Understanding the Function of a PN Junction Diode | Explained by Experts

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SUMMARY

The discussion focuses on the function of a silicon PN junction diode, specifically addressing the barrier potential of approximately 0.7V that exists between the p-type and n-type materials. It clarifies that measuring this potential difference directly across the diode terminals with a voltmeter may yield a reading of zero volts due to contact potentials. The conversation emphasizes the necessity of applying an external voltage equal to the barrier potential to overcome the depletion layer formed by electron-hole recombination. Additionally, it suggests using an oscilloscope for more accurate measurements when the diode is forward biased.

PREREQUISITES
  • Understanding of semiconductor physics, specifically PN junctions
  • Knowledge of barrier potential and depletion layers in diodes
  • Familiarity with measurement tools such as voltmeters and oscilloscopes
  • Concept of contact potentials in metal-semiconductor interfaces
NEXT STEPS
  • Research the principles of semiconductor physics and PN junction behavior
  • Learn about measuring techniques for semiconductor devices using oscilloscopes
  • Explore the effects of contact potentials in metal-semiconductor junctions
  • Study the characteristics and applications of silicon diodes in electronic circuits
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Electronics students, electrical engineers, and anyone interested in semiconductor technology and diode functionality.

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?
 
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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:
laxclarke said:
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?
unplebeian said:
The instrument to measure is a voltmeter, or for more accurate results you can use a scope.

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.
 

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