Chapter 12, Problem 12.46P

(a)

To determine

The value of the quiescent current $I_{DQ1}$ and $I_{DQ2}$ .

Answer to Problem 12.46P

The value of $I_{DQ1}$ is $11.87\text{mA}$ and the value of $I_{DQ2}$ is $26.77\text{mA}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

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

  $V_A=\left(I_{D1}+I_{D2}\right)250\Omega$

The expression for the voltage $V_{GS2}$ is given by,

  $V_{GS2}=\sqrt{\frac{I_{D2}}{K_p}-V_{TP}}$

The expression for $V_{GS2}$ in terms of the drain current of $I_{D1}$ is given by,

  $V_{GS2}=I_{D1}{R}_{D1}$

The expression for the value of the gate to source voltage of the second transistor in terms of the second transistor drain current is given by,

  $V_{GS2}=\sqrt{\frac{I_{D2}}{K_P}}-V_{TP}$

Substitute $I_{D1}{R}_{D1}$ for $V_{GS2}$ in the above equation.

  $I_{D1}{R}_{D1}=\sqrt{\frac{I_{D2}}{K_P}}-V_{TP}$

Substitute $525\Omega$ for $R_{D1}$ and $10\text{mA}/{\text{V}}^2$ for $K_P$ in the above equation.

  $\begin{array}{l}{I}_{D1}\left(525\Omega \right)=\sqrt{\frac{I_{D2}}{\text{mA}}}-V_{TP}\\ I_{D2}=I_{D1}{\left[\left(1.660\right)-3.162\right]}^2\end{array}$ …… (1)

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

  $V_G-V_{GS1}-{\left(I_{D2}\right)}^2\left(250\Omega \right)=I_{D1}\left(750\Omega \right)$

Substitute $7.6\text{V}$ for $V_G$ , $I_{D1}\left(1.660\right)-3.162$ for $I_{D2}$ in the above equation.

  $\begin{array}{l}7.6\text{V}-V_{GS1}-{\left(I_{D1}\left(1.660\right)-3.162\right)}^2\left(250\Omega \right)=I_{D1}\left(750\Omega \right)\\ I_{D1}=10{\left(V_{GS1}-1\right)}^2\end{array}$ …… (2)

From trial and error method the value of the current $I_{D1}$ is $3.98\text{mA}$ .

The expression for the value of the voltage $V_{GS2}$ is given by,

  $V_{SG2}=I_{D1}{R}_{D1}$

Substitute $3.98\text{mA}$ for $I_{D1}$ and $525\Omega$ for $R_{D1}$ in the above equation.

  $\begin{array}{c}{V}_{SG2}=\left(3.98\text{mA}\right)\left(525\Omega \right)\\ =2.0895\text{V}\end{array}$

Substitute $2.0895\text{V}$ for $V_{SG2}$ in equation (2).

  $\begin{array}{c}{I}_{D1}=10{\left(2.0895\text{V}-1\right)}^2\\ =11.87\text{mA}\end{array}$

Substitute $11.87\text{mA}$ for $I_{D1}$ in equation (2).

  $\begin{array}{c}{I}_{D2}=\left(11.87\text{mA}\right){\left[\left(1.660\right)-3.162\right]}^2\\ =26.77\text{mA}\end{array}$

Conclusion:

Therefore, the value of $I_{DQ1}$ is $11.87\text{mA}$ and the value of $I_{DQ2}$ is $26.77\text{mA}$ .

(b)

To determine

To show: The expression for the small signal current gain is $A_i=\frac{-g_{m2}{R}_{D1}}{1+\frac{1}{g_{m1}\left(R_F+R_{D2}\right)}+\frac{g_{m1}{R}_{D1}{R}_{D2}}{R_F+R_{D2}}}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

The small signal equivalent circuit is shown in Figure 2

  

From above the expression for the output current is given by,

  $I_O=g_{m2}{V}_{sg2}$

The expression for the voltage $V_{sg2}$ is given by,

  $V_{sg2}=g_{m1}{V}_{gs1}{R}_{D1}$

Apply KCL at the upper left node.

  $\begin{array}{l}{I}_i+g_{m1}{V}_{gs1}+\frac{V_A+V_{gs1}}{R_F}=0\\ V_A=-R_F\left[I_i+g_{m1}{V}_{gs1}+\frac{V_{gs1}}{R_F}\right]\end{array}$

The expression to determine the value of the $V_{gs1}$ is given by,

  $V_{gs1}=\frac{I_O}{g_{m1}{g}_{m2}{R}_{D1}}$

The expression for the value of the current at the output is given by, $I_O=\frac{V_A}{R_{D2}}+\frac{V_A-\left(-V_{gs1}\right)}{R_F}$

Substitute $-R_F\left[I_i+g_{m1}{V}_{gs1}+\frac{V_{gs1}}{R_F}\right]$ for $V_A$ in the above equation.

