Resistor/Led and Basic Electricity Questions

a7ecorsair

The LED doesn't "draw" voltage from the power supply. In this example, the LED will never be forward biased because the power supply output is too low and therefore the LED will not allow any current flow.

Let's assume you have a variable output power supply that goes from 1 volt to 3 volts and can deliver 500 ma of current.
Hook your 20 ma rated LED up to this PS and start adjusting the voltage upward. There will be no current flow until the LED forward biases. At 2 volts the LED will forward bias and burn up because there will be 500 ma flowing through it.

Khivar

Thanks I understand
I need a small precision in a similar matter but with a different power supply. If I take 2 LR6 1.5V 2600mah battery and hook them up and take a 3V LED 20mA, in this case the LED will be forward biased but what current will it get and will it work ?

mpoulton

Typical LED's have operating voltages from about 1.8 to 3.5V. Voltage is determined directly by the color of the LED. See my earlier post on this, or the wikipedia article on LED's for a decent explanation. A 0.7V LED would be very dark red or infrared and is unusual. A 2V LED would probably be green.

All LED's have a specified operating voltage (discussed in detail in my earlier post), but this does not mean they are intended to be used with a constant voltage supply and thus don't require a resistor. Here's an LED data sheet for a green LED with a rated Vf of 2.1V for example: http://media.digikey.com/pdf/Data%20...ST-C191GKT.pdf

a7ecorsair

Get yourself a variable resistor to work with and use a 12V power supply. Using a 3V supply to illuminate a 3V LED is a poor design.

curiousB

You can make a solid state series resistor that will act as a constant current source using an LM117 voltage regulator and a resistor. For 20mA select R to be approximately 62 Ohms. 56 Ohms is a standard value and that will yield 22mA across a widely varying input voltage up to 30 volts. That might take some of the guess work out of it.