# Detecting Open Circuit in Op Amp circuit

The situation

I have a AD7689 ADC whose voltage input is 0 - 4.096 V. The max. voltage of the measured signal is around 10V, and it never goes below 0.5 V. I'm using a resistor divider, followed by a LMV324 in a voltage follower config to drive the ADC as shown below.

My Question:

In previous applications, I've simply put a 100k pull down at the input pin of the ADC, so when there's an open circuit the ADC will read 0V and this will trigger some routine in software.

Will R3 still work in the circuit below as the default value when an open circuit is detected, or should I put such pull down resistor before the resistor divider?

• R3 serves no purpose here. That node is being driven by the opamp, and the voltage divider at the input also serves as a pull-down in case of an open circuit. Commented Nov 19, 2021 at 20:50
• Thank you, looking back at the circuit after reading your answer has made me feel a bit silly for asking such a question. I appreciate it @brhans Commented Nov 19, 2021 at 20:53

No op-amp needed. All you need is 2 resistors:

simulate this circuit – Schematic created using CircuitLab

R3 protects your input from excessive voltage.

Not the answer are looking for, but:

simulate this circuit – Schematic created using CircuitLab

You should put a limiting resistor to protecct the ADC input from overvoltage, if opamp supply is higher than ADC supply (Don't quite understand what standard value is 4.75V as in your schematics). The capacitor is optional if you want to implement a low pass filter, which you should need it as anti-alias filter. Diodes are built in GPIO.

If the input of the voltage divider is floating, then the output of opamp is the lowest possible voltage that this opamp can output - few milivolts if it's a rail to rail.