Upper limit for resistance values in voltage divider?
The input impedance of your ADC puts an upper limit on the usable voltage divider resistance. The ADC's input impedance appears in parallel with \$R_2\$ and is part of the divider ratio equation:







simulate this circuit &ndash; Schematic created using CircuitLab

(Resistor values are for illustration purposes only. You need to consult your ADC's datasheet to find its input impedance, which may not be purely resistive. Also, your ADC's datasheet may mention other restrictions and/or provide recommended resistor values.)

Without the ADC load the divider equation is

$$V_{\text{out}} = \frac{R_2}{R_1 + R_2}V_{\text{in}}$$

However, the ADC input impedance is actually in parallel with \$R_2\$ so the output voltage is really

$$V_{\text{out}} = \frac{R_2 \parallel R_{\text{ADC}}}{R_1 + R_2 \parallel R_{\text{ADC}}}V_{\text{in}}$$

If \$R_{\text{ADC}} \gg R_2\$ then \$R_{\text{ADC}} \parallel R_2 \approx R_2\$ (\$R_{\text{ADC}}\$ is effectively an open circuit) and can be ignored in the divider equation. However, if \$R_{\text{ADC}} \approx R_2\$ then \$R_{\text{ADC}} \parallel R_2 \approx R_2/2\$ and your actual divider ratio is not what you thought it was. A good rule of thumb is to make sure \$10 \times R_2 \leq R_{\text{load}}\$, where the ADC is \$R_{\text{load}}\$ in this case.

Also, the noise generated by a resistor is proportional to its resistance so you'll see more noise with very high valued resistors.