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I'm a begginer in electronics so I have some very basic questions. I would like to protect an OpAmp power supply rails from overcurrent. But I don't know at what value I should set this protection. The OpAmp is the OPA445. I found this in the datasheet :

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Should I make the protection for 4.2 mA because I'm working in a room where the temperature is regulated at 20°C?

edit: The reason why i need to know the current of the load is that I want to make a protection for the Opamp. I have four of them supplied by two dc/dc controllers in parallel. One for the positive rail and one for the negative rail. (LTC3863 and LTC3864). In order to size the components I need to determine how much current is drawn by the four opamps.

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    \$\begingroup\$ Software circles have a name for this, YAGNI, You Aren't Going to Need It. Overcurrent protection for individual components, especially at this level, is unheard of. \$\endgroup\$ – Neil_UK Apr 14 at 7:25
  • \$\begingroup\$ As a beginner, you will introduce more issues with this overcurrent protection than that it will solve. As you gain more experience, you will learn that this level of protection sis simply not needed and that you should focus your efforts on making a proper design. \$\endgroup\$ – Bimpelrekkie Apr 14 at 9:31
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I would like to protect an OpAmp power supply from overcurrent.

Basically, you can't, because the current drawn by the opamp depends on the load on its output.

It's a high voltage opamp, +/-30V, so if you use a 2kOhm load, and the output is near +30V it will output 15mA into the load. If the output is shorted to ground, it will output about 25mA. Since you have 4 opamps, it's impossible to know if several opamps are providing legit current to loads, or if one is shorted, the supply currents would not be any different.

If the load will be a very high resistance, then the simplest protection option is to put a resistor in the output. You can wrap the feedback around it since the opamp supports differential input voltage equal to the supply voltage. Note this is important, if the opamp datasheet says "maximum differential input voltage 1V" or something like that, that would be exceeded if the output was shorted, and the output is connected to negative input, and the positive input is driven by another circuit. So this needs more resistors to limit current in the inputs.

Another solution would be to use an opamp with proper protection. Check TI search engine, there are several opamps with +/-30V supplies and proper protection, also cheaper than OPA445.

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Before embarking in a 4mA current protection, what would be the damage of such a fault? would your power supply break?

It's interesting that you want to protect your supply, but you usually consider what your supply can give, not what is required in the standard situation (unless you are doing some kind of overload protection, like for motor). So more than protecting the supply (which probably has at least 100mA available) you could be interested in protecting the amplifier (if the load has some chance of fault). Unless you are in the ultra-reliability or safety circuit market you don't really care if your amp draws 4 or 10 milliamps (and even then you usually don't care anyway!)

These curves are mostly for determining the power budget i.e. to size up the power supply. As in, I have 10 amplifiers so I need to decide if I need a 50mA power supply or a 100mA power supply or for how much it will go when supplied by a battery.

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  • \$\begingroup\$ I guess i didn't explain myself well. It is indeed the opamp that i want to protect. I have four opamp "pumping" on this supply. The big plan is to supply those opmap with a protecting circuit before. I have a 0/+30V IN and then two DC/DC converter. One to regulate to +30V and the other to -30V. And I need to know how much curent is drawn be my load (four opamp) in order to set up the regulators (LTC3863 and LTC3864). I know it's a bit overkill to use those but the negative voltage protection has given me a real headache! \$\endgroup\$ – Neeko Apr 14 at 8:34
  • \$\begingroup\$ In that case estimate you maximum temperature, add the quiescent and output current and then some margin. \$\endgroup\$ – Lorenzo Marcantonio Apr 14 at 9:05
  • \$\begingroup\$ the negative voltage protection Oh, now you want voltage protection? But your question is about over current protection. \$\endgroup\$ – Bimpelrekkie Apr 14 at 9:33
  • \$\begingroup\$ @Bimpelrekkie Well, i figured how to do my voltage protection. But i thought my solution needed also current protection. And since I need a current value for both current and voltage protection I asked about current. \$\endgroup\$ – Neeko Apr 14 at 10:44

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