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I'm making an analysis of the error introduced by two different resistors in a voltage divider (and taking the voltage out from the middle of two series resistors).

I'm using this for an NTC reading, which is part of the divider. I know that the maximum error is the sum of the relative errors of the resistors present in the divider. What I don't know, is how to demonstrate this in an expression so that if the divider is made by 10 resistors and not just 2, I can still demonstrate easily the maximum error. So obtaining a graph like this: http://www.electronicproducts.com/Passive_Components/Resistors_and_Potentiometers/Resistive_voltage_dividers.aspx. I know that this could be a dummy question.

Moreover, the NTCs are providing an absolute error in °K which is related to the B factor and the ambient temperature resistance relative errors. Something like this http://product.tdk.com/en/techjournal/tfl/sensor_actuator/NTCG2/images/001e_p1_gla.gif. Should this error be added to the error that I will find from the voltage divider? The advantage, however, is that this error can be offset by calculating it and storing it in a Look-up table, since seems to depend only by the NTC itself and the ambient temperature.

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The maximum error in the voltage output of a divider is not the sum of the errors, as your link shows. It only approaches the sum if one resistor is much bigger than the other.

The maximum error will occur at one extreme or the other for each resistor and for the thermistor. For N parts, there are only 2^N combinations, so it's possible to do an exhaustive search numerically at each data point without any thinking, but it should be obvious which way the parts will affect the output.

If your only requirement is to calculate the maximum error, simply assume each part is at its worst-case tolerance limit for the reading to be high or low, and calculate said error.

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  • \$\begingroup\$ Yes, and I suppose that the case of R1/(R1+R2+R3), will melt R2+R3 to have both the positive half of the error, if R1 is calculated by reducing it by its error. So, if E is the absolute error equal for all resistors, I can find the relative error eps by: (R1+E/(R1+E + R2-E + R3-E))/(R1/(R1+R2+R3)) = eps \$\endgroup\$
    – thexeno
    Sep 25, 2015 at 22:54

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