Chapter 10, Problem 10.1P

To determine

The model parameters for an NMOS transistor operating at different values

Answer to Problem 10.1P

  $\begin{array}{l}{C}_{ox}=8.625\text{ mF}\\ V_{ov}=0.16\text{ V}\\ g_m=2.5\text{ mA/V}\\ \chi \text{= 0}\text{.228}\\ g_{mb}=0.57\text{ mA/V}\\ r_0=50\text{ k}\Omega \\ C_{gs}=23.598\text{ fF}\\ C_{ov}=3.105\text{ fF}\\ C_{sb}=4.24\text{ fF}\\ C_{dB}=3.396\text{ fF}\\ f_T=15.1\text{ GHz}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{I}_D=200\text{ µA}\\ V_{SB}=0.35\text{ V}\\ V_{DS}=0.7\text{ V}\\ W=12\text{ µm}\\ L=0.3\text{ µm}\\ t_{ox}=4\text{ nm}\\ {\mu }_n=450{\text{ cm}}^2/\text{V·s}\\ \gamma =0.5V^{1/2}\\ 2{\varphi }_f=0.85\text{ V}\\ \lambda =0.1{\text{ V}}^{-1}\\ V_0=0.9\text{ V }\\ C_{sb0}=C_{db0}=5\text{ fF}\\ L_{ov}=0.03\text{ µm}\\ g_{mb}=\chi g_m\\ \text{where,}\\ \chi =\frac{\gamma }{2\sqrt{2{\varphi }_f+V_{SB}}}\\ {ϵ}_{ox}=3.45\times {10}^{-11}\text{ F/m}\end{array}$

  

Fig: Given NMOS transistor

Calculation:

The equation for oxide capacitance is given by,

  $C_{ox}\text{=}\frac{{\epsilon }_{ox}}{t_{ox}}$

Plugging the values

  $\begin{array}{c}{C}_{ox}\text{=}\frac{3.45\times {10}^{-11}}{4\text{ nm}}\\ =8.625{\text{ mF/m}}^2\end{array}$

The equation for overdrive voltage is given by,

  $V_{ov}=\sqrt{\frac{2I_D}{{\mu }_n{C}_{ox}\left(\frac{W}{L}\right)}}$

Plugging the values

  $\begin{array}{l}{V}_{ov}=\sqrt{\frac{2\times 200\times {10}^{-6}}{450\times {10}^{-4}\times 8.625\times {10}^{-3}\times \left(\frac{12\times {10}^{-6}}{0.3\times {10}^{-6}}\right)}}\\ V_{ov}=0.16\text{ V}\end{array}$

Here,

  $V_{DS}>V_{ov}$ (saturation region)

The equation for transconductance is given by,

  $g_m=\frac{2I_D}{V_{ov}}$

Plugging the values

  $\begin{array}{c}{g}_m=\frac{2\times 200\times {10}^{-6}}{0.16}\\ =2.5\text{ mA/V}\end{array}$

  $\begin{array}{c}\chi =\frac{0.5}{2\sqrt{0.85+0.35}}\\ =0.228\end{array}$

The equation for the body trans conductance is given by,

  $g_{mb}=\chi g_m$

Plugging the values

  $\begin{array}{c}{g}_{mb}=0.228\times 2.5\\ =0.57\text{ mA/V}\end{array}$

The equation for the output resistance through the transistor is given by,

  $r_0=\frac{1}{\lambda I_D}$

Plugging the values

  $\begin{array}{c}{r}_0=\frac{1}{0.1\times 200\times {10}^{-6}}\\ =50\text{ k}\Omega \end{array}$

  $C_{o\chi }=W{L}_{OV}{C}_{ox}$

Plugging the values

  $\begin{array}{l}{C}_{o\chi }=12\times {10}^{-6}\times 0.03\times {10}^{-6}\times 8.625\times {10}^{-3}\\ C_{o\chi }=3.105\text{ fF}\end{array}$

The equation for gate-source capacitance is given by,

  $C_{gs}=\frac{2}{3}WL{C}_{ox}+C_{ov}$

Plugging the values

  $\begin{array}{l}{C}_{gs}=\frac{2}{3}\times 12\times {10}^{-6}\times 0.3\times {10}^{-6}\times 8.625\times {10}^{-3}+3.105\text{ F}\\ C_{gs}=23.805\text{ fF}\end{array}$

The equation for the source body capacitance is given by,

  $C_{SB}=\frac{C_{sbo}}{\sqrt{1+\frac{V_{SB}}{V_0}}}$

Plugging the values

  $\begin{array}{c}{C}_{Sb}=\frac{5F}{\sqrt{1+\frac{0.35}{0.9}}}\\ =4.24\text{ fF}\end{array}$

The equation for drain body capacitance is given by,

  $C_{dB}=\frac{C_{dbo}}{\sqrt{1+\frac{V_{DS}+V_{SB}}{V_o}}}$

Plugging the values

  $\begin{array}{c}{C}_{dB}=\frac{5\text{f}}{\sqrt{1+\frac{0.7+0.35}{0.9}}}\\ =3.396\text{ fF}\end{array}$

The equation for unity gain frequency is given by,

  $f_T=\frac{g_m}{2\pi (C_{gs}+C_{ov})}$

Plugging the values

  $\begin{array}{c}{f}_T=\frac{2.5}{2\pi (23.598+3.105)}\\ =15.1\text{ GHz}\end{array}$