(i.e. incompatible power supply)
Generally in the market, LEDs are using either 2.8-3.2volt, 12volt, 24-36volt forward. What does that mean?
Simply, it is the required voltage from the power supply to make the LEDs work.
If the voltage supplied is under the LEDs requirement;
1. The LEDs may not light up
2. The LEDs may light up, albeit at a lower brightness or effectiveness as compared to its maximum potential.
If the voltage supplied is over/above the LED requirement;
1. The LEDs may be overcharged and light up brighter than its usual or standard light output.
2. This will reduce the lifespan of the LEDs dramatically, resulting in the LEDs melting or failing quickly. *Do take note not to super charged LEDs without any expertise supervision as it may cause hurt or fire hazard.
So whats the importance of the forward voltage? Shouldn't we be concerned with just the current supplied to the LEDs?
Please refer to the graph below;
Focusing on the curve in the graph, we can see that only when the voltage reaches 1.6v - 1.7v, does the LED start receiving its required current(amphere) to light up. It then quickly reaches 20mA when the voltage reaches 2v. The key lies between 1.5v - 1.6v where the forward voltage is required to open the p-n gate of a diode. A small change in voltage will result in a large effects of the forward current (If).
The key question to ask is; what will be sufficient to push through the pn-gate of a diode?
You can supply voltage from 0v - 1.5v and not result in any noticeable change in the current. But, should you supply a forward voltage much larger than required, it will generate an even larger forward current that not only pushes through the pn-gate, but also melting the LED chip and circuitry quickly due to heat.
A solution to prevent over supply of current is actually to have a current limiting resistor in your circuit after the pn-gate. However, this solution not only cost money, but is also dependant on the space availability within the circuit. That said, the resistor will also heat up resulting in heat dissipation issues. In my view, it is best to ascertain the voltage required by the LED and using appropriate power supply from the beginning.
In conclusion, if i were to supply a 12 volt power to a 5volt LED chipset, there is a high probability that the LED will be burned out shortly.
Above which are purely my opinions in relation to the topic of discussion. Please do not hesitate to post any constructive comments below for future discussions.
- Light-emitting diodes by E.Fred Schubert