Chapter 8, Problem 8.45P

(a)

To determine

The design for class AB enhancement mode MOSFET.

Answer to Problem 8.45P

The design for class AB enhancement mode MOSFET is shown in Figure 2.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The diagram for the class AB output stage using the enhancement mode MOSFETs is with resistance $R_3$ and $R_4$ equals to zero is shown below.

The required diagram is shown in Figure 2

  

Conclusion:

Therefore, the design for class AB enhancement mode MOSFET is shown in Figure 2.

(b)

To determine

The value of $R_1$ and $R_2$ .

Answer to Problem 8.45P

The value of the resistance $R_1$ is $4\text{k}\Omega$ and $R_2$ is $4\text{k}\Omega$ .

Explanation of Solution

Calculation:

The expression to determine the value of the resistance $R_1$ is given by,

  $R_1=\frac{V^{+}-v_{SGpl}}{I_{D1}}$

Substitute $10\text{V}$ for $V^{+}$ , $2\text{V}$ for $v_{SGpl}$ and $2\text{mA}$ in the above equation.

  $\begin{array}{c}{R}_1=\frac{10\text{V}-2\text{V}}{2\text{mA}}\\ =4\text{k}\Omega \end{array}$

The expression for the value of the resistance is given by,

  $R_2=R_1$

Substitute $4\text{k}\Omega$ for $R_1$ in the above equation.

  $R_2=4\text{k}\Omega$

Conclusion:

Therefore, the value of the resistance $R_1$ is $4\text{k}\Omega$ and $R_2$ is $4\text{k}\Omega$ .

(c)

To determine

The value of the current in $M_1$ and $M_2$ .

Answer to Problem 8.45P

The value of the current $I_{D1}$ is $2\text{mA}$ and the value of the current $I_{D2}$ is $2\text{mA}$ .

Explanation of Solution

Calculation:

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

  $v_{GSn3}=\sqrt{\frac{I_{DQ}}{K_{n3}}}+V_{TN}$

Substitute $5\text{mA}$ for $I_{DQ}$ , $5\text{mA}/{\text{V}}^2$ for $K_{n3}$ and $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{v}_{GSn3}=\sqrt{\frac{5\text{mA}}{5\text{mA}/{\text{V}}^2}}+1\text{V}\\ =\text{2}\text{V}\end{array}$

The expression for the gate to source voltage for $M_1$ is given by,

  $v_{SGp1}=v_{GSn3}$

Substitute $\text{2}\text{V}$ for $v_{GSn3}$ in the above equation.

  $v_{SGp1}=\text{2}\text{V}$

The expression for the drain current for transitory $M_1$ is given by,

  $I_{D1}=k_{p1}{\left(v_{SGp1}-\left|V_{TP}\right|\right)}^2$

Substitute $2\text{mA}/{\text{V}}^2$ for $K_{p1}$ , $2\text{V}$ for $v_{SGp1}$ and $1\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{c}{I}_{D1}=\left(2\text{mA}/{\text{V}}^2\right){\left(\left(2\text{V}\right)-\left|1\text{V}\right|\right)}^2\\ =2\text{mA}\end{array}$

The expression for the value of the drain current of transistor $M_2$ is given by,

  $I_{D1}=I_{D2}$

Substitute $2\text{mA}$ for $I_{D1}$ in the above equation.

  $I_{D2}=2\text{mA}$

Conclusion:

Therefore, the value of the current $I_{D1}$ is $2\text{mA}$ and the value of the current $I_{D2}$ is $2\text{mA}$ .

(c)

To determine

The current in each of the transistor, the value of the input voltage $v_I$ and the power delivered to the load,

Answer to Problem 8.45P

The value of the drain current $i_{D3}$ is $0.0233\text{A}$ , the value of the input voltage is $5.014\text{V}$ , the power delivered to the load is $81.67\times {10}^{-3}\text{W}$ , the value of the current through the transistor $M_2$ is $3.188\text{mA}$ and the value of the drain current for $M_1$ is $0.835\text{mA}$ .

