Let you quickly learn the driver matching solution of LED lamps

With the development of the LED industry, major manufacturers have begun to change from miscellaneous to specialized. Engineers only need to match according to the parameters provided by customers. As for how to match? The following will teach you how to design lamps.

        Now let's take an 18" LED tube as an example to analyze the design ideas:

　Let's take the most commonly used 3528 lamp beads as an example. When LED lamp tubes first came out, due to the large price gap, light strip manufacturers all used LED chips with high current consistency and high brightness . The current of this type of chip is generally: 20mA;

        First, calculate the power of a lamp bead: 3528 A lamp bead is composed of 1 chip, voltage: 3.2V; (parallel circuit: voltage remains unchanged, total current is added) If it is a good quality LED chip: the current of a single chip is 0.02A, then the power of a lamp bead is 0.02A×3=0.06W. However, in the actual design of the LED chip on the market: the current of a single chip is designed to be 0.018A (considering the margin), then the power of a lamp bead is 0.018A×3.2v=0.0576w.

　　Second, we must first consider whether the lamp is isolated or non-isolated. Power W calculation = voltage V × current A, so we can first roughly estimate how many lamp beads are needed: 18/0.0576=312.5, about 312 lamp beads

　　Third, we assume that it is isolated, and the voltage is generally less than 36v, that is, 36/3=12 series. Non-isolated is generally set to 24 series. 312/12=26 parallel; 312/24=13 parallel. In this way, we can preliminarily determine two solutions: 12 series 26 parallel isolation and 24 series 13 non-isolated.

　　Fourth, we can choose the driver solution based on this: First, for this kind of engineer, you can communicate directly with the driver factory, ask them to adjust the input power to 18w, and tell them your lamp bead string and parallel method. If you are facing sales, you can tell you that the current and voltage of the LED driver output are as follows: Isolated solution: voltage V=36vI=0.18*26=470MA Non-isolated solution: voltage V=72vI=0.18*13=235MA

　　Fifth, it is worth noting that the above is an ideal design, but in reality we often have to consider a lot when designing, such as cost. First of all, the number of our lamp beads is 312, and the power of the lamp board is already 18w, so the power of the lamp must be greater than 18w. So at this time, we have to consider the efficiency and pf value of the driver: if the power of the power supply is 18w, the pf value is 0.95 and the efficiency is 0.85, then we will design the pcb board according to the power supply.

　　The method is as follows: the power of the light board = 18*0.85 = 15.3w, the number of equal numbers = 15.3w/0.0576 = 265.6, which is about 266. But considering 12 strings, we set the number of lamp beads to 264. This is another two solutions: 12 strings and 22 parallels and 24 strings and 11 parallels. Of course, it can also be 11 strings and 24 parallels and 22 strings and 12 parallels. As long as the isolated voltage is less than 36v, it will be fine. This is also a reverse design idea. Generally, the two are combined.

　　In this way, your work is basically completed, and all that remains is to open the light board of this solution.