SOLUTION

In design, not all specifications are important to meet exactly. For example, the power supply must be exactly 5 V but the output swing can exceed the specifications if necessary for some other reason. Let us begin with the dc current. Both the power dissipation and the slew rate will influence the current. The slew rate requires a current greater than $100 \mu \mathrm{~A}$ while the power dissipation requires a current less than $200 \mu \mathrm{~A}$. Let us compromise with a current of $150 \mu \mathrm{~A}$.



We will first begin with M 3 because the only unknown is $W_3 / L_3$. Solving the upper righthand relationship of providing figure for $W_3 / L_3$, we get



$$

\frac{W_3}{L_3}=\frac{2 I}{K_p^{\prime}\left[V_{D D}-v_{\mathrm{OUT}}(\max )\right]^2}=\frac{2 \cdot 150}{50(1)^2}=6

$$





The upper left-hand relationship of Fig. 5.3-7 gives $W_4 / L_4=W_3 / L_3$ if $I=I_{\text {BIAS }}$. Next, we use the lower right-hand relationship of Fig. 5.3-7 to define $W_1 / L_1$ as



$$

\frac{W_1}{L_1}=\frac{\left(A_\nu \lambda\right)^2 I}{2 K_N^{\prime}}=\frac{(50 \cdot 0.04)^2(150)}{2 \cdot 110}=2.73

$$





To design $W_2 / L_2$, we will first calculate $V_{D S 1}$ (sat) and use the $v_{\text {OUT }}(\min )$ specification to define $V_{D S 2}$ (sat). $V_{D S 1}$ (sat) is given as



$$

V_{D S 1}(\mathrm{sat})=\sqrt{\frac{2 I}{K_N^{\prime}\left(W_1 / L_1\right)}}=\sqrt{\frac{2 \cdot 150}{110 \cdot 2.73}}=0.8 \mathrm{~V}

$$

Subtracting this value from 1.5 V gives $V_{D S 2}(\mathrm{sat})=0.5 \mathrm{~V}$. Therefore, $W_2 / L_2$ is



$$

\frac{W_2}{L_2}=\frac{2 I}{K_N^{\prime} V_{D S 2}(\mathrm{sat})^2}=\frac{2 \cdot 150}{110 \cdot 0.5^2}=5.57

$$





Finally, the lower left-hand relationship of Fig. 5.3-7 gives the value of $V_{G G 2}$ as



$$

\begin{gathered}

V_{G G 2}=V_{D S 1}(\mathrm{sat})+\sqrt{\frac{2 I}{K_N^{\prime}\left(W_2 / L_2\right)}}+V_{T N}=1.0 \mathrm{~V}+0.7 \mathrm{~V}+0.7 \mathrm{~V}=2.4 \mathrm{~V} \\

\frac{W_5}{L_5}=\frac{2 I_{B I A S}}{\left(V_{G G 2}-V_T\right)^2 K_N^{\prime}}=\frac{2.150}{1.7^2(110)}=0.94

\end{gathered}

$$





This example illustrates that by varying the $W / L$ ratios of the transistors, an output voltage range of 2.5 V is achieved over which all transistors stay in saturation.