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I thought I should have some hardware projects on my Github so I decided to start with a fairly simple one. This is a night light that lights up when it's dark enough.
Above is the final schematic for this circuit. There are two parts to it, first is the right side which uses an NPN transistor as a switch. The left side is a voltage divider.
These calculations were done to figure out the required resistance for the LED.
Typically, the forward voltage of an LED is between 1.8 and 3.3 volts
-- Wikipedia
The specific LED I was using has a forward voltage drop of around 2V and needs about 10mA for it to function optimally. Using Ohm's law, we can calculate the required resistance.
$$ R = \frac{V_{power} - V_{forward}}{I} = \frac{6 - 2}{0.01} = 400 $$
I don't have a 400 ohm resistor. 330 is close enough (alternatively 470 would work too but I don't have that either).
When looking at the datasheet for the transistor I used, it says the minimum voltage to turn the base on is 580mV. This means as long as a value greater than 580mV is passing through the base, there will be a current flowing from the collector to the emitter.
The formula for voltage dividers can be found with a quick search.
Above is an image from Wikipedia. Wikipedia also gives us the formula for a voltage divider as
$$ V_{out} = V_{in}\frac{R_2}{R_1 + R_2} $$
In the schematics I drew up, $R_2$ of the voltage divider formula is the photoresistor and $R_1$ from the formula is the 100k resistor. This was determined by knowing the resistance of a photoresistor increases as it gets darker and when it's darker, we want more voltage. Since we want voltage and resistance to be porportional, the photoresistor should be $R_2$. I took a multimeter and measured the value of the photoresistor under normal light conditions. It was about 10k. We know $V_{in}$ is 6V and we want $V_{out}$ to be 0.58V so we now need to solve for the value of $R_1$.
$$V_{out} = V_{in}\frac{R_2}{R_1 + R_2}$$
$$0.58 = 6\frac{10}{10 + R_1}$$
$$5.8 + 0.58R_1 = 60$$
$$0.58R_1 = 54.2$$
$$R_1 \approx 93.4$$
I don't have a 93.4k resistor so I used an 100k resistor.
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