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I am using a sensor that gives 5V output However, the ADC I am using can sense only a 3.3V at absolute maximum. A very simple way of translating a 5V to 3.3V was to use a voltage divider as shown. However, the resistor R2 comes in parallel with ADC equivalent impedance and so the overall ratio of voltage divider changes. How can I solve this problem? P.S I don't know the equivalent impedance of ADC and none is provided in datasheet

One solution I found was to use an op-amp between ADC and R2 as then R2 will be in parallel with op-amp input impedance which is infinite ideally and so problem will be solved. But I don't know if it'll work. Will it work? If not then what is the solution?

Also I am completely open to other ways of translating 5V output to 3.3V ADC input. Thanks

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Which sensor it is? Which ADC it is? What resistor values you use? Knowing the sensor means knowing how much load resistance it can drive, and knowing the ADC means knowing what source resistance it needs to work properly, and knowing the resistances means double-checking if they are suitable or not. \$\endgroup\$
    – Justme
    Oct 14 '21 at 18:44
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    \$\begingroup\$ Your first question should be how to determine the ADC input impedance (and/or leakage current, equally important). Without that (and a bunch of other things like required accuracy) you can't really make an informed decision. An op-amp buffer may or may not be the right approach- it can add additional issues but is the most flexible. Typically SAR ADC input impedance (if lacking a buffer) may depend on sample rate so there may not be a single answer in the datasheet. \$\endgroup\$ Oct 14 '21 at 23:05
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The input impedance of many ADCs is so high that it can be ignored and doesn't affect the result provided the resistors are not too high (ie <10k ohms).

It is often beneficial to put a capacitor from the input of the ADC to ground to provide the charge needed when the ADC performs a sample. A value such as 0.1uF ceramic is suitable if the signal is not rapidly changing. As Kartman comments, it is also useful to filter out high-frequency components such as noise.

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  • \$\begingroup\$ Adding the capacitor also forms a low pass filter which is generally a good idea to get rid of some noise and can go some way to providing a bit more ESD protection. \$\endgroup\$
    – Kartman
    Oct 14 '21 at 22:52

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