Chapter 16, Problem 16.1EP

Consider the NMOS inverter with resistor load in Figure 16.3(a) biased at $V_{DD}=3\text{V}$ . Assume transistor parameters of ${k^{\prime }}_n=100\mu {\text{A/V}}^{\text{2}}$ , $W/L=4$ , and $V_{TN}=0.5\text{V}$ . (a) Find the value of $R_D$ such that ${\upsilon }_O=0.1\text{V}$ when ${\upsilon }_I=3\text{V}$ . (b) Using the results of part (a), determine the maximum current and maximum power dissipation in the inverter. (c) Using the results of part (a), determine the transition point for the driver transistor. (Ans. (a) $R_D=29.6\text{k}\Omega$ ; (b) $i_{D,\max }=0.098\text{mA/V}$ , $P_{D,\max }=0.294\text{mW}$ ;(c) $V_{It}=1.132\text{V}$ , $V_{Ot}=0.632\text{V}$ )

To determine

The value of $R_D$

Answer to Problem 16.1EP

The value of $R_D$ is $29.6k\Omega$ .

Explanation of Solution

Given:

  $V_{DD}=3\text{V}$ .

  $k_n{}^{\text{'}}=100\mu {\text{A/V}}^{\text{2}}$ , $W/L=4$ , and $V_{TN}=0.5\text{V}$

  $v_o=0.1\text{V}$ when $v_1=3\text{V}$ .

Calculation:

Consider the NMOS inverter with resister load biased at $V_{DD}=3\text{V}$ is shown in Figure $1$ .

  

Figure 1

From the circuit,

  $\begin{array}{c}{v}_{DS}=v_o\\ =0.1\text{V}\\ v_{GS}=v_1\\ =3\text{V}\end{array}$

Calculate the value of $V_{GS}-V_{TN}$

  $\begin{array}{c}{V}_{GS}-V_{TN}=3-0.5\\ =2.5\text{V}\end{array}$

Since, $V_{DS}<V_{GS}-V_{TN}$ , the transistor is operating in non-saturation region.

The current drain equation is,

  $\begin{array}{c}{i}_D=K_n\left[2\left(v_{GS}-V_{TN}\right)v_{DS}-v^2{}_{DS}\right]\\ =\left(\frac{k_n{}^{\text{'}}}{2}\right)\left(\frac{W}{L}\right)\left[2\left(v_{GS}-V_{TN}\right)v_{DS}-v^2{}_{DS}\right]\left(\text{Since}{K}_n=\left(\frac{k_n{}^{\text{'}}}{2}\right)\left(\frac{W}{L}\right)\right)\\ =\left(\frac{100\times {10}^{-6}}{2}\right)\left(4\right)\left[2\left(2.5\right)\left(0.1\right)-{0.1}^2\right]\\ =98\text{μA}\end{array}$

Apply Kirchhoff’s voltage law to the circuit.

  $\begin{array}{l}{v}_O=V_{DD}-R_D{I}_D\\ R_D=\frac{V_{DD}-v_O}{I_D}\\ R_D=\frac{3-0.1}{98\times {10}^{-6}}\\ R_D=29.6k\Omega \end{array}$

Conclusion:

Therefore, the value of $R_D$ is $29.6k\Omega$ .

To determine

The maximum current and maximum power dissipation in the inverter.

Answer to Problem 16.1EP

The maximum power transfer is, $P_{D,\max }=0.294\text{mW}$

The maximum drain current is, $i_{D,\max }=0.098\text{mA}$

Explanation of Solution

Given:

  $V_{DD}=3\text{V}$ .

  $k_n{}^{\text{'}}=100\mu {\text{A/V}}^{\text{2}}$ , $W/L=4$ , and $V_{TN}=0.5\text{V}$

Calculation:

Consider Figure 1.

The maximum current is,

  $\begin{array}{c}{i}_{D,\max }=\frac{V_{DD}-v_O}{R_D}\\ =\frac{3-0.1}{29.6\times {10}^3}\\ =0.098\text{mA}\end{array}$

The maximum power transfer is,

  $\begin{array}{c}{P}_{D,\max }=i_{D,max}\times V_{DD}\\ =0.98mA\times 3V\\ =0.294mW\end{array}$

Conclusion:

Therefore, the maximum power transfer is, $P_{D,\max }=0.294\text{mW}$ and the maximum drain current is, $i_{D,\max }=0.098\text{mA}$

To determine

The transition point for the driver transistor.

Answer to Problem 16.1EP

The transition point for the driver transistor is

  $V_{1t}=1.132\text{V}$

  $V_{Ot}=0.632\text{V}$

Explanation of Solution

Given:

  $V_{DD}=3\text{V}$ .

  $k_n{}^{\text{'}}=100\mu {\text{A/V}}^{\text{2}}$ , $W/L=4$ , and $V_{TN}=0.5\text{V}$

Calculation:

Transitionpoints for the driver resistor is

  $\begin{array}{l}{K}_n{R}_D{\left(V_{1t}-V_{TN}\right)}^2+\left(V_{1t}-V_{TN}\right)-V_{DD}=0\\ \left(2\times {10}^{-4}\right)\left(29.6\times {10}^3\right){\left(V_{1t}-0.5\right)}^2+\left(V_{1t}-0.5\right)-3=0\\ 5.92{\left(V_{1t}-0.5\right)}^2+{\left(V_{1t}-0.5\right)}^2=3\\ 5.92\left({\left(V_{1t}\right)}^2-1V_{1t}+0.25\right)+V_{1t}-3.5=0\\ 5.92{\left(V_{1t}\right)}^2-4.92V_{1t}+2.02=0\\ V_{1t}=1.132,-0.3013\end{array}$

As the transition voltage is positive and greater than zero for NMOS, the input transition voltage is $V_{1t}=1.132\text{V}$

The output transition voltage is,

  $\begin{array}{c}{V}_{Ot}=V_{1t}-V_{TN}\\ =1.132-0.5\\ =0.632\text{V}\end{array}$

Conclusion:

Therefore, the transition point are

  $V_{1t}=1.132\text{V}$

  $V_{Ot}=0.63\text{2V}$