How is energy consumed in PN junction?

Discussion in 'Homework Help' started by Heidi L, Nov 27, 2012.

  1. Heidi L

    Thread Starter New Member

    Nov 12, 2012
    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!
  2. crutschow


    Mar 14, 2008
    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.
    Heidi L likes this.