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Question 6.3: For a p-channel transistor with a gate oxide thickness of 10...

For a p-channel transistor with a gate oxide thickness of 10 nm,calculate the boron ion doseFB(B+ions/cm2)F_{B} (B^{+ }ions/cm^{2}) required to reduce VTV_{T} from -1.1 V to – 0.5 V. Assume that the implanted acceptors form a sheet of negative charge just below the Si surface. If, instead of a shallow B implant, it was a much broader distribution, how would the VTV_{T} calculation change? Assuming a boron ion beam current of 10−510^{-5} A, and supposing that the area scanned by the ion beam is  650cm2650 cm^{2}, how long does this implant take?

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Since the B negative ions are assumed to form a sheet of charge just Solution below the gate oxide, they affect the fixed oxide charge term in VFBV_{FB}. Thus,

Ci=ϵid=(3.9)(8.85×10−14)10−6=3.45×10−7F/cm2C_{i} =\frac{\epsilon _{i} }{d} =\frac{(3.9)(8.85\times 10^{-14})}{10^{-6}} = 3.45 \times 10-7 F/cm^{2}

 

–0.5=−1.1+qFBCi– 0.5 = -1.1 +\frac{qF_{B}}{C_{i}}

 

FB=3.45×10−71.6×10−19(0.6)=1.3×1012cm−2F_{B} =\frac{3.45\times 10^{-7}}{1.6 \times 10^{-19}} (0.6) = 1.3\times 10^{12} cm^{-2}

If the B distribution is deeper, we cannot assume it to be a sheet charge in VFBV_{FB}. If it is approximately constant over the maximum depletion width, we should instead change the substrate doping term in the expression for VTV_{T}.Otherwise, the calculation has to be done by numerically solving Poisson’s equation in the depletion region, with a varying doping concentration corresponding to the B distribution, to obtain the voltage drop in the semiconductor. For a beam current of 10 μA scanned over a 650ـcm² target area,

10−5(C/s)650cm2t(s)=1.3×1012(ions/cm2)×1.6×10−19(C/ion)\frac{10^{-5}(C/s)}{650 cm^{2}} t(s) = 1.3 \times 10^{12} (ions/cm^{2}) \times 1.6 \times 10^{-19} (C/ion)

The implant time is t = 13.5 s.

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