A light emitting diode (LED) transfers electrical current into light. A typical LED works well with two volts (2V) applied across the device. Any more, and the current becomes dangerously high and can damage the part. The relationship between the voltage and the current is plotted below. Some example points are:

Consider the circuit in Figure 0.4.2. The voltage supply is nine volts (V_in = 9V), and the LED voltage will be around 2V (V_led = 2V). This means there is 7V across the resistor.

The following analysis is based on Ohms law (once it is arranged to solve for resistance):

Also, this analysis is based on Kirchoff's current law, which states the current flowing into a junction must equal the current flowing out. Since only two devices are attached at the V_led node, the resistor and the LED, the current flowing out of the resistor must be equal to the current flowing into the LED.

If we want to set 1mA through the LED (to provide a nice light), and since the current through the LED is equal to the current through the resistor, then the resistance can be calculated.

This is an approximation; because the voltage drop across the diode will not necessarily be exactly 2V, the voltage across the resistor is not exactly 7V. However, the voltage difference will probably not change the current significantly. For example, based on the number above, at 1mA the voltage drop across the diode is actually 1.9V, which means the voltage across the resistor is really 7.1V. However, our approximation gets us within a couple of percent of the exact answer:

Put 1.8V across the LED, and 15uA will go through the diode

Put 1.9V across the LED, and 1mA will go through the diode

Put 2.1V across the LED, and 6A will go through the diode

Put 1.9V across the LED, and 1mA will go through the diode

Put 2.1V across the LED, and 6A will go through the diode

Figure 0.4.1: LED voltage-current relationship

At 15uA, the LED light will be barely visible. At 1mA, the LED light will be reasonably bright. At 6A, the LED will release smoke and never work again. One way to look at this is that over a broad range of current, the voltage will be about 2V.Consider the circuit in Figure 0.4.2. The voltage supply is nine volts (V_in = 9V), and the LED voltage will be around 2V (V_led = 2V). This means there is 7V across the resistor.

Figure 0.4.2: LED circuit

The following analysis is based on Ohms law (once it is arranged to solve for resistance):

Also, this analysis is based on Kirchoff's current law, which states the current flowing into a junction must equal the current flowing out. Since only two devices are attached at the V_led node, the resistor and the LED, the current flowing out of the resistor must be equal to the current flowing into the LED.

If we want to set 1mA through the LED (to provide a nice light), and since the current through the LED is equal to the current through the resistor, then the resistance can be calculated.

This is an approximation; because the voltage drop across the diode will not necessarily be exactly 2V, the voltage across the resistor is not exactly 7V. However, the voltage difference will probably not change the current significantly. For example, based on the number above, at 1mA the voltage drop across the diode is actually 1.9V, which means the voltage across the resistor is really 7.1V. However, our approximation gets us within a couple of percent of the exact answer:

worst case offset

Technically, then, we would like a 7.1K resistor to set 1.9V across the LED. However, before one gets too pedantic, remember that:

Technically, then, we would like a 7.1K resistor to set 1.9V across the LED. However, before one gets too pedantic, remember that:

- They don't actually make 7.1k (or even 7k) resistors. The closest "standard" values would be 6.8k or 7.5k. It is possible to get closer with precision resistors or combinations of resistors, but...
- The 1mA is really just an order of magnitude set for power consumption- there is nothing magical about hitting that current level exactly right, especially since...
- The actually voltage of a "9V" battery will be between 7V - 10V depending on how old the battery is, so being precise is probably moot.