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Hi, I am measuring voltage on the voltage divider

Divider is powered from the same power supply as ADS1115.  AIN0 connected to VDD, AIN1 to the voltage divider.

Theory: easy calculation gives me: VDD = 5V , R1=1k, R2=10k, V = 0.455

Practical measurement:

VDD 5V; measured 0.447V; expected 0.455V So far so good.

VDD 4.7064; Measured 0.3980V; expected: 0.4228V

Where is the problem?

• Hi Petr,

Welcome to the E2E forum! It is easier to see what is happening with raw data instead of calculated data along with the register settings being used. It is also helpful to see a series of continuous data to see how noise may be affecting the result. All that said, I think I know what is going on.

The ADS1115 is a switched capacitor input that is considered as an impedance. But this impedance is based on an average current. The actual current draw from sampling the input is dependent on the input voltage applied. So what is really happening is you have a current divider at the junction of R2, R1 and AIN1. Figure 26 in the ADS1115 datasheet on page 16 shows a simplified version of the input circuit. More current will flow depending on the actual input voltage relative to the reset voltage of 0.7V.

The 10k resistor is limiting the current, which in turn is split. It is for this reason that we recommend that input filter resistor values be limited to 1k or less. You should see improvement by lowering the resistor values (R1 = 1k, R2,= 100), and adding a capacitor across R1 to act as a charge reservoir for the input.

Best regards,
Bob B
• Hi Bob,