Close

More on artificial lighting

A project log for Growing vegetables in sealed containers

I want to work out whether or not it is possible to grow vegetables indoors in sealed, airtight containers.

will.stevenswill.stevens 04/13/2018 at 21:010 Comments

In an earlier log entry I made about artificial lighting, I wasn’t sure how efficient the LEDs I was using were, and so I didn’t really know how much light the plant that was under them was receiving.

A lot of LED datasheets don’t contain data directly about the efficiency of the LED – i.e. how much light energy the LED outputs per unit of electrical energy.

I found this useful post about how to calculate/estimate the efficiency of an LED from the datasheet parameters: https://electronics.stackexchange.com/questions/325949/how-can-i-estimate-the-optical-power-that-a-single-color-led-generates

For example, the red LEDs I had been using for previous experiments is a Cree C503B-RCS 624nm Red 5mm dome LED.

The forward voltage is 2.1V, so with a forward current of 20mA the LED consumes 42mW

The luminous intensity is 6.6cd. The 50% power angle of the LED is 30 degrees, so lets say that the intensity is on average 4.95cd within the 30 degree angle, and ignore anything outside of this.

The solid angle is 2*pi*(1 – cos 15 degrees) = 0.214 steradians, so the luminous flux of the LED is 4.95*0.214 = 1.06 lumens.

The luminousity function at 624nm is 0.333 (obtained from http://www.ies.org/definitions/table-134-definitions/table-134-definitive-values-of-the-special-luminous-efficiency-function-for-photopic-vision-v/ ). The radiant flux is therefore 1.06 / (683*0.333) = 4.66mW, so the efficiency of the LED is 4.66/42 = 11 percent.

Realising that this LED wasn’t very efficient, and also can’t tolerate sustained currents of more than about 30mA, I ordered some more efficient LEDs that can take a higher current: the OSRAM OSLON Signal 120 LJ CKBP JZKZ 25-1-35. The datasheet gives the luminous intensity, but it also gives the luminous flux, and in one place it gives a figure for efficiency, so I can use these to check my estimates. The calculations below all assume a forward current of 350mA because that’s what the datasheet parameters are based on, but when I use this LED I’ll actually be using less current than this.

The forward voltage is 2.15V, so with a forward current of 350mA the LED consumes 753mW

The luminous intensity is typically 21.8cd at 350mA. The 50% power angle of the LED is 125 degrees, so lets say that the intensity is on average 16.4cd within the 125 degree angle

The solid angle is 2*pi*(1-cos 62.5 degrees) = 3.38 steradians, so the luminous flux of the LED is 16.4*3.38 = 55.4 lumens.

The data sheet doesn’t give the typical value for the luminous flux, but it gives a minimum and maximum value, and the average of these is 66 lumens, so my estimate is low (perhaps because I haven’t accounted for anything outside the 125 degree angle) but not too far off.

The luminosity function at 625nm is 0.321, the radiant flux is therefore 66 / (683*0.321) = 301mW, and the efficiency of the LED is 301/753 = 40 percent.

In one place in the datasheet (under text for the “Electrical thermal resistance junction / solder point”) an efficiency figure of 38% is mentioned, so my calculation is not far from that.

Discussions