4.1, 4.5-4.6: MOSFET Layout and Capacitances

4.2

We know that:

thus, using ${ϵ}_{ox}=345fFc{m}^{-1}$:

4.3

Using:

so using $\epsilon =103.545\cdot {10}^{-14}Fc{m}^{-1}$ and $q=e=1.6\cdot {10}^{-19}C$ we get:

4.65

We know that:

where $\epsilon =345fFc{m}^{-1}$, so plug in our values of $W=10\mu m$ and $L=0.25\mu m$:

4.67

Use:

where since $C_{ox}^{″}=\frac{{\epsilon }_{ox}}{t_{ox}}=345nFc{m}^{-2}$ then:

4.70

We are given that $C_{ox}^{″}=\frac{{\epsilon }_{ox}}{t_{ox}}=23nFc{m}^{-2}$, and $C_{GSO}=C_{GDO}=20pF/m$ and the $L=2\mathrm{\Lambda }=1\mu m$, and $W=10\mathrm{\Lambda }=5\mu m$, so $W=5\cdot {10}^{-4}cm$ and $L={10}^{-4}cm$.

a

For triode:

b

For saturation:

c

For cutoff:

4.72

Consider the following:

If $\mathrm{\Lambda }=0.5\mu m$ then $L=2\mathrm{\Lambda }=1\mu m$ and $W=10\mu m=5\mu m$. We are also given that $N_D={10}^{16}c{m}^{-3}$, and $N_A={10}^{20}c{m}^{-3}$. We also have $C_{JSW}=C_J\cdot 5\cdot {10}^{-4}c{m}^{-1}$:

via the symmetry of the diagram here $A_S=A_D$. Furthermore:

We know from EE 306 that:

plugging in values gives ${\varphi }_o=V_T\ln \left(\frac{N_A{N}_D}{n_i^2}\right)=0.935V$:

so then:

thus:

And by the symmetry of this problem, $C_{SB}=C_{DB}=\boxed{{\displaystyle 26.1fF}}$.

6.11: NMOS Inverter-Dynamic Characteristics

6.136

We'll want to derive:

Adding the delays of each inverter from two round trips we get the result. For the first trip, since $N$ is given and required to be odd, then we add:

For the second run we get the opposite order of delays, as we are flipped:

Adding them together:

6.139

Given:

Given that we have a load of $C=0.5pF$. For a resistive load we use:

Thus:

Thus: