Calculate the voltage for a diode

In summary: 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? If you mean I= Is (eeV/kT -1), then you would need to find V that will maintain the same current I.
  • #1
Lavace
62
0

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!
 
Physics news on Phys.org
  • #2
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:
 
  • #3


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!
 
  • #4
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:
 
  • #5


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!
 
  • #6
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:
 
  • #7


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
 
  • #8


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?
 

What is a diode?

A diode is an electronic component that allows current to flow in one direction but blocks it in the opposite direction.

How does a diode work?

A diode is made up of a semiconductor material, typically silicon, with two terminals attached to it. When a forward voltage is applied across the diode, it allows current to flow through it. When a reverse voltage is applied, it blocks the flow of current.

Why is it important to calculate the voltage for a diode?

Calculating the voltage for a diode is important because it helps determine the appropriate amount of forward voltage needed to turn the diode on and allow current to flow through it.

What factors affect the voltage for a diode?

The voltage for a diode is affected by the type of semiconductor material used, the dopants added to the material, and the temperature at which it is operating.

How do you calculate the voltage for a diode?

The voltage for a diode can be calculated using Ohm's law, which states that voltage equals current multiplied by resistance. In the case of a diode, the resistance is known as the diode's "forward resistance" or "dynamic resistance."

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Electrical Engineering
Replies
32
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
788
  • Other Physics Topics
Replies
3
Views
3K
  • Advanced Physics Homework Help
Replies
5
Views
742
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
979
Back
Top