To make high-efficiency LED grow lights, it is very important to choose high-efficiency LED grow light diodes. The brands Koray is using include: Osram , Samsung , Seoul, Toyoda, etc.

For the same brand and same model, with different brightness levels, Lux/BIN is very different. Especially with different current and power, the PPE varies greatly.

Let’s take a look at the following chart

As we can see:

The PPE at 65mA 2.71V is 3.02umol/J

Driven at 100mA current, PPE is 2.89umol/J

Driven at 150mA current, PPE is up to 2.72umol/J

Osram red diodes have a relatively high PPE reaching to 4umol/J, which is a very high efficiency.

So how many LED grow light diodes are needed in a grow light for a PPE of 2.7umol/J?

If it's a 430W LED grow light(like Koray G4000), and the efficiency is required to reach 2.7umol/J,

then the PPF should be 430*2.7=1161umol/S,

using the Samsung LM301H SK, PPE 2.89umol/J, 100mA*2.85V=0.285W,

the required quantities of diodes are 1400+pcs.

However, the efficiency of Osram red diodes is much higher than the white light. If we add some red diodes, the number of white diodes can be reduced. The specific numbers, as well as the PPE, the light attenuation of the lens or the protective measures, etc. is needed to be considered.

Koray G4000UR LED grow light has 72 Osram Red/IR diodes, PPE is designed to meet 2.7+umol/J.