Diode Equation and Saturation Current

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SUMMARY

The discussion centers on the diode equation I(V)=I_{sat}e^{\frac{Ve}{nKT}}-I_{sat}, specifically addressing the conditions under which the saturation current (I_{sat}) becomes negligible. Participants confirm that when the voltage (V) significantly exceeds the product of the ideality factor (n) and thermal voltage (V_{T}), the second term of the equation becomes insignificant. With initial conditions of V > 0.1V, kT approximately 0.025 eV, and n approximately 2, it is established that the saturation current is primarily observable in reverse bias conditions, typically in the range of nA to µA, due to the low recombination rates of minority carriers.

PREREQUISITES
  • Understanding of the diode equation and its components
  • Knowledge of thermal voltage (V_{T}) and its significance
  • Familiarity with forward and reverse bias conditions in diodes
  • Basic concepts of carrier recombination in semiconductor physics
NEXT STEPS
  • Study the derivation and implications of the diode equation in detail
  • Explore the concept of thermal voltage (V_{T}) and its calculation
  • Investigate the effects of minority carrier recombination on diode performance
  • Learn about avalanche breakdown and its relation to reverse bias conditions
USEFUL FOR

Electrical engineers, physics students, and anyone involved in semiconductor device design or analysis will benefit from this discussion, particularly those focusing on diode behavior and characteristics in various bias conditions.

DeldotB
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Good day all:

For the diode equation: I(V)=I_{sat}e^{\frac{Ve}{nKT}}-I_{sat}

I also know that the second term in the equation is negligible if: V> > nV_{T}

Vt is thermal voltage
n is ...just a factor related to the quality of the material

I have these initial conditions:
V> 0.1V\, \, \, ,kT\approx .025 eV\, \, \, n\approx 2

Its easy to show that the my inital conditions lead to a V much greater that n*thermal voltage.

My question is: Why is the saturation Current negligible? I know it has something to do with recombination rate but can anyone give me a simple explanation? So many websites just say that the order relation is true but give no evidence as to why.
 
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In forward bias, the Is component is tiny. You only see it in reverse bias before avalanche breakdown...

http://www.electrotechservices.com/electronics/images/diode_graph.jpg
diode_graph.jpg

Does that help?
 
Because most of the current is in the forward direction and the reverse current, due to the minority carriers, is low. Typically in the order of nA to uA.
 

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