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Say we use a 500 ohm resistor to convert current signal from 4-20 ml to 2-10 volts for an analog input rated for 0-10v.
What happens if I put a 1/4 watt and the current was high enough to burn the resistor, does that effectively protect the analog input?
The current loop drive will have some maximum voltage (typically 24 volts) that it regulates to get the desired current flow. With the 500 ohm resistor, when it's trying to maintain a current of 20mA through the resistor, its output voltage will be 10 volts – simple Ohm's Law, v = I × r.
What happens if the 500 ohm resistor burns up? The current driver will do everything it can to force 20mA through the now-open circuit. It will increase its output voltage to the maximum, in the attempt to make 20mA flow. Now you have 24 volts (depending on the driver) across your 10v input.
If the resistor goes open circuit, the full loop supply voltage would appear across the analog input.
So, no.
That's the last part of the loop that should fail.
Also allow for loop current being present when the power to the ADC or MCU is not; it generally needs some divider or RC filter to the device inputs, plus a diode clamp.
A series resistor (e.g. 10kΩ) with a Schottky diode clamp to the analog input power supply should protect the analog input from any type of input faults.
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