I studied the transport of majority carriers across the grain boundaries in polysilicon both experimentally and theoretically. In spite of the importance of this work no researcher looked at it.
Newly, this subject matter became alive again because of the micro crystalline and polycrysatline solar cells. I noticed this as young researchers ask questions related to this issue on the research gate.
Therefore i decided to put this question and add my answer which is my paper: PROPERTIES OF DIFFUSED RESISTORS IN SOLAR-GRADE SEMICRYSTALLINE SILICON A. EL-EMAWY, A. ZEKRY, M. EL-KOOSY and H. FIKRY Faculty of Engineering, Ain Shams University, 1 Elsarayat Street, Abbasia, Cairo, Egypt (Received 20 January 1987; in revised form 19 February 1987)
with the Abstract-The transport properties of diffused p +-layers in n-type semicrystalline solar-grade silicon have been examined. It has been found that the grain boundary slightly increases the resistance of the majority carrier flow. A simple method is proposed to determine the average grain size, the barrier height and the effective trapping density at the grain boundary. Resistors made of these diffused layers show anomalous behaviour concerning their voltage and temperature dependence. This anomaly is caused by the soft 2-V characteristics of the isolation p-n junction.
For more information please follow the link: Article A. Elemawy, A. Zekry, M. El-Koosy and H.F. Ragai, “Propertie...
Dear Colleagues,
The grain boundaries in materials because of its poycrysatline nature affect much its electrical transport properties. It is so that the grain boundaries interrupts the path of the mobile charge carriers and thereby increases the resistance imparted by the material for the motion of the mobile carriers when ever they cross the boundary they meet a larger resistance. So in general one can model the resistance Rt of the material by two resistances: the grain bulk resistance Rbulk which is the usual volume resistance of the material and the grain boundaries resistance Rgb. Then Rt= Rbulk +Rgb
As usual the bulk resistance can be expressed by
Rbulk= L/sigma A, with L is the length of the bulk, A its cross sectional area and sigma its conductivity. The conductivity sigma depends on majority charge carriers under thermal equilibrium. A assuming n-type semicondcutor with electron concentration no then sigma= q un n0 where q is the electronic charge, and un is the electron mobility.
At the As the grain boundaries are surfaces of the bulk grains then these surfaces will contain interface states Nt in the energy gap which can trap the majority electrons in them and cause a depletion layer adjacent to the grain boundary. These depleted layers will be positively charged by the donor ions. As a consequence a dipole layer is formed at the grain boundaries causing the formation of a potential energy barrier for the flow the electrons across the grain boundaries. Only the electrons in the bulk that have thermal energy greater than the energy barrier will be able to cross the barrier. This imparts a large impedance for the low of the electrons across the grain boundaries. Assuming the resisatnce of one grain boundary is Rgbs and the number of boundaries in the path of the electrons is eqaul to N, then the total grain boundary resistance will be
Rgb= RgbsN
In the next post i will speak moe about Rgbs.
Best wishes
Dear Colleagues,
with reference to the paper above, the current through the grains depends on the voltage dropped on the grain boundaries Vgbt and can be expresses by
I = Is sinh (qVgbt/2N kT),
with the reverse saturation current Is = Aq n(0) (kT/2pi mn,*)^0.5 exp (--qVB/kT)
It can be expressed in the reduced form Is= Is0 exp ( - qVB/kT)
where n(0) is the majority carrier concebtrationin the center of the grain, kT is the thermal energy mn^* is the effective mass of electrons, VB is the barrier height at thermal equilibrium. g is the number of grain boundaries in the path of the current.
It is clear that the current at the grain boundaries varies nonlinearly with the voltage drop at the grain boundary Vgbt. It is more or less an exponential relation ship. The current depends largely on the barrier height VB at the grain boundaries and decreases exponentially with VB. So as VB inn-creases the I decreases appreciabley.
Best wishes
To complete the analysis of the effect of grain boundaries on the majority carrier transport it is required only to express the barrier height VB under thermal equilibrium in terms of the doping concentration in the grain ND and assuming that the grain size Lg is greater than double the depletion region width wd such that Lg> 2 wd. Under this condition there will be only partial depletion of the grain around the grain boundary.
Assuming also that the trap density at the surface of the grain is Nt, then these traps will capture neighboring electrons from the bulk of the grain where they become negatively charged leaving the adjacent region positively charged by the donor ions. At equilibrium the number of charged traps will be Ntc and the width of the space charge region will be Wd,
From the neutrality conditions Ntc= Wd ND, therefore wd depends on the charged trap density which comprises all traps lying under the Fermi level.The electric field associated with the ND ions will have a linear distribution from zero at the bulk edge of the space charge region and maximum at the surface of the grain.
It follows that the potential barrier can be expressed as in the case single sided p-n junction VB= q NDWd^2/2 epsilon= q Ntc( Ntc/2ND epsilon), with Ntc= 10^12 /cm^2, ND=10^17/cm^3, q= 1.6x10^-19, epsilon= 12x8.88x10^-14 F/cm^2,
VB= .75 V. Such high potential barrier will impart high impedance to the flow of minority carrier. The consequence of this high barrier is that the resistance will be dominated by the grain boundary. Increasing the doping to 10^18/cm^3 will decrease the barrier to 75 mV rendering the barrier very small and therefore the grain boundaries will have a much smaller effects on the resistance of the polycrysatlline material.
Best wishes
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