Chapter 12, Problem 12.8TYU

To determine

The magnitude of the current gain and the relationship between $R_F,R_{E2}$ and current gain using computer simulation.

Answer to Problem 12.8TYU

The relation between the current gain with the change in resistance $R_F$ and $R_{E2}$ is depicted by the expression $A_i=\frac{\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{F\text{}}B\right\}}{\frac{-1}{g_{m2}}\left[\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right]\left\{\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{AC+g_{m1}}\right\}\left[\left(AB\right)-\frac{1}{R_F}\right]\right\}}$ and the plot for the relation is shown in Figure 6.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Mark the nodes and redraw the circuit.

The given diagram is shown in Figure 2

  

Figure 2

Mark the values and draw the PSpice circuit for the above circuit.

The required circuit is shown in Figure 3

  

Figure 3

The snip for the drop box of the internal parameters of the transistor is shown in Figure 4

  

Figure 4

The simulation settings to estimate the magnitude of the current gain as the value of $R_{E2}$ is varied between $0.4\text{k}\Omega$ and $1.6\text{k}\Omega$ is shown in Figure 5

  

Figure 5

Then left click on the trace option then select add trace and type “ABS(l(RL)L(li))” then command in the trace magnitude of the current gain as the emitter resistance varies between $0.4\text{k}\Omega$ and $1.6\text{k}\Omega$ . the required relation is shown in Figure 6

  

Figure 6

By KCL the expression for the current $I_i$ is given by,

  $\begin{array}{l}{I}_i=\frac{V_{\pi 1}}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}}+\frac{V_{\pi 1}-V_{e2}}{R_F}\\ V_{e2}=V_{\pi 1}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)-I_i{R}_F\end{array}$

The expression for the node voltage is given by,

  $V_{C1}=V_{\pi 2}+V_{e2}$

Apply KCL at node $V_{C1}$ .

  $\begin{array}{l}{g}_{m1}{V}_{\pi 1}+\frac{V_{C1}}{R_{C1}\left|\right|R_{B2}}+\frac{V_{\pi 2}}{r_{\pi 2}}=0\\ g_{m1}{V}_{\pi 1}+\frac{V_{\pi 2}+V_{e2}}{R_{C1}\left|\right|R_{B2}}+\frac{V_{\pi 2}}{r_{\pi 2}}=0\\ g_{m1}{V}_{\pi 1}+V_{\pi 2}\left(\frac{1}{R_{C1}\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)+\frac{V_{e2}}{R_{C12}\left|\right|R_{B2}}=0\end{array}$

Substitute $V_{\pi 1}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)-I_i{R}_F$ for $V_{e2}$ in the above equation.

  $\begin{array}{l}{g}_{m1}{V}_{\pi 1}+V_{\pi 2}\left(\frac{1}{R_{C1}\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)+\frac{1}{R_{C12}\left|\right|R_{B2}}\left[V_{\pi 1}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)-I_i{R}_F\right]=0\\ V_{\pi 1}=\frac{I_i{R}_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)-V_{\pi 2}\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)\right]}\end{array}$

The expression for the output current is given by,

  $\begin{array}{l}{I}_O=-\left(g_{m2}{V}_{\pi 2}\right)\left(\frac{R_{C2}}{R_{C2}+R_L}\right)\\ V_{\pi 2}=\frac{-I_O}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\end{array}$

Apply KCL at node $V_{e2}$ .

  $\begin{array}{l}\frac{V_{\pi 2}}{r_{\pi 2}}+g_{m2}{V}_{\pi 2}=\frac{V_{e2}}{R_{E2}}+\frac{V_{e2}-V_{\pi 1}}{R_F}\\ \left(\frac{1}{r_{\pi 2}}+g_{m2}\right)V_{\pi 2}+\frac{V_{\pi 1}}{R_F}=\left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)V_{e2}\end{array}$

Substitute $V_{\pi 1}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)-I_i{R}_F$ for $V_{e2}$ in the above equation.

  $\begin{array}{l}\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)V_{\pi 2}+\frac{V_{\pi 1}}{R_F}=\left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)\left(V_{\pi 1}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)-I_i{R}_F\right)\\ \left(\frac{1}{r_{\pi 2}}+g_{m2}\right)V_{\pi 2}+I_i{R}_F\left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)=V_{\pi 1}\left\{\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)\right]-\frac{1}{R_F}\right\}\end{array}$

Substitute $\frac{I_i{R}_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)-V_{\pi 2}\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)\right]}$ for $V_{\pi 1}$ in the above equation.

  $\left[\begin{array}{l}\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)V_{\pi 2}\\ +I_i{R}_F\left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)\end{array}\right]=\frac{\left[\begin{array}{l}{I}_i{R}_F\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\\ -V_{\pi 2}\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)\end{array}\right]}{\left[g_{m1}+\left[\begin{array}{l}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\\ \left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)\end{array}\right]\right]}\left\{\left[\begin{array}{l}\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\\ \left(\frac{1}{R_{E2}}+\frac{1}{R_F}\right)\end{array}\right]-\frac{1}{R_F}\right\}$

Consider $A=\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}$ , $B=\frac{1}{R_{E2}}+\frac{1}{R_F}$ , $C=\frac{1}{R_{C1}\left|\right|R_{B2}}$ and $D=\frac{1}{R_{C1}\left|\right|R_{B2}\left|\right|r_{\pi 2}}$ in the above equation. So the equation is,

  $\left[\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)V_{\pi 2}+I_i{R}_{F}B\right]=\frac{\left[I_i{R}_{F}C-V_{\pi 2}D\right]}{\left[g_{m1}+AC\right]}\left\{AB-\frac{1}{R_F}\right\}$

  $V_{\pi 2}\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]\right\}=I_i\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{F}B\right\}$

  $\left[\frac{-I_O}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right]\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]\right\}=I_i\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{FB}B\right\}$

  $\frac{I_O}{I_i}=\frac{\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{F\text{}}B\right\}}{\frac{-1}{g_{m2}}\left[\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right]\left\{\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{AC+g_{m1}}\right\}\left[\left(AB\right)-\frac{1}{R_F}\right]\right\}}$

Thus, the expression for the small signal current gain is,

  $A_i=\frac{\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{F\text{}}B\right\}}{\frac{-1}{g_{m2}}\left[\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right]\left\{\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{AC+g_{m1}}\right\}\left[\left(AB\right)-\frac{1}{R_F}\right]\right\}}$

Conclusion:

Therefore, the relation between the current gain with the change in resistance $R_F$ and $R_{E2}$ is depicted by the expression $A_i=\frac{\left\{\frac{R_{F}C}{g_{m1}+AC}\left[AB-\frac{1}{R_F}\right]-R_{F\text{}}B\right\}}{\frac{-1}{g_{m2}}\left[\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right]\left\{\left\{\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{D}{AC+g_{m1}}\right\}\left[\left(AB\right)-\frac{1}{R_F}\right]\right\}}$ and the plot for the relation is shown in Figure 6