In general, ideality factor ''n'' could be calculated by:
n=(V2-V1)/ln(I2/I1)
where voltage and current are taken from the dark I-V curve.
For simple PN Si diode, it is close to 2.
In my case NPNN Si diode and PNN Si diode, however, the ideality factor is less than 1 (0.05-0.2).
I could not find any explanation or research where the ideality factor could be less than 1.
Any suggestions or comments?
It looks like in formula you missing factor of e/kT = 26mV (room temperature).
Basically you have to take your voltage in 26mV 'units'
DEAR Vytautas Kavaliūnas ,
There is strong evidence that it may be falsely calculated as hinted also by Nicolay.
We have
I=I0 exp V/nVt
Assuming two currents I1 and I2, at V1 and V2, then
V2-V1= nVt ln (I2/I1)
Then
n=(V2-V1)/ Vt ln (I2/I1),
So, you have to divide your values by Vt= 26 mv at room temperature
So dividing your values by Vt, one gets
2 to 8
These values are greater than one.
Best wishes
Ideality factor "n" is an empirical parameter showing the deviation of a measured I(V) junction characteristics and the "ideal" Shockley equation I = Is (e^(qV/kT) - 1) leading to I = Is (e^(qV/nkT) - 1). Parasitic resistance to access through the semiconductor to the active zone of a junction is the main cause of this non-ideality factor in high-injection mode (high current density). Carrier recombination can also introduce non-idealities at low-current density. However, all these two phenomena lead to an ideality factor n > 1.
A parameter that can be confused/mixed with the ideality factor is the effective temperature of the junction Te. Indeed, the diode equation can be written I = Is (e^(qV/nkTe) - 1) taking into account possible increase of the junction temperature regarding ambient temperature. Again this neglected phenomenon is usually interpreted as an ideality factor > 1 because of Te > Tambiant (due to dissipations) considered as T.
However, supposing that you cooldown your junction (?), and assume T = 300 K or any higher temperature than the junction Te itself : your first impression could be that the ideality factor is lower than 1, but in fact due to an over-estimation of Te.
You have to consider the 26 mV factor and divide your result by this constant. Hope you will get the right value.
Thank you all for the comments. Yes, by mistake I did not add e/kT to the equation and another mistake, I did the calculation in Volts, and not mV. After this, it seems that the ideality factor at a certain voltage is 1.5-2 which is decent.
Thank you all for a prompt response and explanation.
I calculated the ideality factor of natural dye sensitized solar cell in the dark from the log (I) versus V/thermal voltage and i obtained value of (1.2). The same cell tested under various lights. From log(I) versus V+IRs/thermal voltage yielded lower ideality factor around 0.82 under LED and 0.72 under tungestin bulb . Is it possible ? @ abdelhalim abdelnaby zekry
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