breadboards and accurate voltage measurments

Question

I am trying to setup a thermistor(PT1000) to work with arduino. For that I used voltage divider and 5V pin as the voltage source. I was getting very unstable results so I removed the thermistor and used 2 1kOhm resistors instead.

I was still getting results ranging from very close to true value of 1kOhm to 1.2kOhm. For reference, 4 Ohms are equivalent temperature change of 1 degree celsius.

I measured the voltage drop of each resistor and was surprised that it is far from equivalent. Check the video of full setup and the closeup wiring of the voltage divider - https://easyupload.io/5fctxm

I cannot explain the 0.3V voltage drop difference between two resistors of the same rating (resistance of 1000 Ohm +-0.5% was verified by multimeter) in any other way but that the breadboard contacts and jump wires are not suitable for accurate voltage measurement. Can someone with more experience confirm if my breadboard is broken or is such behaviour intended.

Answer 1

In order to assess the stability of analog readings on a breadboard, I did an experiment very similar to yours. I plugged two 1 kΩ resistors on a breadboard, connected in series between my Arduino's GND and +5V. I connected the middle point to A0 and made an histogram of one million consecutive calls to analogRead(A0).

Here is my test code:

const int histo_length = 64;
const long histo_count = 1000000;
long histo[histo_length];
int histo_start;

void setup() {
    Serial.begin(9600);

    // Center the histogram on the average reading.
    int sum = 0;
    for (int i = 0; i < 16; i++)
        sum += analogRead(A0);
    int average = sum / 16;
    histo_start = average - histo_length/2;

    // Fill the histogram.
    Serial.println("Ouliers:");
    for (long i = 0; i < histo_count; i++) {
        int x = analogRead(A0);
        if (x >= histo_start && x < histo_start + histo_length)
            histo[x - histo_start]++;
        else
            Serial.println(x);  // print oulier
    }

    // Print it out.
    Serial.println("Histogram:");
    for (int i = 0; i < histo_length; i++) {
        Serial.print(histo_start + i);
        Serial.print(" ");
        Serial.println(histo[i]);
    }
}

void loop() {}

Note that the full histogram would not fit in the memory of my Uno, hence the trick of printing out the outliers instead of recording them.

And here is the resulting histogram:

508 0
509 39386
510 960567
511 47
512 0

All other histogram bins were empty (no outliers).

As you can see, 96% of the calls returned 510. Most of those that did not return 510 returned 509, and less than 0.005% of them returned 511. The RMS dispersion of the readings is about 0.04 ADC steps.

I can hardly imagine anything more stable than this.

Answer 2

Schematics even for such a question are indispensable.

Assuming from the following:

I cannot explain the 0.3V voltage drop difference between two resistors of the same rating

... means the voltage across each of 2 different resistors both labeled 1K Ohms in series with the same 2.1K Ohm resistor connected to 5V and 0V is different by 0.3V.

^{simulate this circuit – Schematic created using CircuitLab}

If the two 1K Ohm resistors are assumed to be 0.5% (5th color band green) and the 2.1K Ohm resistor is perfect as well as the 5V potential. Then a 1K Ohm resistor of, say, 0.5% less impedance would be 995 ohms. And another, say, 0.5% more impedance would be 1005 ohms. The voltage across the 1st in this resistor divider circuit would be 1005 x ( 5 / (1005 + 2100)) or 1.62V. And the 2nd would be 995 x (5 / (995 + 2100)) or 1.61. This is a difference of 0.01V. That is less then the measured difference between the two different 1K Ohm resistors given the maximum possible percent error of 0.5%.

So the the 0.3V difference between voltage readings of the two different 1K Ohm resisters are difficult to account for. Perhaps cleaning the connections may improve the measurements.