Chapter 16, Problem 16.3EP

(a)

To determine

The output voltage $v_o$

Answer to Problem 16.3EP

The output voltage $v_o$ is $v_o=0.0414\text{V}$

Explanation of Solution

Given:

Power supply voltage, $V_{DD}=3\text{V}$

Intrinsic trans conductance parameter, $k_n{}^{\text{'}}=100\times {10}^{-6}{\text{A/V}}^{\text{2}}$

Device parameter for driver transistor, $V_{TND}=0.4\text{V}$

Device parameter for load transistor, $V_{TNL}=-0.8\text{V}$

Aspect ratio of driver transistor, ${\left(\frac{W}{L}\right)}_D=6$

Aspect ratio of load transistor, ${\left(\frac{W}{L}\right)}_L=2$

When input voltage, $v_I=3\text{V}$

Calculation:

Given depletion-load NMOS inverter:

  

For the NMOS inverter with Depletion load, the output voltage equation is given by

  $\frac{K_D}{K_L}\left[2\left(v_I-V_{TND}\right)v_o-v_o{}^2\right]=\left(-V_{TNL}{}^2\right)$

The parameters $K_D$ and $K_L$

  $\begin{array}{l}{K}_D=\left(\frac{k_n{}^{\text{'}}}{2}\right){\left(\frac{W}{L}\right)}_D\\ K_D=\left(\frac{100\times {10}^{-6}}{2}\right)\times \left(6\right)\\ K_D=3\times {10}^{-4}A/V^2\\ K_L=\left(\frac{k_n{}^{\text{'}}}{2}\right){\left(\frac{W}{L}\right)}_L\\ =\left(\frac{100\times {10}^{-6}}{2}\right)\times \left(2\right)\\ K_L=1\times {10}^{-4}A/V^2\end{array}$

Now substituting all the values in the above voltage equation,

  $\begin{array}{l}\frac{3\times {10}^{-4}}{1\times {10}^{-4}}\left[2\left(3-0.4\right)v_o-v_o{}^2\right]={\left(-0.8\right)}^2\\ 3\left[2\left(2.6\right)v_o-v_o{}^2\right]=0.64\\ 5.2v_o-v_o{}^2=\frac{0.64}{3}\\ 5.2v_o-v_o{}^2=0.2133\\ v_o{}^2-5.2v_o+0.2133=0\end{array}$

On comparing the above equation with quadratic equation $\left(a{x}^2+bx+c=0\right)$ whose roots are

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ v_o=\frac{-\left(-5.2\right)\pm \sqrt{\left(-5.2\right)-4\left(1\right)\left(0.2133\right)}}{2\left(1\right)}\\ =\frac{5.2\pm \sqrt{26.1868}}{2}\\ v_o=\frac{5.2\pm 5.1173}{2}\\ v_o=\frac{5.2+5.1173}{2},\frac{5.2-5.1173}{2}\\ v_o=\frac{10.3173}{2},\frac{0.0827}{2}\\ v_o=5.15865,0.04135\end{array}$

The output voltage cannot be greater than $V_{DD}=3\text{V}$ so the output voltage is $v_o=0.0414\text{V}$

Conclusion:

Therefore, the output voltage $v_o$ is $v_o=0.0414\text{V}$

(b)

To determine

The maximum current and maximum power dissipation in the inverter.

Answer to Problem 16.3EP

The maximum current and maximum power dissipation in the inverter are

  $\begin{array}{l}{i}_{D,max}=0.064\text{mA}\\ P_{D,\max }=0.192\text{mW}\end{array}$

Explanation of Solution

Given:

Power supply voltage, $V_{DD}=3\text{V}$

Intrinsic trans conductance parameter, $k_n{}^{\text{'}}=100\times {10}^{-6}{\text{A/V}}^{\text{2}}$

Device parameter for driver transistor, $V_{TND}=0.4\text{V}$

Device parameter for load transistor, $V_{TNL}=-0.8\text{V}$

Aspect ratio of driver transistor, ${\left(\frac{W}{L}\right)}_D=6$

Aspect ratio of load transistor, ${\left(\frac{W}{L}\right)}_L=2$

Calculation:

