zachtheterrible
Active Member
Just wanna make sure that there's nothing more to this before I go blowing up my 1W LED, if I have 4.5 volts, and I have a 1W LED (330mA, 3.3v rated), I just use ohm's law to calculate a resistor of 13 ohms, correct?
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stevez said:Zach-for future reference (so you only have to remember Ohm's Law) look at the LED/resistor this way.
You have 4.5 volts supply and an LED that is rated at 3.3 V/330 ma. You'd like 330 ma of current to flow when 4.5 volts is applied to the LED/resistor.
In a series circuit the flow of current is the same thru all components so in this case it will be 330 ma.
In a series circuit the total voltage drop across both elements is the total applied voltage - in this case it's 4.5 volts. Each component in the series circuit will have a voltage drop across it and the sum of the voltage drops across each component will total the applied voltage. You've said that 330 ma will flow when the voltage drop across the LED is 3.3 volts. That means that the voltage drop across the resistor is 1.2 volts. Since we said that the current flow thru all elements is the same (in a series circuit) then we know the resistor has 330 ma flowing - 1.2 volts and 330 ma is a 3.64 ohm resistor.
This may seem trivial compared to just cranking thru the formula but I though it might help to understand.