Hi, I was reading Sedra & Smith and came across this line in the pn-diode section:

"When the pn junction terminals are left open-circuited, the voltage measured between them will be zero. That is, the voltage across the depletion region does not appear between the junction terminals. This is because of the contact voltages existing at the metal-semiconductor junctions at the terminals, which counter and exactly balance the barrier voltage. If this were not the case, we would have been able to draw energy from the isolated pn junction, which would clearly violate the principle of conservation of energy."

I am having trouble understanding how the ~0.7 barrier voltage is countered by the metal contacts. Anyone lend a hand?

Thanks,
JP

 

bump to the top!

 

When charges migrate across the junction and create the depletion region, think about where those charges came from and what that does.

 

When charges migrate across the junction and create the depletion region, think about where those charges came from and what that does.

I understand that the depletion regions on each side of the junction have stationary charges left in the material that are the opposite sign of the majority carriers. But what does the metal-semiconductor junction have to do with it?

 

You have metal on both ends which does not like an electric field existing within it and electrons can flow easily to/from the metal at the metal-semiconductor interface. If you had 0.7V of potential between the two, electrons will flow off one interface and onto the other until a countering potential is set up that exactly matches the one from the p-n junction.

 

You have metal on both ends which does not like an electric field existing within it and electrons can flow easily to/from the metal at the metal-semiconductor interface. If you had 0.7V of potential between the two, electrons will flow off one interface and onto the other until a countering potential is set up that exactly matches the one from the p-n junction.

Isn't this the case only when the metal is incasing the entire diode and there must be a net field of zero within? This case is for metal on either side, no?

 

This might be of help .... see section 4 [page 16] of the pdf attachment.

Although I'm not sure the potential graph entirely makes sense. For instance another site gives typical electric potentials for metal [gold] Si semiconductor [N or P type] interfaces as ....

\(\Phi_{mp}=0.34V\)

and

\(\Phi_{mn}=0.8V\)

See ...

http://ecee.colorado.edu/~bart/book/book/chapter3/ch3_2.htm

So the metal-to-N type interface has the higher potential. The graph on page 17 of the attachment seems to indicate the opposite.