# Pn junction diode

1. Jan 14, 2008

### laxclarke

Say we have a silcon pn junction (diode) - i.e., a block of p-type on
left, attached to a block of n-type semiconductor on right:

anode ------[ p | n ]------ cathode

Because of diffusion we get a barrier potention at the junction, which
makes the n-side/cathode 0.7V higher than the p-side/anode:

~~~~~~~~ - 0.7V +

anode ------[ p | n ]------ cathode (*)

1) Now is there any way to "measure" this potential difference right
from the diode using some instrument?

Now to get ride of the depletion layer (barrier) we need an "opposite"
external voltage equal in magnitude to the 0.7V shown in (*) :

~~~~~~~ (barrier potential)
~~~~~~~ - 0.7V +
anode ------[ p | n ]------ cathode

~~~~~~~ + 0.7V -
~~~~~~~ (external voltage)

2) Now, why isn't the resulting voltage of the diode 0V (sum of
barrier and external)? How come we only measure the external 0.7V
using a voltmeter when the diode is forward biased?

3) Is there anything wrong with the thought process I've outlined
above?

2. Jan 14, 2008

### unplebeian

The instrument to measure is a voltmeter, or for more accurate results you can use a scope (only when the diode is on).

Correction: Because of diffusion the electron-hole paris recombine and a depletion layer is formed. To overcome this depletion layer an external potential (generally 0.7V for Silicon) is needed, hence you measure 0.7V and not 0V.

Once the depletion layer is formed it prevents the further recombination of holes and electrons.

Last edited: Jan 14, 2008
3. Jan 14, 2008

### ranger

Please note that it is not as simple as placing a voltmeter across the terminals of the diode and measuring the barrier potential, Vb. In fact one would expect a reading of zero volts. Give some thought to contact potentials. Remember that we have metal-metal and metal-semiconductor contacts.