Chapter 12, Problem 12.9TYU

To determine

To show: The variation in magnitude of the input resistance $R_{if}$ as the feedback resistance varies.

To explain: The influence of $R_{if}$ on $R_F$ .

Answer to Problem 12.9TYU

The influence of $R_F$ on $R_{if}$ is determined by the relation $\frac{R_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)+\left(\frac{A_{if}}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$ and the plot for the variation in the input resistance when the feedback resistance is varied between $5\text{k}\Omega$ and $50\text{k}\Omega$ is shown in Figure 6.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Feedback resistor value is varied between $5\text{k}\Omega$ and $50\text{k}\Omega$

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 $R_F$ is varied between $5\text{k}\Omega$ and $50\text{k}\Omega$ is shown in Figure 5

  

Figure 5

Then left click on the trace option then select add trace and type “ABS(V(vi)l(li))” then command in the trace magnitude of the resistance $R_{if}$ as the feedback resistance varies between $5\text{k}\Omega$ and $50\text{k}\Omega$ , 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.

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

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,

  $V_{\pi 1}=\frac{I_i{R}_{F}C-V_{\pi 2}D}{g_{m1}+\left[AC\right]}$

The expression for the output current is given by,

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

Substitute $\frac{-I_O}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)$ for $V_{\pi 2}$ in equation (1).

  $V_{\pi 1}=\frac{I_i{R}_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)-\left(\frac{-I_O}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$

Substitute $A_{if}{I}_i$ for $I_O$ in the above equation.

  $V_{\pi 1}=\frac{I_i{R}_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)+\left(\frac{A_{if}{I}_i}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$

  $\frac{V_{\pi 1}}{I_i}=\frac{R_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)+\left(\frac{A_{if}}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$

  $R_{if}=\frac{R_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)+\left(\frac{A_{if}}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$

Conclusion:

Therefore, the influence of $R_F$ on $R_{if}$ is determined by the relation $\frac{R_F\left(\frac{1}{R_{C12}\left|\right|R_{B2}}\right)+\left(\frac{A_{if}}{g_{m2}}\left(\frac{R_{C2}+R_L}{R_{C2}}\right)\right)\left(\frac{1}{R_C\left|\right|R_{B2}\left|\right|r_{\pi 2}}\right)}{g_{m1}+\left[\left(\frac{R_F}{R_s\left|\right|R_{B1}\left|\right|r_{\pi 1}\left|\right|R_F}\right)\left(\frac{1}{R_{C1}\left|\right|R_{B2}}\right)\right]}$ and the plot for the variation in the input resistance when the feedback resistance is varied between $5\text{k}\Omega$ and $50\text{k}\Omega$ is shown in Figure 6