- #1
laxclarke
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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?
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?