Why Does Current Rise Exponentially in a Diode After Reaching +0.6V?

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SUMMARY

The exponential rise of current in a diode after reaching approximately +0.6V is primarily due to the microscopic effects related to the semiconductor's band structure and the behavior of electrons following Fermi-Dirac statistics. This phenomenon is not significantly influenced by macroscopic factors such as internal resistance, inductance, or capacitance. The built-in potential of 0.6V to 0.7V is characteristic of silicon diodes, while other types, such as Schottky diodes and LEDs, exhibit different threshold voltages. Understanding the junction theory is essential for comprehending the exponential current-voltage relationship in diodes.

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so I've seen this graph all over the internet, but nowhere seems to explain why the current after +0.6v rises exponentially and not just straight up. is it mainly due to the internal resistance of the diode or is there significant inductance/capacitance in there as well. what's the dominating effect(s)?

I'm guessing at that voltages significantly higher than Vf its just the internal resistance (r=1/gradient), but the curve leading up to Vf, what's causing it?


http://www.yegopto.co.uk/media_yegOpto/image/400%20x%20400%20images/volt_current%20graph.jpg


also, what's a typical internal resistance for a power diode?

thanks
 
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No, it is a microscopic effect. It has to do with the shape of the bands (the density of states) in the semiconductor; you can't really think of it in terms of "macroscopic" variables like capacitance etc.
It is not too difficult to derive the correct expression using a band model and you should be able to find that derivation in just about any book on semiconductor physics.

The fact that is rises exponentially is essentially just a consequence of the electrons obeying Fermi-Dirac statistics.
 
No the exponential relationship is not due to series resistance. It's true that bulk resistivity in the neutral regions can have a significant effect in some diodes at high current levels (making the I/V characteristic less steep than predicted by the exponential relationship) but no the exponential relation is not due to resistance. It's a lot more complicated than that. Start with this for example : http://books.google.com.au/books?id...resnum=2&ved=0CA8Q6AEwAQ#v=onepage&q=&f=false
 
The 0.6 volts is called the built-in potential and depends on the type of semiconductor material in the diode.
The current–voltage curve is exponential. In a normal silicon diode at rated currents, the arbitrary “cut-in” voltage is defined as 0.6 to 0.7 volts. The value is different for other diode types — Schottky diodes can be rated as low as 0.2 V and red or blue light-emitting diodes (LEDs) can have values of 1.4 V and 4.0 V respectively.
http://en.wikipedia.org/wiki/Diode#Semiconductor_diodes"
And as f95toli implies, you would need to study the junction theory to see why the curve is exponential.
 
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thanks guys.

not so simple question after all then, it's going to take me a long while to understand all that.
 
Look at this URL
http://www.allaboutcircuits.com/vol_3/chpt_3/1.html
and scroll down to the diode equation. kT at room temperature is about 26 millivolts.
For some reason, I cannot open Wiki URL in post #4.
Bob S
 

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