Question 6.1: An n^+ -polysilicon-gate n-channel MOS transistor is made on...
An n^{+}-polysilicon-gate n-channel MOS transistor is made on a p-type Si substrate with N_{a} = 5 × 10^{15} cm^{-3}. The SiO_{2} thickness is 100 Å in the gateregion, and the effective interface charge Q_{i} is 4 ×10^{10} q C>cm^{2}. Find C_{i} and C_{min} on the C–V characteristics, and find W_{m}, V_{FB}, and V_{T}.
\phi _{F} =\frac{kT}{q}\ln \frac{N_{a} }{n_{i} }=0.0259\ln \frac{5 \times 10^{15}}{1.5 \times 10^{10}} = 0.329 eV
W_{m} =2\left[\frac{\epsilon _{s}\phi _{F} }{q N_{a} } \right]^{1/2 }=2\left[\frac{11.8 \times 8.85 \times 10^{-14} \times 0.329}{1.6 \times 10^{-19} \times 5 \times 10^{15}} \right]^{1/2 } =4.15 \times 10^{-5} cm = 0.415 \mu m
From Fig. 6–17,\Phi _{ms}\cong – 0.95 V,and we have
Q_{i} = 4 \times 10^{10} \times 1.6 \times 10^{-19} = 6.4 \times 10^{-9} C/cm^{2}
C_{i} =\frac{\epsilon_{i}}{d} =\frac{3.9 \times 8.85 \times 10^{-14}}{0.1\times 10^{-5}} =3.45\times 10^{-7}F/cm^{2}
V_{FB} =\Phi _{ms} -Q_{i} /C_{i} = -0.95 – 6.4 \times 10^{-9}/3.45 \times 10^{-7} = -0.969 V
Q_{d} =-qN_{a} W_{m} =-1.6 \times 10^{-19}\times 5 \times 10^{15} \times 4.15 \times 10^{-5}=-3.32\times 10^{-8} C/cm^{2}
V_{T} =V_{FB} -\frac{Q_{d} }{C_{i} } +2h_{F}=-0.969+\frac{3.32 \times 10^{-8}}{3.45 \times 10^{-7}} +0.658 = -0.215 V
C_{d} =\frac{\epsilon _{s}}{W_{m}} =\frac{11.8 \times 8.85 \times 10^{-14}}{4.15 \times 10^{-5}} =2.5 \times 10^{-8} F/cm
C_{min} =\frac{C_{i}C_{d} }{C_{i}+C_{d}} =\frac{3.45 \times 10^{-7} \times 2.5 \times 10^{-8}}{3.45 \times 10^{-7} + 2.5 \times 10^{-8}} =2.33 \times 10^{8} F/cm^{2}
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