The video is very informative and pretty good for people who haven't worked with LED's or building circuits before. BTW, the original link (led.linear1.org) does appear to be dead, but check the description because an updated link is provided (
https://ledcalculator.net/). The ability to suggest LED voltages and currents is particularly good for those who have cannibalized LEDs from somewhere or gotten some without documentation as to their operating specs.
I'd say there's a couple of things that could use clarification...particularly for those that have never used a breadboard or familiar with the way you measure electronic properties (volts, millivolts, amps, ohms...etc).
Breadboards are actually wired internally in order to facilitate the "lego" like construction of circuits. The breadboard has 4 general regions of connections. One on each edge runs the length of the board. Typically at one end of the board you plug your source voltages for the circuits. The 2 middle regions are wired by rows perpendicular to the long edge. This allows you to plug in individual elements into the same "wire" so to speak. Because of the way holes are connected, there is a variety of ways to wire things up and it will work such as the way shown in the video. In fact if you think about that underlying structure you will see how that circuit worked.
Normally though for clarity one thinks of breadboard circuits running across the width of the board so often you will see examples where the circuits straddle the middle with the positive end running along one side and negative to the other. A more detailed explanation can be found at
https://www.sciencebuddies.org/science- ... breadboard (or if this link is dead just search on "how to use a breadboard").
Another thing that may be confusing for the novice is the explanation at the end of how one goes from .435 to 470 Ohms as the best match, and that's because the .435 is not in Ohms. One of the most problematic things when working on low power circuits is that current is never (at least should never) be a substantial value in Amps (1 Amp could kill you). Instead as noted the equation is being divided by milliamps (mA) or 1/1000Amps. 1 Ohm represents a resistor that will drop 1Volt per 1A, or 1V/1A. Understandably mixing the power of measurements can easily throw your calculations off. I recommend converting everything to Ohms, Volts, and Amps before doing any electrical equations to keep things straight rather than trying to compensate after the fact, such as in this case multiplying the result by 1000.*
So a better way to do the equation is turning the 20mA measure into .02A, then dividing 8.7V by .02A which yields 435Ω (Ω is the symbol for Ohm). In addition to make sure you are doing things correctly always put in the values you are actually measuring in the equation as you write it.
So instead of writing R=(12-3.3)/20, write R= (12V-3.3V/20mA), or as I suggested R=(12V-3.3V)/.02A**, that way you make sure you haven't mixed things up either by item (sticking in current where the Voltage goes) or by factor (confusing milliamps for amps). To give you a sense of how important that is, if you had substituted a .5Ω where the 470Ω resistor should have gone you would have certainly blown the LED.
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Postscript notes:
*Converting to different measurement unit values is fairly easy as it is just multiplication or division by powers of 10. Keep in mind the prefixes involved such as milli always refers to 1/1000, whereas kilo means 1000x. In practice LED circuits will work with Volts and Ohms, but only in current be measured frequently as "milliamps" as they are typically very low power type things.
** A bit of a math postscript. Leaving the measurement unit in is also a nice way to check physics equations. Recall I said 1Ω =1V/1A. Also recall that in math multiplying anything by 1 is the same as it not being there, so we can write the same equation as Ω =V/A. If one takes the equation as shown we could get a result of 8.7V/.02A = 8.7(1V) / .02 (1A) = 435 (V/A), then substitute Ω for the (V/A) so you get 435Ω. If you had left it in milliamps you would get 8.7V/20mA =.435(V/mA)...note (V/mA) is not 1Ω , that tells you the math may not be quite right yet (i.e. multiply by 1000 to convert mA value to A...math to show that requires complex fractions and left as a problem for the student
)
