How to Compensate Voltage for Temperature Change in a Diode?

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Homework Statement


I'm trying to calculate what voltage change will compensate for a temperature change of 1 degree for a diode.
I have set up the circuit, taken down results and plotted the graph (its non-linear). So I re-plotted it using natural logs of I, and got a straight line with a gradient. Now I'm lost as to how to answer this question.


Homework Equations


I= Is x (e^(eV/kT) -1)
It's the characteristic equation for V vs I for a diode.


The Attempt at a Solution


I took natural logs to get:
ln I = lnIs x (eV/kT - 0)

Then said the gradient is equal to e/kt.
But then where do I go from there?

A little push in the right direction would be great!

Thanks in Advance!
 
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Lavace said:
I= Is x (e^(eV/kT) -1)
It's the characteristic equation for V vs I for a diode.

I took natural logs to get:
ln I = lnIs x (eV/kT - 0)

Hi Lavace! :smile:

If you mean I= Is (eeV/kT -1),

then I/Is + 1 = eeV/kT:smile:
 


tiny-tim said:
Hi Lavace! :smile:

If you mean I= Is (eeV/kT -1),

then I/Is + 1 = eeV/kT:smile:

But then I'd end up with ln(I - Is) + 0 = eV/kT

Then the gradient is equal to e/kT, so how do we calculate what voltage change will compensate for a temperature change of 1 degree for a diode for this? I have a value of the gradient as well!
 
Lavace said:
But then I'd end up with ln(I - Is) + 0 = eV/kT …

No … ln(I/Is + 1) is not ln(I/Is) + log(1), is it? :smile:
 


tiny-tim said:
No … ln(I/Is + 1) is not ln(I/Is) + log(1), is it? :smile:

Sorry to make it so slow but where des this come into finding the voltage change for 1 degree?

Thanks for the advice!
 
Lavace said:
Sorry to make it so slow but where des this come into finding the voltage change for 1 degree?

oooh, that's electronics, isn't it? :confused:

i'm just here for the maths. :redface:
 


ot sure if this helps or hinders.
But there is Wien's Law that allows you to determine Peak wavelength for a given temperature in Kevin
ie

<br /> \lambda _{peak} T = 2.90 \times 10^{ - 3} meters.Kelvin<br />

The energy gap is the energy required by a semiconductor needed before starting to conduct, the energy is related to the wavelength of the EM wave.

So

<br /> E_g = hf = \frac{{hc}}{\lambda }<br />

where c is the speed of light
 


Lavace said:

Homework Statement


I'm trying to calculate what voltage change will compensate for a temperature change of 1 degree for a diode.

It sounds like you want to maintain the same current I, when there is a temperature change, by changing V.

I= Is x (e^(eV/kT) -1)

Okay, so if I is to not change, then the right-hand-side of this equation must also not change. If T changes, for example from 20 C to 21 C, what must happen to V in order to keep this expression at the same value?
 
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