Calculating η for a Diode: A Puzzling Experiment

[jono90one]

Homework Statement

I have been doing an experiment, and I needed to calculate the value of η for a diode.
The value should be between 0.5 and 1.0...
The value for alpha obtained is 20.

Homework Equations

αV=ηeV/kT
Where k is Boltzmann’s constant, eV is the electron volt, T is temperature.
I=I(0)(e^αV) (neglecting the minus 1)

The Attempt at a Solution

I don't really understand how η can be a constant seeming V is a variable. The values of the temperature where never recorded (never asked for in the experiment), so do i assume T = 298K? What voltage do i choose to associate that with? the average?
As by doing that i get 0.33 for η, outside the acceptable range where the range of voltages taken where between 0.8 and 0.5 (so i let v=0.65).
Even if i choose v=0.8, it's still less than 0.5 because v would have to be greater than 1.

The values we were asked to gather were between 0.5 and 0.8 v anyway, so that cannot be wrong. Alpha is defiantly correct as i drew it by hand and plotted it on origin. So currently either 0.33 is correct despite not being in the range, or I’m using the wrong values of V in my calculation (or T).

Hope you can advice me :)

 

[gneill]

αV=ηeV/kT
Where k is Boltzmann’s constant, eV is the electron volt, T is temperature.
I=I(0)(e^αV) (neglecting the minus 1)

The Attempt at a Solution

I don't really understand how η can be a constant seeming V is a variable. The values of the temperature where never recorded (never asked for in the experiment), so do i assume T = 298K? What voltage do i choose to associate that with? the average?
As by doing that i get 0.33 for η, outside the acceptable range where the range of voltages taken where between 0.8 and 0.5 (so i let v=0.65).
Even if i choose v=0.8, it's still less than 0.5 because v would have to be greater than 1.

I think you may be confusing the diode voltage V with the unit V. In your expression for α:

αV = ηeV/kT

the V is the unit [V], yielding the value for α to be : α = ηeV/(kTV)

The volt unit is in the denominator of the expression so that when α multiplies the diode voltage (let's call it V_d), the result is unitless. This is required for the parameter of the exponential function to be unitless.

Your expression for the current is then

Id = I_s e^αV_d

The value of α should be approximately η * 38.68 V^-1

300K is a commonly used temperature for semiconductor testing.

 

[jono90one]

Ohh i see where i went wrong lol. I now get 0.518 for the value of eta :D

Thanks!