I did a simple op-amp based 100X gain single ended circuit. I tried to order 100 IC's and manufactured them. But found that out of 100 only 60 devices have same and expected baseline. What could be the reason for this variation and how do i improve my yield? (PS: I use TLV9061) All resistors used are SMD type 0603 1% tolerance resistors. Also sensor that i would be using is a capacitive load of 220pF, will this op-amp be good for that. enter image description here Test case: If the vsensor is not connected, the output of the op amp should be 0.33 as an ideal condition. I considered OK range as 0.3 to 0.4 . Only 60 of them were marked OK. Rest all are out of range, may be 0.2V, 3.0V, 2.08V etc

  • \$\begingroup\$ You need to post your circuit. \$\endgroup\$ – Andy aka Dec 27 '18 at 14:07
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    \$\begingroup\$ In what way did those 40 units not perform as you expected? What input signal are you applying? This question is going to get close as "Unclear what you're asking" soon without some more effort on your part ... \$\endgroup\$ – brhans Dec 27 '18 at 14:23
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    \$\begingroup\$ Did you design the PCB in MS Paint as well? Surely you have a proper schematic? There is a CircuitLab button on the editor toolbar and you can create a proper schematic with that. The inverting and non-inverting inputs are not marked on your doodle. \$\endgroup\$ – Transistor Dec 27 '18 at 14:41
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    \$\begingroup\$ Image of the PCB assembly (both sides) might tell us something. Another possibility is that you've been given fake op-amps. \$\endgroup\$ – pjc50 Dec 27 '18 at 14:44
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    \$\begingroup\$ When I consider Vos, your output (for a working condition) can range from 172mV to 495mV. You may have instability on some items. Did you check the output with a scope? \$\endgroup\$ – Peter Smith Dec 27 '18 at 15:21

Look at section 8.10 in the data sheet. The part you have chosen to use has an input offset voltage of \$\pm 0.3\mathrm{mV}\$ typical, \$\pm 2\mathrm{mV}\$ worst case. Your output offset voltage in an amplifier like this is your non-inverting gain times the input offset voltage -- so you're looking at between \$\pm 30\mathrm{mV}\$ and \$\pm 200\mathrm{mV}\$ of offset.

I think that you're lucky to have a yield of 60%, although the picture may be more grim if you did your measurements over any sort of a temperature range.

On the bright side, with input bias currents in the pA and gain-setting resistors in the k\$\Omega\$, you don't need to worry about offsets from the input current + your mismatched resistors, or from input current bias.

You need to build some offset correction into your circuit, and run the numbers (or do a lot of testing) to verify that the drift over temperature will be acceptable, or you need to choose a part that has significantly better voltage offsets without too much input bias current or input bias current offset.

  • \$\begingroup\$ In selecting opamps, all those that can't operate from a 3.3V supply should be culled - I wonder if OP did this step? And what about rail-to-rail? \$\endgroup\$ – glen_geek Dec 27 '18 at 17:28
  • \$\begingroup\$ @glen_geek. OK, fine. The OP needs to check all the parameters to find an op-amp that fits. Ah, the sublime joy of sorting through all 50,000 available part numbers to find the one that works -- and hoping that it's still in production, of course. \$\endgroup\$ – TimWescott Dec 27 '18 at 21:20
  • \$\begingroup\$ Grin: I can tell you've been there and done that. And I've been nasty enough to require students to do that exercise (thank goodness for "parametric search"). \$\endgroup\$ – glen_geek Dec 27 '18 at 21:24
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    \$\begingroup\$ @glen_geek. Been there, done that. I actually enjoy it, even if I always end up sliding into software jobs to pay the bills. \$\endgroup\$ – TimWescott Dec 27 '18 at 21:28

What could be the reason for this variation

When operating at high gains in DC-coupled situations, input offset voltage can be a problem. The worst input offset voltage given in table 8.10 the datasheet is 2mV, which when multiplied by your gain of 100 means a potential output offset of 200mV, this would explain an output voltage in the range of 0.13V to 0.53V, it doesn't explain why you were getting voltages in the volts though.

However that is not the whole story.

Firstly lets re-read table 8.10 in the datasheet carefully and in particular pay attention to what it doesn't tell us. In particular at the head of the table we find the statement (bold mine).

For VS (Total Supply Voltage) = (V+) – (V–) = 1.8 V to 5.5 Vat TA = 25°C, RL = 10 kΩ connected to VS / 2, VCM = VS / 2, and VOUT = VS / 2 (unless otherwise noted)

In other words the parameters in that table only apply when you are in the middle of the amplifiers common mode voltage range. Anywhere else and it may be worse, how much worse? they don't guarantee that!

However they do provide a "typical performance characteristics" graph "figure 4" which implies that this amplifier maintains it's offset performance at the negative rail (but not the positive rail).

So unfortunately I don't have a good explanation for the behavior you are seeing, only several possibilities.

  1. The parts you have are not the parts you think you have. Either due to a mistake or outright counterfeiting. Have you checked the markings on the devices against the expected part markings in the "packaging" section of the datasheet?
  2. You got unlucky and your parts offset performance at low common mode voltages is much worse than "typical".
  3. The lower supply voltage significantly effects the performance of the part (I think this is unlikely)

how do i improve my yield?

First ask yourself if you really need to DC couple, if you can get away with AC coupling then offsets (both voltages and currents) become far less of an issue.

Secondly read data sheets very carefully with an eye to what they are not telling you.

Thirdly always buy your parts from reputable sources.

Fourthly try to avoid operating near the rails unless you really have to. Even if the amplifier is advertised as rail-rail.

But ultimately there is always a risk that unknown unknowns that did not show up with the first couple of prototypes will bite you when you try to scale up manufacturing.


Transistors are fickle things. They have a specification but also a min/max specification. The min/max is what gets you. Op-amps are made out of transistors so they inherit the same qualities.

The min/max is in a Gaussian distribution. Most will be around the norm but there will be outliers. I think you were lucky to get even a 60% yield.

There are 3 things you can do.

  1. Select very accurate components. Resistors that are 0.1% or better and an Op-amp who's offset voltage is within the parameters. This way even if the offset is off a bit it will still be within tolerance after the 100x gain. (depending on the bandwidth requirements a chopper amp might work)
  2. Get an Op-amp that has an offset trimmer pin. Manually trim the offset with a potentiometer.
  3. Add trimmer potentiometers to the gain and offset circuits and manually trim it that way.

Option 1 is expensive. Option 2 can adjust the offset but the gain may be askew. It also requires someone to trim the pot. Option 3 can fix issues with both the offset and gain but requires someone to trim 2 pots.

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    \$\begingroup\$ Your remark about transistors is illogical. OP amps are designed with transistors in a way that cancels out the instability. Even the absolute most accurate OP amps on the planet is built from transistors. And how will going to 0.1% resistors help in this design? He already says that an error of even 10% is fine. Finally, min/max are the actual limit, if you have outliers outside those then the batch is faulty. \$\endgroup\$ – pipe Dec 27 '18 at 15:34
  • \$\begingroup\$ @pipe If the instability was canceled out then there would never be an offset. The most accurate op-amps are binned, they are selected through testing. The selection process basically finds the op-amps that have the most well matched and balanced transistors. The parts that didn't make the cut are just sold under a different part number with a wider tolerance. How did you get an error of 10%? The full range is roughly 0v to 3v so with a range of 0.3 to 0.4 is an error of 3%. \$\endgroup\$ – vini_i Dec 27 '18 at 16:22

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