Resistance Change In Voltage Divider

**EDIT: ** In Comment #2 and #28 you can find the solution to the problem I described in the topic. Also, there's another problem in this topic which you can read it from comment #10, the solution to that problem is in the comment #26.

Hello,
I'm trying to use an ADS1015 and a voltage divider to measure the resistance of a PT100 temperature sensor.
Here's my circuit:

And this is how it looks on Breadboard (consider the potentiometer as the PT100):

My code works fine, my problem is when I connect the PT100 to the Breadboard, or lets say it better, when I connect the +5V or GND wires to the divider, the PT100 resistance drops about 5/10 Ohms,
How do I measure the resistance? I disconnect the power and measure the PT100 resistance, so the resistance drop happens when no voltage is applied to the circuit. Also when I disconnect the ADS1015 module from breadboard, the resistance is correct one again, which is basically just the same thing as disconnecting the wires.

Setrik_aZ:
when I connect the +5V or GND wires to the divider, the PT100 resistance drops about 5/10 Ohms,

Could you please clarify the meaning of 5/10 ohms? Does that mean 0.5 ohm? Or does it mean 5 to 10 ohm?

might be your wiring, try to connect only 2 wire.

Southpark:
Could you please clarify the meaning of 5/10 ohms? Does that mean 0.5 ohm? Or does it mean 5 to 10 ohm?

5 to 10 Ohms

rzk:
might be your wiring, try to connect only 2 wire.

Thanks it helped, I reduced the wires to 3 (+5, GND, Vout), but its still changing a bit when I connect the wires, about 1 Ohm, which is important because it means 2 degrees, is there a way to make it zero? can it be because of the breadboard?

Hi there! 1 Ohm is relatively small ..... relatively. See the resistor R1? That has a 5 percent uncertainty about it. So for R1 being 200 ohm, the uncertainty in resistance is plus or minus 10 ohm already.

Southpark:
Hi there! 1 Ohm is relatively small ..... relatively. See the resistor R1? That has a 5 percent uncertainty about it. So for R1 being 200 ohm, the uncertainty in resistance is plus or minus 10 ohm already.

No, the R1 is actually a Multi-turn potentiometer which I adjusted it on exactly 200 Ohm.
I mean its not THAT important in my application now that I think, but if there's a way to fix it, I will, if there's not, it doesn't matter.

The problem with breadboard is that the contacts are not as good as soldered connections and I wouldn't be surprised to get more than 1 Ohm variation just from the breadboard, regardless of anything else. I mean, you might not but I wouldn't trust it.

You cannot measure resistance reliably in-circuit, as you then have the resistance of the circuit in parallel
with the resistor in question.

Have a google of pt100 circuits , 4wire lead compensation .

Any resistance you have in your leads and connections adds to that of the pt100 and creates errors .

Setrik_aZ:
when I connect the +5V or GND wires to the divider, the PT100 resistance drops about 5 to 10 ohms.
How do I measure the resistance?

One quick question is - how did YOU measure the resistance? (ie. how did you determine that 5 to 10 ohm drop in resistance?).

One way - with the existing circuit - is to use the arduino to determine the voltage of your analog pin … for example, the analog pin (analog to digital conversion) voltage were measured by the arduino in a digital range of 0 to 1023, and let’s say the arduino gave a result of digital level 450. Then the associated voltage would be 450 * (1/1024) * 5V, which is approximately 2.197 volt.

That allows you to use the 200 ohm (with 5% uncertainty) resistor to estimate the current flowing through the ‘resistor’ of your sensor - which happens to be in series, such 2.197/200 = 0.011 A.

Knowing the voltage difference across the PT100 will be 5V - 2.197V = 2.803V ------- and knowing the current through the PT100 is 0.011A, allows you to use ohms law to estimate the PT100 resistance.

