Chapter 10, Problem 73P

If the input impedance is defined as Z_in = V_s/I_s, find the input impedance of the op amp circuit in Fig. 10.116 when R₁ = 10 kΩ, R₂ = 20 kΩ, C₁ = 10 nF, C₂ = 20 nF, and ω = 5000 rad/s.

Figure 10.116

To determine

Calculate the input impedance $\left(Z_{\text{in}}\right)$ of the op amp circuit in Figure 10.116.

Answer to Problem 73P

The value of input impedance $\left(Z_{\text{in}}\right)$ in the given op amp circuit is $21.21\angle -45°\text{k}\Omega$.

Explanation of Solution

Given data:

Refer to Figure 10.116 in the textbook for the op amp circuit.

The values of resistance $R_1$ and $R_2$ are $10\text{k}\Omega$ and $20\text{k}\Omega$ respectively.

The values of capacitance $C_1$ and $C_2$ are $10\text{nF}$ and $20\text{nF}$ respectively.

The value of angular frequency $\omega$ is $5000\frac{\text{rad}}{\text{s}}$.

Formula used:

Write the expression to calculate impedance of the capacitor.

$Z_C=\frac{1}{j\omega C}$ (1)

Here,

$\omega$ is the angular frequency, and

$C$ is the value of capacitor.

Write the formula for input impedance of the given op amp circuit.

$Z_{\text{in}}=\frac{V_s}{I_s}$ (2)

Here,

$V_s$ is the source voltage, and

$I_s$ is the source current.

Calculation:

Substitute $5000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $10\text{nF}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(5000\frac{\text{rad}}{\text{s}}\right)\left(10\text{nF}\right)}\\ =\frac{1}{j\left(5000\frac{\text{rad}}{\text{s}}\right)\left(10\times {10}^{-9}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{nF}=1\times {10}^{-9}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j20\text{k}\Omega \end{array}$

Substitute $5000\frac{\text{rad}}{\text{s}}$ for $\omega$ and $20\text{nF}$ for $C$ in equation (1) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(5000\frac{\text{rad}}{\text{s}}\right)\left(20\text{nF}\right)}\\ =\frac{1}{j\left(5000\frac{\text{rad}}{\text{s}}\right)\left(20\times {10}^{-9}\frac{\text{s}}{\Omega }\right)}\left\{\begin{array}{l}\because 1\text{nF}=1\times {10}^{-9}\text{F}\\ 1\text{F}=1\frac{\text{s}}{\Omega }\end{array}\right\}\\ =-j10\text{k}\Omega \end{array}$

The frequency domain representation of given circuit with the node voltage is shown in Figure 1.

Apply Kirchhoff’s current law at node $V_1$.

$\frac{V_1-V_s}{10\text{k}\Omega }+\frac{V_1-V_o}{-j20\text{k}\Omega }+\frac{V_1-V_2}{20\text{k}\Omega }=0$ (3)

According to the properties of an ideal op amp, the voltage appear across the input terminals of the op amp is zero.

The voltage at the inverting terminal will be appear at the non-inverting input terminal. Since the output voltage is fed back to the inverting terminal, the voltage $V_2=V_o$.

$V_2=V_o$ (4)

Substitute equation (4) in (3).

$\begin{array}{l}\frac{V_1-V_s}{10}+\frac{V_1-V_o}{-j20}+\frac{V_1-V_o}{20}=0\\ \frac{V_1}{10}-\frac{V_s}{10}-\frac{V_1}{j20}+\frac{V_o}{j20}+\frac{V_1}{20}-\frac{V_o}{20}=0\\ \left(\frac{1}{10}-\frac{1}{j20}+\frac{1}{20}\right)V_1-\left(\frac{1}{20}-\frac{1}{j20}\right)V_o=\frac{V_s}{10}\\ \left(0.1+j0.05+0.05\right)V_1-\left(0.05+j0.05\right)V_o=0.1V_s\end{array}$

Reduce the equation as follows.

$\left(0.15+j0.05\right)V_1-\left(0.05+j0.05\right)V_o=0.1V_s$

$\left(3+j\right)V_1-\left(1+j\right)V_o=2V_s$ (5)

Apply Kirchhoff’s current law at node $V_2$.

$\frac{V_2-V_1}{20\text{kΩ}}+\frac{V_2-0}{-j10\text{kΩ}}=0$ (6)

Substitute equation (5) and (7).

$\begin{array}{l}\frac{V_o-V_1}{20}-\frac{V_o-0}{j10}=0\\ \frac{V_o-V_1}{2}=\frac{V_o}{j1}\\ \frac{V_1}{2}=\frac{V_o}{2}-\frac{V_o}{j1}\end{array}$

Simplify the equation as follows.

$\begin{array}{l}0.5V_1=\left(0.5+j1\right)V_o\\ V_1=\frac{0.5+j1}{0.5}{V}_o\end{array}$

$V_1=\left(1+j2\right)V_o$ (7)

Substitute equation (7) in (5).

$\begin{array}{l}\left(3+j\right)\left(1+j2\right)V_o-\left(1+j\right)V_o=2V_s\\ \left(1+j7\right)V_o-\left(1+j\right)V_o=2V_s\\ \left(j6\right)V_o=2V_s\end{array}$

Simplify the above equation as follows.

$V_o=\frac{2}{\left(j6\right)}{V}_s$

$V_o=-j\frac{1}{3}{V}_s$ (8)

Substitute equation (8) in (7).

$V_1=\left(1+j2\right)\left(-j\frac{1}{3}{V}_s\right)$

$V_1=\left(\frac{2}{3}-j\frac{1}{3}\right)V_s$ (9)

From Figure 1, write the expression for current $I_s$.

$\begin{array}{c}{I}_s=\frac{V_s-V_1}{10\text{k}\Omega }\\ =\frac{\left(\frac{1+j}{3}\right)V_s}{10\text{k}\Omega }\end{array}$

$\frac{I_s}{V_s}=\frac{1+j}{30\text{k}\Omega }$ (10)

Substitute equation (10) in (2).

$\begin{array}{c}{Z}_{\text{in}}=\frac{30\text{k}\Omega }{1+j}\\ =15-j15\text{k}\Omega \\ =21.21\angle -45°\text{k}\Omega \end{array}$

Conclusion:

Thus, the value of input impedance $\left(Z_{\text{in}}\right)$ in the given op amp circuit is $21.21\angle -45°\text{k}\Omega$