Maximum current is

  $\begin{array}{c}{i}_{D,\max }=K_L{\left(-V_{TNL}\right)}^2\\ =\left(1\times {10}^{-4}\right){\left(-0.8\right)}^2\\ =\left(1\times {10}^{-4}\right)\left(0.64\right)\\ i_{D,max}=0.064\text{mA}\end{array}$

Maximum Power dissipated in the inverter is

  $\begin{array}{l}{P}_{D,max}=i_{D,\max }\times V_{DD}\\ P_{D,max}=0.064\text{mA}\times 3\text{V}\\ P_{D,\max }=0.192\text{mW}\end{array}$

Conclusion:

Therefore, maximum current and maximum power dissipation in the inverter are

  $\begin{array}{l}{i}_{D,max}=0.064\text{mA}\\ P_{D,\max }=0.19\text{2mW}\end{array}$

(c)

To determine

The transition points for the driver and load transistors.

Answer to Problem 16.3EP

The transition points for the driver and load transistors are

  $\begin{array}{l}{V}_{It}=0.862\text{V},V_{Ot}=0.462\text{V}\\ V_{It}=0.862\text{V},V_{Ot}=2.2\text{V}\end{array}$

Explanation of Solution

Given:

Power supply voltage, $V_{DD}=3\text{V}$

Intrinsic trans conductance parameter, $k_n{}^{\text{'}}=100\times {10}^{-6}{\text{A/V}}^{\text{2}}$

Device parameter for driver transistor, $V_{TND}=0.4\text{V}$

Device parameter for load transistor, $V_{TNL}=-0.8\text{V}$

Aspect ratio of driver transistor, ${\left(\frac{W}{L}\right)}_D=6$

Aspect ratio of load transistor, ${\left(\frac{W}{L}\right)}_L=2$

When input voltage, $v_I=3\text{V}$

Calculation:

Transition points for the driver transistor

  $\begin{array}{c}{\left(\frac{W}{L}\right)}_D{\left(v_{It}-V_{TND}\right)}^2={\left(\frac{W}{L}\right)}_L{\left(-V_{TNL}\right)}^2\\ 6{\left(v_{It}-0.4\right)}^2=2{\left(0.8\right)}^2\\ \sqrt{6{\left(v_{It}-0.4\right)}^2}=\sqrt{2{\left(0.8\right)}^2}\\ \left(v_{It}-0.4\right)=\frac{\left(0.8\right)\sqrt{2}}{\sqrt{6}}\\ v_{It}-0.4=\left(0.8\right)\left(0.577\right)\\ v_{It}-0.4=0.4616\\ v_{It}=0.4616+0.4\\ v_{It}=0.8616\\ v_{It}=0.862\text{V}\end{array}$

Output transition point is

  $\begin{array}{c}{V}_{ot}=V_{It}-V_{TND}\\ =0.862-0.4\\ V_{ot}=0.462\text{V}\end{array}$

Transition points for the load transistor

  $\begin{array}{l}{\left(\frac{W}{L}\right)}_D{\left(v_{It}-V_{TND}\right)}^2={\left(\frac{W}{L}\right)}_L{\left(-V_{TNL}\right)}^2\\ 6{\left(v_{It}-0.4\right)}^2=2{\left(0.8\right)}^2\\ \sqrt{6{\left(v_{It}-0.4\right)}^2}=\sqrt{2{\left(0.8\right)}^2}\\ \left(v_{It}-0.4\right)=\frac{\left(0.8\right)\sqrt{2}}{\sqrt{6}}\\ v_{It}-0.4=\left(0.8\right)\left(0.577\right)\\ v_{It}-0.4=0.4616\\ v_{It}=0.4616+0.4\\ v_{It}=0.8616\\ v_{It}=0.862V\end{array}$

Output transistor point is

  $\begin{array}{c}{V}_{ot}=V_{It}+V_{TND}\\ =3+\left(-0.8\right)\\ V_{ot}=2.2\text{V}\end{array}$

Conclusion:

The transition points for the driver and load transistors are

  $\begin{array}{l}{V}_{It}=0.862\text{V},V_{Ot}=0.462\text{V}\\ V_{It}=0.862\text{V},V_{Ot}=2.2\text{V}\end{array}$