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There's something about the LED formula that I don't understand.

It says that if I have one LED with a forward voltage of 3V and a forward current of 20mA then if I want to drive my circuit with a 6V battery, the formula to determine the resistor's value for the LED is:

$$\frac{(6-3)V}{0.02A} = 150\Omega$$

In a circuit with the elements connected in series the amperage is the same anywhere.

So by using the formula differently:

$$\frac{6V}{150\Omega} = 0.04A$$

So does the LED is actually driven by 0.04A?

The formula is of course correct, but I don't understand where my logic is mistaken. Can someone help me?

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  • \$\begingroup\$ In your first calculation you allowed for the 3 volt drop across the LED, but you did not allow for the LED voltage drop in the second calculation. \$\endgroup\$ – Peter Bennett Nov 13 '14 at 19:01
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If you applied less than 3 volts across the LED, virtually no current would flow so, it has to be assumed that there is about 3 volts across the LED and that this causes a current of about 20m to flow. If there is 3 volts across the LED and the power supply is 6 volts, there MUST be 3 volts across the resistor. Given that there is expected to be 20mA flowing, it's a simple case of ohms law to calculate R: -

R = 3 volts / 20mA = 150 ohms.

Take a look how a 2V LED might conduct: -

enter image description here

This is just a picture I took from the web. Below about 1.7 volts it hardly conducts any current and at 2 volts it's taking 20mA. As you can see, you can almost assume that at approximately 2 volts the current could be anywhere between 5mA and 45mA. For this particular LED, from a 6 volt supply, there would be 4 volts across the resistor with about 20mA flowing and this would lead to a resistor value of 200 ohms.

Below is another way of looking at how the series resistor alters the overall impedance of the circuit. Again, this is for a 2 volt LED put in series with a 100 ohm resistor: -

enter image description here

With 100 ohms in series the net resistance of the two components dictates that at about 4 volts applied, the current is about 17mA.

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  • \$\begingroup\$ Hi Andy. Do you mean that the LED will sink 20ma no matter what? Then could we assume that the total resistance in the 6v driven circuit is: 6 / 0.02 = 300 omh? \$\endgroup\$ – Mr Bonjour Nov 13 '14 at 12:52
  • \$\begingroup\$ I've modified my answer with another worked example. The method is approximate but without working out a non-linear formula, the answer you get from the simplified method is better than the likely spread of tolerances on several LEDs in a batch. \$\endgroup\$ – Andy aka Nov 13 '14 at 12:56
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So if I use the formula diferently: 6v/150ohm = 0.04a.

In the above, you're evidently assuming that there is 6V across the 150 ohm resistor. Remember that Ohm's law relates the voltage across and current through the resistor:

$$v_R = R \cdot i_R$$

So, if you want to solve for the current through the resistor, you divide the voltage across the resistor by the resistance.

Since you have a 6V battery and since there is about 3V across the LED over a wide range of current, you can approximate the voltage across the resistor as

$$v_R \approx 6V - 3V$$

Then, to correctly apply Ohm's law, write:

$$v_R = 3V = R \cdot i_R $$

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