Chapter 11, Problem 11.80P

(a)

To determine

The small signal parameters for each of the transistor and the value of the composite transconductance for the given specifications.

Answer to Problem 11.80P

The value of the small signal parameters are $g_{m2}=9.5517\text{mA}/\text{V}$ , $r_{\pi 2}=15.7\text{k}\Omega$ , $g_{m1}=448.69\text{μA}/\text{V}$ and the value of the composite transconductance is $g_m^C=8.416\text{mA}/\text{V}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The expression to determine the value of the emitter current of the second transistor is calculated as,

  $\begin{array}{c}{I}_{E2}=I_{Bias2}-I_{Bias1}\\ =0.50\text{mA}-0.25\text{mA}\\ =0.25\text{mA}\end{array}$

The expression to determine the value of the current $I_{C2}$ is given by,

  $I_{C2}=\left(\frac{\beta }{1+\beta }\right)I_{E2}$

Substitute $150$ for $\beta$ and $0.25\text{mA}$ for $I_{E2}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=\left(\frac{150}{1+150}\right)0.25\text{mA}\\ =248.344\text{μA}\end{array}$

The value of the transconductance $g_{m2}$ is calculated as,

  $\begin{array}{c}{g}_{m2}=\frac{I_{C2}}{0.026\text{V}}\\ =\frac{248.344\text{μA}}{0.026\text{V}}\\ =9.5517\text{mA}/\text{V}\end{array}$

The expression to determine the value of the small signal resistance is given by,

  $r_{\pi 2}=\frac{\beta }{g_{m2}}$

Substitute $150$ for $\beta$ and $9.5517\text{mA}/\text{V}$ for $g_m$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{150}{9.5517\text{mA}/\text{V}}\\ =15.7\text{k}\Omega \end{array}$

The value of the drain current $I_{D1}$ is given by,

  $\begin{array}{c}{I}_{D1}=I_{Bias2}-I_{C2}\\ =0.50\text{mA}-248.344\text{μA}\\ =\text{251}\text{.66}\text{μA}\end{array}$

The value of the transconductance $g_{m1}$ is given by,

  $g_{m1}=2\sqrt{K_n{I}_{D1}}$

Substitute $0.2\text{mA}/{\text{V}}^2$ for $k_n$ and $251.66\text{μA}$ for $I_{D1}$ in the above equation.

  $\begin{array}{c}{g}_{m1}=2\sqrt{\left(0.2\text{mA}/{\text{V}}^2\right)\left(251.66\text{μA}\right)}\\ =448.69\text{μA}/\text{V}\end{array}$

The expression to determine the value of the composite transconductance is given by,

  $g_m^C=\frac{g_{m1}\left(1+g_{m2}{r}_{\pi }\right)}{1+g_{m1}{r}_{\pi }}$

Substitute $448.69\text{μA}/\text{V}$ for $g_{m1}$ , $9.5517\text{mA}/\text{V}$ for $g_{m2}$ and $15.7\text{k}\Omega$ for $r_{\pi }$ in the above equation.

  $\begin{array}{c}{g}_m^C=\frac{448.69\text{μA}/\text{V}\left(1+\left(9.5517\text{mA}/\text{V}\right)\left(15.7\text{k}\Omega \right)\right)}{1+\left(448.69\text{μA}/\text{V}\right)15.7\text{k}\Omega }\\ =8.416\text{mA}/\text{V}\end{array}$

Conclusion:

Therefore, the value of the small signal parameters are $g_{m2}=9.5517\text{mA}/\text{V}$ , $r_{\pi 2}=15.7\text{k}\Omega$ , $g_{m1}=448.69\text{μA}/\text{V}$ and the value of the composite transconductance is $g_m^C=8.416\text{mA}/\text{V}$ .

(b)

To determine

The small signal parameters for each of the transistor and the value of the composite transconductance for the given specifications.

Answer to Problem 11.80P

The value of the small signal parameters are $g_{m2}=17.193\text{mA}/\text{V}$ , $r_{\pi 2}=8.724\text{k}\Omega$ , $g_{m1}=205.87\text{μA}/\text{V}$ and the value of the composite transconductance is $g_m^C=11.087\text{mA}/\text{V}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The expression to determine the value of the emitter current of the second transistor is calculated as,

  $\begin{array}{c}{I}_{E2}=I_{Bias2}-I_{Bias1}\\ =0.50\text{mA}-0.05\text{mA}\\ =0.45\text{mA}\end{array}$

The expression to determine the value of the current $I_{C2}$ is given by,

  $I_{C2}=\left(\frac{\beta }{1+\beta }\right)I_{E2}$

Substitute $150$ for $\beta$ and $0.45\text{mA}$ for $I_{E2}$ in the above equation.

  $\begin{array}{c}{I}_{C2}=\left(\frac{150}{1+150}\right)0.45\text{mA}\\ =447.02\text{μA}\end{array}$

The value of the transconductance $g_{m2}$ is calculated as,

  $\begin{array}{c}{g}_{m2}=\frac{I_{C2}}{0.026\text{V}}\\ =\frac{447.02\text{μA}}{0.026\text{V}}\\ =17.193\text{mA}/\text{V}\end{array}$

The expression to determine the value of the small signal resistance is given by,

  $r_{\pi 2}=\frac{\beta }{g_m}$

Substitute $150$ for $\beta$ and $17.193\text{mA}/\text{V}$ for $g_{m2}$ in the above equation.

  $\begin{array}{c}{r}_{\pi 2}=\frac{150}{17.193\text{mA}/\text{V}}\\ =8.724\text{k}\Omega \end{array}$

The value of the drain current $I_{D1}$ is given by,

  $\begin{array}{c}{I}_{D1}=I_{Bias2}-I_{C2}\\ =0.50\text{mA}-447.02\text{μA}\\ =\text{52}\text{.98}\text{μA}\end{array}$

The value of the transconductance $g_{m1}$ is given by,

  $g_{m1}=2\sqrt{K_n{I}_{D1}}$

Substitute $0.2\text{mA}/{\text{V}}^2$ for $k_n$ and $\text{52}\text{.98}\text{μA}$ for $I_{D1}$ in the above equation.

  $\begin{array}{c}{g}_{m1}=2\sqrt{\left(0.2\text{mA}/{\text{V}}^2\right)\left(\text{52}\text{.98}\text{μA}\right)}\\ =205.87\text{μA}/\text{V}\end{array}$

The expression to determine the value of the composite transconductance is given by,

  $g_m^C=\frac{g_{m1}\left(1+g_{m2}{r}_{\pi }\right)}{1+g_{m1}{r}_{\pi }}$

Substitute $205.87\text{μA}/\text{V}$ for $g_{m1}$ , $17.193\text{mA}/\text{V}$ for $g_{m2}$ and $15.7\text{k}\Omega$ for $r_{\pi }$ in the above equation.

  $\begin{array}{c}{g}_m^C=\frac{205.87\text{μA}/\text{V}\left(1+\left(17.193\text{mA}/\text{V}\right)\left(8.724\text{k}\Omega \right)\right)}{1+\left(205.87\text{μA}/\text{V}\right)15.7\text{k}\Omega }\\ =\frac{0.031}{2.796}\\ =11.087\text{mA}/\text{V}\end{array}$

Conclusion:

Therefore, the value of the small signal parameters are $g_{m2}=17.193\text{mA}/\text{V}$ , $r_{\pi 2}=8.724\text{k}\Omega$ , $g_{m1}=205.87\text{μA}/\text{V}$ and the value of the composite transconductance is $g_m^C=11.087\text{mA}/\text{V}$ .