Chapter 7, Problem 7.69P

For the PMOS common−source circuit shown in Figure P769, the transistor parameters are: $V_{TP}=-2\text{V}$ , $K_P{\text{=1mA/V}}^2$ , $\lambda =0$ , $C_{gs}=15\text{pF}$ , and $C_{gd}=3\text{pF}$ . (a) Determine the upper 3dB frequency. (b) What is the equivalent Miller capacitance? State any assumptions or approximations that you make. (c) Find the midband voltage gain.

Figure P7.69

To determine

The upper $3\text{dB}$ frequency

Answer to Problem 7.69P

The upper $3\text{dB}$ frequency is $10.4\text{MHz}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Calculate the gate voltage $V_G$

  $\begin{array}{l}{V}_G=\left(\frac{R_2}{R_1+R_2}\right)[10-(-10)]-V_{DD}\\ =\left(\frac{R_2}{R_1+R_2}\right)[20]-V_{DD}\end{array}$

Substitute $22\times {10}^3$ for $R_2,8\times {10}^3$ for $R_1,$ and 10 for $V_{DD}$

  $V_G=\left(\frac{22}{22+8}\right)20-10$

  $V_G=4.67\text{V}$

The expression for drain current $I_D$ is,

  $I_D=\frac{V_{CC}-V_{SG}-V_G}{R_S}$

Substitute 10 for $V_{CC},0.5\times {10}^3$ for $R_S$ and 4.67 for $V_G$

  $I_D=\frac{10-V_{SG}-4.67}{0.5\times {10}^3}\mathrm{...........}(1)$

Consider the another expression for drain current $I_D$

  $I_D={10}^{-3}{\left(V_{SG}-2\right)}^2\dots \dots (2)$

Equate equations (1) and (2)

  $\frac{10-V_{SG}-4.67}{0.5\times {10}^3}={10}^{-3}{\left(V_{\text{SG}}-2\right)}^2$

  $10-V_{SG}-4.67=0.5\left(V_{SG}^2-4V_{SG}+4\right)$

  $20-2V_{SG}-9.34=V_{SG}^2-4V_{SG}+4$

  $V_{SG}^2-2V_{SG}-6.66=0$

Calculate source-to-gate voltage $V_{\text{SG}}$

  $V_{SG}=\frac{2\pm \sqrt{4-4(1)(-6.66)}}{2(1)}$

  $=3.77\text{V}$

Calculate transconductance $g_m$

  $g_m=2K_p\left(V_{SG}+V_{TP}\right)$

Substitute ${10}^{-3}$ for $K_P,3.77$ for $V_{SC}$ and -2 for $V_{TP}$

  $g_m=2\left({10}^{-3}\right)(3.77-2)$

  $g_m=3.54\text{mA}/\text{V}$

Calculate the Miller Capacitance.

  $C_M=C_{Sd}\left[1+g_{=}\left(R_D\parallel R_L\right)\right]$

Substitute $3\times {10}^{-12}$ for $C_{PD},3.54\times {10}^{-3}$ for $g_m,2\times {10}^3$ for $R_D,$ and $5\times {10}^3$ for $R_L$

  $\begin{array}{l}{C}_M=3\times {10}^{-12}\left[1+3.54\times {10}^{-3}\left(2\times {10}^3\parallel 5\times {10}^3\right)\right]\\ =3\times {10}^{-12}\left[1+3.54\times {10}^{-3}\left(\frac{2\times 5\times {10}^3}{2+5}\right)\right]\\ =18.2\text{pF}\end{array}$

Calculate the time constant $\tau$

  $\tau =R_{eq}\left(C_{gs}+C_M\right)\mathrm{............}(3)$

Where

  $R_{eq}=R_i\parallel R_1\parallel R_2$

  $R_{eq}=0.5\parallel 8\parallel 22$

  $R_{eq}=0.461\text{kΩ}$

Recall equation (3).

  $\tau =R_{eq}\left(C_{gs}+C_M\right)$

Substitute $0.461\times {10}^3$ for $R_{eq},15\times {10}^{-12}$ for $C_{gs},$ and $18.2\times {10}^{-12}$ for ${\text{C}}_{\text{M}}$

  $\tau =\left(0.461\times {10}^3\right)(15+18.2)\times {10}^{-12}$

  $\tau =15.30\text{ns}$

Calculate the upper 3 dB frequency $f_B$

  $\begin{array}{l}{f}_B=\frac{1}{2\pi \tau }\hfill \\ \text{Substitute}15.3\times {10}^{-9}\text{for}\tau \hfill \\ f_B=\frac{1}{2\pi \left(15.30\times {10}^{-9}\right)}\hfill \\ =10.4\text{MHz}\hfill \end{array}$

Therefore, the upper $3\text{dB}$ frequency is $10.4\text{MHz}$

To determine

The value of the Miller capacitance

Answer to Problem 7.69P

The value of the Miller capacitance is $18.2\text{pF}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Calculate the gate voltage $V_G$

