# 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
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0
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!

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2. ### crutschow Expert

Mar 14, 2008
21,340
6,118
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.

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