  $I_O=-R_F\left[I_i+g_{m1}{V}_{gs1}+\frac{V_{gs1}}{R_F}\right]\left(\frac{1}{R_{D2}}+\frac{1}{R_F}\right)+\frac{V_{gs1}}{R_F}$

Substitute $\frac{I_O}{g_{m1}{g}_{m2}{R}_{D1}}$ for $V_{gs1}$ in the above equation.

  $\begin{array}{l}{I}_O=-R_F{I}_i\left(\frac{1}{R_{D2}}+\frac{1}{R_F}\right)-R_F{V}_{gs1}\left[\left(g_{m1}+\frac{1}{R_F}\right)\left(\frac{1}{R_{D2}}+\frac{1}{R_F}\right)-\frac{1}{{\left(R_F\right)}^2}\right]\\ I_O\left[1+\frac{R_F}{g_{m1}{g}_{m2}{R}_{D1}}\left(\frac{g_{m1}}{R_{D2}}+\frac{g_{m1}}{R_F}+\frac{1}{R_F{R}_{D2}}\right)\right]=I_i\left(1+\frac{R_F}{R_{D2}}\right)\\ \frac{I_O}{I_i}=\frac{-g_{m2}{R}_{D1}}{1+\frac{1}{g_{m1}\left(R_F+R_{D2}\right)}+\frac{g_{m1}{R}_{D1}{R}_{D2}}{R_F+R_{D2}}}\\ A_i=\frac{-g_{m2}{R}_{D1}}{1+\frac{1}{g_{m1}\left(R_F+R_{D2}\right)}+\frac{g_{m1}{R}_{D1}{R}_{D2}}{R_F+R_{D2}}}\end{array}$

Conclusion:

Therefore, the expression for the current gain is $A_i=\frac{-g_{m2}{R}_{D1}}{1+\frac{1}{g_{m1}\left(R_F+R_{D2}\right)}+\frac{g_{m1}{R}_{D1}{R}_{D2}}{R_F+R_{D2}}}$ .

(c)

To determine

The value of the current gain.

Answer to Problem 12.46P

The value of the current gain is $-2.33$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Calculation:

The expression to determine the value of the $g_{m1}$ is given by,

  $g_{m1}=2\sqrt{K_n{I}_{D1}}$

Substitute $10\text{mA}/{\text{V}}^2$ for $K_n$ and $3.98\text{mA}$ for $I_{D1}$ in the above equation.

  $\begin{array}{c}{g}_{m1}=2\sqrt{\left(10\text{mA}/{\text{V}}^2\right)\left(3.98\text{mA}\right)}\\ =12.62\text{mA}/\text{V}\end{array}$

The expression to determine the value of the $g_{m2}$ is given by,

  $g_{m2}=2\sqrt{K_P{I}_{D2}}$

Substitute $10\text{mA}/{\text{V}}^2$ for $K_P$ and $11.87\text{mA}$ for $I_{D2}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=2\sqrt{\left(10\text{mA}/{\text{V}}^2\right)\left(11.87\text{mA}\right)}\\ =21.79\text{mA}/\text{V}\end{array}$

The expression for the current gain is given by,

  $A_i=\frac{-g_{m2}{R}_{D1}}{1+\frac{1}{g_{m1}\left(R_F+R_{D2}\right)}+\frac{g_{m1}{R}_{D1}{R}_{D2}}{R_F+R_{D2}}}$

Substitute $500\Omega$ for $R_F$ , $525\Omega$ for $R_{D1}$ , $250\Omega$ for $R_{D2}$ , $12.62\text{mA}/\text{V}$ for $g_{m1}$ and $21.79\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{c}{A}_i=\frac{-\left(21.79\text{mA}/\text{V}\right)\left(525\Omega \right)}{1+\frac{1}{\left(21.79\text{mA}/\text{V}\right)\left(500\Omega +250\Omega \right)}+\frac{\left(12.62\text{mA}/\text{V}\right)\left(525\Omega \right)\left(250\Omega \right)}{500\Omega +250\Omega }}\\ =-2.33\end{array}$

Conclusion:

Therefore, the value of the current gain is $-2.33$ .