Hi community,
I just started to learn electronics with Arduino. I have a question about diodes, but before, please, take a look at the circuit below:
If I measure with a tester the voltage between the anode and the cathod I get a -1.60V value, but I expect 0V in this configuration.
What's wrong? Forgive me if it sounds like a stupid question! 
LEDs are also photodiodes, and generate a small current and voltage when exposed to light, similar to PV panels. That is why you get a voltage reading when the switch is open. Overview here.
LEDs are color sensitive, responding best to light that is a bit bluer than the emission color.
Use the diagonal terminals on that switch.
If I measure with a tester the voltage between the anode and the cathod I get a -1.60V value, but I expect 0V in this configuration.
With the LED on? That's about right for an LED.
A regular silicon diode (with current through it) has about a 0.7V "forward voltage drop" across it. LEDs have a higher forward voltage.
The diode (or LED) has high resistance below it's rated voltage. Above its rated forward voltage, the resistance drops dramatically and current flows freely. So with a resistor in series, the LED voltage "falls into place" and it remains constant with the remaining voltage dropped across the resistor.
The resistor limits the current. Without the resistor the total resistance is very low (at 5V) and you'd get excessive current that could blow the LED or your Arduino, etc.
You can calculate the current (through both the LED and resistor) by applying Ohm's Law to the known resistance and the known voltage across the resistor.
larryd:
Use the diagonal terminals on that switch.
Or pull the switch out, turn 90 degrees and replace.
And if you really want to get an earful about LEDs, check out my Blog:
http://www.sinistercircuits.com/Blog/LEDs_LightemWithoutSmokinem/index.php
larryd:
Use the diagonal terminals on that switch.
You would do well to read a datasheet or check with a continuity tester before incorrectly assuming
the pin-out of the push-button - you have a permanent short wiring it that way.
Frankly. there is far too little information in the original posting to know what is happening. 
If the OP comes back to explain more, it may be worth discussing.
On the posted photo, it is stated that the switch is open, so the LED is connected only to ground at one pin.
The text indicates that the OP measures a DC voltage of ~1.6V across the LED.
Only photodiode action explains all those observations (reply #1).
Thank you guys, sorry for the late reply but I had some problems with my internet connection.
@jremington - You are right: the pushbutton is open, therefore the circuit is open. In this configuration, I measured a negative DC voltage across the LED.
As I wrote earlier, I expect a 0V value. Am I wrong? :-\
Must be working under a bright light there!
Must be working under a bright light there!
The open circuit voltage of a photodiode is roughly constant, independent of light intensity (except for total darkness, of course). However, the short circuit current is proportional to the intensity.
jremington:
The open circuit voltage of a photodiode is roughly constant, independent of light intensity (except for total darkness, of course). However, the short circuit current is proportional to the intensity.
Ah, yes...but it's not exactly open circuit, is it. There is the meter's internal impedance. I gave it a try with my best mmeter and got at most around 0.2V
But, that was under fairly dim lighting, and I had to press the LED right up against the fluorescent bulb on my desk lamp [the rest of the room lighting is all LED].
So, I agree with Paul__B, must be some bright lighting -- or a different explanation.
Thank you guys, just an update. I tried to measure the voltage into a dark room but I obtained the same value as previous tests (below the scenario):
Then I measured the voltage in direct polarization and it was as expected:
Some kind of enigma! :-\
Anyway, I will make further tests, thank you!
And what does your meter indicate with nothing connected to the test clips?
Paul__B:
And what does your meter indicate with nothing connected to the test clips?
It indicates 0V (as expected). The meter works fine 
Show us a image of the meter dial and test lead placement when you are measuring the voltage across the LED in the circuit with the open switch.
BTW, in the same LED/switch circuit, what do you measure across the open push button switch?
larryd:
Show us a image of the meter dial and test lead placement when you are measuring the voltage across the LED in the circuit with the open switch.BTW, in the same LED/switch circuit, what do you measure across the open push button switch?
Hi larryd, below the images:
Do you live under high voltage power lines or near a radio/TV broadcast antenna? I.e., maybe your meter probe wires are picking up some sort of RFI/EMF that is being rectified by the LED.
Well done and documented experiments! Congratulations, few people are able to do so.
Sadly I don't understand this phenomenon either but I can confirm it exists. Not only LEDs but all PN junctions do this (i.e. "normal" diodes, BJTs). When you connect one lead to something and leave the other open a voltage "magically" appears over the device. I believe it has something to do with AC coupling between mains, power source and you - holding the test leads (the isolated part) make the voltage higher. I think only the long wires of testing leads and you touching them makes this significant so I postponed further experiments to "later".
