Totally new to electronics but my son wants to learn so we endeavoring to do so together. I hope someone can validate or correct my question regarding sizing or resistor for a single led, as well as the impact of changing the resistor.
Power supply = 5.15V / 2.5A
LED = 3.0Vf / 20mA
As I am trying to fully grasp Ohms law, I came up with the following:
V=IR
I=0.02A ( from the LED spec)
V=2.15V (5.15 from the supply - 3.0 drop from LED spec)
So R=0.02/2.15
R=107.5 Ohms
Common sizes of resistors near this seem to be 110 and 120 Ohm, and I'd probably go with the 120 to give a little safety margin given its an unregulated supply and may fluctuate.
Question 1 - is my thought process correct so far?
Now, I don't have any resistors of that size on hand but I do have a 210 Ohm on hand. I confused myself trying to figure out what this would do. If the resistance is now fixed, but I am not sure which would change - voltage or current. My gut says voltage would stay at 2.15 and it is current that would change. So that means the I would now solve for I:
I=V/R
I=2.15/210
I=10.24mA
Running almost half the current would result in a much dimmer led is my assumption and the voltage would not be changed at all due to the different resistor size.
Question 2 - do I have this part right?
Thanks for putting up with some very basic questions.
Power supply = 5.15V / 2.5A
LED = 3.0Vf / 20mA
As I am trying to fully grasp Ohms law, I came up with the following:
V=IR
I=0.02A ( from the LED spec)
V=2.15V (5.15 from the supply - 3.0 drop from LED spec)
So R=0.02/2.15
R=107.5 Ohms
Common sizes of resistors near this seem to be 110 and 120 Ohm, and I'd probably go with the 120 to give a little safety margin given its an unregulated supply and may fluctuate.
Question 1 - is my thought process correct so far?
Now, I don't have any resistors of that size on hand but I do have a 210 Ohm on hand. I confused myself trying to figure out what this would do. If the resistance is now fixed, but I am not sure which would change - voltage or current. My gut says voltage would stay at 2.15 and it is current that would change. So that means the I would now solve for I:
I=V/R
I=2.15/210
I=10.24mA
Running almost half the current would result in a much dimmer led is my assumption and the voltage would not be changed at all due to the different resistor size.
Question 2 - do I have this part right?
Thanks for putting up with some very basic questions.