Chapter 16, Problem D16.11P

(a)

To determine

The design parameter for the circuit.

Answer to Problem D16.11P

The required value of the width to length ratio of the driver is $5.0408$ and of the load is $2.47$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression to determine the power dissipated in the circuit is given by,

  $P_D=i_D{V}_{DD}$

Substitute $1.8\text{V}$ for $V_{DD}$ and $80\text{μW}$ for $P_D$ in the above equation.

  $\begin{array}{l}80\text{μW}=i_D\left(1.8\text{V}\right)\\ i_D=44.44\text{μA}\end{array}$

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

  $i_D=\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_L{\left(-V_{TNL}\right)}^2$

Substitute $44.44\text{μA}$ for $i_D$ , $100\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{l}44.44\text{μA}=\left(\frac{100\text{μA}/{\text{V}}^2}{2}\right){\left(\frac{W}{L}\right)}_L{\left(-\left(-0.6\text{V}\right)\right)}^2\\ {\left(\frac{W}{L}\right)}_L=2.47\end{array}$

The expression to determine the value of the width to length ratio of the driver and the load transistor is given by,

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

Substitute $1.8\text{V}$ for $v_I$ , $0.3\text{V}$ for $V_{TND}$ , $0.06\text{V}$ for $v_O$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{c}\frac{K_D}{K_L}=\frac{{\left[-\left(-0.6\text{V}\right)\right]}^2}{\left[2\left(1.8\text{V}-0.3\text{V}\right)\left(0.06\text{V}\right)-{\left(0.06\text{V}\right)}^2\right]}\\ =2.0408\end{array}$

The expression for the ratio of the width to length ratio of driver to transistor is given by,

  $\frac{K_D}{K_L}=\frac{{\left(\frac{W}{L}\right)}_D}{{\left(\frac{W}{L}\right)}_L}$

Substitute $2.0408$ for $\frac{K_D}{K_L}$ and $2.47$ for ${\left(\frac{W}{L}\right)}_L$ in the above equation.

  $\begin{array}{l}2.0408=\frac{{\left(\frac{W}{L}\right)}_D}{2.47}\\ {\left(\frac{W}{L}\right)}_D=5.0408\end{array}$

Conclusion:

Therefore, the required value of the width to length ratio of the driver is $5.0408$ and of the load is $2.47$ .

(b)

To determine

The transition for the driver and the load transistor.

Answer to Problem D16.11P

The value of the input transition point or the load is $0.720\text{V}$ and the value of the output transition point is $1.2\text{V}$ . In case of the driver the output transition point is $0.720\text{V}$ and the value of the input transition point is $0.420\text{V}$ .

Explanation of Solution

Calculation:

For load.

The expression to determine the value of the output transition point is given by,

  $v_{Ot}=V_{DD}+V_{TNL}$

Substitute $1.8\text{V}$ for $V_{DD}$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{c}{v}_{Ot}=1.8\text{V}+\left(-0.6\text{V}\right)\\ =1.2\text{V}\end{array}$

The expression to determine the input transition point is given by,

  $\sqrt{\frac{K_D}{K_L}}\left(v_{It}-V_{TND}\right)=\left(-V_{TNL}\right)$

Substitute $2.0408$ for $\frac{K_D}{K_L}$ , $0.3\text{V}$ for $V_{TND}$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{l}\sqrt{2.0408}\left(v_{It}-0.3\text{V}\right)=\left(-\left(-0.6\text{V}\right)\right)\\ v_{It}=0.720\text{V}\end{array}$

For driver.

The expression to determine the input transition point is given by,

  $\sqrt{\frac{K_D}{K_L}}\left(v_{It}-V_{TND}\right)=\left(-V_{TNL}\right)$

Substitute $2.0408$ for $\frac{K_D}{K_L}$ , $0.3\text{V}$ for $V_{TND}$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{l}\sqrt{2.0408}\left(v_{It}-0.3\text{V}\right)=\left(-\left(-0.6\text{V}\right)\right)\\ v_{It}=0.720\text{V}\end{array}$

The expression to determine the value of the output transition point is given by,

  $v_{Ot}=v_{It}-V_{TND}$

Substitute $0.720\text{V}$ for $v_{It}$ and $0.3\text{V}$ for $V_{TND}$ in the above equation.

  $\begin{array}{c}{v}_{Ot}=0.720\text{V}-0.3\text{V}\\ =0.420\text{V}\end{array}$

  $v_{It}$

Conclusion:

Therefore, the value of the input transition point or the load is $0.720\text{V}$ and the value of the output transition point is $1.2\text{V}$ . In case of the driver the output transition point is $0.720\text{V}$ and the value of the input transition point is $0.420\text{V}$ .

(c)

To determine

The value of maximum power dissipation in the inverter and the output voltage for the given input.

Answer to Problem D16.11P

The value of the output voltage is $0.03\text{V}$ and the power dissipated in the circuit is $80\text{μW}$ .

Explanation of Solution

Calculation:

The power dissipation is same even if the width to length ratio of the driver is doubled and is given by,

  $P_D=80\text{μW}$

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

  $i_D=\left(\frac{{k^{\prime }}_n}{2}\right){\left(\frac{W}{L}\right)}_L{\left(-V_{TNL}\right)}^2$

Substitute $44.44\text{μA}$ for $i_D$ , $100\text{μA}/{\text{V}}^2$ for ${k^{\prime }}_n$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{l}44.44\text{μA}=\left(\frac{100\text{μA}/{\text{V}}^2}{2}\right){\left(\frac{W}{L}\right)}_L{\left(-\left(-0.6\text{V}\right)\right)}^2\\ {\left(\frac{W}{L}\right)}_L=2.47\end{array}$

The expression for the width to length ratio of the driver is given by,

  ${\left(\frac{W}{L}\right)}_D=2{\left(\frac{W}{L}\right)}_L$

Substitute $2.47$ for ${\left(\frac{W}{L}\right)}_L$ in the above equation.

  $\begin{array}{c}{\left(\frac{W}{L}\right)}_D=2\left(2.47\right)\\ =\mathrm{10..816}\end{array}$

The expression for the ratio of the width to length ratio of driver to transistor is given by,

  $\frac{K_D}{K_L}=\frac{{\left(\frac{W}{L}\right)}_D}{{\left(\frac{W}{L}\right)}_L}$

Substitute $\mathrm{10..816}$ for ${\left(\frac{W}{L}\right)}_D$ and $2.47$ for ${\left(\frac{W}{L}\right)}_L$ in the above equation.

  $\begin{array}{c}\frac{K_D}{K_L}=\frac{\mathrm{10..816}}{2.47}\\ =4.081\end{array}$

The expression to determine the value of the output voltage is given by,

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

Substitute $1.8\text{V}$ for $v_I$ , $0.3\text{V}$ for $V_{TND}$ , $4.081$ for $\frac{K_D}{K_L}$ and $-0.6\text{V}$ for $V_{TNL}$ in the above equation.

  $\begin{array}{l}4.081=\frac{{\left[-\left(-0.6\text{V}\right)\right]}^2}{\left[2\left(1.8\text{V}-0.3\text{V}\right)\left(v_O\right)-{\left(v_O\right)}^2\right]}\\ 4.081\left[3v_O-v_O^2\right]=0.36\\ v_O=2.97\text{V},0.03\text{V}\end{array}$

Conclusion:

Therefore, the value of the output voltage is $0.03\text{V}$ and the power dissipated in the circuit is $80\text{μW}$ .