I have Cu2ZnSnS4 films on soda lime glass and Silicium wafer. I choose Si wafer for correct measurement of elemental analysis. In EDS analysis how can I ignore (or remove) Si from spectrum? Is it reliable measurement for other metals' composition. Any other reliable and robust method will be appreciated (not XRF beacuse I have no XRF facility in our lab).
Commonly this is not a big deal. I am sure that each commertial software is able to exclude single elements. Typically this is "necessary" for carbon, but you can often define any element. At least for the bruker software this works. Another option is to define regions in your spectrum which should not be considered during analysis. If you define such region covering the Si peak it cannot be detected anymore.
1. You can use low accelerating voltages (no signal from deeper layers).
2. You can use another substrate, such as plastic.
I will ignore Si and find composition without Si. Mr Viladimir which voltage value you recommend just for the above elements.
It is depend upon thickness of your sample. You can check penetration depth of electron in your sample by using CASINO software tool.
www.probesoftware.com/Technical.html
I would like to mention that Si K-emission lines do not overlap with any other EDX-signal from Zn, Cu, Sn and S! So it should not be a problem to separate the different elemental contributions!
All the best, Dirk
Hi Dirk Luetzenkirchen-Hecht .
I deposited V,Nb,Mo,Ta,W on Si wafer. The at% of the elements vary with the change in acceleration voltage . About 5 at% difference was observed for 20 and 30 KV. I tried excluding Si peaks. I am confused about which voltage I should take.
How do I explain the change in result?
Do not forget that you will change the penetration depth of the electron beam when changing the primary energy - if you have an inhomogeneous sample, this will definitely affect the quantitative analysis, as any software usually assumes that the sample is homogeneous ... So, be careful. depending on the thickness of your layer, I would try to reduce the excitation energy, but Nb and Mo can be difficult to separate by only considering the L-emissions at around 2-3 keV photon energy. You have to "play" a bit to come to consistent results. Good luck
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