Resistance measure

[Berroa Ferenc]

First of „Happy New Year” to ALL of you.

Please, help me understand an electrical and mathematical problem.

I’m working in a remote input/output module, and when I measure an unknow resistance with a digital ohmmeter I get 17% greater value than reading it with the ADC of the microcontroller.



I use a potential divider to know the resistance in this way:



5V---|Reference resistance|---ADC---|Unknow Resistance|---GND

Reference resistance = 10K



The mathematical equation I use is:



V_ADC = (ADC value * (5V / 1023))



Resistance = ((V_ADC / 5V) * Reference resistance) / (1 - (V_ADC / 5V))



I would like to know your opinion about this. The digital multimeter is an standard one.

The resistance I’m measuring is 1K with 20% tolerance.
The other problem is that resitance reading is not linear because the non linearity of the ADC reading. Any one can give me an idea?

Thank you in advance.

Cordially: Berroa Ferenc.

 

[Nigel Goodwin]

You're measuring it wrong - feed the resistor from a constant current source, and simply measure the voltage across it - this gives a linear reading, with no maths required (and the correct way round).

 

[Berroa Ferenc]

You're measuring it wrong - feed the resistor from a constant current source, and simply measure the voltage across it - this gives a linear reading, with no maths required (and the correct way round).

Thank you Nigel !!

 

[dougy83]

5V---|Reference resistance|---ADC---|Unknow Resistance|---GND
Reference resistance = 10K

Resistance = ((V_ADC / 5V) * Reference resistance) / (1 - (V_ADC / 5V))

If 'A' is the voltage at the ADC input, and 'R' is the unknown,

A/5 = R / (10k + R)
5/A = (10k + R) / R = 10k/R + 1
R*(5/A - 1) = 10k
R = 10k / (A/5 - 1), which seems a little different to you final equation.