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I have a circuit powered by 12V that accepts an input signal of 0V to 10V. The circuit shall act as a buffer to the signal, with an added LED indicator. In certain circumstances, the circuit itself may not be powered and in this case, it shall pass on the unbuffered signal, and use the power from the signal itself to drive the LED.

This is what I've come up with:

circuit

Now it works beautifully both in the simulation and in real life, but I'm wondering whether there is a chance of damaging the opamp. When the opamp is unpowered (Vsupply = 0V), the inverting inputs see a much higher voltage than the supply, and internal clamping diodes will not work. Also, what about the outputs, will they be safe if a voltage is applied?

I thought about current limiting, but I don't really see how I could do that in this situation. I'm open to a different opamp type that will withstand this type of abuse. I know the OPA121 does, but unfortunately it's not available at my chosen PCBA manufacturer.

Is there any change to the circuit that would make it safe with an LM358, or are there other types of opamp that are better suited?

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  • \$\begingroup\$ Why don't you just let the input signal drive the LED directly and forget about the op-amps? I'm sure you have a reason why you shouldn't but, it's not clear what that is. \$\endgroup\$
    – Andy aka
    Commented Mar 25 at 14:31
  • \$\begingroup\$ The signal may be coming from a high output impedance in some cases (it comes from external gear which I have no control over), so there may be a non-significant loss of signal amplitude when it's driving the LED directly. The user would be advised to power my device in such cases. Other users could leave my device unpowered, since there signal is strong enough. \$\endgroup\$
    – one_three_three_seven
    Commented Mar 25 at 14:53

1 Answer 1

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To determine this, take a look at the Absolute Maximum Ratings in the datasheet. enter image description here

When the maximums are specified directly in volts, you can usually drive those inputs without VCC. When there are ESD diodes, the ratings will be relative to VCC/GND.

You do have to be careful that you don't end up with the new version of the chip where ESD diodes were added to the inputs or another change was made. Hobby grade board assemblers do not always provide the exact part specified in their inventory.

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  • \$\begingroup\$ Great, thanks! I didn't know that. I assumed that these input voltage ratings were always assuming that Vs >= Vi. So you're saying that judging from the datasheet those LM variants would be fine with the above circuit? I'm not worried about different versions, my fab house specifies the exact variant including the datasheet. \$\endgroup\$
    – one_three_three_seven
    Commented Mar 25 at 14:50
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    \$\begingroup\$ That will (usually) be stated (Vs>=Vi >= GND) in the absolute maximums. For this part, you can also look at the internal schematic and simulate the internal transistor structure under those conditions to see what happens. \$\endgroup\$
    – Dean Franks
    Commented Mar 25 at 17:16
  • \$\begingroup\$ Perfect. Thanks a ton! \$\endgroup\$
    – one_three_three_seven
    Commented Mar 25 at 18:38

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