How to calculate the PPF of your grow light.

Hi RIU!

I'm new here and was hoping some of you LED experts might help me with something.

I was watching a YouTube vid (

) by Growmau5 (great LED YouTuber btw as I'm sure you all know) on the Vero 29 COB's and how efficient they are in Micromoles/per Watt and how you can use this to calculate the 'Photosynthetic Photon Flux' (PPF) and 'Photosynthetic Photon Flux Density' (PPFD) in a given grow space based on your light setup and the area your growing in to then work out the number of micromoles/per watt/ per second that set up will give you. Growmau5's explanation on how you calculate the PPF just didn't add up for me. Now I'm sure he's quite correct and I'm missing something but...

As I understand it you calculate the PPF by using the total input wattage of the driver x the COB's efficacy but this doesn't make sense to me.

Why is the input wattage used to calculate the PPF and not the total wattage being used by the actual COB's? My brain (which is usually wrong to be fair) wants to use the total wattage that the COB's are utilising instead of the total input wattage of the driver as isn't this the actual amount of energy being output by the COB's not the total input wattage.

To give an example using the input wattage of a driver as in the video:

Using six Vero 29 SE D's at 1400ma and 37.6v with a 240v Maxwell HLG-240H-C1400 driver with an input wattage of 250.6w then PPF would equal (Input wattage of the driver x Efficacy of the COB(1.91 micromoles per watt)) 235.56w x 1.91 = 449.92 PPF.

Now if you removed all the COB's except one the calculation would remain the same as the input wattage of the driver and the efficacy of the COB are unchanged giving you the same answer even though only one COB is now being used, which has to be wrong?

I must be missing something as it seems to me that the formula for PPF given in the video can only give you the maximum PPF that a light can put out when all of the input wattage is utilised. Does the efficacy number change if the input wattage of the driver is not fully utilised?

Now using the actual wattage used by the COB's instead of the input wattage of the driver I get these answers with the same scenario:

Six Vero 29 SE D's at 1400ma and 37.6v with a 240v Maxwell HLG-240H-C1400 driver (Output wattage of 6 COB's x COB efficacy) = 225.6w x 1.91 = 430.9 PPF. Which is pretty close to the answer given above.

Now the same formula with just one COB: (Output wattage of one COB x COB Efficacy) 51.38w x 1.91 = 98.14 PPF. This makes way more sense to me but I'm struggling to believe that Growmau5 has this wrong and believe I must have missed something.

Any helpful comments would be greatly appreciated.

PPF is the total amount of photons of light energy in umoles that a light source produces. The efficacy(or efficiency) of a luminaire is based on the conversion of energy in joules/sec(also called watts) to photons of PPF so the total amount of photons for an array powered by a driver would be the product of the efficacy and the power consumed by the array. The amount of photons can be calculated based on the actual energy consumed BY the chip so the power loss from driver efficiency isn't part of that calculation.
So, if you know the efficacy of your chip @ your drive current then you can extrapolate the PPF from that.

Mick The Lick said:

Output wattage of 6 COB's x COB efficacy) = 225.6w x 1.91 = 430.9 PPF.

Click to expand...

Of what value is PPF? It's got nothing to do with horticulture. Flux is a lighting parameter.

Why would you need to convert the output at each wavelength from photometric lumens to radiometric watts. Then convert the watts to Moles? What does that get you?

Converting PPF to PPFD is nearly impossible.

The radiometric SPD in the data sheet looks like this:



In photometric lux it looks like this.



After each lux wavelength above is converted to PPFD it looks like this.



Once you have the PPF it's pretty much useless. what you need is PPFD which is the amount of light in µmol/m²/s that reaches the plants. PPF will only give how many photons leave the LED each second.

Lumens and PPF are typically measured using an integrating sphere.

What is being measured is the flux which is number of photons that exit the LED each second.



To get the irradiance (lux, W/m²/s, or PPFD µmol/m²/s) which is the number of photons that will illuminate an area at some specified distance (i.e. the height of the grow light from the canopy) each second.

The formula to convert PPF to PPFD is PPF / Area

But the Area is not the canopy. It's 4⋅π⋅r² steridians which is the surface of a sphere.
r is the radius of the sphere.

If the LED is 3,000 lumens the lux = 3000 ÷ 4⋅π⋅r².
Use 1 meter for the radius r.
3000 ÷ 12.566 = 238.7 lux over an area of 12.566 m²

Then things start getting difficult. Then you have to estimate the percentage of photons exiting the LED reach the canopy. To get that you have to know the height of the grow light and Radiation Pattern (in Bridgelux terminology) from the datasheet.

If you were to convert the Radiation Pattern which is spherical like:



Which needs to be converted to a flat earth by extending the distance with the Inverse Square Law


For a Vero 29 that looks like:

The two inset images are the spherical Radiation Pattern and flattened pattern.



Then you have to take the spherical surface from way above and flatten it like flatten a globe to a flat map.




You probably don't need me to go on. I can if you please.

GrowLightResearch said:

That is not how it works.

Click to expand...

thats exactly how it works.

the 1.91 umol/J is a direct measurement and already includes all the math which you provided to convert a given spectrum to PAR.

all the extra info you provide about irradiance patterns is extraneous to his question of determining PPF. nobody asked about PPFD

CobKits said:

thats exactly how it works.

Click to expand...

Sorry I was wrong, I misinterpreted the conversion factor.

CobKits said:

nobody asked about PPFD

Click to expand...

What use is PPF?
Plants need PPFD not PPF.

CoBs still suck compared to strips.

GrowLightResearch said:

What use is PPF?
Plants need PPFD not PPF

Click to expand...

Thanks for the reply and sorry for my general ignorance, I'm very new to all this.

From Growmau5's vid he used the PPF to calculate what I assume is the average PPFD being output from the lighting array by dividing it by the area of the grow room/tent as you set out in your post. While this obviously doesn't give you the exact spread of PPFD for a given area which I think is what you were going into and looks pretty tricky, My limited understanding does suggest that it gives you a ballpark figure to work with which might be helpful in deciding things like whether to supplement your space with CO2 and the like.

GBAUTO said:

PPF is the total amount of photons of light energy in umoles that a light source produces. The efficacy(or efficiency) of a luminaire is based on the conversion of energy in joules/sec(also called watts) to photons of PPF so the total amount of photons for an array powered by a driver would be the product of the efficacy and the power consumed by the array. The amount of photons can be calculated based on the actual energy consumed BY the chip so the power loss from driver efficiency isn't part of that calculation.
So, if you know the efficacy of your chip @ your drive current then you can extrapolate the PPF from that.

Click to expand...

Thanks, GBAUTO. So I'm right in thinking that the input wattage of the driver isn't relevant and what you need is the total wattage actually being used of the LED's/luminaire and it's efficacy?