Chapter 16, Problem 16.43P

(a)

To determine

To show: The resistance of the NMOS device is equal to $\frac{1}{\left[\left({k^{\prime }}_n{\left(\frac{W}{L}\right)}_n\right)\left(V_{DD}-V_{TN}\right)\right]}$ when $v_I\cong 0$ and resistance of PMOS is $\frac{1}{\left[\left({k^{\prime }}_P{\left(\frac{W}{L}\right)}_P\right)\left(V_{DD}-V_{TP}\right)\right]}$ .

Explanation of Solution

Calculation:

Consider the case when the input voltage is,

  $v_I=V_{DD}$

The expression for the conduction parameter of the NMOS transistor is given by,

  $K_N=\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_n$

The expression for the drain current of the NMOS transistor when the transistor is biased in the non-saturation is given by,

  $i_{DN}=K_N\left[2\left(v_I-V_{TN}\right)v_{DSN}-v_{DSN}^2\right]$

The expression to determine the resistance of the NMOS transistor is given by,

  $\frac{1}{r_{ds}}=\frac{d{i}_N}{d{v}_{DSN}}$

Substitute $K_N\left[2\left(v_I-V_{TN}\right)v_{DSN}-v_{DSN}^2\right]$ for $i_{DN}$ in the above equation.

  $\begin{array}{c}\frac{1}{r_{ds}}=\frac{d\left(K_N\left[2\left(v_I-V_{TN}\right)v_{DSN}-v_{DSN}^2\right]\right)}{d{v}_{DSN}}\\ \frac{1}{r_{ds}}=2K_N\left(v_I-V_{TN}\right)-K_N{V}_{DSN}\\ r_{ds}=\frac{1}{2K_N\left(v_I-V_{TN}\right)-K_N{V}_{DSN}}\end{array}$

Substitute $0\text{V}$ for $v_{DSN}$ , $\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_n$ for $K_N$ and $V_{DD}$ for $v_I$ in the above equation.

  $\begin{array}{c}{r}_{ds}=\frac{1}{2\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_n\left(V_{DD}-V_{TN}\right)-\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_n\left(0\text{V}\right)}\\ =\frac{1}{{k^{\prime }}_n{\left(\frac{W}{L}\right)}_n\left(V_{DD}-V_{TN}\right)}\end{array}$

The expression for the current in the NMOS transistor is given by,

  $i_{DP}=K_P\left[2\left(v_I+V_{TP}\right)v_{SDP}-{\left(v_{SDP}\right)}^2\right]$

The expression for the conduction parameter of the NMOS transistor is given by,

  $K_P=\left(\frac{{k^{\prime }}_p}{2}\right){\left(\frac{W}{L}\right)}_p$

The expression for the resistance of the PMOS device is given by,

  $\frac{1}{r_{SD}}=\frac{d{i}_{DP}}{d{v}_{SDP}}$

Substitute $K_P\left[2\left(v_I+V_{TP}\right)v_{SDP}-{\left(v_{SDP}\right)}^2\right]$ for $i_{DP}$ in the above equation.

  $\begin{array}{c}\frac{1}{r_{SD}}=\frac{d}{d{v}_{SDP}}\left[K_P\left[2\left(v_I+V_{TP}\right)v_{SDP}-{\left(v_{SDP}\right)}^2\right]\right]\\ r_{SD}=\frac{1}{2K_P\left(v_I+V_{TP}\right)-2K_P{v}_{SDP}}\end{array}$

Substitute $K_P\left[2\left(v_I+V_{TP}\right)v_{SDP}-{\left(v_{SDP}\right)}^2\right]$ for $i_{DP}$ in the above equation.

  $r_{SD}=\frac{1}{2K_P\left(v_I+V_{TP}\right)v_{SDP}-2K_P{v}_{SDP}}$

Substitute $\left(\frac{{k^{\prime }}_p}{2}\right){\left(\frac{W}{L}\right)}_p$ for $K_P$ , $V_{DD}$ for $v_I$ and $0$ for $v_{SDP}$ in the above equation.

