We are making Si solar cells (n-type emitter, p-type base), where the rear contact (with base) is made by depositing Aluminum. The purpose is to create a p+ BSF layer and the rear contact altogether by depositing 800 nm Aluminum. To ensure that Al diffuses to the bottom of the base and creates a BSF layer, the samples are put in furnace (1000 degC for 10 min) after Al deposition, and then cooled down. Front contact is 500 nm Ag. The resulting cells give good Voc and Jsc, but the fill factor is consistently low (
First think about physics (material- or device-related reasons for the low fill factor) and only after that about technology (processing steps that lead to the material or deivce properties obtained). Therefore, start your reasoning by looking at the shape of your current–voltage characteristic (dark and illuminated) and think about a possible influence of shunt resistance and series resistance (cf. http://pveducation.org/pvcdrom/solar-cell-operation/impact-of-both-resistances). Also, think about a possible deviation from the superposition principle (after comparing the dark with the illuminated current–voltage characteristic; cf. https://www.researchgate.net/publication/3070940_Application_of_the_superposition_principle_to_solar-cell_analysis). About lateral inhomogeneities: How large are your samples? Would it make sense to investigate them by (lock-in) thermography or electroluminescence?
Article Application of the superposition principle to solar-cell analysis
Dr. Wagner, thanks for your reply. My cells are 2 cm*2cm. The pveducation resource suggests that there is impact of both shunt and series resistance behind the low fill factor. An almost identical J-V curve is obtained with Rs = 7.5 ohm*cm2 and Rsh = 65 ohm*cm2. The cells give uniform resistivity across the surface, so I am not sure whether lateral inhomogeneities are there that caused the low FF. Do you have insights on possible reason behind the Rs and Rsh obtained?
Well, Rs = 7.5 ohm*cm2 for a cell of 2 cm*2cm means an absolute Rs of 1.875 ohm, which is three orders of magnitude too large (compared to industrial standards) for a regular operation.
I suggest to separately investigate front and back contact resistance by "structuring" the contacts for sending current laterally through either region. By this I mean to create isolated spots, e.g. on the Al back side, and measure the resistance between them; on the front side, you could make a TLM measurement (cf. https://en.wikipedia.org/wiki/Transmission_line_measurement and http://tuttle.merc.iastate.edu/ee432/topics/metals/tlm_measurements.pdf) using the grid fingers (which you cut off from the bus bar).
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