Light-Emitting Diode (LED)

A LED or Light-Emitting Diode is a semiconductor that emits light when there is a specific voltage and current apply across the two terminals of the LED. This is a process called electroluminescence. The symbol of a LED is the same as a regular diode except that there is two or more arrows pointing away from the diode. The arrows are used to indicate that the diode is a light-emitting diode. The positive side of the diode is called the anode and the negative side is called the cathode.

The light-emitting diode needs a specific positive voltage between its anode and cathode to generate lights. This voltage is called the forward voltage (Vf) and will vary depending of the LED model and color. If you exceed the forward voltage between the anode and cathode, the LED will start drawing a lot of current. The LED will burn and can explodes in some cases. The LED has a forward current (If) flowing through it when a specific voltage is applied. In all applications, this current needs to be control so it doesn’t exceed the maximum forward current supported by the LED. If the maximum forward current is exceeded, the LED will burn and can explodes in some cases.

Light-Emitting Diode have a polarity and it is important to respect this polarity to avoid causing damage to the LED. They can withstand a small reverse voltage but they generally have a much lower maximum reverse voltage than a regular diode. This reverse voltage limit is called the breakdown voltage. If you exceed this limit, a high current will flows and the LED will be damage. The LED could burn or even explode in some cases.

Forward current, forward voltage and relative intensity

The forward voltage is the voltage necessary between the anode and the cathode for the diode to be active. If a voltage lower than the forward voltage is applied, the LED won’t work or the light will be dim if you are near the forward voltage. There will be a current called the forward current that will flow through the LED when it is active. In figure 3, you can see the current flowing through the light-emitting diode vs the forward voltage applied between the anode and cathode. This curve is for the QTLP690C LED from Fairchild Semiconductor. The forward voltage is specify at 2V in the datasheet and we can see that over 2V, the curve is linear which means the diode is completely active. Below 2V, it starts to conducts at around 1.75V and will light up a bit but it is not a linear curve. Every model have different curves and you should look in the datasheet of your LED model for this information.

In Figure 4, we have the relative intensity of the same LED model vs the forward current (mA). We can see that the intensity of the LED is mostly linear with the current except below 2mA or less. All light-emitting diode will have theses curve in their datasheet with the information on the maximum current they can support, the forward voltage needed and much more. You should always verify the datasheet of your LED before using it to make sure that you use your LED within its limit. For this specific model (QTLP690C), the maximum continuous forward current supported is 30mA (except for the yellow LED which supports a maximum of 25mA). It is way better to control a light-emitting diode with a current source than a voltage source. A small variation in the forward voltage would break the LED. If we look at figure 3, a voltage of 2.1V applied between the anode and cathode would most likely break the diode since we are over the maximum rating for the forward current of this model. This is only 100mV which is not a lot. Directly applying a voltage to the terminal of a light-emitting diode is not recommended since a small variation of the voltage source could destroy the diode.

Do note that you can exceed the 30mA limit if you are applying a pulse since the 30mA limit is for a continuous current. In most datasheet, the length of the pulse and the frequency will be specify with a maximum current that can be apply for the specific pulse. Exceeding this limit will cause the LED to burn or explodes. Also, the maximum continuous current is going to lower with the ambient temperature and the de-rating curve can be found in the datasheet of your component. It is important to verify this if you are going to operate a LED at high ambient temperature.

LED internal resistance

Light-emitting diode is not a perfect component. The LED has a small internal resistance. For most applications, this resistance can be ignore but in some cases it is important since it will have an impact on how a circuit will behave. In figure 5, we have a simplify model of the internal component of a LED. When a current is flowing through the LED, there will be a voltage drop across the resistor and across the diode.

In most datasheet, the internal resistance is not included and you will need to calculated its value based on the graph showing the forward current versus the forward voltage. To calculated its value, we need to work with the linear section of this graph. The voltage drop across the diode will be a fixed value when working in the linear section. This means that the forward current increasing with the voltage is actually cause by the internal resistance.

To calculate the internal resistance, we will need two points from figure 6. Our first point is going to be at 2.0V and our second point is going to be at around 2.25V since the currents at these voltage are easier to read. Point 1 : 2.0V at 20mA and Point 2 : 2.25V at 60mA. Since we know the voltage drop across the diode doesn’t change with current, the extra voltage will be added to the resistor terminals. For a forward current increase of 40mA, we have an additional voltage drop at the resistor of 0.25V. We just need Ohm’s law to find the internal resistance of the LED. Below, we have the formula that you can use for calculating the internal resistance. Voltage divided by current gives us the internal resistance of the LED.

$Rint=\cfrac{V2-V1}{I2-I1}=\cfrac{2.25V-2.0V}{60mA-20mA}=6.25\Omega$

The internal resistance is important when you are working with a voltage source that match exactly the forward voltage of the LED since you are most likely gonna connect the LED directly to the voltage source without an external resistor. Without an external resistor to limit the current, the current is limited by the internal resistance of the LED. A small change of voltage at your voltage source will cause a massive current to flow into your LED which could potentially destroy your LED if the current exceed the absolute maximum ratings of your light-emitting diode.

For example, if we have a 2V voltage source that is not reliable and sometimes output 2.1V for a little while. We will have an increase of 100mV across our LED. We know that our internal resistor is 6.25 ohms. We can calculate the additional amount of current that is gonna flow into our LED with Ohm’s law.

$I=\cfrac{V}{R}=\cfrac{0.1V}{6.25\Omega=16mA}$

An additional 16mA is going to flow into the LED but this is a big problem. We already had 20mA at 2.0V and now we have 36mA at 2.1V. Do note that this information can be found without doing any calculation, you can look at the curve in figure 6. In the datasheet of the QTLP690C, it is written that it supports a maximum continuous forward current of 30mA. This mean that if the 2.1V stays for long enough, it could potentially destroy this LED.

The solutions to this problem : We have to use a different model of LED with a lower forward voltage than the voltage source or use a higher voltage source so we can use an external resistor to limit the current in both cases. The external resistor should be a couple hundreds of Ohms to protect against a small change in the voltage source. This will protect the LED from a massive current that could flow into the ligh-emitting diode due to a small voltage change at the source. By picking a resistor with a couple hundreds of Ohms, the internal resistance of the LED can be safely ignore since a difference of a couple of Ohms won’t have any major impact on the amount of current flowing into the LED.