Electronic – Model for Diode Shows Voltage Increase not Drop

depletion-regiondiodespn-junctionsemiconductors

It's well-known that if a diode is forward biased at a sufficient voltage, there is a voltage drop across the diode, i.e. it dissipates power.

However, I was reading Wikipedia's article on the depletion region of the diode (aka PN junction), and it occurred to me the electric field, E, as labeled below, is set up by immobilized ions and causes the voltage, V, as indicated to be positive:

enter image description here

I've also drawn the symbol of the diode in the above.

This result is definitely what I would expect, because if a battery (and resistor) is connected to the diode in the usual, forward-biased way, there should be voltage drop across the diode and the voltage V (as labeled) should be negative.

Where did I go wrong?

Best Answer

You understanding is not wrong, but incomplete. You basically showed how a diode is "built", but not how it is used.

schematic

simulate this circuit – Schematic created using CircuitLab

If you have a circuit like the one above, for instance, the 3 V supply must overcome the built-in voltage of your diode (the one you drew in your picture) to generate an electric field (directed from anode to cathode) strong enough to push current from anode to cathode.