Simple transistor

Boki =E5=AF=AB=E9=81=93=EF=BC=9A

Because the Vf was increased?

Diodes are not perfect... If you push enough current through them, the voltage drop rises. Check out the I vs V curve of the diode. Take a look at:

Yes, of course, but how is the voltage at very high current status..

P connects 3.3V, N connects to ground. ?

no, no condition here, just to know why....

...and you've probably let the magic smoke out of the device...

Taking a typical 1N914B diode as an example, and assuming a simple square-law model, the data sheet gives a maximum forward voltage of 1V at 100mA which makes the model:

I = (100mA/1V^2)*Vf

So you'd need to push about 960mA across the junction. This is not something that you'd want to do on a continuous basis! But you could get away with it for short duration pulses (less than a mSec).

If "Boki" is measuring 3.1V DC across the diode, it's not a diode anymore.

If P is going directly to 3.3V and N is going directly to ground, with no resistor in series with it, the diode will be destroyed (an open circuit, as John said). Brian

About 59-60mV per decade at room temp for ideal diodes, based on the differential of the simple EM equation.... which would produce a great deal of current, cranked up to 3.3V. Impossible amounts, if there wasn't another effect operating.

There is a resistive component that can vary a lot from diode type to type. For example, the 1N4148 exhibits about 0.6 ohms (see the Rs figure in a Spice model.) That will limit the current, but not so much as to survive 3.3V.

It's probably open or nearly so.

Jon

??????? are you sure?

I connect P to 3.3V, and then N nothing else but directly to ground...

I think it will get a large current and then broken ... ( but I think that open is not at beginning)

Best regards, Boki.

"it's not a diode anymore."

hahha...

Except some of those. ;) But the OP wrote "PN diode" for whatever that may have meant.

Jon

LED diodes? :)