Chapter 11, Problem 11.54P

(a)

To determine

To show: The expression for the value of the current $\frac{i_{d2}}{I_Q}$ and $\frac{i_{d1}}{I_Q}$ :

  $\frac{i_{D2}}{I_Q}=\frac{1}{2}+\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$

  $\frac{i_{D1}}{I_Q}=\frac{1}{2}+\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Figure 1

Calculation:

The expression for the current $i_{D1}$ is given by,

  $i_{D1}=I_{DSS}{\left(1-\frac{v_{GS1}}{V_P}\right)}^2$

The expression for the current $i_{D2}$ is given by,

  $i_{D2}=I_{DSS}{\left(1-\frac{v_{GS2}}{V_P}\right)}^2$

Both the JFET are matching so,

  $\begin{array}{l}\sqrt{i_{D1}}-\sqrt{i_{D2}}=\sqrt{I_{DSS}}\left(1-\frac{v_{GS2}}{V_P}\right)+\sqrt{I_{DSS}}\left(1-\frac{v_{GS1}}{V_P}\right)\\ \sqrt{i_{D1}}-\sqrt{i_{D2}}=\sqrt{I_{DSS}}\left(\frac{1}{-V_P}\right)\left(v_{GS1}-v_{GS2}\right)\end{array}$

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

  $v_d=\left(v_{GS1}-v_{GS2}\right)$

So the difference of current is,

  $\sqrt{i_{D1}}-\sqrt{i_{D2}}=\sqrt{I_{DSS}}\left(\frac{1}{-V_P}\right)v_d$

The value of the sum of the drain current is given by,

  $i_{D2}=i_Q-i_{D1}$

So,

  $\sqrt{i_{D1}}-\sqrt{i_Q-i_{D1}}=\sqrt{I_{DSS}}\left(\frac{1}{-V_P}\right)v_d$

Square both the sides of the above equation.

  $\begin{array}{l}{\left(\sqrt{i_{D1}}-\sqrt{i_Q-i_{D1}}\right)}^2=I_{DSS}{\left(\frac{1}{-V_P}\right)}^2{\left(v_d\right)}^2\\ i_Q-I_{DSS}\left(\frac{1}{-V_P}\right)\left(I_Q-I_{DSS}{\left(\frac{1}{-V_P}\right)}^2{v}_d^2\right)\end{array}$

Again square both the sides of the equation.

  $\begin{array}{l}{i}_{D1}{I}_Q-i_{D1}^2=\frac{1}{4}{\left(I_Q-i_{DSS}{\left(\frac{1}{-V_P}\right)}^2{v}_d^2\right)}^2\\ i_{D1}^2-i_{D1}{I}_Q+\frac{1}{4}{\left(I_Q-i_{DSS}{\left(\frac{1}{-V_P}\right)}^2{v}_d^2\right)}^2=0\\ i_{D1}=\frac{I_Q\pm \sqrt{I_Q{}^2-\left(I_Q{}^2+i_{DSS}^2\right){\left(\frac{1}{-V_P}\right)}^4{v}_d^4-2I_Q{i}_{DSS}{\left(\frac{1}{-V_P}\right)}^2{v}_d^2}}{2}\\ i_{D1}=\frac{I_Q}{2}\pm \frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)v_d\sqrt{2\frac{I_{DSS}}{I_Q}-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\end{array}$

Solve further as,

  $\frac{i_{D1}}{I_Q}=\frac{1}{2}+\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$

The expression for the current $i_{D2}$ is given by,

  $i_{D2}=I_Q-i_{D1}$

  $\begin{array}{l}{i}_{D2}=I_Q-\left[\frac{I_Q}{2}+\frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)v_d\sqrt{2\frac{I_{DSS}}{I_Q}-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\right]\\ i_{D2}=I_Q-\frac{1}{2}{I}_Q-\frac{1}{2}\left(\frac{1}{-V_P}\right)v_d\sqrt{2\frac{I_{DSS}}{I_Q}-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\\ \frac{i_{D2}}{I_Q}=\frac{1}{2}-\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\end{array}$

Conclusion:

Therefore, the expression for current ratios are $\frac{i_{D2}}{I_Q}=\frac{1}{2}-\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$ and $\frac{i_{D1}}{I_Q}=\frac{1}{2}+\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$ .

(b)

To determine

To show: The base current is switched to other transistor when $\left|v_d\right|=\left|v_P\right|\sqrt{\frac{I_Q}{I_{DSS}}}$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Figure 1

Calculation:

The diagram for the normalized dc transfer characteristics of MOSFET differential amplifier as a function of differential input voltage is shown in Figure 2

  

Figure 2

The expression to derive the expression for the differential input voltage is given by,

  $\begin{array}{l}\frac{i_{D1}}{I_Q}=\frac{1}{2}\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\\ 0.5=\frac{1}{2}\left(\frac{1}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\\ v_d=V_P\sqrt{2\left(\frac{I_Q}{I_{DSS}}\right)}\end{array}$

Therefore, the expression for the normalized differential voltage is given by,

  $\begin{array}{l}{v^{\prime }}_d=V_P\sqrt{2\left(\frac{I_Q}{I_{D_{SS}}}\right)}\\ \frac{v_d}{\sqrt{2}}=V_P\sqrt{\left(\frac{I_Q}{I_{D_{SS}}}\right)}\\ \left|v_d\right|=\left|V_P\right|\sqrt{\left(\frac{I_Q}{I_{D_{SS}}}\right)}\end{array}$

Conclusion:

Therefore, the required expression is $\left|v_d\right|=\left|V_P\right|\sqrt{\left(\frac{I_Q}{I_{D_{SS}}}\right)}$ .

(c)

To determine

To show: The expression for the maximum forward transconductance is

  $\left(\frac{1}{-V_P}\right)\sqrt{\frac{2I_{DSS}{I}_Q}{2}}$ .

Explanation of Solution

Calculation:

The expression for the current $i_{D1}$ is given by,

  $i_{D1}=\frac{I_Q}{2}+\left(\frac{I_Q}{-2V_P}\right)v_d\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}$

Differentiate both the sides of the equation.

  $\frac{d{i}_{D1}}{d{v}_d}=\left[\begin{array}{l}0+\frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)v_d\cdot \left(\frac{1}{2\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}}\right)+\\ \sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{v_d}{V_P}\right)}^2}\frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)\end{array}\right]$

  ${\frac{d{i}_{D1}}{d{v}_d}|}_{v_d=0}=\left[\begin{array}{l}0+\frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)\left(0\right)\cdot \left(\frac{1}{2\sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{0}{V_P}\right)}^2}}\right)+\\ \sqrt{2\left(\frac{I_{DSS}}{I_Q}\right)-{\left(\frac{I_{DSS}}{I_Q}\right)}^2{\left(\frac{0}{V_P}\right)}^2}\frac{1}{2}{I}_Q\left(\frac{1}{-V_P}\right)\end{array}\right]$

  ${\frac{d{i}_{D1}}{d{v}_d}|}_{v_d=0}=\left(\frac{1}{-V_P}\right)\sqrt{\frac{2\left(\frac{I_{DSS}{I}_Q{}^2}{I_Q}\right)}{4}}$

  ${\frac{d{i}_{D1}}{d{v}_d}|}_{v_d=0}=\left(\frac{1}{-V_P}\right)\sqrt{\frac{2\left(\frac{I_{DSS}{I}_Q{}^2}{I_Q}\right)}{4}}$

Conclusion:

Therefore, the expression for conductance is $\left(\frac{1}{-V_P}\right)\sqrt{\frac{2I_{DSS}{I}_Q}{2}}$ .