I am using a number of single-supply op-amps (OPA4344) in a circuit, and am using one of them to supply a VCC/2 value for a virtual ground to the + side of several other op-amps.

VCC is +5 volts. When I first power up the board, I get 2.5v from the output, but after awhile the output jumps to around 4.5 volts and stays there until I power off and back on again.

enter image description here

I read here that:

Due to the strong (i.e., unity gain) feedback and certain non-ideal characteristics of real operational amplifiers, this feedback system is prone to have poor stability margins. Consequently, the system may be unstable when connected to sufficiently capacitive loads. In these cases, a lag compensation network (e.g., connecting the load to the voltage follower through a resistor) can be used to restore stability.

As you can see, I am already using a resistor on the output. The datasheet for the 4344 (referenced earlier) claims the op-amp is "unit gain stable."

Is there something else that can be causing the instability? Do I need a separate resistor for each output (currently I the + inputs of three op-amps tied to VOUT).

  • \$\begingroup\$ What happens if you just use a single op amp? \$\endgroup\$ Nov 15, 2011 at 20:11
  • \$\begingroup\$ I believe I tried that, and got the same result, but will repeat that experiment tonight. (I have this particular device at home and am on-site right now.) BTW, I forgot to mention this circuit is laid out on a PCB, and is not on a breadboard. \$\endgroup\$
    – tcrosley
    Nov 15, 2011 at 20:15
  • \$\begingroup\$ Maybe that opamp is still just a little twitchy at unity gain and acts up depending on load? Can you try something basic like a LM324 and see if it makes a difference? Those get used a lot in voltage follower applications. Should be cheaper too. \$\endgroup\$ Nov 15, 2011 at 20:28
  • 1
    \$\begingroup\$ What are your other 3 opamps in the package doing? If at least one of them is "not happy" it can affect the whole package (where not happy = something seriously wrong, depends on the opamp) \$\endgroup\$
    – Jason S
    Nov 15, 2011 at 20:29
  • 2
    \$\begingroup\$ I added capacitors per your suggestion and also replaced the chip. The voltage on the + lead was correct. \$\endgroup\$
    – tcrosley
    May 18, 2015 at 21:51

3 Answers 3


VCC is +5 volts. When I first power up the board, I get 2.5v from the output, but after awhile the output jumps to around 4.5 volts and stays there until I power off and back on again.

At first I thought this sounds like a case of phase inversion outside the common-mode input range (which for the OPA344 is -0.1V to (Vcc - 1.5V = 3.5V in your case). It's rarer these days, but some op-amps exhibit gain reversal when outside their common-mode range, causing an effective latch-up condition. For an op-amp with phase inversion, as long as you stay within the common-mode range, you should be fine, but if it ever strays outside, there's no guarantee that it will operate properly.

But the OPA334 datasheet says this:

The OPA334 and OPA335 series op amps are unity-gain stable and free from unexpected output phase reversal. They use auto-zeroing techniques to provide low offset voltage and very low drift over time and temperature.

So at this point we're left with a couple of things to try, assuming you can reproduce this problem easily.

  1. Check all the opamp pin voltages with an oscilloscope. Make sure Vcc and Vss are what you expect, and check to see if the + pin of the op-amp is the 2.5V that you expect.

  2. Add a capacitor (100-1000pF) between op-amp + and ground. You should be doing this anyway to keep the impedance of the voltage divider node low at high frequencies so it does not pick up noise. If this fixes the problem, you may be running into RF rectification (If this is the case, I'm surprised, but it's possible.) where the op-amp behaves linearly with low-frequency signals, but behaves nonlinearly like a rectifier with high-frequency signals and turns AC into a DC bias.

  3. Add a bypass capacitor across the op-amp supply. (supply noise shouldn't make that much of a difference, but you never know)

  4. Replace the op-amp with another of the same model -- the one on the board could be damaged.

If all still looks good, then you've got quite a stumper.


Place a resistor in series with the feedback you have provided. In other words, remove the zero resistance lead from output to negative input and place a low value resistor, say about 2.2k. Hopefully it will solve the problem.

This resistor along with input capacitance of the amplifier forms a sort of compensation that ensures that high frequency oscillations are blocked. If high frequency gain is reduced, Barkhausean criterion cannot be satisfied and hence no oscillations. These oscillations are at such a high frequency that they end up rectified by nonlinearities in the op-amp and the resulting DC gives the effect you see.


I ran into a similar problem with various op-amp such as AD712 in particular when I simply tried to build a voltage follower circuit with it. Note that the AD712 is unity-gain stable per its own datasheet:

AC performance
Settles to ±0.01% in 1.0 μs
16 V/μs minimum slew rate (AD712J)
3 MHz minimum unity-gain bandwidth (AD712J)

The datasheet first page third paragraph states:

The combination of excellent noise performance and low input current also make the AD712 useful for photo diode preamps. Common-mode rejection of 88 dB and open-loop gain of 400 V/mV ensure 12-bit performance even in high speed unity-gain buffer circuits.

The datasheet even suggests a unity-gain follower circuit and displays its performance on page 9!
Alas! Implementing the same circuit gave me a ~5 V oscillation at ~2.8 MHz when the op-amp non-inverting input was grounded. When the non-inverting input was let free, the output would quickly drift to the negative rail.

The only solution that stabilized the output was to add a fairly small resistor (50 Ω) between the output and the op-amp inverting input as suggested by Dr.V.S.V.Mani in another answer. It is not clear to me though how this solution affects the circuit performance or its output impedance.

  • 1
    \$\begingroup\$ The fact that an OpAmp is unity-gain stable doesn't mean you can provide any type of load (e.g. capacitive). Here's why a small resistor can be needed at the output, you can find it explained in most application notes on capacitive load drives, and even in datasheets themselves sometimes. \$\endgroup\$
    – LuC
    May 11, 2023 at 10:07
  • 1
    \$\begingroup\$ The AD712 datasheet has a section named "driving a large capacitive load". This can affect precision and settling time, of course, so in specific cases, you would use a part that fits them better \$\endgroup\$
    – LuC
    May 11, 2023 at 10:14
  • \$\begingroup\$ C.J - Hi, Are you trying to ask the implied question in the last sentence of your post? if so, you're in the wrong place, as this area is just for answers, not for questions. That implied question is liable for removal, in order to stop your post turning into a question. || That question would be suitable for asking as a new question, with a link back to the answer you mention for context. || It seems you are new here, so please see the tour & help center for details of how Stack Exchange differs from typical forums (including its Q&A format). Thanks. \$\endgroup\$
    – SamGibson
    May 11, 2023 at 21:26
  • \$\begingroup\$ Thank you for helping out @LuC. Capacitive load was indeed my suspicion. I did my measurement with about a meter of coaxial cable so it might be this. The AD712 datasheet section you mention shows some sort of an integrator. Am I right? I am not sure I understand it right. Also, Horowitz Hill The Art of Electronics mentions the topic in section 4.6.2 so maybe I should try some of the solutions they suggest. \$\endgroup\$
    – C.J
    May 11, 2023 at 21:39
  • \$\begingroup\$ @SamGibson that last sentence is a warning that I do not have a full grasp of the possible consequences of that solution. \$\endgroup\$
    – C.J
    May 11, 2023 at 21:44

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