How to Compensate Voltage for Temperature Change in a Diode?

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To calculate the voltage change that compensates for a 1-degree temperature change in a diode, the characteristic equation I = Is x (e^(eV/kT) - 1) is used. By taking the natural logarithm, the relationship between current and voltage can be linearized, revealing that the gradient corresponds to e/kT. To maintain constant current despite temperature changes, the voltage must adjust accordingly, as the right side of the equation must remain constant. The discussion emphasizes understanding how voltage and temperature interact to keep current stable in diode applications. This analysis is crucial for accurate diode circuit design and temperature compensation.
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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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