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Temperature effect on diode function

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flaxwalter

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Hi, I have a question about the diode equation function. If I graph vd versus id for a given temperature, it returns me a characteristic curve, but when I try with a higher temperature and I graph the equation, I find a curve more to the right, however the electronics books say that at a higher temperature the curve must go towards the left and viceversa. The equation I am graphing is: Id = 1 * 10 ^ -12 * {exp [(1.6 * 10 ^ -19 * Vd) / (1.38 * 10 ^ -23 * Kelvin Temp)] -1}.
What am I doing wrong?
Thank you.
 
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Vf goes down, Rs goes up, forward voltage tempco depends
on whether you're at high or low current.
 

thank you, dick_freebird, (I dont understand what is tempco.) I mean if we only graph the equation, results what I said, I am only refering to graphing, If 1.- I do write 273 Kelvin, I get a curve. 2.- Then if I do write 298 Kelvin, I get another curve, the second curve appears more to the right than the first. Ok, electronic books say otherwise. Is there any missing in my equation? Or am I using the wrong equation?
 

At higher current levels, ohmic resistance is a greater proportion of total impedance (instead of the PN junction). This causes the volt-vs-Ampere curve to become more linear, so it ramps less abruptly upward. The appearance is to flatten the curve and make it extend toward the right more than upward.

By 'curve moving to the left' implies that less voltage is needed to produce same Ampere level.

- - - Updated - - -

This goes along with the idea of thermal runaway in semiconductors.
 

The equation in post #1 (apparently Shockley equation) hides the fact that also the saturation current Is is temperature dependent which results in an overall negative temperature coefficient of Vd at constant diode current. In other words, graph Vd versus Id is shifted to the right with increasing temperature.

I never saw a text book stating otherwise.
 

Thanks, BraththeRad and FvM. This is what I mean:
curva1.png
Yes, it is Shockley eq. I am trying to understand what you say. And I think that if Shockley eq. is a general equation, it should behave in an understable way. I am doing a simple graph, setting values of Temp like I said before, first less T, and then more T, and I get the opposite to the book image I posted here.
 

Your curve is Id versus Vd, opposite to what previously stated.

The behavior is mainly caused by Is temperature dependency in Shockley equation.
 

Ok, let's see what happens with the function. I only graphed it using Geogebra:
analisis 2 de la funcion Id con deslizador.jpg
Edit: Oh, I dont know why this .gif does not animate.
I will try to put it in a web page.
But meanwhile I will post two pictures:
pict 1 de gif.png
pict 2 de gif.png
Shockley equation is a general equation, so I think it does not matter what happens with Is or the dc resistance in a given point, It is simply the graphing of Shockley eq.
And in some books I have read de opposite behavior. That is my big question. I thank you for your attention and comments.
 
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At const Is, the graph as shown by the equation in post #1, moves to the right, and not to the left. (with increase in temp).

As mentioned in post #7, the Is is actually temp dependent. This is not seen in the graph.

But this is not the complete story. Just to confirm, you plot for I vs Vd for negative voltages.

The text book figure you quote does not say that it is a plot of the Shockley equation (it is a more realistic curve).

How does the saturation current depend on T; dig more and see the result.
 
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Here you have my graph with Geogebra, also with negative Vd values:
look: -242 Celsius, (if I do Temp < 273.15 Celsius, the graph inverts to the right, but this is no the question, just saying)
pict 3 de gif.png
then I graph to 25 Celsius:
pict 4 de gif.png
And to Vd<= -14000 V:
pict 5 de gif.png
(Insist: this graph is made by me with Geogebra)

The text book figure you quote does not say that it is a plot of the Shockley equation (it is a more realistic curve).
mmm... ok, ok... This may be the reason. Good point. Thank you.

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I guess what they told me is the same. I will investigate more.
 

look: -242 Celsius, (if I do Temp < 273.15 Celsius, the graph inverts to the right, but this is no the question, just saying)

I do not understand what you mean by "the graph inverts to the right"; if the temp is less than 0C.

You should expand the plot close to zero. The nature of the plot is I(Vd)+Is=Is*exp(alpha*Vd/T)

Is basically is diffusion controlled; and diffusion is also strongly temp dependent (with some exponential terms).

Sorry, I do not have the equation right on hand. But anyway you will always get a nice smooth exponential curve for I(Vd) vs Vd at a given temp.

The equation does not consider zener breakdown either.
 

I do not understand what you mean by "the graph inverts to the right"; if the temp is less than 0C..
Sorry, I meant "if I do Temp < - 273.15". Watch how the curve turns right:
pict 6 de gif.png

you can also tabulate the graph using Excel and with this equation:
=1*10^-12*(EXP( (1.6 * 10^-19 * A1) / (1.38 * 10^-23 * (B1+273.15))) -1)
In A1 cell : 0 And then lets say a range with 0 to 0.9 down
Then in B2 cell a Temperature, lets say 25
And so on.

you will always get a nice smooth exponential curve for I(Vd) vs Vd at a given temp..
Yes, that is what I have done for example using Geogebra and animating the graph.

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Here is the interactive graph made by me in Geogebra, you only have to open this link and you will be able to move the slider to watch the shockley eq. for a range of Temperatures:


https://www.geogebra.org/graphing/pxdzmcbq
 
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Sorry, I meant "if I do Temp < - 273.15"

It is not possible to go below absolute zero. Hence T (in degrees Kelvin; absolute temperature) cannot be negative.

Under some non-equilibrium conditions (lasers during lasing state) T can be negative but that is a different story.

Most diodes will not work at temperatures close to absolute zero.

Basically the graph is y=A*exp(B*x) where A and B are constants but depend on temperature. (A and B are usually positive in this case)

If the absolute temp becomes negative (less than -273C), then the graph flips over. Why? B*x becomes -B*x which is same as B*(-x).

Or you are asking something else??

By the way, the current can be considered as produced by carriers that are in turn produced by the voltage (similar to a discharge tube). These processes give rise to the familiar exponential graph.
 
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It is not possible to go below absolute zero. Hence T (in degrees Kelvin; absolute temperature) cannot be negative.
Yes I know, my comment was only mathematically, analizing the function.
That's the reason I said in post #10
flaxwalter said:
"look: ...the graph inverts to the right, but this is no the question, just saying")

Thank you for your comments.
 
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In other words, when you increase the temperature of a material, the movement of the electrons inside the material also increases, which results more possibility for conductivity. Forward Bias of PN Junction Diode: As you increase the temperature, the intrinsic carrier concentration increases.

[Moderator action: deleted business weblink]
 
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While thermally-generated carriers may increase, thermal
effects such as scattering drive mobility reduction for a
net increase in bulk resistivity as temperature increases.
Lifetime effects (conductivity modulation) that matter
at higher currents, driving down resistivity, also are
temperature-modulated (balance between generation
rate and recombination rate, region by region).

But at the bottom of it all is this: textbooks are not
without errors, neither is our understanding of them.
 

the movement of the electrons inside the material also increases, which results more possibility for conductivity...

Rather the opposite.

At a fixed carrier concentration, increase in temperature will cause more collisions and greater dissipation and higher resistance and lower conductivity.

However, with increase in temperature, the carrier concentration increases exponentially and this causes increase in conductivity. This effect has larger effect compared to the increase in resistance due to dissipation.
 
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