0
\$\begingroup\$

I am having a bit of a hard time understanding the voltage read in the voltage divider that is being an input for an ADC to an NXP microcontroller, S32K118. Below is the schematic of the circuit.

schematic of the circuit

Where P1 and R1 represent an analogic temperature sensor, just for hardware debugging purposes.

The designed and the real values of each component are represented in the table below:

Component Design Real
P1 10k Ohms 9.4k Ohms
R1 4.7k Ohms 4.4k Ohms
R2 15k Ohms 13.4k Ohms
R3 22k Ohms 21.9k Ohms
C1 100n F 100n F

According to the MCU (S32Kxx) datasheet, the ADC input impedance equivalency diagram is shown below:

ADC input impedance equivalency

As the signal from the sensor is updated at a very slow rate, I consider the signal being DC, therefore no current flows into the ADC pin and the voltage at the node (R1,R2 & R3) is the same as in the node (R3 & C1).

Then, calculating the voltage divider in both cases, which resistance of the potentiometer was changed from the minimum to the maximum, using the real values, I got the following values:

$$ Vout = \frac{(P1 + R1)}{(P1 + R1) + R2}{Vin} $$

V_{in} = 5V

Vout_{min} = 1.23V

Vout_{max} = 2.45V

However, the values that I am reading with the multimeter are:

Vout_{min} = 0.22V

Vout_{max} = 0.65V

That makes no sense, I have checked all the tracks and components, and they are in order.

Am I missing some point that is leading to misunderstanding something?

\$\endgroup\$
4
  • \$\begingroup\$ It's not clear where \$V_{IN}\$ is. Please label it on your schematic. \$\endgroup\$ Commented Jun 4 at 11:13
  • \$\begingroup\$ Measure the voltages at each point. My guess is you’ve set the adc pin as gpio output. \$\endgroup\$
    – Kartman
    Commented Jun 4 at 11:13
  • \$\begingroup\$ @Kartman the voltage in both nodes is the same, indicating no current is flowing in the R3 resistor. \$\endgroup\$ Commented Jun 4 at 11:38
  • \$\begingroup\$ I see four nodes. Measure and annotate the voltages at each. It has to be something fundamental. \$\endgroup\$
    – Kartman
    Commented Jun 4 at 13:13

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.