"the resistor will block a percentage of the voltage in a circuit"
This is a useless general statement. It might apply for a specific circuit but otherwise ignore it. Perhaps it is the output of Google Translate.
usually we talk about drawing current and dropping voltage
if an article says 'blocking' it is just plain odd usage. A capacitor blocks DC.
I am surprised your LED survived with a 100 ohm resistorm as there was roughly 95mA flowing through it. The usual LED specifies a max of 30mA [continuous]. [Yes, I know high-power LEDs can handle much more].
I look at it this way. A low value resistor is more about dropping voltage but as the value is increased it is more abouting limiting current. View attachment 101784
Try for yourself. I used Excel. A 10V power supply, with two resistors. Vary the value of one resistor - I varied R2. At a low value R2 is dropping voltage [and limiting current], but at higher values it is more about limiting current.
But the voltage rating of the LED is used in the calculation of the resistor value.
The resistor value is the voltage across it divided by the current you need.
No.Exactly, at this point I need the LED's amps to figure out the resistor. I.e. for the 12v supply and the LED being 20mA, I need 600 Ohms resistor. Correct? Rated voltage of LED should concern me as far as it does not exceed the 12v, so it's not underpowered. Is this train of thought correct?
No.
You are incorrect. The voltage rating of the LED is also supposed to be used in the calculation of the resistor value.
The LED has a voltage across it. The LED is in series with the resistor so the resistor has voltage across it that is LESS than the supply voltage. The current is determined by Ohm's Law to be the voltage across the resistor divided by the resistor value.
If the LED is a white one with a voltage range of 3.0V to 4.0V then the maximum voltage across the resistor is (12V - 3V=) 9V. Then the resistor value is (9V/20mA=) 450 ohms. 470 ohms is the nearest standard value then the current is (9V/470 ohms=) 19.2mA. If the LED voltage is actually 4.0V then the voltage across the resistor is (12V - 4V=) 8V and the current will be (8V/470 ohms=) 17.0mA which will look almost as bright as if it had 19.2mA.
The range of LED voltage affects the current more if the supply voltage is much lower like 5V.
Then the maximum voltage across the resistor is (5V - 3V=) 2V. Then the resistor value is (2V/20mA=) 100 ohms. If the LED voltage is actually 4V then the voltage across the resistor is (5V - 4V=) 1V and the current will be (1V/100 ohms=) 10.0 mA which will look a little dimmed.
If it's no trouble, would you explain the process of how would I go about connecting four 3v LEDs (20mA) in series with 12v supply?
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