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## Q. 13.1

Analysis of an Inverting Amplifier

Figure 13.6 shows a version of the inverting amplifier that can have high gain magnitude without resorting to as wide a range of resistor values as is needed in the standard inverter configuration. Derive an expression for the voltage gain under the ideal op amp assumption. Also, find the input impedance and output impedance.
Evaluate the results for $R_1 = R_3 = 1 \text{ k} \Omega \text{ and } R_2 = R_4 = 10 \text{ k}\Omega$. Then, consider the standard inverter configuration of Figure 13.5 with $R_1 = 1 \text{ k}\Omega$, and find the value of $R_2$ required to achieve the same gain. ## Verified Solution

Generally, if a network of resistors is connected between the inverting input and the output, negative feedback exists.

First, we verify that negative feedback is present. Assume a positive value for $v_i$, which results in a negative output voltage of very large magnitude. Part of this negative voltage is returned through the resistor network and opposes the original input voltage. Thus, we conclude that negative feedback is present.
Next, we assume the conditions of the summing point constraint:

$v_i \text{ and }i_i=0$

Then, we apply Kirchhoff’s current law, Kirchhoff’s voltage law, and Ohm’s law to analyze the circuit. To begin, we notice that $v_{\text{in}}$ appears across $R_1 (\text{ because } v_i = 0)$.
Hence, we can write

$i_1=\frac{v_{\text{in}}}{R_1} \quad \quad \quad (13.8)$

Next, we apply Kirchhoff’s current law to the node at the right hand end of $R_1$ , to obtain

$i_2=i_1 \quad \quad \quad (13.9)$

(Here, we have used the fact that $i_i = 0$.)
Writing a voltage equation around the loop through $v_i, R_2 , \text{ and } R_3$, we obtain

$R_2i_2=R_3i_3 \quad \quad \quad (13.10)$

(In writing this equation, we have used the fact that $v_i = 0$.) Applying Kirchhoff’s current law at the top end of $R_3$ yields

$i_4=i_2+i_3 \quad \quad \quad (13.11)$

Writing a voltage equation for the loop containing $v_o, R_4 , \text{ and } R_3$ gives

$v_o=-R_4i_4-R_3i_3 \quad \quad \quad (13.12)$

Next, we use substitution to eliminate the current variables $(i_1 , i_2 , i_3 , \text{ and } i_4 )$ and obtain an equation relating the output voltage to the input voltage. First, from Equations 13.8 and 13.9, we obtain

$i_2=\frac{v_{\text{in}}}{R_1} \quad \quad \quad (13.13)$

Then, we use Equation 13.13 to substitute for $i_2$ in Equation 13.10 and rearrange terms to obtain

$i_3=v_{\text{in}}\frac{R_2}{R_1R_3} \quad \quad \quad (13.14)$

Using Equations 13.13 and 13.14 to substitute for $i_2 \text{ and } i_3$ in Equation 13.11, we find that

$i_4=v_{\text{in}}\left(\frac{1}{R_1}+\frac{R_2}{R_1R_3} \right) \quad \quad \quad (13.15)$

Finally, using Equations 13.14 and 13.15 to substitute into 13.12, we obtain

$v_o=-v_{\text{in}}\left(\frac{R_2}{R_1}+\frac{R_4}{R_1}+\frac{R_2R_4}{R_1R_3} \right) \quad \quad \quad (13.16)$

Therefore, the voltage gain of the circuit is

$A_v=\frac{v_o}{v_{\text{in}}}=-\left(\frac{R_2}{R_1}+\frac{R_4}{R_1}+\frac{R_2R_4}{R_1R_3} \right) \quad \quad \quad (13.17)$

The input impedance is obtained from Equation 13.8:

$R_{\text{in}}=\frac{v_{\text{in}}}{i_1}=R_1 \quad \quad \quad (13.18)$

Inspection of Equation 13.16 shows that the output voltage is independent of the load resistance. Thus, the output appears as an ideal voltage source to the load. In other words, the output impedance of the amplifier is zero.
Evaluating the voltage gain for the resistor values given $(R_1 = R_3 = 1 \text{ k} \Omega \text{ and } R_2 = R_4 = 10 \text{ k}\Omega)$ yields

$A_v=-120$

In the basic inverter circuit of Figure 13.5, the voltage gain is given by Equation 13.5, which states that

$A_v=-\frac{R_2}{R_1}$

Therefore, to achieve a voltage gain of 120 with $R_1 = 1 \text{ k}\Omega, \text{ we need } R_2 = 120 \text{ k}\Omega$.
Notice that a resistance ratio of 120:1 is required for the basic inverter, whereas the circuit of Figure 13.6 has a ratio of only 10:1. Sometimes, there are significant practical advantages in keeping the ratio of resistances in a circuit as close to unity as possible. Then, the circuit of Figure 13.6 is preferable to the basic inverter shown in Figure 13.5. 