  $\begin{array}{l}{V}_G=\left(\frac{R_2}{R_1+R_2}\right)[10-(-10)]-V_{DD}\\ =\left(\frac{R_2}{R_1+R_2}\right)[20]-V_{DD}\end{array}$

Substitute $22\times {10}^3$ for $R_2,8\times {10}^3$ for $R_1,$ and 10 for $V_{DD}$

  $V_G=\left(\frac{22}{22+8}\right)20-10$

  $V_G=4.67\text{V}$

The expression for drain current $I_D$ is,

  $I_D=\frac{V_{CC}-V_{SG}-V_G}{R_S}$

Substitute 10 for $V_{CC},0.5\times {10}^3$ for $R_S$ and 4.67 for $V_G$

  $I_D=\frac{10-V_{SG}-4.67}{0.5\times {10}^3}\mathrm{...........}(1)$

Consider the another expression for drain current $I_D$

  $I_D=K_P{\left(V_{SG}+V_{TP}\right)}^2$

Substitute ${10}^{-3}$ for $K_P$ and $-2$ for $V_{TP}$

  $I_D={10}^{-3}{\left(V_{SG}-2\right)}^2\dots \dots (2)$

Equate equations (1) and (2)

  $\frac{10-V_{SG}-4.67}{0.5\times {10}^3}={10}^{-3}{\left(V_{\text{SG}}-2\right)}^2$

  $10-V_{SG}-4.67=0.5\left(V_{SG}^2-4V_{SG}+4\right)$

  $20-2V_{SG}-9.34=V_{SG}^2-4V_{SG}+4$

  $V_{SG}^2-2V_{SG}-6.66=0$

Calculate source-to-gate voltage $V_{\text{SG}}$

  $V_{SG}=\frac{2\pm \sqrt{4-4(1)(-6.66)}}{2(1)}$

  $=3.77\text{V}$

Calculate transconductance $g_m$

  $g_m=2K_p\left(V_{SG}+V_{TP}\right)$

Substitute ${10}^{-3}$ for $K_P,3.77$ for $V_{SC}$ and -2 for $V_{TP}$

  $g_m=2\left({10}^{-3}\right)(3.77-2)$

  $g_m=3.54\text{mA}/\text{V}$

Calculate the Miller Capacitance.

  $C_M=C_{gd}\left[1+g_m\left(R_D\parallel R_L\right)\right]$

Substitute $3\times {10}^{-12}$ for $C_{PD},3.54\times {10}^{-3}$ for $g_m,2\times {10}^3$ for $R_D,$ and $5\times {10}^3$ for $R_L$

  $\begin{array}{l}{C}_M=3\times {10}^{-12}\left[1+3.54\times {10}^{-3}\left(2\times {10}^3\parallel 5\times {10}^3\right)\right]\\ =3\times {10}^{-12}\left[1+3.54\times {10}^{-3}\left(\frac{2\times 5\times {10}^3}{2+5}\right)\right]\\ =18.2\text{pF}\end{array}$

The value of the Miller capacitance is $18.2\text{pF}$

To determine

The mid-band voltage gain.

Answer to Problem 7.69P

The mid-band voltage gain is $-4.66$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Draw the small-signal model of the circuit shown in figure

  

Determine the expression for output voltage $v_o$

Apply the voltage division principle.

  $v_o=g_m{V}_{sg}\left(R_D\parallel R_L\right)\dots \dots \text{(4)}$

Determine source-to-gate voltage $V_{gs}$

  $V_{\text{sg}}=-\left[\frac{\left(R_1\parallel R_2\right)}{\left(R_1\parallel R_2\right)+R_i}\right]v_i$

Substitute $8\times {10}^3$ for $R_1,22\times {10}^3$ for $R_2,$ and $0.5\times {10}^3$ for $R_i$

  $\begin{array}{l}{V}_{\text{sg}}=-\left[\frac{\left(8\times {10}^3\parallel 22\times {10}^3\right)}{\left(8\times {10}^3\parallel 22\times {10}^3\right)+0.5\times {10}^3}\right]v_i\\ =-\left[\frac{\left(5.8666\times {10}^3\right)}{\left(5.8666\times {10}^3\right)+0.5\times {10}^3}\right]v_i\\ =-0.9215v_i\end{array}$

Substitute $0.9215v_i$ for $V_{sg}$ in equation (4)

  $v_o=g_m\left(-0.9215v_i\right)\left(R_D\parallel R_L\right)$

  $\frac{v_o}{v_i}=-g_m(0.9215)\left(R_D\parallel R_L\right)$

Therefore, the expression for the mid-band voltage gain is,

  $A_v=-g_m(0.9215)\left(R_D\parallel R_L\right)$

Substitute $3.54\times {10}^{-3}$ for $3.54\times {10}^{-3}$ for $g_m,2\times {10}^3$ for $R_D,$ and $5\times {10}^3$ for $R_L$

  $\begin{array}{l}{A}_v=-3.54\times {10}^{-3}(0.9215)\left(2\times {10}^3\parallel 5\times {10}^3\right)\\ =-4.66\end{array}$