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I have these circuits: enter image description here Their goal is to protect the +vcc and -vcc pins of an opamp. They limit the voltage to +30 V (-30 V) and 25 mA in normal conditons.

The 1200 ohm resistors act like 25 mA load at 30 V.

When there is a short on one rail, this circuit drops the Vout voltage. Which means that the supply isn't "symetric" anymore. And i don't think thats good ( more about that here and here). So what I would like is that when a short happens on one rail the other one is also shutdown until the fault is adressed and cleared. I was thinking of some sort of voltage monitor that would trigger a BJT or MOSFET on one rail when the votlage on the other rail is not high enough. But the negative rail makes thing a bit more complicated for me.

OPA445 datasheet

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  • \$\begingroup\$ Hmm. Maybe use two window comparators (one for each supply), then sum (AND) their outputs and use that signal to turn on/off both outputs. \$\endgroup\$
    – rdtsc
    Commented Apr 20, 2021 at 14:16
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    \$\begingroup\$ I am trying to understand what you mean by "protecting" the op-amp from undervoltage. Does something bad happen if the voltage is too small? What if a signal is applied while the power is turned off? Wouldn't you rather turn off the signal if the voltage is too low? I'm just trying to understand what you are aiming at. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Apr 20, 2021 at 14:41
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    \$\begingroup\$ Q5&Q6 and Q8&Q7 are undervoltage protection because my opamp is rated for +/-10 to +/- 45 V. Why all this complexity? All other circuit desigers including myself would simply make a +/- 30 V supply and feed the opamps with that, DONE. Even though I've commented on your previous questions about this that you don't need this, you're still "protecting". Explain us WHY? \$\endgroup\$
    – Bimpelrekkie
    Commented Apr 20, 2021 at 14:50
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    \$\begingroup\$ I'm not trying to make you angry. Oh, I am not angry at all! :-) I am just challenging your design choices so that we can have a discussion and maybe someone will learn something (you or me, if you have good arguments). opamp datasheet that indicates +/-10 to +/-45 supply votlage And that means that you should design your circuit such that the supply is between +/-10 to +/-45 V. If you use voltage regulators to provide +/- 30 V you would be done. There is really no need for (extra) protection. \$\endgroup\$
    – Bimpelrekkie
    Commented Apr 20, 2021 at 15:24
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    \$\begingroup\$ The datasheet (which I added to your question) in the table "Electrical Characteristics" in the row "Operating Voltage Range" in the column "Min", gives the value +/-10V. This means that proper operation is not guaranteed if the supply voltage falls below that level. However, the IC will not be damaged by operating simply with a low supply voltage. [There may be other issues, like input signals exceeding the supply rails, but barring these other issues, there won't be damage]. So what are the negative consequences you are trying to prevent with cutting off a low supply voltage? \$\endgroup\$
    – Math Keeps Me Busy
    Commented Apr 20, 2021 at 17:10

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