But why when I connect a volt-meter across the whole diode will I not see this built in potential?
Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop).
Edit
Maybe a more visual explanation would help? Think of the Fermi-level as the height of water in two buckets. The first bucket on the left has a very low water level, the second bucket on the right has a very high water level. P-doped semiconductor have a Fermi-level close to the valence band so this is represented by the left bucket. N-dopded semiconductors have a Fermi-level very close to the conduction band, this is represented by the very high water level in the right bucket. We now connect the buckets together with a pipe and the water level equilibrates.
| | |~~~~~~~~| | | | |
| | | | | | | |
| | | | ---> |~~~~~~~~ ========= ~~~~~~~~|
| | | | | | | |
|~~~~~~~~| | | | | | |
|________| |________| |________| |________|
Now obviously semiconductor band structure is not a bucket and water is not an electron gas but this illustrates the basic process. Let's try and stretch the metaphor a little... Before we connected the buckets they were both electrically neutral, this implies that by allowing the exchange of water we should expect an internal build-in electric field and associate potential.
But notice, that the "voltage drop" across the buckets is just the potential energy difference between the left and the right sides, which is zero because both waters levels have the same height.
So back to your question; we know the pn-junction has a built-in potential, so why can't we measure that? Well this is the potential required to lower the n-side band such that the Fermi levels of both sides can align. The internal potential is equal to the difference of Fermi-levels (water heights) before alignment.
The applied voltage and the voltage inside the pn junction are opposite to each other in direction, so the energy barrier gets reduced so more electrons can cross the depletion zone and thus increasing the current.
Best Answer
You are right that there are both drift and diffusion currents in the pn-diode. However, due to current continuity in the one-dimensional diode model, the total current (electrons and holes) can be described as the sum of the minority diffusion currents in the quasi-neutral p- and n-regions at the respective depletion zone boundaries. This corresponds to the so-called Shockley model of the diode, which works pretty well at not to high voltages when recombination/generation in the depletion zone can be neglected.