What Do e and V Represent in the PN Junction Current Equation?

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What are e and V in exponent of the equation I = I0(eeV/kT - 1)? is it really one variable "eV", as in electron volts, or is e just 2.718... and V for volts?
 
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Two; e is the charge on an electron, V is the potential.
 
It's confusing. The first e is just the base of the natural logarithms (2.718...), but the e in the exponent is the charge on the electron. I prefer to write it as:
e^\frac{qV}{kT} to eliminate this confusion. Note that if you measure kT in electron volts (eV), then the electron charges cancel, and you can just use the voltage V directly. Note that at room temperature, kT is .026 eV, so you can write:
I = I_0 e^\frac{V}{.026}
 
Thanks!
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?

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