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 doseF_{B} (B^{+ }ions/cm^{2}) required to reduce V_{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 V_{T} calculation change? Assuming a boron ion beam current of 10^{-5} A, and supposing that the area scanned by the ion beam is 650 cm^{2}, how long does this implant take?
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 V_{FB}. Thus,
C_{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 +\frac{qF_{B}}{C_{i}}
F_{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 V_{FB}. If it is approximately constant over the maximum depletion width, we should instead change the substrate doping term in the expression for V_{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,
\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.