Chapter 11, Problem 11.68P

Consider the diff-amp in Figure P11.68. The PMOS parameters are: $K_p=$ $80\mu \text{A}/{\text{V}}^2,{\lambda }_p=0.02{\text{V}}^{-1},V_{TP}=-2\text{V}.$ The NMOS parameters are: $K_n=80\mu \text{A}/{\text{V}}^2,{\lambda }_n=0.015{\text{V}}^{-1},V_{TN}=+2\text{V}.$ (a) Determine the open-circuit differential-mode voltage gain. (b) Compare this value to the gain obtained when $R_1=0.$ (c) What is the output resistance of the diff-amp for parts (a) and (b)?

To determine

The open circuit differential mode voltage gain.

Answer to Problem 11.68P

The value of the differential mode voltage gain is $56.06$ .

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Figure 1

Calculation:

From above, the expression for the source to gate voltage is given by,

  $V_{sg4}\cong I_x{R}_1$

Apply KCL at node $V_1$ .

  $I_x+g_{m4}{V}_{g4}=\frac{V_x-V_{sg4}}{r_{o4}}$

Substitute $I_x{R}_1$ for $V_{sg4}$ in the above equation.

  $\begin{array}{l}{I}_x+g_{m4}{V}_{g4}=\frac{V_x-I_x{R}_1}{r_{o4}}\\ \frac{V_x}{I_x}=r_{o4}\left[1+g_{m4}+\frac{R_1}{r_{O4}}\right]\\ R_O=r_{o4}\left[1+g_{m4}+\frac{R_1}{r_{O4}}\right]\end{array}$

The expression for the transconductance of the second transistor is evaluated as,

  $g_{m2}=2\sqrt{\left(K_n\right)\frac{I_{DQ}}{2}}$

Substitute $80\text{μA}/{\text{V}}^2$ for $K_n$ and $0.2\text{mA}$ for $I_{DQ}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=2\sqrt{\left(80\text{μA}/{\text{V}}^2\right)\frac{0.2\text{mA}}{2}}\\ =0.197\text{mA}/\text{V}\end{array}$

The internal output resistance of the second transistor is given by,

  $r_{O2}=\frac{1}{{\lambda }_n\left(\frac{I_{DQ}}{2}\right)}$

Substitute $0.015{\text{V}}^{-1}$ for ${\lambda }_n$ and $0.2\text{mA}$ for $I_Q$ in the above equation.

  $\begin{array}{c}{r}_{O2}=\frac{1}{\left(0.015{\text{V}}^{-1}\right)\left(\frac{0.2\text{mA}}{2}\right)}\\ =667\text{k}\Omega \end{array}$

The expression for the transconductance of the fourth transistor is evaluated as,

  $g_{m4}=2\sqrt{\left(K_n\right)\frac{I_{DQ}}{2}}$

Substitute $80\text{μA}/{\text{V}}^2$ for $K_n$ and $0.2\text{mA}$ for $I_{DQ}$ in the above equation.

  $\begin{array}{c}{g}_{m4}=2\sqrt{\left(80\text{μA}/{\text{V}}^2\right)\frac{0.2\text{mA}}{2}}\\ =0.179\text{mA}/\text{V}\end{array}$

The internal output resistance of the fourth transistor is given by,

  $r_{O4}=\frac{1}{{\lambda }_p\left(\frac{I_{DQ}}{2}\right)}$

Substitute $0.02{\text{V}}^{-1}$ for ${\lambda }_p$ and $0.2\text{mA}$ for $I_Q$ in the above equation.

  $\begin{array}{c}{r}_{O4}=\frac{1}{\left(0.02{\text{V}}^{-1}\right)\left(\frac{0.2\text{mA}}{2}\right)}\\ =500\text{k}\Omega \end{array}$

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

  $R_O=r_{O4}\left[1+g_m{R}_1+\frac{R_1}{r_{O4}}\right]$

Substitute $500\text{k}\Omega$ for $r_{O4}$ , $0.179\text{mA}/\text{V}$ for $g_{m4}$ and $1\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{R}_O=\left(500\text{k}\Omega \right)\left[1+\left(0.179\text{mA}/\text{V}\right)\left(1\text{k}\Omega \right)+\frac{1\text{k}\Omega }{500\text{k}\Omega }\right]\\ =590.5\text{k}\Omega \end{array}$

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

  $A_d=g_{m2}\left(r_{O2}\left|\right|R_O\right)$

Substitute $0.179\text{mA}/\text{V}$ for $g_{m2}$ , $667\text{k}\Omega$ for $r_{O2}$ and $590.5\text{k}\Omega$ for $R_O$ in the above equation.

