Capacitor on sensor output purpose

Question

I'm currently using a demo-board from sparkfun for quick prototyping of a hall-effect sensor application. The specific chip I'm using is the ACS723-5AB, a Hall-Effect sensor that can measure between -5A and +5A.

Now that I'm done using the demo-board from sparkfun I'm looking to just directly use the sensor chip in my final circuit design.

One of the improvements I wanted to make was to add a low-pass filter on the sensor output for smoothing the signal a bit by adding a simple RC filter, but when I look at the circuit-design of the SparkFun board, they don't use an RC filter but just a capacitor to ground on the output line.

I can't imagine anything drawing current from the sensor output, so decoupling doesn't seem likely. It doesn't really create a filter without a resistor (as far as I know?)

My question is: What is the purpose of capacitor C2 on Viout?

link to circuit diagram of daughterboard

Link to sensor datasheet

Answer 1

It doesn't really create a filter without a resistor (as far as I know?)

It probably does because it's likely that the output amplifier will have some output resistance that isn't trivially small. The data sheet tell us two things here: -

• Output Resistive Load from VIOUT to GND is 4.7 kΩ minimum
• Output Capacitance Load from VIOUT to GND is 10 nF maximum

This fairly convinces me that the output impedance is not insignificant else why have such a high minimum load AND why have such a high limit for output capacitance; most op-amps would probably go unstable with a load capacitance as high as 10 nF so, I expect that there is an internal output resistor.

What is the purpose of capacitor C2 on Viout?

An extra means of reducing high frequency output noise.