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It's a pretty simple circuit. I'm measuring a voltage with an ADC, which has a max input of 5V. The signal I'm measuring goes to the drain of a F12N10L MOSFET. The MOSFET's source is connected to ground, and it's gate to the output of the comparator.

The comparator will have the reference voltage (5V from a regulator) and the signal I'm measuring as inputs. The idea is, than when that when the signal is bigger than the reference voltage, the MOSFET turns on and the ADC will read ground.

The MOSFET works fine with 5V on gate (10V max limit). My supply on the whole circuit will be 9V, which is why I think I should use a comparator with 5V output. Since I am a newbie, I don't know if there is such a thing (comparator that is powered by 9V but has 5V output).

The question is, what comparator should I use? What is commonly used? Are there any problems with my circuit?

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Most low speed comparators have open drain outputs - that is, they drive low, but not high. You can exploit this by tying the gate of the FET to 5V via a resistor. When the comparator turns on, it will pull the gate low, switching the FET off.

Source impedance is a concern here. Your ADC probably works fine with a high impedance source, but when you turn on the FET, it will cause a voltage drop if your source is high impedance, which will reduce the voltage measured by the comparator, which will turn the FET back off, and so forth. Hysteresis fixes this, but only if you know what the source's range of impedances is likely to be. Also, because the source impedance determines the current through the LED, it may not light up appreciably in an overvoltage condition.

A simple way to ensure your inputs don't exceed the rails is the same method used in ICs: a series resistor such as R1 (though a higher value would be ideal), followed by a pair of reverse biased diodes to VCC and GND. If the input voltage goes out of this range, the protection diodes will conduct, keeping the voltage within a diode drop of the rails.

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  • \$\begingroup\$ I plan to pass the Signal through a OpAmp follower, thus having a low impedance source. About the diode solution: I don't understand where the current will go if I have an overvoltage situation. Into my 5V source? How does that work? Doesn't it damage it? Hm... My solution doesn't offer protection for negative Signal values. :( Could you please give a short explanation to how the diode solution works (maybe a quick schematic), and what diodes should I use? Thank you very much. \$\endgroup\$
    – Dragos Puri
    Apr 6, 2015 at 9:53
  • \$\begingroup\$ In that case, why not have the opamp scale its output voltage down to a safe range for your ADC? Alternately, use a pair of zeners in your feedback path to clip the output voltage, as described here: ecircuitcenter.com/Circuits/op_limiter1/op_limiter1.htm \$\endgroup\$
    – Nick Johnson
    Apr 6, 2015 at 9:56

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