Question

A temperature sensor is built with a semiconductor diode. Assume the bandsap of the semiconductor is 1 eV, independent of temperature. The voltage across the diode was 0.6 V when the temperature was 300K. When the diode is brought into intimate contact with a foreign object, the voltage across the dlode changes to 0.65 V. What is the temperature of the foreign object? HTML

Answer 1

The forward voltage , also called barrier voltage V_D is given by

                                      $V_{D} = \frac{k\hspace{1mm}T}{q} \hspace{1mm}\ln\left ( \frac{N_{A}\hspace{0.5mm}N_{D}}{n_{i}^{2}} \right )$

Since enegrgy gap is independent of temperature, we can write the expression for intrinsic carrier concentration as

                                        $n_{i} = \sqrt{N_{c}\hspace{1mm}N_{v}}\hspace{1mm}e^{-E_{g}/kT}$

So that

                                             $V_{D} = \frac{k\hspace{1mm}T}{q} \hspace{1mm}\ln\left ( \frac{N_{A}\hspace{0.5mm}N_{D}}{N_{c}\hspace{1mm}N_{v}\hspace{1mm}e^{-2E_{g}/kT}} \right )$

At T = 300K, For E_g = 1eV , we have V_D = 0.6V, So we have

              $0.6 = \frac{1.38*10^{-23}\hspace{1mm}*300}{1.9*10^{-19}} \hspace{1mm}\ln\left ( \frac{N_{A}\hspace{0.5mm}N_{D}}{N_{c}\hspace{1mm}N_{v}\hspace{1mm}e^{-(2*1.6*10^{-19}/1.38*10^{-23}*300)}} \right )$

Or

                                    $23.177 = \ln\left ( \frac{N_{A}\hspace{0.5mm}N_{D}}{N_{c}\hspace{1mm}N_{v}\hspace{1mm}e^{-(2*1.6*10^{-19}/1.38*10^{-23}*300)}} \right )$

Or

                                                                $\frac{N_{A}N_{D}}{N_{C}N_{V}} = e^{-54.08}$

So the expression for forward voltage drop is given by

                            $V_{D} = \frac{k\hspace{1mm}T}{q} \hspace{1mm}\ln\left ( \frac{e^{-54.08}}{\hspace{1mm}e^{-2E_{g}/kT}} \right )$

Now we will plot V_D vs T using matlab as shown below:

k = 1.38*1e-23;
q = 1.6*1e-19;
Eg = 1.6*1e-19;
T = (250:0.1:350);
Vd = k*T/q.*log(exp(-54.08)./(exp(-2*Eg./(k*T))));
plot(T,Vd);
grid on;
xlabel('Temperture');
ylabel('V_{D}(V)');

So we find that V_D changes to 0.65V at T = 289.4K.