# Decreasing and reading output of a sensor (0 - 10 V) using a voltage divider - Output impedance

I have a pressure sensor giving an output between 0 - 10 V.

I want to read and save the value using an arduino, so I am using a voltage divider (https://learn.sparkfun.com/tutorials/voltage-dividers) to scale it down to 0 - 5 V. Since I want to halve the voltage, I am using the same resistance twice.

I have noticed that the voltage changes according to the value of the chosen resistance pair. I assume this is due to the output impedance of the sensor.

On the sensor there is a label reading "OUT 0...10 V RL>10K". I assume this is the impedance of the output.

Does this mean that one should use a pair of at least 10K Ohm resistance for the voltage divider?

Info on the sensor = https://www.hubacontrol.com/fileadmin/user_upload/domain1/Produkte/Product_catalogue.pdf - Page 109 (Sensor type 692)

• A link to the data sheeet of the sensor is required if guesses are to be avoided. Commented Jul 19, 2018 at 14:49
• @Andyaka I linked information from the manufacturer. Unfortunately I do not have additional datasheets. Commented Jul 19, 2018 at 14:56
• @Andyaka im pretty sure he meant page 109 of the 206 page document Commented Jul 19, 2018 at 15:04
• @Andyaka to be fair, it only took about 3 seconds on my pc Commented Jul 19, 2018 at 15:09
• @Andyaka well... I feel like providing a link to a datasheet isnt really particularly lazy. While it would be nice to have an excerpt picture, I think providing a link to the catalogue with the page number is perfectly reasonable Commented Jul 19, 2018 at 15:12

Looking at the datasheet, it does look like it needs a load resistance of more than 10kOhm.

The total load resistance of your potential divider will depend on two things: the value of your resistors, and the input resistance of whatever you're connecting to the divider output.

Since you're connecting the divider output directly to the arduino ADC, and the ADC input resistance is very high, the only thing you really have to worry about is the resistor value.

The total resistance of your divider will therefore just be R1+R2, so you need to make sure that this adds up to >10kOhm. I would make the total resistance >20kOhm just in case.

• "I would make the total resistance >20kOhm just in case." So two 10K would be enough, right? Commented Jul 19, 2018 at 15:17
• @Keine yes, that would be fine Commented Jul 19, 2018 at 15:18

according to the attached datasheet, you are not using a sensor and this is a pressure transmitter, that you use it in 3 wire and output voltage is change between 0 to 10 V. 3 wire transmitter always have an output current 4-20mA or 0-20mA , or have a output voltage : 0-5 volt, 1-5 volt or 0-10 volt . In this transmitter they used a current supply for 2 wire to produce 4-20 mA, and for 3 wire system they used a resistor(load) , and convert the current to voltage. When you use resistors for voltage dividing, actually you are change the load(output resistor).

• I'm not sure thats true, if you look at the current consumption, for three wire mode it's <5mA, so it cant be using a 500 ohm resistor to generate 10V Commented Jul 19, 2018 at 15:10
• @Rashid what are you talking about? On page 109 is says very clearly Current consumption < 5 mA Commented Jul 19, 2018 at 15:16
• @Rashid Also, it says the output goes down to 0V, which wouldnt be possible if it were just a 4-20ma device with a 500ohm resistor, since the minimum voltage would be 2V Commented Jul 19, 2018 at 15:20
• @Rashid yes, it cant produce these voltages or currents without power, but my point is that the "three wire mode" is not just the "two wire mode" with an internal 500 ohm resistor. They are separate operating modes Commented Jul 19, 2018 at 15:22
• I see you edited your answer, but it's still incorrect, saying that it's just a 4-20ma device with a resistor to convert it to a voltage output device. It's not. Commented Jul 19, 2018 at 15:42