  $\begin{array}{c}{A}_d=\left(0.179\text{mA}/\text{V}\right)\left(\left(667\text{k}\Omega \right)\left|\right|\left(590.5\text{k}\Omega \right)\right)\\ =0.179\text{mA}/\text{V}\left(\frac{\left(667\text{k}\Omega \right)\left(590.5\text{k}\Omega \right)}{\left(667\text{k}\Omega \right)+\left(590.5\text{k}\Omega \right)}\right)\\ =56.06\end{array}$

Conclusion:

Therefore, the value of the open circuit differential mode voltage gain is $56.06$ .

To determine

To compare: The value of the open circuit differential mode voltage gain for the given changes with the value obtained in part (a).

Answer to Problem 11.68P

The value of the differential voltage gain is $51.15$ . This indicates, that when $R_1=0$ then there will be decrease in gain value.

Explanation of Solution

Given:

The given circuit is shown in Figure 1

  

Figure 1

The value of resistance $R_1=0$

Calculation:

When $R_1=0$ then $R_o=r_{o4}=500\text{k}\Omega$

The expression for the open circuit differential mode voltage gain is given by,

  $A_d=g_{m2}\left(r_{o2}\left|\right|r_{O4}\right)$

Substitute $0.179\text{mA}/\text{V}$ for $g_{m2}$ , $667\text{k}\Omega$ for $r_{O2}$ and $500\text{k}\Omega$ for $r_{O4}$ in the above equation.

  $\begin{array}{c}{A}_d=\left(0.179\text{mA}/\text{V}\right)\left(\left(667\text{k}\Omega \right)\left|\right|\left(500\text{k}\Omega \right)\right)\\ =0.179\text{mA}/\text{V}\left(\frac{\left(667\text{k}\Omega \right)\left(500\text{k}\Omega \right)}{\left(667\text{k}\Omega \right)+\left(500\text{k}\Omega \right)}\right)\\ =51.15\end{array}$

When the resistance $R_1=0$ the gain will decrease.

Conclusion:

Therefore, the value of the differential voltage gain is $51.15$ .

To determine

The value of the output resistance of the differential amplifier for part (a) and (b).

Answer to Problem 11.68P

The differential output resistance of the amplifier for part (a) is $313\text{k}\Omega$ and part (b) is $286\text{k}\Omega$ .

Explanation of Solution

The given circuit is shown in Figure 1

  

Figure 1

Calculation:

The output resistance for the differential amplifier for the open circuit differential voltage gain is calculated as,

  $R_{od}=r_{o1}\left|\right|R_O$

Substitute $667\text{k}\Omega$ for $r_{O2}$ and $590.5\text{k}\Omega$ for $R_O$ in the above equation.

  $\begin{array}{c}{R}_{od}=667\text{k}\Omega \left|\right|590.5\text{k}\Omega \\ =\frac{\left(667\text{k}\Omega \right)\left(590.5\text{k}\Omega \right)}{\left(667\text{k}\Omega \right)+\left(590.5\text{k}\Omega \right)}\\ =313\text{k}\Omega \end{array}$

The output resistance for the differential amplifier when $R_1=0$ is calculated as,

  $R_{od}=r_{o2}\left|\right|r_{o4}$

Substitute $667\text{k}\Omega$ for $r_{O2}$ and $500\text{k}\Omega$ for $r_{o4}$ in the above equation.

  $\begin{array}{c}{R}_{od}=667\text{k}\Omega \left|\right|500\text{k}\Omega \\ =\frac{\left(667\text{k}\Omega \right)\left(500\text{k}\Omega \right)}{\left(667\text{k}\Omega \right)+\left(500\text{k}\Omega \right)}\\ =286\text{k}\Omega \end{array}$

Conclusion:

Therefore, the differential output resistance of the amplifier is $313\text{k}\Omega$ and when the value for $R_1=0$ is $286\text{k}\Omega$ .