How to Compensate Voltage for Temperature Change in a Diode?

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Homework Help Overview

The discussion revolves around calculating the voltage change required to compensate for a temperature change of 1 degree in a diode. The context involves the diode's characteristic equation relating current and voltage, specifically focusing on how temperature affects this relationship.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of natural logarithms to linearize the diode equation and explore the implications of the gradient in relation to temperature changes. Questions arise about the correct interpretation of logarithmic transformations and their relevance to finding the voltage change.

Discussion Status

The conversation is ongoing, with participants offering insights and clarifications about the mathematical relationships involved. Some guidance has been provided regarding the interpretation of the gradient, but there is no explicit consensus on how to proceed with the calculation of the voltage change.

Contextual Notes

There are indications of confusion regarding the logarithmic manipulation of the diode equation, and participants are questioning the assumptions made in their calculations. The discussion also touches on the broader context of electronics and the relationship between temperature and voltage in semiconductor behavior.

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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

[tex] \lambda _{peak} T = 2.90 \times 10^{ - 3} meters.Kelvin[/tex]

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

[tex] E_g = hf = \frac{{hc}}{\lambda }[/tex]

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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