Electronic – Internal difference in a diode

I think, the answer is relatively simple. Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction? Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured. With other words: You cannot use the diffusion voltage to drive any current through an external load.

I think, there is a small confusion on your side regarding the meaning of "barrier voltage".

Caused by a diffusion process across the pn-junction there is a "barrier" consisting of a potential difference (called voltage). This "built-in voltage" causes the diffusion to stop at a certain equlibrium. Hence, "barrier voltage" and "built-in voltage" are two different expressions for the same quantity.

For a current to exist through the device there must be an external voltage which is able to reduce this barrier. Thus, this voltage further supports the diffusion process and allows a exponentially rising current. This effect is visible already at an external forward voltage of app. 0.1 Volt).