# How is energy consumed in a PN junction?

In the schematic above, there's a PN junction with a source $V_d$, the barrier potential $V_J$, contact potentials $aV_{J0}$ on the P-side and (1-a)$V_{J0}$ on the N-side. $V_{J0}$ indicates the barrier potential of the PN junction with no source connected, $V_J$ the barrier potential with this source present and 'a' is a fraction in $aV_{J0}$.

How can we interpret the whole circuit in the point of view of erergy-providing and energy-consuming?

1) There are voltage drops $aV_{J0}$, (1-a)$V_{J0}$. Does that mean a test positive unit charge will "loose" energy while going from the (+) side to the (-) side? If so, in what means does the test charge get it's energy? What kind of form of energy is it transferred to?

2) Across the depletion region from P to N-side, there is an electric field pointing to the P-side. I guess there must be a reverse electric field acting on a test positive charge for it to pass through the build-in field. Where does that reverse electric field come from?

3) If the external source $V_d$ > $V_{J0}$(the barrier potential of the PN junction with no source connected), will there still be a barrier voltage inside the PN junction?

Thank you!