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通过蓝光发光二极管(LED)芯片激发红、绿两种荧光粉制作白光LED,首先保持两种荧光粉的配比不变,依次从2700~13000 K改变相关色温,发现在某一色温段时,显色指数达到最高。然后依次改变两种荧光粉的配比,重复试验,发现不同的荧光粉配比,达到最高显色指数所对应的色温段不同。试验结果表明,通过合理匹配红、绿荧光粉和硅胶三者之间的比例,可以实现在2700~13000 K之间的任一色温区,显色指数均能达到90以上,在4000 K以下的低色温区,显色指数可达到96。基于此,通过选择和匹配LED蓝光芯片、荧光粉的激发、发射波长,以及它们之间的比例关系,可以实现在任意色温段使显色指数最大化的白光LED光谱设计。
Through the blue light-emitting diode (LED) chip excitation red and green phosphors to produce two white LED, first of all keep the ratio of the two phosphors unchanged, followed by 2700 ~ 13000 K to change the relevant color temperature, found in a color temperature segment, Color rendering index reached the highest. Then, the proportions of the two phosphors are changed in turn, and the test is repeated, and it is found that the different color phosphors have different color temperature ranges corresponding to the highest color rendering index. The experimental results show that by matching the ratio of red, green phosphor and silica gel reasonably, any color temperature range from 2700 to 13000 K can be achieved, and the color rendering index can all reach above 90 and below 4000 K Low color temperature, color rendering index can reach 96. Based on this, by selecting and matching the LED blue chip, the excitation of the phosphor, the emission wavelength, and the ratio between them, it is possible to achieve a white LED spectrum design that maximizes the color rendering index at any color temperature.