  $\begin{array}{c}{r}_{SD}=\frac{1}{2\left(\frac{{k^{\prime }}_p}{2}\right){\left(\frac{W}{L}\right)}_p\left(V_{DD}+V_{TP}\right)\left(0\right)-2\left(\frac{{k^{\prime }}_p}{2}\right){\left(\frac{W}{L}\right)}_p\left(0\right)}\\ =\frac{1}{\left[\left({k^{\prime }}_P{\left(\frac{W}{L}\right)}_P\right)\left(V_{DD}+V_{TP}\right)\right]}\end{array}$

Conclusion:

Therefore, the resistance of the NMOS device is $\frac{1}{{k^{\prime }}_n{\left(\frac{W}{L}\right)}_n\left(V_{DD}-V_{TN}\right)}$ and PMOS transistor is $\frac{1}{\left[\left({k^{\prime }}_P{\left(\frac{W}{L}\right)}_P\right)\left(V_{DD}+V_{TP}\right)\right]}$ .

(b)

To determine

The value of the maximum current that the NMOS transistor can sink and the current that the PMOS can source.

Answer to Problem 16.43P

The maximum value of the current of that NMOS device can sink is $0.336\text{mA}$ and the current that it can source is $0.336\text{mA}$ and the maximum current that PMOS can source is $0.168\text{mA}$ .S

Explanation of Solution

Calculation:

The expression for the resistance off the NMOS transistor is given by,

  $r_{ds}=\frac{1}{{k^{\prime }}_n{\left(\frac{W}{L}\right)}_n\left(V_{DD}-V_{TN}\right)}$

Substitute $80\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ , $2$ for ${\left(\frac{W}{L}\right)}_n$ , $5\text{V}$ for $V_{DD}$ and $0.8\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{r}_{ds}=\frac{1}{80\text{μA}/{\text{V}}^2\left(2\right)\left(5\text{V}-0.8\text{V}\right)}\\ =1.488\times {10}^{-3}\text{M}\Omega \end{array}$

The conversion from $1\text{M}\Omega$ into $\text{k}\Omega$ is given by,

  $1\text{M}\Omega ={10}^3\text{k}\Omega$

The conversion from $1.488\times {10}^{-3}\text{M}\Omega$ into $\text{k}\Omega$ is given by,

  $1.488\times {10}^{-3}\text{M}\Omega =1.488\text{k}\Omega$

The conversion from $2.9761\times {10}^{-3}\text{M}\Omega$ into $\text{k}\Omega$ is given by,

  $2.9761\times {10}^{-3}\text{M}\Omega =2.9761\text{k}\Omega$

The expression to determine the value of the current through the NMOS is given by,

  $i_{DN}\left(\max \right)=\frac{v_{DSN}}{r_{DS}}$

Substitute $0.5\text{V}$ for $v_{DSN}$ and $1.488\text{k}\Omega$ for $r_{DS}$ in the above equation.

  $\begin{array}{c}{i}_{DN}\left(\max \right)=\frac{0.5\text{V}}{1.488\text{k}\Omega }\\ =0.336\text{mA}\end{array}$

The expression for the resistance of the PMOS transistor is given by,

  $r_{SD}=\frac{1}{\left[\left({k^{\prime }}_P{\left(\frac{W}{L}\right)}_P\right)\left(V_{DD}+V_{TP}\right)\right]}$

Substitute $2$ for ${\left(\frac{W}{L}\right)}_P$ , $40\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_P$ , $5\text{V}$ for $V_{DD}$ and $-0.8\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{c}{r}_{SD}=\frac{1}{\left[\left(40\text{μA}/{\text{V}}^2\right)\left(2\right)\left(5\text{V}+\left(-0.8\text{V}\right)\right)\right]}\\ =2.9761\times {10}^{-3}\text{M}\Omega \\ =2.9761\text{k}\Omega \end{array}$

The expression for the maximum value of the current through the PMOS transistor is given by,

  $i_{DP}\left(\max \right)=\frac{v_{DSP}}{r_{SD}}$

Substitute $0.5\text{V}$ for $v_{DSP}$ and $2.9761\text{k}\Omega$ for $r_{DS}$ in the above equation.

  $\begin{array}{c}{i}_{DP}\left(\max \right)=\frac{0.5\text{V}}{2.9761\text{k}\Omega }\\ =0.168\text{mA}\end{array}$

Conclusion:

Therefore, the maximum value of the current of that NMOS device can sink is $0.336\text{mA}$ and the current that it can source is $0.336\text{mA}$ and the maximum current that PMOS can source is $0.168\text{mA}$ .