Day Three - Dusk-to-Dawn Light & BJTs 3/7/17

Hey guys, today we first had our amazing hot dog experiment.

I think the purpose of this experiment is to let us know that the hot dog will burn out eventually, and as the hot dog burn out, there is no more current going through, which implies there is current going through the LED as well.

Secondly, we had some calculation on finding the equivalent resistance of a circuit, and we also use KVL and KCL to find some values with some unknowns.

Here is how we calculate the equivalent resistance. We dissect the circuit into three parts, and we calculate the resistance of each part.

Thirdly, we had our experiment of the light sensor. As we change the light around the photocell, we are changing the resistance of the photocell and thus the current. Please note that there is an inverse relationship between the resistance of the photocell and the light surrounding the photocell.

Since we have resistances of 0.9k ohms and 10k ohms, we use these data for our calculation instead of 5k and 20k.

Note: this is the calculation we have using KVL; however, in our case, we are using 0.9k and 10k ohms instead of 5k and 20k ohms. 

By using 0.9k (exposed to light) and 10k (covered by hand) ohms resistors for the photocell in the calculation, we get Vb = 0.4128v for 0.9k resistor and Vb = 2.5v for 10k resistor.

In our measurement, Vb = 0.4v when exposed to light and Vb = 1.92v when covered by hand.

For the first situation (exposed to light), the voltage difference is only 0.0128v, we can conclude it is accurate from that; however, for the second situation, the voltage difference is 0.58v, which is around 25% of difference, and we conclude that it is inaccurate because this significant percentage difference can be resulting from the different lighting environments during the measurement of the photocell and the actual implementation of the circuit.

Here is a video of the functioning circuit.

Summary:

On day three, we learned to use Bipolar Junction Transistors, which is a current controlled current source in our light sensor device. As the photocell is covered by hand, the resistance of the photocell increases to 10k, and the increase in resistance induces a smaller current flowing through the 10k resistor and the photocell; in addition, as smaller current flow through the 10k resistor, the voltage drop across the 10k resistor is half of the voltage source 5v, which implies there 5 voltage drop across the base and the ground. Since the diode inside the BJT requires 0.7 voltage difference between base and emitter in order to let the current flow from the collector to the emitter, this sufficient voltage controlled by current will induce another current flowing through the transistor, and this is the whole idea of BJT.