Chapter 4, Problem 4.78P

For the p−channel JFET source−follower circuit in Figure P4.78, the transistor parameters are: $I_{DSS}=2\text{mA}$ , $V_P=+1.75\text{V}$ , and $\lambda =0$ . (a) Determine $I_{DQ}$ and $V_{SDQ}$ . (b) Determine the small−signal gains $A_{\upsilon }={\upsilon }_o/{\upsilon }_i$ and $A_i=i_o/i_i$ . (c) Determine the maximum symmetrical swing in the output voltage.

Figure P4.78

To determine

The value of the $I_{DQ}$ and $V_{SDQ}$ .

Answer to Problem 4.78P

The value of the drain current $I_{DSQ}$ is $1\text{mA}$ and $V_{SDQ}$ is $5\text{V}$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

Figure 1

Calculation:

The value of the voltage across the resistance $R_1$ is given by,

  $\begin{array}{c}{V}_{R1}=\frac{90\text{k}\Omega }{90\text{k}\Omega +110\text{k}\Omega }\left(10\text{V}\right)\\ =4.5\text{V}\end{array}$

The expression to determine the value of the voltage $V_{R1}$ is given by,

  $V_{R1}=V_G-\left(5\text{k}\Omega \right)I_Q$

Substitute $4.5\text{V}$ for $V_{R1}$ in the above equation.

  $\begin{array}{l}4.5\text{V}=V_G-\left(5\text{k}\Omega \right)I_Q\\ I_Q=\frac{4.5\text{V}-V_{SG}}{5\text{k}\Omega }\end{array}$

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

  $I_D=\frac{4.5\text{V}+V_{SG}}{5\text{k}\Omega }$

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

  $I_D=I_{DSS}{\left(1-\frac{V_{GS}}{V_P}\right)}^2$ ….. (1)

Substitute $2\text{mA}$ for $I_{DSS}$ , $\frac{4.5\text{V}-V_{SG}}{5\text{k}\Omega }$ for $I_D$ and $1.75\text{V}$ for $V_P$ in the above equation.

  $\begin{array}{l}\frac{4.5\text{V}-V_{SG}}{5\text{k}\Omega }=\left(2\text{mA}\right){\left(1-\frac{V_{GS}}{1.75\text{V}}\right)}^2\\ V_{GS}=0.51\text{V}\end{array}$

Substitute $2\text{mA}$ for $I_{DSS}$ , $0.51\text{V}$ for $V_{GS}$ and $1.75\text{V}$ for $V_P$ in equation (1).

  $\begin{array}{c}{I}_D=\left(2\text{mA}\right){\left(1-\frac{0.51\text{V}}{1.75\text{V}}\right)}^2\\ =1\text{mA}\end{array}$

Mark the values and redraw the circuit.

The required diagram is shown in Figure 2

  

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

  $V_{SDQ}=10\text{V}-I_D\left(5\text{mA}\right)$

Substitute $1\text{mA}$ for $I_D$ in the above equation.

  $\begin{array}{c}{V}_{SDQ}=10\text{V}-\left(1\text{mA}\right)\left(5\text{mA}\right)\\ =5\text{V}\end{array}$

Conclusion:

Therefore, the value of the drain current $I_{DSQ}$ is $1\text{mA}$ and $V_{SDQ}$ is $5\text{V}$ .

To determine

The current gain and the voltage gain of the circuit.

Answer to Problem 4.78P

The value of the current gain of the circuit is $4.18$ and the voltage gain is $0.844$ .

Explanation of Solution

Calculation:

The expression for conductance is given by,

  $g_m=\frac{2I_{DSS}}{\left|V_P\right|}\left(1-\frac{V_{GS}}{V_P}\right)$

Substitute $2\text{mA}$ for $I_{DSS}$ , $1.75\text{V}$ for $V_P$ and $0.51\text{V}$ for $V_{GS}$ in the above equation.

  $\begin{array}{c}{g}_m=\frac{2\left(2\text{mA}\right)}{\left|1.75\text{V}\right|}\left(1-\frac{0.51\text{V}}{1.75\text{V}}\right)\\ =1.62\text{mA}/\text{V}\end{array}$

The expression to determine the voltage gain is given by,

  $A_v=\frac{g_m\left(R_S\left|\right|R_L\right)}{1+g_m\left(R_S\left|\right|R_L\right)}$

Substitute $1.62\text{mA}/\text{V}$ for

  $g_m$ , $5\text{k}\Omega$ for $R_L$ and $10\text{k}\Omega$ for $R_S$ in the above equation.

  $\begin{array}{c}{A}_v=\frac{\left(1.62\text{mA}/\text{V}\right)\left(\left(5\text{k}\Omega \right)\left|\right|\left(10\text{k}\Omega \right)\right)}{1+\left(1.62\text{mA}/\text{V}\right)\left(\left(5\text{k}\Omega \right)\left|\right|\left(10\text{k}\Omega \right)\right)}\\ =0.844\end{array}$

Mark the value and draw the small signal circuit.

The required diagram is shown in Figure 3

  

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

  $R_i=\frac{R_1{R}_2}{R_1+R_2}$

Substitute $110\text{k}\Omega$ for $R_2$ and $90\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{R}_i=\frac{\left(90\text{k}\Omega \right)\left(110\text{k}\Omega \right)}{\left(90\text{k}\Omega \right)+\left(110\text{k}\Omega \right)}\\ =49.5\text{k}\Omega \end{array}$

The expression to determine the value of the current gain is given b, $A_i=A_v\left(\frac{R_i}{R_L}\right)$

Substitute $0.844$ for $A_v$ , $110\text{k}\Omega$ for $R_L$ and $49.5\text{k}\Omega$ for $R_i$ in the above equation.

  $\begin{array}{c}{A}_i=\left(0.844\right)\left(\frac{49.5\text{k}\Omega }{110\text{k}\Omega }\right)\\ =4.18\end{array}$

Conclusion:

Therefore, the value of the current gain of the circuit is $4.18$ and the voltage gain is $0.844$ .

To determine

The value of the maximum symmetrical swing in the output voltage.

Answer to Problem 4.78P

The value of the maximum voltage swing is $6.66\text{V}$ .

Explanation of Solution

Calculation:

The expression to determine the maximum voltage swing is given by,

  $v_{sd}=\left(\frac{R_S{R}_L}{R_S+R_L}\right)I_D$

Substitute $1\text{mA}$ for $I_D$ , $10\text{k}\Omega$ for $R_L$ and $5\text{k}\Omega$ for $R_S$ in the above equation.

  $\begin{array}{c}{v}_{sd}=\left(\frac{\left(10\text{k}\Omega \right)\left(5\text{k}\Omega \right)}{\left(10\text{k}\Omega \right)+\left(5\text{k}\Omega \right)}\right)\left(1\text{mA}\right)\\ =3.33\text{V}\end{array}$

The maximum voltage swing is twice the source to drain voltage which is $6.66\text{V}$ .

Conclusion:

Therefore, the value of the maximum voltage swing is $6.66\text{V}$ .