# measure resistance of thermistor without second resistor

Hello, just wondering if is it possible to do that without second resistor ? I've made some calculation fix me if I am wrong :

Lets say aref is 1.04
lets say that known resistance is 11.320 k within current aref, and measured voltage after thermistor is 0.66

so the resistance on adc=0.51 voltage(bypassed how to convert ADC value to voltage) would be :

a=(1.04*11.320)/0.66

R=a-(aadc/aref)=17.837 - (17.8370.51/1.04) = 9.09

does it works ? AREF is the reference voltage for the ADC, it must be solid. By default its connected to Vcc internal
to the chip.

I'm not really sure what you're saying, you claim no second resistor, but what's this 11.32k?

11.32kit is known resistance of thermistor, usually it is 10k in 25C.

if is it possible to do that without second resistor

To estimate the thermistor resistance, you need some way to measure the current through it, and the Arduino cannot do that directly.

No, you can't measure a resistance with a voltage sensing ADC unless you have a voltage
divider with a second resistor, or a constant current circuit.

jremington:
To estimate the thermistor resistance, you need some way to measure the current through it, and the Arduino cannot do that directly.

Right, but aref are known and it is stable, otherwise you would not get stable results without stable aref. So in my case aref are known, and it is stable(It is Internal aref on my board). When you measure resistance and voltage within resistance(11.32 and 0.66 in my case), you can calculate rest.

MarkT:
No, you can't measure a resistance with a voltage sensing ADC unless you have a voltage
divider with a second resistor, or a constant current circuit.

hmm, docs says that there is internal 32k resistor when you use internal/external aref ? If you would not have constant current, and external aref, you would not measure it any way, even with second resistor, because you have to know the voltage right ?

When you measure resistance

How will you do that?

. When you measure resistance and voltage within resistance(11.32 and 0.66 in my case), you can calculate rest.

No you can't. That sentence simply dosn't make sense. What is "resistance and voltage within resistance"? it is gobbledygook.

I think you are fundamentally misunderstanding the role of the reference voltage in an D/A converter. It is the voltage that is used as the reference ( hence the name ) for the unknown voltage input. Show us a wiring diagram showing how you think you are going to wire it up. Maybe that will give us an insight into where your thinking is going wrong. If so we can correct you.

you would not measure it any way, even with second resistor, because you have to know the voltage right ?

Wrong, you measure the voltage, you can only measure voltage with the D/A, you can not measure current, you can not measure resistance you can only measure voltage. You can work out resistance only from the voltage drop across a known resistance / unknown resistance potential divider.

If no current flows through a resistor then there is no voltage dropped across it. As the input impedance of the analogue input is so high, virtually no current will flow through it. So a resistance connected with one end to a voltage and the other end to an analogue input results in the analogue measuring the voltage the other end of the resistor is connected to. What ever the value of the resistance is, the voltage measured will always be the same.

ihitmani:
Right, but aref are known and it is stable, otherwise you would not get stable results without stable aref. So in my case aref are known, and it is stable(It is Internal aref on my board). When you measure resistance and voltage within resistance(11.32 and 0.66 in my case), you can calculate rest.
hmm, docs says that there is internal 32k resistor when you use internal/external aref ? If you would not have constant current, and external aref, you would not measure it any way, even with second resistor, because you have to know the voltage right ?

AREF is the reference for the ADC, you must not take any current from it or it will no longer be a known
voltage, so the ADC will produce nonsense. It needs to be a known, stable voltage, and free from noise.

To measure resistance with a voltmeter, you either need a voltage source and second resistor, or a current
source. What is the problem with that anyway?

I suspect the OP is confusing the ADC's 32k reference resistance mentioned in the datasheet with the internal pull-up resistors on the GPIO pins. The latter could be used to form a voltage divider with the thermistor, to "measure the resistance of thermistor" without a second (external) resistor (I'm not recommending that approach, as it has several problems). But without a schematic, who knows what the OP is thinking?

does it works ? No.

Lets say aref is 1.04
lets say that known resistance is 11.320 k within current aref, and measured voltage after thermistor is 0.66

Where are you getting this "after thermistor" voltage? You don't have an after because your measuring circuit only contains the thermistor.

Draw a schematic of the circuit you are thinking of, and we'll be able to tell you exactly what you're misunderstanding and why it won't work. Without it we can only speculate about what mistake you're making.