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A co-worker and I were a little bored at work the last week and stumbled our way into playing around with basic circuits. Long story short we have made several simple series and parallel circuit diagrams and we practice working out the math and voltage drops across resistors and what not.

A question we cannot seem to be definitively answered is this; What is the relationship of Voltage Droppedacrossthe LED and the LEDs rated Forward Voltage? Lets say we have a simple circuit:

9V DC Power Source > 350Ω Resistor > Single Red LED with Vf of 2.5V

First we calculated (based on the data sheet of the LED stating 20mA for I) that we would need a 325Ω resistor. (I used a 350Ω above to get the current under the LED rating, 19.444mA specifically).

This gives us a voltage drop of 6.806V at the resistor and 2.194V at the LED.

What is the relationship of Voltage Droppedacrossthe LED and the LEDs rated Forward Voltage?

In addition:

Does the Vd across the LED being LESS than the Vfratingof the LED have anything to do with, well anything?

What would happen if Vd wasmorethan Vf? Would the LED just burn itself out?

Does Vdneedto be less than Vf?

Thanks in advance for what seem to be some simple questions that we're having difficulty getting straight answers to.

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# Forward Voltage / Dropped Voltage / LEDs

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