Hello,
I hope this is the correct forum; I'm new. So, I have a 7.4v/2800mA/20w battery I salvaged from a portable DVD player that went kaput. I wanted to test the yellow wire (the battery has three wires: red, yellow, and black), which Co-Pilot said was likely connected to a thermistor. Here's my setup:

The issue is, I can't run the code to test the resistance because the battery is powering the Arduino, so the Arduino (it's an Uno, btw) won't connect to the computer, which was why I added the diode (IN5819), but it's still won't connect.
Thanks in advance for any help.

 

hi thes,
Welcome to AAC.
A 7.4V battery should be connected to Vin, not +5V on the Uno.
E

Added:

won't connect to the computer,

I suspect the UNO is now damaged.

 

Hello Eric,
Thanks for the prompt reply. Aha! Yes, it works when put into the VIN, and - fortunately - the Arduino doesn't appear to be damaged. Hopefully a mistake I won't repeat.

 

Hi thes.
That's good news.
Do you know how to connect the Thermistor?
E

 

Are you sure it's a thermistor? The yellow wire might just be a voltage-monitoring connection for a battery-management system.

 

So, I'm not sure it's a thermistor; that was what Co-Pilot suggested and what I was testing for. However, the voltage stayed the same even when applying some light heat.

 

Hi these,
What is that measured voltage?
E

 

Eric,
This was the code I used:


void setup() {
Serial.begin(9600);
}
void loop() {
int raw = analogRead(A0);
float voltage = raw * (5.0 / 1023.0);
Serial.println(voltage);
delay(1000);
}

And the reading was 2.87 over and over.

 

hi these,
Can you measure that 'raw/wire voltage with a voltmeter, just to compare with the ADC, in case that first wrong Vin connection has damaged the UNO?
E

 

It's giving me 8.2v, and the battery should be about full charge. I'm not sure how much voltage would have dropped over the circuit as is.

 

hi these,
Sorry I was not clear, I mean the voltage on the Yellow wire as it comes out of the battery, no other connection.
E

 

Gotcha. It looks like 4.4 maybe 4.5.

 

Hi these,
That looks as though it could be the centre wire between two fully charged 3.7V Li-ion cells batteries, which for ~7.4V.
The old battery charger would be then able to monitor the two cells during charging.

As it actually measures about 4.4v and the ADC says about 2.87v, I would suggest you try a new UNO..
E

 

Okay, will do. Thanks for your help. I've got an off-brand one attached to something else I'm not using, so I'll fetch that and give it a go.

 

Hi these,
Using your Sketch, it works OK.
This shows 3 different ADC voltage inputs 3.7v, 5v and 0v.
E

void setup() {

Serial.begin(9600);

}

void loop() {

int raw = analogRead(A0);

float voltage = raw * (5.0 / 1023.0);

Serial.println(voltage);

delay(1000);

}

3.71
3.72
3.71
3.71
3.71
3.81

5.00
5.00
5.00
5.00

0.00
0.00
0.00
0.00

 

So, I'm not sure it's a thermistor; that was what Co-Pilot suggested and what I was testing for. However, the voltage stayed the same even when applying some light heat.

Another example of why it's not a good idea to trust AI.

 

Yeah, learning electronics using AI as a backup has been 70% reliable, which is not great. It's reluctant to tell me when I'm wrong and it's too flattering, which could breed overconfidence. But it's been good for snippets of code, finding resources, and some basic brainstorming, at least.

 

It's reluctant to tell me when I'm wrong

It does at least accept criticism graciously when you point out its errors.

 

The problem with Artificial Intelligence at this state of development, is that it has no real intelligence (i.e. understanding) of the subject of interest.
It just picks statements from it vast memory that seems to pertain to your question or problem.
That's why it works okay for boiler-plate type answers, but some of its replies are rather off-the-wall when real understanding is needed..

 

You should be able to test for the presence of a thermistor with a multimeter.
Check if there are two of the wires with no voltage across them.
Then check the resistance between those two wires.
A thermistor will have several kΩ of resistance.