Chapter 4, Problem 4.70P

Consider the circuit shown in Figure P4.70. The transistor parameters are $V_{TP1}=-0.4\text{V}$ , $V_{TP2}=0.4\text{V}$ , ${\left(W/L\right)}_L=20$ , ${\left(W/L\right)}_L=80$ , ${k^{\prime }}_p=40\mu {\text{A/V}}^{\text{2}}$ , ${k^{\prime }}_n=100\mu {\text{A/V}}^{\text{2}}$ , and ${\lambda }_1={\lambda }_2=0$ . Let $R_{\text{in}}=200\text{k}\Omega$ . (a) Design the circuit such that $I_{DQ1}=0.1\text{mA}$ , $I_{DQ2}=0.3\text{mA}$ , $V_{SDQ1}=1.0\text{V}$ , and $V_{SDQ2}=2.0\text{V}$ . The voltage across $R_{S1}$ is to be 0.6 V. (b) Determine the small−signal voltage gain $A_{\upsilon }={\upsilon }_o,{\upsilon }_i$ . (c) Find the small−signal output resistance $R_o$ .

Figure P4.70

To determine

The design parameters of a multistage transistor circuit to meet the specifications.

Answer to Problem 4.70P

The values are:

  $R_{S2}=4.42\text{ k}\Omega$

  $R_{S1}=6\text{ k}\Omega$

  $R_1=342.86\text{ k}\Omega$

  $R_2=479.99\text{ k}\Omega$

  $R_{D1}=20\text{ k}\Omega$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TP1}=-0.4;V_{TN2}=0.4\text{ V, }{\left(\frac{W}{L}\right)}_1=20,{\left(\frac{W}{L}\right)}_2=80,k_p^{\text{'}}\text{=40 }\mu \text{A}/{\text{V}}^{\text{2}},k_{n_1}^{\text{'}}\text{=100 }\text{mA}/{\text{V}}^{\text{2}}\text{, }\\ {\lambda }_1\text{=}{\lambda }_2\text{=0, }{R}_{in}=200\text{ k}\Omega ,I_{DQ1}=0.1\text{ mA, }{I}_{DQ2}=0.3\text{ mA, }{V}_{SDQ1}=1\text{ V, }{V}_{DSQ2}=2\text{ V, }\\ V_{RS1}=0.6\text{ V}\end{array}$

The given circuit is shown below.

  

Calculation:

The value of source resistor of transistor 1 is

  $\begin{array}{c}{R}_{S1}=\frac{V_{RS1}}{I_{DQ1}}\\ \text{=}\frac{0.6}{0.1}\text{ k}\Omega \\ R_S=6\text{ k}\Omega \end{array}$

The equation of drain current in M₁ transistor,

  $\begin{array}{l}{I}_{DQ1}=\frac{k_p^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{\left(\text{}{V}_{SGQ1}+V_{TP1}\right)}^2\\ 0.1=\frac{0.04}{2}\times 20{\left(\text{}{V}_{SGQ1}-0.4\right)}^2\\ V_{SGQ1}=0.9\text{ V}\end{array}$

The equation of drain current in M₂ transistor,

  $\begin{array}{l}{I}_{DQ2}=\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_n{\left(\text{}{V}_{SGQ1}-V_{TN1}\right)}^2\\ 0.3=\frac{0.1}{2}\times 80{\left(\text{}{V}_{SGQ1}-0.4\right)}^2\\ V_{SGQ2}=0.6739\text{ V}\end{array}$ ’

The gate voltage of the M₁ transistor is

  $\begin{array}{c}{V}_{G1}=V^{+}-V_{RS1}-V_{GSQ1}\\ =1.8-0.6-0.9\\ =0.3\text{ V}\end{array}$

Also, the gate voltage of the M₁ transistor is

  $\begin{array}{c}{V}_{G1}=\frac{R_2}{R_1+R_2}\left(V^{+}-V^{-}\right)+V^{-}\\ =\frac{R_{in}}{R_1}{V}_{DD}\because R_{in}=\frac{R_1{R}_2}{R_1+R_2}\\ 0.3=\frac{200\times {10}^3}{R_1}\left(3.6\right)-1.8\\ R_1=342.86\text{k}\Omega \end{array}$

Hence,

  $200=\frac{342.86R_2}{342.86+R_2}$

Which yields

  $R_2=479.99\text{ k}\Omega$

From a KVL equation around the drain to source loop,

  $\begin{array}{l}{V}^{+}=V_{SDQ1}+I_{DQ1}\left(R_{S1}+R_{D1}\right)+V^{-}\\ 1.8=1+0.6+\left(0.1\times {10}^{-3}\right)R_{D1}-1.8\\ R_{D1}=20\text{k}\Omega \end{array}$

Now,

  $\begin{array}{l}{V}_{S2}=V_{D1}-V_{GSQ2}\\ V_{S2}=V^{-}+I_{DQ1}{R}_{D1}-V_{GSQ2}\\ V_{S2}=-1.8+0.1\times {10}^{-3}\times 20\times {10}^3-0.6739\\ V_{S2}=-0.4739\text{ V}\end{array}$

Then,

  $\begin{array}{c}{R}_{S2}=\frac{V_{S2}-V^{-}}{I_{DQ2}}\\ =\frac{-0.4739+1.8}{0.3\times {10}^3}\\ R_{S2}=4.42\text{ k}\Omega \end{array}$

To determine

The small-signal voltage gain of the given circuit.

