At 80°F the contact potential for a given p-n junction is 0.026 V. If the temperature is raised to 180°F, the increase (mV) in the contact potential will be:   ________________

The equation for the contact potential, also known as the built-in potential, for a p-n junction is given to us on pg. 381 of the FE reference handbook and is as follows:

From the equation, we can see that the contact potential is linear based on temperature (in Kelvin). Let's rewrite the equation to make this clearer:

$V_0=\left(T\right)\left(\frac{k}{q}\ln \frac{N_a{N}_d}{{n_i}^2}\right)$

Since in the equation T is represented in Kelvin (K), we want to convert our values.

$$\begin{gathered} °K = (°F+459.67)*\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right) \\ °K = (80+459.67)*\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right) = 299.82 \\ °K = (180+459.67)*\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right) = 355.37 \end{gathered}$$

Since the only values that will be changing in our equation are the contact potential $V_0$ and temperature $T$,  we can plug in our given values and solve for the constants:

$$\begin{gathered} V_0=\left(T\right)\left(\frac{k}{q}\ln \frac{N_a{N}_d}{{n_i}^2}\right) \\ .026 = (299.82)\left(\frac{k}{q}\ln \frac{N_a{N}_d}{{n_i}^2}\right) \\ .0000867 =\frac{k}{q}\ln \frac{N_a{N}_d}{{n_i}^2} \end{gathered}$$

We now have a numerical value for the part of the equation we know will remain constant while the temperature changes. We can now plug in our new temperature, along with this numerical value we found in our previous step, to find the new contact potential:

$$\begin{gathered} V_0=\left(T\right)\left(\frac{k}{q}\ln \frac{N_a{N}_d}{{n_i}^2}\right) \\ {V}_0=(355.37)(.0000867) \\ {V}_0=.03081V=30.81mV \end{gathered}$$

Now, taking the delta between the new and original contact potentials:

$∆V_0=30.81mV - 26mV=4.81mV$

Therefore, the increase in contact potential will be approximately 4.81mV.