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I have this circuit which is part of a bigger schematic used for capacitance measurement. This circuit should output 7.5V as the opamp is configured as a repeater and the input voltage is 7.5V (15V divided by 2).

I don't exactly know the role of R3 and why its value is 500kohm.

D2 is one of the 2 back-to-back diodes that are placed in the real design.

The problem is that if I short D2 the output goes up to 13V and stays there even if I remove the short from D2. The circuit enters this state even if I randomly touch R3 or D2 using tweezers or even finger, not necessarily shorting D2. The output reverts to the normal state if I briefly short one of the inputs to the ground.

I implemented the below schematic on a breadboard to try to replicate the issue and I did it, the issue is still there so it's not influenced by other circuits on my board.

I also have some older boards with the same design and same opamp part number, but a different batch I guess .... the problem started appearing some weeks ago and I didn't have this problem before, 90% of my boards have this issue now.

  • If I remove D2 the problem disappears
  • If I replace the opamp with an older one but same part number the problem disappears
  • If I replace R3 with 100kohm the problem disappears

enter image description here

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    \$\begingroup\$ What have you got hung on the output of the thing? It is jfet input, but 500k is still a big ask. It is not going to take much leakage in that diode to potentially make this go sideways if there is a bit of offset voltage in the wrong direction. \$\endgroup\$
    – Dan Mills
    Commented Jun 12, 2023 at 15:10
  • \$\begingroup\$ I agree with @DanMills. You may have some positive feedback through the leakage of the diode that acts to keep the op-amp in the saturated state once it gets there. The 500 K resistor seems unnecessarily high. \$\endgroup\$
    – John D
    Commented Jun 12, 2023 at 16:27

2 Answers 2

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the problem started appearing some weeks ago

If I replace the opamp with an older one but same part number the problem disappears

If the parts aren't coming from DigiKey or another authorized Renesas distributor, you possibly got fakes. But it's not a great design anyway.

  • D2 should probably be a BJT B-E diode instead, since it can have a much lower leakage than the diode shown. It was placed there to supposedly prevent phase reversal. Obviously, it doesn't work.

  • CA3140 has a phase reversal that it gets stuck in. That's the effect you're observing.

Use a better op-amp (or, in some ways, a worse op-amp - CA3140 is over-specialized here).

If I replace R3 with 100kohm the problem disappears

This points to diode leakage being a problem.

If I remove D2 the problem disappears

Again, diode leakage...

I have this circuit which is part of a bigger schematic used for capacitance measurement

This is a rather poor-man's circuit. CA3140 is really not the best part to use. It's unnecessarily noisy, and its MOS inputs are not needed here. LM741 or LM358 would be better suited if you insist on using parts almost half a century old. If you're OK with going back to the 80s, then TL07x/08x will be a much better alternative and still dirt cheap.

If someone asked me to redesign this using chips of similar vintage as CA3140, I'd do something like:

schematic

simulate this circuit – Schematic created using CircuitLab

OA1 can be pretty much any jellybean legacy wide supply range op-amp - TL071, TL081, LM741, LM358. Q1-R3 are a class A output follower that isolates the op-amp from the capacitive load.

If the load is mostly 15V-referenced, i.e. sources current into the OUT node, instead of sinking it to GND, then the transistor should be a PNP instead:

schematic

simulate this circuit

Or just use a TLE2426, if price and availability are no object.

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From the datasheet, page 8:

"Moreover, some current limiting resistance should be provided between the inverting input and the output when the CA3140 is used as a unity gain voltage follower. This resistance prevents the possibility of extremely large input signal transients from forcing a signal through the input protection network and directly driving the internal constant current source which could result in positive feedback via the output terminal. A 3.9kΩ resistor is sufficient."

Probably the diode makes this worse.

The diode and resistor might be there just in case some transitory input voltage is outside opamp supply voltage.

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