For a university project that is due for tomorrow (!) I have a question concerning the use of voltage dividers on the Arduino's analog inputs. A prototype I'm using provides a variable voltage output through a voltage divider. I have the basic prototype rigged up, but the instance I connect the voltage divider's to an analog input on my Arduino Diecimila, the voltage dividers' outputs drop to zero immediately. I've tried using just one, basic voltage divider (10k R1 and 10k R2, meaning 5/2=2.5V output), but the same problem occurs: a nice and steady 2.5 output, but as soon as I connect it to an Audrino analog input pin, it drops to zero. I've also tried connecting a 10K resistor between the divider's output and the analog input to prevent any current being drained from the divider's output, but the problem is still there.

I hope you can help me with this problem.

Thanks,

Sebastian

You don't say where the voltage is coming from that goes through your divider. If it is separate from the Arduino you must also connect the ground of the voltage source to the ground of the arduino.

Also
Make sure the software has not initialises the analogue input pins to be an output.

basic voltage divider (10k R1 and 10k R2, meaning 5/2=2.5V output), but the same problem occurs: a nice and steady 2.5 output, but as soon as I connect it to an Audrino analog input pin, it drops to zero.

Assuming it is wired up correctly and the software is right then maybe the Audrino pin is damaged. Have you tried it on the other analogue input pins?

I have the same situation, with a 22k/22k divider input.
Checking the resistance between that ADC pin and ground it's very small (~1 KOhm), whereas a healthy ADC pin shows .7 MOhm.

Not sure what you mean by healthy. If you move your potential divider to a previously healthy pin does that show up as unhealthy?

Anyway you can't measure an arduino pin like that, a resistance meter puts a current through it's two probes so you are forcing current through your pins. Is this with the arduino powered or not? Either way unfortunately it is not much of a test.