Ideality factor - schotkky diode

[hananl]

Hi,
everywhere I look for the equation describing the current of a diode, I get
I=I0exp(qvb/nkt-1)

n=1 for ideal case. for non-ideal diode you got values between 1-2.

in order to calculate n you have the equation: n=1+kt/(2*q*phi)
(phi=contact barrier height)

anyone knows how to derive this equation for n?

thanks.

 

[es1]

http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_4.htm

See section 4.4.4.

n is just something that is added there because we recognize the equations we use to model a pn junction (see everything before section 4.4.4) are not perfect, there are important second order effects. Adding a fudge factor is an easy way to increase the accuracy of the model. Of course you don't have to do that. There are equations to express the additional effects, but engineers rarely need that kind of accuracy so we don't bother.

AFAIK, n is always found experimentally via curve fitting and it is usually pretty close to one. The diodes I am using now, for example, have n=1.08. BJTs wired as diodes can have n quite small, say less than 1.01. Diodes inside ASICs typically have larger n's due to dimensions and smaller currents.

 

[hananl]

Hi,
I know that there are deviations from the ideal model, also I undestand why (due to generation-recombination currents). So I can understand why for certain materil its n value is bigger(smaller) than another semiconductor.
But I DO need to know how get this equation. infact I need to prove it since I am making a lab report, and need to show this relation but don't know how to approch it.
thanks.

 

[es1]

Well, section 4.4.2 in the above link derives the ideal equation based on quasi-fermi energies (see section 2.11.5). Then just say this equation I derived is missing second order effects so I am slapping in a constant factor in the exponential to account for it.

I am not sure how you are supposed to "prove" it (especially when we know it's wrong) via experiment since the whole point is to fit the equation to the data not vice versa. Maybe solve for one diode and then use it to predict the current of a set of diodes?

Proofs are for math class, not engineering. :)

 

[marcusl]

I also understand non-ideality factors to be empirical corrections. You can look in Henisch, Semiconductor Contacts (1984) for a discussion relating to Schottky barriers. He discusses tunneling and other effects, and gives numerous references for further reading.
Edit: He is careful to point out that it is purely empirical in nature.

 

[hananl]

es1,
It seems resonable to add a constant, so I don't understand how I was told that:
n=1+kt/(2*q*phi) ... ?

 

[es1]

I am not sure either.

My guess is it's based on the assumption that the error is exclusively based on the recombination current which is dominated by the electron-hole pairs with the minimum amount of energy necessary to recombine. The barrier height is q(phi-vb) so if they meet in the middle the energy is half that. So then the left side of the equation becomes I multiplied by some error term which is portional to e^(-(phi-vb)/2Vt). Then make an assumption about I for a given vb and solve for n.

Just a guess based on reverse engineering your equation for n. Also, I didn't plug it in so the math might not work out in the end.

 

[hananl]

Hi,
I let it go, but anyhow, if you are interested:
take the term for Js, the current for schottky diode:
http://ece-www.colorado.edu/~bart/book/book/chapter3/ch3_4.htm section 3.4.1 the first equation. we would like to have something like
J=J'*exp(qvb/kt-1)
However, if we take J' as the part in section 3.4.1 that's only before the exp, that's not good, since we have a depndace on Va (the voltage we apply on the jucntion) and we would like to have J'=const. so that's where n came from. we assum Va<<phi. also we use : x^0.5=exp^(0.5lnx)
we assume j'=the whole part before the exp, excluding the Va. using this ln "trick", and presenting ln with a taylor seriouse, my instructor told me, than eventually you get n=1+kt/(2q*phi). than you'll have: J=J'exp(qvb/nkt -1).

thanks es1.

Hananl