Answer to Problem 4.70P

  $A_v=-2.13$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TP1}=-0.4;V_{TN2}=0.4\text{ V, }{\left(\frac{W}{L}\right)}_1=20,{\left(\frac{W}{L}\right)}_2=80,k_p^{\text{'}}\text{=40 }\mu \text{A}/{\text{V}}^{\text{2}},k_{n_1}^{\text{'}}\text{=100 }\text{mA}/{\text{V}}^{\text{2}}\text{, }\\ {\lambda }_1\text{=}{\lambda }_2\text{=0, }{R}_{in}=200\text{ k}\Omega ,I_{DQ1}=0.1\text{ mA, }{I}_{DQ2}=0.3\text{ mA, }{V}_{SDQ1}=1\text{ V, }{V}_{DSQ2}=2\text{ V, }\\ V_{RS1}=0.6\text{ V}\end{array}$

The given circuit is shown below.

  

Calculation:

Draw the small-signal equivalent circuit as below.

  

The transconductance of transistors is,

  $\begin{array}{c}{g}_{m1}=2\sqrt{\frac{k_p^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_1{I}_{DQ1}}\\ =2\sqrt{0.02\times 20\times 0.1}\\ =\text{0}\text{.4 mA}/\text{V}\\ g_{m2}=2\sqrt{\frac{k_n^{\text{'}}}{2}{\left(\frac{W}{L}\right)}_2{I}_{DQ2}}\\ =2\sqrt{0.05\times 80\times 0.3}\\ =\text{2}\text{.191 mA}/\text{V}\end{array}$

The output voltage of the transistor 2,

  $V_O=g_{m2}{V}_{gs2}{R}_{S2}$

The output voltage of the first stage or the input to the second stage is

  $V_{O1}=g_{m1}{V}_{sg1}{R}_{D1}=V_{gs2}+g_{m2}{V}_{gs2}{R}_{s2}$

The input voltage of the first stage,

  $V_i=-\left(V_{sg1}+g_{m1}{V}_{sg1}{R}_{s1}\right)$

The voltage gain of the circuit

  $\begin{array}{c}{A}_v=\frac{g_{m1}{V}_{sg1}{R}_{D1}}{-\left(V_{sg1}+g_{m1}{V}_{sg1}{R}_{s1}\right)}\times \frac{g_{m2}{V}_{gs2}{R}_{S2}}{V_{gs2}+g_{m2}{V}_{gs2}{R}_{s2}}\\ A_v=\frac{-g_{m1}{R}_{D1}}{\left(1+g_{m1}{R}_{s1}\right)}\times \frac{g_{m2}{R}_{S2}}{\left(1+g_{m2}{R}_{s2}\right)}\\ A_v=\frac{-0.4\times 20}{\left(1+0.4\times 6\right)}\times \frac{2.191\times 4.42}{\left(1+2.191\times 4.42\right)}\\ =-2.13\end{array}$

To determine

The small-signal output resistance of the given circuit.

Answer to Problem 4.70P

  $R_O=413.69\text{ k}\Omega$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}{V}_{TP1}=-0.4;V_{TN2}=0.4\text{ V, }{\left(\frac{W}{L}\right)}_1=20,{\left(\frac{W}{L}\right)}_2=80,k_p^{\text{'}}\text{=40 }\mu \text{A}/{\text{V}}^{\text{2}},k_{n_1}^{\text{'}}\text{=100 }\text{mA}/{\text{V}}^{\text{2}}\text{, }\\ {\lambda }_1\text{=}{\lambda }_2\text{=0, }{R}_{in}=200\text{ k}\Omega ,I_{DQ1}=0.1\text{ mA, }{I}_{DQ2}=0.3\text{ mA, }{V}_{SDQ1}=1\text{ V, }{V}_{DSQ2}=2\text{ V, }\\ V_{RS1}=0.6\text{ V}\end{array}$

The given circuit is shown below.

  

Calculation:

Draw the small-signal equivalent circuit as below.

  

The small-signal output resistance

  $R_O=\frac{1}{g_{m2}}\parallel R_{S2}$

Then from part (a) and (b), $R_{S2}=4.42\text{ k}\Omega$ , $g_{m2}=\text{2}\text{.191 mA}/\text{V}$

  $R_O=\frac{1}{2.191}\parallel 4.42\text{ k}\Omega$

  $R_O=413.69\text{ k}\Omega$