps that just can't generate the type of fast transients this is supposed to help with. I have used this approach on opamps operating at a few hundred MHz, and on microwave amplifiers. Even then, I didn't actually measure it making an improvement, it was just defensive in nature. Unlike resistors, the ferrite chips are very unlikely to hurt anything.

\$\endgroup\$
3
  • \$\begingroup\$ Hahaha --- oh, my incompetence with PCB design!! :-) Turns out that for placement reasons, I was putting the resistors close to the power supply, and then taking a long (say, 4 or 5 cm) trace to the target op-amps. The capacitors are right next to the op-amps power-in pins, but the problem you describe is still present in my design as it is now. Fortunately, my layout is not finished yet, so I will certainly make the changes. Actually, since we're talking up to about 5 or 10MHz operation, I'll probably remove the resistors. \$\endgroup\$
    – Cal-linux
    Dec 19 '16 at 13:41
  • \$\begingroup\$ Also, I'm quite surprised by this: "long power leads whose inductance can form a LC resoantor with your bypass caps. This is a parallel LC resonance". I had always understood that the parasitic inductance of the traces was in series with the trace. Is it not? Your comment suggests that it is inductance between the trace and GND? Why / how would the parasitic inductance go from the trace to GND? \$\endgroup\$
    – Cal-linux
    Dec 19 '16 at 13:46
  • 1
    \$\begingroup\$ Not in parallel with the PCB trace, in parallel with the capacitor as seen by the opamp. Imagine a current pulse originating at the opamp supply pin. It has two parallel paths to ground -- one through the bypass capacitor, and one through the PCB trace back to the power supply, and then to ground via the power supply capacitors. Those two parallel paths can form a resonant circuit that has high impedance. \$\endgroup\$
    – Evan
    Dec 19 '16 at 18:23
1
\$\begingroup\$

The latter. You are unnecessarily introducing additional impedance in your supply rails.

...by placing a resistor in series I not only make the path to GND low-impedance...

Well, the path to ground doesn't improve, it's simply unaffected, while the impedance between the supply and load is increased. The net effect is that the overall supply impedance is increased.

Also, consider your "low-pass filter": $$\frac{1}{2\pi\cdot 1\Omega \cdot 100\text{nF}} \simeq 1.59 \text{MHz}$$

If you're worried about 1+ MHz noise in your supply rails, you need to solve the problem by fixing your supply, not by adding "filters" at your loads.

So, you should keep the capacitors there, decoupling capacitors are good, but the resistors are non-sensical.

Do you actually have a situation where you have high-frequency noise coupled in through the supply rails? Generally, cleaning up the supply rails by improved regulation and selecting op-amps with higher PSRR is the fix. Also, better PCB layout.

To counter interaction between op-amps within the circuit, good decoupling and layout should be sufficient, but for high-performance applications, you can use separate regulators for each section of the circuit.

The only situation where I have seen RC filtering on supply rails is in tube-amps, which isolate supply stages to prevent feedback through the rails, and also smooth the 60Hz from the rails.