V/I = R … which is 2.803/0.011 = 254.79 ohm for the PT100 in this example.

when I connect the +5V or GND wires to the divider, the PT100 resistance drops about 5/10 Ohms

You are driving 16.67mA through the RTD, probably causing it to self heat. Since you are measuring the voltage across the resistor instead of the RTD, higher RTD resistance means lower voltage across the resistor..

Southpark:
One quick question is - how did YOU measure the resistance? (ie. how did you determine that 5 to 10 ohm drop in resistance?).

I'm not using the Arduino yet, I measured the resistance with my multimeter after turning of the power.

One way - with the existing circuit - is to use the arduino to determine the voltage of your analog pin ...... for example, the analog pin (analog to digital conversion) voltage were measured by the arduino in a digital range of 0 to 1023, and let's say the arduino gave a result of digital level 450. Then the associated voltage would be 450 * (1/1024) * 5V, which is approximately 2.197 volt.

That allows you to use the 200 ohm (with 5% uncertainty) resistor to estimate the current flowing through the 'resistor' of your sensor - which happens to be in series, such 2.197/200 = 0.011 A.

Knowing the voltage difference across the PT100 will be 5V - 2.197V = 2.803V ------- and knowing the current through the PT100 is 0.011A, allows you to use ohms law to estimate the PT100 resistance.

Thanks you, I was about to post another topic about measuring the unknown resistor n a voltage divider since I cant do it with the equation in Wikipedia which is

If I understood you correctly, what you said is an alternative for this?! English is not my main language so I'm a little confused.
If yes, can you please write the equation for me?
If no, can you explain it more please?
And also, I want to know if a Wheatstone bridge would help me to get more stable and more accurate measurement?

JCA34F:
You are driving 16.67mA through the RTD, probably causing it to self heat. Since you are measuring the voltage across the resistor instead of the RTD, higher RTD resistance means lower voltage across the resistor..

Yes I'm measuring the voltage across the fixed resistor because other way if I disconnect the sensor from the circuit, 5V will be on the ADC pin, and I was thinking about increasing the ADS1015 gain for more resolution, and that way the 5V would destroy the ADS1015 so I decided to measure the fixed resistor voltage.

I'm now confused, I didn't knew that a current would flow though the voltage divider!
Can you give me a YouTube video or a page or whatever to read? or explain it here? I watched few videos but they don't talk about what I need to know, and as I said the equation in Wikipedia gives me a wrong value.

For example, the PT100 is about 110 Ohms in 25°C, so:

R1 = 110 (measured with multimeter)
R2 = 200 (measured with multimeter)

Vout = (200/310)*5 = 3.22V

But when I read the voltage with my multimeter and the ADC, its 2.97

Another example:

R1 = Unknown
R2 = 200
Vout = 2.97

R1 = 200 ( 5 / 2.97 - 1 ) = 136.7 Ohm

Which is wrong and if I measure the R1 with multimeter, its 110 Ohm.

so you're comparing resistance value of your multimeter with adc calculation, right?
here is the missing point
ADC depend on Voltage input, while your main supply component only from arduino 5V
arduino supply can drop to below 4.8 v when connecting more than 1 component, and it's normal
the things you can do

  • adjust your equation
  • compare the temperature result with real Thermometer

thx

The equation is correct and so is your maths.

If you are not measuring 3V22 or close then something is not as you think it is.

Is there anything connected to Vout other than your multimeter?

Are you assuming 5V supply or are you actually measuring 5V?

Is your meter calibrated?

I'm now confused, I didn't knew that a current would flow though the voltage divider!

You need to understand Ohm's Law and Kirchhoff's circuit laws

rzk:
so you're comparing resistance value of your multimeter with adc calculation, right?
here is the missing point
ADC depend on Voltage input, while your main supply component only from arduino 5V
arduino supply can drop to below 4.8 v when connecting more than 1 component, and it's normal
the things you can do

  • adjust your equation
  • compare the temperature result with real Thermometer

thx

Thanks I will try that and update the topic