5
\$\begingroup\$

Diode voltage changes with temperature according to the Shockley diode equation.

I understand this , but what I don't understand is why diodes have a negative temperature coefficient (why voltage decreases with temperature.)

The temperature (T) is in the numerator of the equation so it should increase.

diode voltage equation

\$\endgroup\$
7
  • \$\begingroup\$ Is is an exponential function of temperature. \$\endgroup\$ Jun 3, 2018 at 2:26
  • \$\begingroup\$ Is has some dependence in temperature but it have little effect in Vd , Not effective as the Term "T" in this equstion \$\endgroup\$
    – isam
    Jun 3, 2018 at 2:50
  • \$\begingroup\$ @isam What you are missing in that equation is the temperature dependence of \$I_\text{S}\$. It's huge. So large in fact, that it changes the sign. See my writing here. The approximate equation looks like this: $$I_S\left(T\right)=I_{S_{T_{nom}}}\cdot\left[\left(\frac{T}{T_{nom}}\right)^{3}e^{\frac{E_g}{k}\cdot\frac{T-T_{nom}}{T\cdot T_{nom}}}\right]$$ The power of 3 is approximate and is sometimes treated as an "adjustable" parameter. But close. \$\endgroup\$
    – jonk
    Jun 3, 2018 at 3:00
  • \$\begingroup\$ @jonk well that change a lot , this is the first time i see this equation , most articles i read just write schockley equation alone , it looks like a lot of physics is involved so they stay floating \$\endgroup\$
    – isam
    Jun 3, 2018 at 3:39
  • \$\begingroup\$ I think my link provides references. But you are right. It's in more esoteric published papers where you find these details. You asked a good question. I'll have to +1 for that. \$\endgroup\$
    – jonk
    Jun 3, 2018 at 13:06

1 Answer 1

7
\$\begingroup\$

Preceding the question

As I'm sure you can find repeated in many places, the simple Shockley diode equation is:

$$I_\text{D}=I_{\text{SAT}\left(T\right)}\left(e^\frac{V_\text{D}}{\eta \:V_T}-1\right), \text{where }V_T=\frac{k\: T}{q}$$

The emission coefficient, \$\eta\$, is usually taken to be 1, by default. The saturation current \$I_{\text{SAT}\left(T\right)}\$ is effectively a \$y\$-axis intercept point found by projecting the log-chart plot of diode current vs diode voltage. The -1 term in the equation removes this offset and brings the results into expectation where the diode voltage is \$0\:\text{V}\$.

Solving for \$V_\text{D}\$:

$$V_\text{D}=\eta\:V_T\:\operatorname{ln}\left(\frac{I_\text{D}} {I_{\text{SAT}\left(T\right)}}+1\right)\approx \frac{k\: T}{q}\:\operatorname{ln}\left(\frac{I_\text{D}} {I_{\text{SAT}\left(T\right)}}\right)$$

Just as you write.

Your question is about why, if this is all there is, that there the sign of the derivative of diode voltage \$V_\text{D}\$ with respect to temperature \$T\$ is negative, rather than positive.

It's a good question.

Answer

The missing element is that the commonly found Shockley equation is short-hand. (I've been hinting about it, above, with the use of \$I_{\text{SAT}\left(T\right)}\$, earlier.)

As it turns out, the temperature dependence of \$I_{\text{SAT}\left(T\right)}\$ is huge. The approximate equation looks like this:

$$I_{\text{SAT}\left(T\right)}=I_{\text{SAT}\left(T_\text{nom}\right)}\cdot\left[\left(\frac{T}{T_\text{nom}}\right)^{3}e^{\frac{E_g}{k}\cdot\left(\frac{1}{T_\text{nom}}-\frac{1}{T}\right)}\right]$$

\$E_g\$ is the effective energy gap (in eV) and \$k\$ is Boltzmann's constant (in appropriate units.) \$T_\text{nom}\$ is the temperature at which the equation was calibrated, of course, and \$I_{\text{SAT}\left(T_\text{nom}\right)}\$ is the extrapolated saturation current at that calibration temperature.

This formula heavily depends upon fundamental thermodynamics theory and the Boltzmann factor, which you can easily look up and is above represented by the factor: \$e^{\frac{E_g}{k}\cdot\left(\frac{1}{T_\text{nom}}-\frac{1}{T}\right)}\$. It's based on the simple ratio of the numbers of states at different temperatures; really no more complex than fair dice used in elementary probability theory. Perhaps the best introduction to the Boltzmann factor is C. Kittel, "Thermal Physics", John Wiley & Sons, 1969, chapters 1-6 in particular.

The power of 3 used in the equation above is actually a problem, because of the temperature dependence of diffusivity, \$\frac{k T}{q} \mu_T\$. And even that, itself, ignores the bandgap narrowing caused by heavy doping. In practice, the power of 3 is itself turned into a model parameter.

The power of 3 is approximate and is sometimes treated as an "adjustable" parameter. But close.

Summary

Once you realize the fuller form of the Shockley equation, you are now able to work out the details about why the temperature variation is so different -- not only in magnitude, but also in sign. The thermal voltage component would normally imply a positive sign. But the \$I_{\text{SAT}\left(T\right)}\$ factor overwhelms this response and changes the sign, itself.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.