Chapter 4, Problem 4.67P

The circuit in Figure P4.67 is a simplified ac equivalent circuit of a folded cascode amplifier. The transistor parameters are $\left|V_{TN}\right|=\left|V_{TP}\right|=0.5\text{V}$ , $K_n=K_p=2{\text{mA/V}}^{\text{2}}$ and ${\lambda }_n={\lambda }_p=0.1{\text{V}}^{-1}$ . Assume the current source $2I_Q=200\mu \text{A}$ is ideal and the resistance looking into the current source $I_Q=100\mu \text{A}$ is $50\text{k}\Omega$ . Determine the (a) small−signal parameters of each transistor, (b) small−signal voltage gain, and (c) output resistance $R_o$ .

Figure P4.67

To determine

The small-signal parameters of each transistors.

Answer to Problem 4.67P

  $r_{o1}=100k\Omega$ , $r_{o2}=100\text{ k}\Omega$ , $g_{m1}=0.8944\text{mA}/\text{V}$ , $g_{m2}=0.8944\text{mA}/\text{V}$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}\left|V_{TN}\right|=\left|V_{TP}\right|=0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p={\lambda }_n=0.1{\text{ V}}^{-1},K_n=K_p=2\text{ mA}/{\text{V}}^2\text{,}{I}_Q=200\mu \text{A },\\ r_o=50\text{ k}\Omega \end{array}$

The given circuit is shown below.

  

Calculation:

The small-signal output resistance of transistor M₁

  $r_{o1}=\frac{1}{{\lambda }_n{I}_Q}=\frac{1}{0.1\times 0.1}\text{ k}\Omega$

  $r_{o1}=100k\Omega$

Similarly, the small-signal output resistance of transistors M₂ and M₃

  $r_{o2}=\frac{1}{{\lambda }_p{I}_Q}=\frac{1}{0.1\times 0.1}$

  $r_{o2}=100\text{ k}\Omega$

Then the transconductance the transistor M₁ ,

  $\begin{array}{l}{g}_{m1}=2\sqrt{K_n{I}_Q}\\ g_{m1}=2\sqrt{2\left(0.1\right)}\end{array}$

  $g_{m1}=0.8944\text{mA}/\text{V}$

Similarly,

  $\begin{array}{l}{g}_{m2}=2\sqrt{K_p{I}_Q}\\ g_2=2\sqrt{2\left(0.1\right)}\end{array}$

  $g_{m2}=0.8944\text{mA}/\text{V}$

To determine

The small-signal voltage gain.

Answer to Problem 4.67P

  $A_v=-44.02$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}\left|V_{TN}\right|=\left|V_{TP}\right|=0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p={\lambda }_n=0.1{\text{ V}}^{-1},K_n=K_p=2\text{ mA}/{\text{V}}^2\text{,}{I}_Q=200\mu \text{A },\\ r_o=50\text{ k}\Omega \end{array}$

The given circuit is shown below.

  

Calculation:

The small signal equivalent circuit is in the below figure.

  

From part (a),

  $r_{o1}=100k\Omega$ , $r_{o2}=100\text{ k}\Omega$ , $g_{m1}=0.8944\text{mA}/\text{V}$ , $g_{m2}=0.8944\text{mA}/\text{V}$

Considering the small signal circuit, add the currents at the output node of the circuit,

  $\begin{array}{l}\frac{V_o}{r_o}+\frac{V_o-V_{sg2}}{r_{o2}}=g_{m2}{V}_{sg2}\\ V_o\left(\frac{1}{r_o}+\frac{1}{r_{o2}}\right)=V_{sg2}\left(g_{m2}+\frac{1}{r_{o2}}\right)\\ V_o\left(\frac{1}{50}+\frac{1}{100}\right)=V_{sg2}\left(0.8944+\frac{1}{100}\right)\\ V_{sg2}=\frac{3V_o}{90.44}\to (1)\end{array}$

Summing the currents at output of the transistor 1

  $\begin{array}{l}{g}_m{V}_i+\frac{V_{sg2}}{r_{o1}}+g_{m2}{V}_{sg2}+\frac{V_{sg2}-V_o}{r_{o2}}=0\\ V_{sg2}\left(\frac{1}{r_{o1}}+g_{m2}+\frac{1}{r_{o2}}\right)-\frac{V_o}{r_{o2}}=-g_{m1}{V}_i\end{array}$

Substituting the equation (1),

  $\begin{array}{l}\frac{3V_o}{90.44}\left(\frac{1}{100}+0.8944+\frac{1}{100}\right)-\frac{V_o}{100}=-0.8944V_i\\ V_o\left(-90.44+268.2+6\right)=-0.8944\times 90.44\times 100V_i\\ \frac{V_o}{V_i}=-44.02\end{array}$

Then the small-signal voltage

  $A_v=-44.02$

To determine

The output resistance of the circuit.

Answer to Problem 4.67P

  $R_o=49.73\text{ k}\Omega$

Explanation of Solution

Given Information:

The given values are:

  $\begin{array}{l}\left|V_{TN}\right|=\left|V_{TP}\right|=0.5\text{ V,}{\lambda }_n=0.05{\text{ V}}^{-1},{\lambda }_p={\lambda }_n=0.1{\text{ V}}^{-1},K_n=K_p=2\text{ mA}/{\text{V}}^2\text{,}{I}_Q=200\mu \text{A },\\ r_o=50\text{ k}\Omega \end{array}$

The given ircuit is shown below.

  

Calculation:

The small signal equivalent circuit for determining the output resistance is shown in the below figure.

  

To find the output resistance, find the relationship between $I_x{\text{ and V}}_x$ .

From part (a).

  $r_{o1}=100k\Omega$ , $r_{o2}=100\text{ k}\Omega$ , $g_{m1}=0.8944\text{mA}/\text{V}$ , $g_{m2}=0.8944\text{mA}/\text{V}$

Adding the currents at the output node,

  $\begin{array}{l}{I}_x=\frac{V_x}{r_o}\text{+}\frac{V_x-V_{sg2}}{r_{o2}}-g_{m2}{\text{V}}_{gs2}\\ I_x=V_x\left(\frac{1}{r_o}\text{+}\frac{1}{r_{o2}}\right)-\left(g_{m2}+\frac{1}{r_{o2}}\right){\text{V}}_{gs2}\\ I_x=V_x\left(\frac{1}{50}\text{+}\frac{1}{100}\right)-\left(0.8944+\frac{1}{100}\right){\text{V}}_{gs2}\to (1)\end{array}$

Also summing the currents at the input side.

  $\begin{array}{l}\frac{V_{gs2}}{r_{o1}}\text{+}{g}_{m2}{V}_{gs2}+\frac{V_{sg2}-V_x}{r_{o2}}=0\\ V_{gs2}\left(\frac{1}{r_{o1}}+g_{m2}+\frac{1}{r_{o2}}\right)=\frac{V_x}{r_{o2}}\\ V_{gs2}\left(\frac{1}{100}+0.8944+\frac{1}{100}\right)=\frac{V_x}{100}\\ V_{gs2}=\frac{V_x}{91.44}\end{array}$

Then substituting in equation (1), we get

  $\begin{array}{l}{I}_x=V_x\left(\frac{1}{50}\text{+}\frac{1}{100}\right)-\left(0.8944+\frac{1}{100}\right)\frac{V_x}{91.44}\\ \frac{V_x}{I_x}=\frac{91.44\times 100}{91.44\left(3\right)-90.44}=49.73\text{ k}\Omega \end{array}$

Then the output resistance is

  $R_o=49.73\text{ k}\Omega$