Determine the metal–semiconductor work function difference, Φms, for a given MOS system and semiconductor doping. For an aluminum–silicon dioxide junction, Φm′ = 3.20 V and, for a silicon–silicon dioxide junction, χ′ = 3.25 V. We may assume that Eg = 1.12 V. Let the p-type doping be Na = 10^15 cm

Question

Determine the metal-semiconductor work function difference, [latex]\phi_{m s}[/latex], for a given MOS system and semiconductor doping.

For an aluminum-silicon dioxide junction, [latex]\phi_{m}^{\prime}=3.20 \mathrm{~V}[/latex] and, for a silicon-silicon dioxide junction, [latex]\chi^{\prime}=3.25 \mathrm{~V}[/latex]. We may assume that [latex]E_{g}=1.12 \mathrm{~V}[/latex]. Let the p-type doping be [latex]N_{a}=10^{15} \mathrm{~cm}^{-3}[/latex].

Accepted Answer

For silicon at [latex]T=300 \mathrm{~K}[/latex], we may calculate [latex]\phi_{f p}[/latex] as

[latex]\phi_{f p}=V_{t} \ln \left(\frac{N_{a}}{n_{j}} ight)=(0.0259) \ln \left(\frac{10^{15}}{1.5 imes 10^{10}} ight)=0.288 \mathrm{~V}[/latex]

Then the metal-semiconductor work function difference is

[latex]\phi_{m s}=\phi_{m}^{\prime}-\left(\chi^{\prime}+\frac{E_{g}}{2 e}+\phi_{f p} ight)=3.20-(3.25+0.560+0.288)[/latex]

or

[latex]\phi_{m s}=-0.898 \mathrm{~V}[/latex]

Comment

The value of [latex]\phi_{m s}[/latex] will become more negative as the doping of the p-type substrate increases.

Question 10.2: Determine the metal–semiconductor work function difference, ...

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