\$\endgroup\$
6
  • \$\begingroup\$ I do not agree with (or do not understand) your comment that "On the contrary, you make the path to ground higher impedance than a wire". I may have misspoken when I said "add a resistor in series" --- I was thinking in series from the power supply to the power-in pin of the op-amp, but now that I think about it, "in series" is completely wrong. Is this why you wrote that comment? If you look at the schematic with the original post, why would the path to GND have high impedance? \$\endgroup\$
    – Cal-linux
    Dec 19 '16 at 5:14
  • \$\begingroup\$ @Cal-linux Sorry, I misread what you said, but my point was that adding the resistor increases the overall impedance of the supply. The impedance between the load and ground remains unchanged, while the impedance between the load and supply is increased. I will edit my post. \$\endgroup\$
    – uint128_t
    Dec 19 '16 at 5:16
  • \$\begingroup\$ Regarding the 1MHz "low-pass" filter --- again maybe a "poor choice of terminology on my part; I'm not using this as a low-pass filter for any noise that's coming from the power supply. The resistor is supposed to "enhance" the effect of the decoupling capacitor, which provides a "short-to-GND" at high frequencies to prevent noise to escape the op-amp; the resistor additionally provides high impedance to everywhere else, so that noise does not propagate to the rest of the circuit. That's how I visualize it --- does it still make no sense? \$\endgroup\$
    – Cal-linux
    Dec 19 '16 at 5:18
  • \$\begingroup\$ The resistor doesn't improve the supply impedance, but yes, you are correct, it does increase the isolation between the op-amp and the rest of the circuit. However, my point is that this is not a good way to increase isolation: if you do in fact need better power supply isolation, there are better ways to do that. \$\endgroup\$
    – uint128_t
    Dec 19 '16 at 5:19
  • 1
    \$\begingroup\$ LPF equation is not correct. 2*pi is missing from the LHS denominator, so the answer is not 10MHz but 1.59MHz. \$\endgroup\$
    – user207421
    Dec 19 '16 at 19:58
0
\$\begingroup\$

I recommend the resistors. They indeed force isolation, at ALL frequencies, between the various opamps. Thus a final opamp, driving the ADC with the sample/hold surges, needing to provide 1Million current surges per second, does not blatantly pollute the entire Analog VDD. And the initial OpAmp, with poor PSRR up at the 1MegaHertz region, does not have a reason to ---- given 0dB PSRR ---- simply copy the VDD ripple right on top of the signal and then amplify.

The VDD tree must be designed. Using resistors in the branches of the tree, is the first step.

Besides, 100 ohms and 100uF becomes a 16Hz rolloff, cutting 60Hz ripple by 12 dB and 120Hz ripple by 18dB, on top of what any LDO achieved.

And what are we to use, below 1MHz where the Beads promise to start to be useful? The resistors work at all frequencies.

I used the resistors (10 Ohms and 10uF) on each opamp (3 per channel) VDD, of a 4 channel IR camera. First pass, we achieved KT-noise-limited performance at the 12 bit level. Even tho the system used switching regulators (remote location), there were no beatnotes in the images. Having seen chroma beatnotes at -60dBc in NTSC systems, to see no beatnotes in a 12-bit (15 bit with auto-ranging) system, aka 90dB, validates the RESISTOR_IN_VDD line.

The automatic dampening is a plus.

\$\endgroup\$
0
\$\begingroup\$

This can be looked at from several different angles.

  • PSRR and supply noise

PSRR of opamps decreases at high frequencies. The effect is usually first order. A first order RC lowpass on the supplies will compensate for it. If your supply has HF noise on it, filtering can help. If the noise comes from a DC-DC though, a single filter can be enough for all opamps.

  • Self-generated noise

Say your opamp drives enough current into the load that its output stage enters class-AB. Now, it draws halfwave-rectified current from its power pins. If this is allowed to couple into the signal, harmonic distortion will increase, and you don't want that.

If your opamp is high-speed and drives a load with HF signal, then its supply current will have lots of HF in it too.

This can couple either through the supply rails (disturbing other opamps), or via crosstalk if the supply traces carrying harmonic current couple into signal lines.

Adding a local cap and a resistor or ferrite bead makes sure the nonlinear and HF currents drawn by the opamp stay inside a tight local loop and don't contaminate the supply.

Both caps (V+ and V-) should have their GND pins at the same spot on the ground plane, to ensure the current going into GND isn't the nonlinear, halfwave rectified class-AB current, but rather the current drawn by the load.

  • Settling

If there is not enough damping in your decoupling network (say, you got lots of 100nF caps connected by slightly inductive traces) then there will be resonance.

Add an opamp which processes fast signals with steep edges. You'd like it to settle quickly once it has finished slewing, but that won't work, because it just pulled a current spike from the supply... and you'll have to wait for the supply caps to stop ringing to get your 0.1% settling time on the output of your opamp.

In this case, a little bit of resistive damping works wonders.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.