For a p-channel transistor with a gate oxide thickness of 10 nm, calculate the boron ion dose FB(B^+ ions/cm^2) required to reduce VT 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, instea of a shallow B implant,it was a

Question

For a p-channel transistor with a gate oxide thickness of 10 nm,calculate the boron ion dose[latex]F_{B} (B^{+ }ions/cm^{2}) [/latex]required to reduce [latex]V_{T}[/latex] 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 [latex]V_{T}[/latex] calculation change? Assuming a boron ion beam current of [latex]10^{-5}[/latex] A, and supposing that the area scanned by the ion beam is [latex]650 cm^{2}[/latex], how long does this implant take?

Accepted Answer

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 [latex]V_{FB}[/latex]. Thus,

[latex]C_{i} =\frac{\epsilon _{i} }{d} =\frac{(3.9)(8.85 imes 10^{-14})}{10^{-6}} = 3.45 imes 10-7 F/cm^{2}[/latex]

[latex]- 0.5 = -1.1 +\frac{qF_{B}}{C_{i}}[/latex]

[latex]F_{B} =\frac{3.45 imes 10^{-7}}{1.6 imes 10^{-19}} (0.6) = 1.3 imes 10^{12} cm^{-2}[/latex]

If the B distribution is deeper, we cannot assume it to be a sheet charge in [latex]V_{FB}[/latex]. If it is approximately constant over the maximum depletion width, we should instead change the substrate doping term in the expression for [latex]V_{T}[/latex].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,

[latex]\frac{10^{-5}(C/s)}{650 cm^{2}} t(s) = 1.3 imes 10^{12} (ions/cm^{2}) imes 1.6 imes 10^{-19} (C/ion)[/latex]

The implant time is t = 13.5 s.

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