# So here\’s some mathematics for a slow Sunday afternoon

So, what I want to know is, what will be the demand for gallium, germanium and indium if multi-junction solar cells become the industry standard?

I don\’t need accurate figures: to somewhere around an order of magnitude is just fine.

First question is why can\’t I work it out myself? Well, partly because my brain huuurts when I try to do such things but more exactly, because I have absolutely no idea how much of each metal by weight goes into such solar cells.

I can\’t see any information with a quick Google. It\’s certainly not the weight of the solar cell: we\’re only using thin films here. Further, I\’ve absolutely no idea, to even two orders of magnitude, what is the weight of a solar cell nor how many acres makes how many tonnes etc.

But what I\’d like to know is, imagine that the world rolls out, over the coming decades, GaAs, Ge and InP based solar cells, the multi-junction kind (as to why they would they can get over 40% efficiency which solves quite a lot of larger problems). Imagine further that this becomes the major renewable (in concept they\’re more efficient than wind etc).

So, how many tonnes per year are needed of Ga, Ge and In? (As and P are in bounteous supply, even if not as yet at the purities required).

I already know how much of each is currently produced: I want to see what the rise in production might have to be.

Anyone?

## 5 thoughts on “So here\’s some mathematics for a slow Sunday afternoon”

1. I can’t see how they are more efficient than wind in the true sense as they won’t produce at night. Of course if you had enough batteries that might help.

2. Solar PV: not as piss-poor as wind, not as good as gas.

DK

3. Industry standard? What industry, once the subsidies are withdrawn?
So your answer is: zero (except from applications we haven’t thought of yet).

4. I make it 3.19g/m2, or 669.9t of GaA to replace the entire installed global PV with the new multi-layer stuff.

(1) Films – say 4 x 150nm thick film per panel. Density of GaA = 5.32g/cm3.

1cm/600nm = 16,667. (5.32/16,667) x 10,000 to get m2 = 3.19g/m2

(2) Wiki gives 2009 global installed PV as 21,000MW. Say PV produces 100W/m2 – UK mid range – this gives 210,000,000m2 installed

x 3.19 g/m2 = 669.9t

I could be quite wrong …