Explanation of Solution

Calculation:

The expression for the current through $M_4$ is given by,

  $I_{D4}=0\text{A}$

The expression for the current through $M_3$ is given by,

  $i_{D3}=\frac{v_O}{R_L}$

Substitute $3.5\text{V}$ for $v_O$ and $150\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{i}_{D3}=\frac{3.5\text{V}}{150\Omega }\\ =0.0233\text{A}\end{array}$

The conversion from $\text{A}$ into $\text{mA}$ is given by,

  $1\text{A}={10}^3\text{mA}$

The conversion from $0.0233\text{A}$ into $\text{mA}$ is given by,

  $0.0233\text{A}=23.3\text{mA}$

The expression for the value of the gate to source voltage for $M_{n3}$ is given by,

  $v_{GSn3}=\sqrt{\frac{i_{D3}}{K_{n3}}}+V_{TN}$

Substitute $23.3\text{mA}$ for $i_{D3}$ , $5\text{mA}/{\text{V}}^2$ for $K_{n3}$ and $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{v}_{GSn3}=\sqrt{\frac{23.3\text{mA}}{5\text{mA}/{\text{V}}^2}}+1\text{V}\\ =\text{3}\text{.16}\text{V}\end{array}$

Apply KVL in at the output terminals.

  $R_1{I}_{D1}+v_{GSn3}+v_O=V^{+}$

Substitute $10\text{V}$ for $V^{+}$ , $3.5\text{V}$ for $v_O$ , $3.16\text{V}$

  $v_{GSn3}$ and $4\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{l}\left(4\text{k}\Omega \right)I_{D1}+3.16\text{V}+3.5\text{V}=10\text{V}\\ I_{D1}=0.835\text{mA}\end{array}$

The expression for the value of the gate to source voltage for $M_{p1}$ is given by,

  $v_{GSp1}=\sqrt{\frac{I_{D1}}{K_{p1}}}+\left|V_{TN}\right|$

Substitute $0.835\text{mA}$ for $I_{D1}$ , $2\text{mA}/{\text{V}}^2$ for $K_{p1}$ and $-1\text{V}$ for $V_{TP}$ in the above equation.

  $\begin{array}{c}{v}_{GSp1}=\sqrt{\frac{0.835\text{mA}}{2\text{mA}/{\text{V}}^2}}+\left|-1\text{V}\right|\\ =\text{1}\text{.646}\text{V}\end{array}$

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

  $v_I=v_O+v_{GSn2}-v_{SGp1}$

Substitute $316\text{V}$ for $v_{GSn3}$ , $3.5\text{V}$ for $v_{}$ and $1.646\text{V}$ for $v_{SGp1}$ in the above equation.

  $\begin{array}{c}{v}_I=3.5\text{V}+316\text{V}-\text{1}\text{.646}\text{V}\\ =5.014\text{V}\end{array}$

Apply KVL at the input terminals.

  $v_I-v_{GSn2}-R_2{i}_{D2}-V^{-}=0$

Substitute $5.014\text{V}$ for $v_I$ , $2\text{mA}/{\text{V}}^2$ for $K_{n2}$ , $4\text{k}\Omega$ for $R_2$ , $-10\text{V}$ for $V^{-}$ and $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{l}5.014\text{V}-v_{GSn2}-\left(4\text{k}\Omega \right)\left(2\text{mA}/{\text{V}}^2\right)-\left(-10\text{V}\right)=0\\ v_{GSn2}=2.2625\text{V}\end{array}$

The expression to determine the value of the drain current for $M_2$ is given by,.

  $I_{D2}=K_{n2}{\left(v_{GSn2}-V_{TN}\right)}^2$

Substitute $2\text{mA}/{\text{V}}^2$ for $K_{n2}$ , $2.2625\text{V}$ for $v_{GSn2}$ and $1\text{V}$ for $V_{TN}$ in the above equation.

  $\begin{array}{c}{I}_{D2}=2\text{mA}/{\text{V}}^2{\left(2.2625\text{V}-1\text{V}\right)}^2\\ =3.188\text{mA}\end{array}$

The expression for the power delivered to the load is given by,

  $P_L=\frac{v_O^2}{R_L}$

Substitute $3.5\text{V}$ for $v_O$ and $150\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}{P}_L=\frac{{\left(3.5\text{V}\right)}^2}{150\Omega }\\ =81.67\times {10}^{-3}\text{W}\end{array}$

Conclusion:

Therefore, the value of the drain current $i_{D3}$ is $0.0233\text{A}$ , the value of the input voltage is $5.014\text{V}$ , the power delivered to the load is $81.67\times {10}^{-3}\text{W}$ , the value of the current through the transistor $M_2$ is $3.188\text{mA}$ and the value of the drain current for $M_1$ is $0.835\text{mA}$ .