Electronic – Matched biasing transistor compensates for temperature changes

If a 0 is applied to a P gate then think of it as a short-circuit. If a 1 is applied to a P gate then think of it as an open-circuit.

Each input, A and B are both going into 1 P gate and 1 N gate each. Now let's walk through the truth table.

A=0, B=0. Both P gates are short circuits and both N gates are open circuits. Therefore C is tied directly to the upper "1" voltage and is completely separated by an open circuit from the bottom "0" voltage, C=1.

A=0, B=1. The P gate B touches is an open-circuit and the other P gate is a short-circuit. C does not have a direct path to the upper "1" anymore. Moreover, the N gate that B is tied to is now a short-circuit which will tie C directly to the lower "0" voltage, C=0.

A=1, B=0. Basically the same as the previous state. You should be able to work it out yourself.

A=1, B=1. See if you can work it out yourself to see why C will equal 0.

Your very first equation is wrong. If you check your link, it should be

Vbe = (kT/q) ln(ic/is)

When you correct your math, the temperature for your circuit is just about 300 K. Actually, assuming a current ratio of 100 gives 302.45 K, but that isn't quite right, and I'm too lazy to solve for the exact values. Plus, I doubt that your experimental error would support any more precision.

And, just as bit of advice, when you are working with measurements which have 3 significant digits, reporting a computation to 16 significant digits suggests very strongly that you have no idea what the limitations of your measurements are. Just a tip.