1) Where and searched

Before we start with the calculation, we must first be clear on what we are looking for and what we already know.

We know...

...what voltage we want in our circuit. This is usually the voltage provided by the transformer unless there are other voltage limiting components or circuits connected in series.

In this example, the operating voltage: V_B = 14 volts

We know...

...forward voltage. The forward voltage is found in the data sheets or in the catalog of the store which sold the LED. Below the conduction voltage, the LED has near infinite resistance. This means that once the forward voltage is reached, the LED has small resistance and has a very high current. Unless the power is limited artificially, this would cause a short-circuit and damage the LED. The simplest way to do this is by mounting a series resistor. Since the current in a series circuit is is equal everywhere this can easily be done.

In this example, the forward voltage of the LED: V_(F) = 1.6 volts

We know...

...forward current of the LED. 

The forward current is the current required by the light emitting diode to be lit. We can obtain this from the data sheets or from the dealer catalog. 

In this example, the forward current of the LED: I_F = 20 mA, ie 0.02 amps

We are looking for ...

...the resistor R_V, which is necessary to protect high current of the LED. Since the LED is a semiconductor, the current without a series resistor would destroy the diode. A semiconductor device has both conductors and non-conductors. Up to the amount of the forward voltage, the diode is a non-conductor having an approximately infinite resistance. Above the forward voltage, the diode is a conductor having an approximately infinitely small resistance. For this reason, it is important to limit the bias resistor R_V to the maximum possible current.

Clearly summarized the whole thing looks like this:

Where:

Operating voltage V_B = 14 V
Forward voltage of the LED V_(F) = 1.6 V
Forward current of the LED: I_F = 20 mA (0.02 A)

Wanted:
Resistor R_V

2)to shred voltage

In a series circuit, total voltage is the sum of individual voltages. 

We need to find out the power to be "destroyed."

For this we simply pull on the fed operating voltage V_B from the voltage required by the LED for its operation. 

In this example, the operating voltage V_B is 14 volts, and the LED forward voltage V_F is 1.6 volt:

voltage V_R, which falls on the series resistor = voltage V_B, at which the circuit is powered - minus voltage V_F , which drops across the diode.

Or in mathematical terms:

V_R = V_B - V_F

V_R = 14 V - 1.6 V

V_R = 12.4 V

3) calculate resistor

The current I_F, at which the diode is to be operated is 20 mA, or 0.02 amps. This figure is obtained from the data sheet of the diode or from the dealer catalog. Since the current in a series circuit is equal everywhere, this same current flows through the resistor R_V.

We have calculated the voltage V_R that falls on the series resistor R_V in the previous step (see point 2). In our example it is 12.4 volts.

Because we now know the voltage V and the current I, we can calculate the resistance R easily with the help of Ohm's law:

resistance R_{V_calculated} = voltage V_R to the resistance - divided by - forward current I_FLED

or in mathematical terms:

R_{V_calculated} = V_R / I_F
R_{V_calculated} = 12.4 V / 0.02 A
R_{V_calculated} = 620 V / A = 620 Ω

Back To The Math: Why are 620 volts / amps = 620 ohms?

We use again Ohm's Law:

Volts = Ohms * Amps
V = R * I | / I (formula change to "R")

Ohms = volts / amps
R = V / I | For "V" we can write "R * I", as we have seen above

R = (R * I) / I | "I" is shortened out, remains:

R = R -> So R = ohms (and not "Volt / Amp").

A mathematical miracle?

No, just Ohm's Law :-)

4) used to determine resistor

In step 3, we have calculated the series resistor as "620 Ω". But you will not find a resistor that has a value of 620 Ω. And before you start to solder several individual resistances in series to add their values, know that it is customarily to use the next higher resistance. In this case, a 680-Ω-resistor. In LED is this small difference does not matter. 

Since we expect 680 Ω flows, we have to calculate the operating current flows I_B and not the forward current I_F (given to us by the light emitting diode).

We calculate this new power again with Ohm's law:

Voltage across the resistor VR(2)/index] = used resistor R_V * Operating current I_B

V_{R_used} = R_{V_used} * I_B | / R_{V_used} (according to formula I_B change)

I_B = U_{R_used} / R_{V_used}

I_B = 12.4 V / 680 Ω

I_B = 0.0182352 V / Ω
= 0.0182352 V / (V / A)
= 0.0182352 A
= 18.23 mA

The light emitting diode according to the manufacturer has a forward current I_F of 20 mA. A slightly higher resistor flows in the circuit, but the actual current is 18.23 mA. For the diode this has no disadvantages. But this value is of interest to those who intend to do greater things with the circuit or may want to hang several of these circuits in a row. For larger circuits, with equivocal results, it may be worthwhile to connect several resistors in series to get the most accurate resistance value.

Depending on the selected tolerance class of resistance (usually recognizable by the last ring of the color coding), this statement can also be calculated by x % with a lower or higher resistance value "R_{V_used}". In our calculation tool these values ​​are also calculated. In this "lesson" we did not include it for the sake of simplicity. 

5) dissipation from resistor determine

Where voltage and current flows, there is power. Power is energy. Energy can not be destroyed but only transformed. For our series resistor means: "This thing is hot!"

Thus the resistance is not destroyed, you must choose the resistance of the appropriate class. For this purpose, we have to figure out once that power falls to this resistance at all.

And since it is also looking impatiently around the corner ... our "Ohm's Law": P = V * I

Loss P_V = voltage across the series resistor V_{R_used} * Operating current I_B

P_V = V_{R_used} * I_B

P_V = 12.4 V * 0.01823 A

P_V = 0.226052 VA
= 0.226052 W
= 226.05 mW

The resistance to be used must therefore be designed for at least a power of 226.05 mW. 

Pretty complicated and extensive, is not it?

But it is actually not a problem to calculate resistors for LEDs because with our tool, you simply need to enter the operating voltage, forward voltage and forward current of the LED and get back the results in less than 1 second. To start our tool, simply click on the tab Calculator.