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uaefame

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Hello everyone,

I started learning about LED, thing i learned are: not all LED have same forward voltage drop and it change according to current input.

For example Yellow LED 5mm has a voltage drop of 2.1V at 20mA at 1 intensity.

Q1> Is my calculation for selecting resistor for LED is true or wrong.

R=(5V-2.1V)/20mA=145ohms
I will select a near value resistor = 150ohms

Q2> I was wandering about the different and the use of these:
1- degree beam angle for IR LED
2- degree angle of half intensity for IR LED
 
In your first queation the LED current will be 19.3mA only if the 5V is actually 5.00V, the 150 ohms is actually 150.0 ohms and if the LED's forward voltage is actually 2.1V.

In your second question the angles are the same.
A "bright" cheap LED is an ordinary cheap old LED that has focussing into a very narrow angle. A good modern LED is bright over a wide angle.
 
You equations are absolutely correct. The only "errors" are really nothing but hair-splitting arguments. You end up having to choose a standard value resistor, so you must consider if you just want to get what we call a "ball park" value (close value) or if the value of current you're calculating for is the absolute maximum value that the LED can take. If the former, you can choose the nearest standard value of resistor after the calculation is made. For the latter, choose the next-higher standard value for the calculation to insure that the actual current does not exceed the maximum value.

With high-brightness LEDs, using a current that's quite a bit less that the maximum value can be power saving and LED-saving without sacrificing much light output.

The color of the LED is probably one of the greatest influences on the forward voltage drop because the die chemistry is different, much like the difference between germanium and silicon diode forward drops. When you're dealing with multiple-color LEDs, the forward voltage is going to change depending upon which color is selected, so you may have to allow for that.

Dean
 
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