In the schematic attached, 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!

 

I'm not sure about your electric field questions so I'll let that for someone else.

The energy dissipated in a PN junction is similar the power dissipated in a resistor and equals V * I. The external voltage source provides the energy which is dissipated as heat in the PN junction, causing the junction temperature to rise. Thus, high current rectifiers must be mounted on a heat sink to keep them from overheating.