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I am learning basic electronics with iCircuit. I have build a very simple circuit with just a 5V battery and a LED connected in series, as seen in the image attached.

There is an option called "Fwd Voltage @ 1A" which I do not understand. The normal forward voltage drop for LED is 1.7V - then what does it mean if I enter "1.7V" in this field? Do I automatically add enough resistance in the circuit to limit the voltage across the LED to be 1.7V?

How do I adjust the "Fwd Voltage @ 1A" in real circuitry, or is that something rated/fixed on the LED?

screenshot from iCircuit

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  • \$\begingroup\$ This circuit may work in in simulation but in the real world I would have a resistor in series with the LED. \$\endgroup\$
    – JIm Dearden
    Commented Feb 3, 2014 at 20:34

3 Answers 3

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For an actual diode, you would look up the forward voltage drop on its data sheet, like this one, for example. The value is given in the table of electrical characteristics; also provided is a graph showing how the forward drop varies with the forward current.

Given the desired brightness, you would determine (Figure 3) the corresponding forward current, and look up (Figure 2) the forward voltage drop at this level. Knowing the supply voltage, you can select a series resistor that will provide the required forward current.

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  • \$\begingroup\$ So if take your datasheet for example, to reach intensity of 2 would require 40mA DC forward current, which corresponds to approximately 2.3V forward voltage. If I am supply power from an arduino's 5V, I should select a resistor that can absorb 5V-2.3V = 1.7V? If the resistor is higher than 1.7V then LED would appear dimmer, and if vice versa, brighter? \$\endgroup\$
    – KMC
    Commented Feb 4, 2014 at 7:32
  • \$\begingroup\$ Basically, yes. But first, 5V - 2.3V = 2.7V, so if I may rephrase the question, you would need a resistor that has 2.7V across it when 40mA is flowing through it: a resistance of 68 ohms. The power dissipated in the resistor will be voltage times current, or 108mW, so the resistor must be at least a 1/8 W; 1/4 W would be a better choice. \$\endgroup\$
    – user28910
    Commented Feb 4, 2014 at 12:37
  • \$\begingroup\$ A few additional notes: 1) The graph in the datasheet is for brightness relative to brightness at 20mA. There is a table further down giving brightness (and its variation) at 20mA. 2) If you're using the Arduino to turn on or off the LED (rather than just connecting it across the 5V supply), you need to check and make sure that the output you're using to drive it is capable of sourcing or sinking the 40mA. This info should be available on its datasheet. 3) 40mA is going to be rather bright; for indoor use as an indicator, 10mA should be more than enough. \$\endgroup\$
    – user28910
    Commented Feb 4, 2014 at 12:50
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Like any diode, the relationship between current and voltage in an LED is not linear, then it is common that the manufacturer specifies a certain voltage at a certain current, in tabular form, or by means of a graph.
In the real circuit, the design parameter is the required brightness. With this value, the current required is determined. The voltage drop over the diode is set based on the current level.

For an ordinary LED, 10 mA is a normal operating current.

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  • \$\begingroup\$ For most small leds 20mA is the typical operating current, not 10mA. \$\endgroup\$
    – Passerby
    Commented Feb 3, 2014 at 23:15
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The forward voltage is a a characteristic of the LED - not something you can change. The forward voltage depends mostly on the colour of the LED.

To calculate the current-limiting resistor, you subtract the forward voltage from the supply voltage, and use the resulting voltage in Ohm's Law to calculate the resistance.

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