# What Thermistor Ω for the Mega 2560?

I just read from Input Impedance of A0-A5 that

whatever you connect to the A/D have an output impedance of 10k or less for best accuracy

However in Using a Thermistor they're using a 10k resistance + a 10k (25°C) thermistor connected in series.
Thus the resulting resistance should be roughly between 42k (0°C) and 11k (100°C).
Both Ω being way above 10k.

Wouldn't it be better to not add the 10k resistance (so the range would only be between 32k and 1k)?

You need to create a potential divider, or you would get no change in voltage with temperature - the resistors are chosen to give the widest range of voltage over the temperature range of interest .
The effective impedance presented to the input is the thermistor in parallel with the 10k resistor ( not in series) as the power supply impedance can be thought of as zero . ( draw it with + and - shorted together).

Ok thanks, parallel made more sense (as we stay <= 10k) but the the article said “we’ll do that by adding another resistor and connecting them in series” ; my eyes maybe

The resistors are in series between the two power rails, but in parallel when viewed from the analog input (to charge/discharge the ADC's holding cap the current flows through both resistors).

Your eyes are fine. The resistors are connected in series and the arduino A/D input is connected to the junction of the two resistors.

The resistance of a resistor is not directly measurable. However the relationship between resistance, current and voltage is known.

Resistance = Voltage / Current

So if we know the Voltage and the current we can calculate the resistance of your thermistor. The 10k mentions helps create the current needed for this measurement. The interaction of the Thermistor and 10k resistor is called a resistance divider. Google "understanding resistance dividers" for a better understanding.

The correct term is "resistor divider" (not "resistance divider").

Simple and powerful concept, indeed, thanks :).

Just to clarify , looks like my post created some confusion - you need resistors in series to create a voltage divider , one arm of which is your sensor .

As far as the resistance seen at the analog input looking outwards is concerned , it sees a resistor to 0v and one to the supply voltage . The impedance of your power supply between 0v and supply can be regarded as zero ohms . So for resistance calculations ( only !) the equivalent circuit looks like those connections are connected together . Draw it out and do the sums.

Weird concept , but there you go !

Suggest you google resistance dividers and have a look at:

inokashira:
The correct term is "resistor divider" (not "resistance divider").

Simple and powerful concept, indeed, thanks :).

Well actually its properly resistIVE divider, like capacitive divider and capacitive dropper.
The meaning is clear anyway, don't worry about it.

The generic term is voltage divider or impedance divider (try saying impedive divider!).

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