In the circuit shown, the open loop gain of the operational amplifier is A₀ = 10⁵. What is the voltage gain of the circuit? (Round off to two decimal places)

The correct answer is –20.00.

Analysis of the Circuit:
Based on the provided circuit diagram:

• The non-inverting terminal ($+$) of the operational amplifier is connected directly to ground, so $V_{+}=0V$.
• The input voltage $V_{\mathrm{in}}=100\mathrm{mV}$ is connected to the inverting terminal ($-$) through an input resistor $R_1=5\mathrm{k\Omega }$.
• A feedback resistor $R_f=100\mathrm{k\Omega }$ is connected between the output and the inverting terminal.
• The open-loop gain is given as $A_0={10}^5$, input resistance $R_{\mathrm{in}}=\infty \Omega$, and output resistance $R_{\mathrm{out}}=0\Omega$.

Step-by-Step Derivation of Voltage Gain:

Let $V_{-}$ be the voltage at the inverting node. The output voltage $V_{\mathrm{out}}$ of the operational amplifier is related to its differential input voltage by the open-loop gain: $V_{\mathrm{out}}=A_0\left(V_{+}-V_{-}\right)$

Substituting $V_{+}=0$ into the expression gives: $V_{\mathrm{out}}=A_0\left(0-V_{-}\right)\Rightarrow V_{-}=-\frac{V_{\mathrm{out}}}{A_0}$

Since the input resistance $R_{\mathrm{in}}$ is infinite, no current enters the inverting terminal. Applying Kirchhoff's Current Law (KCL) at the inverting node: $\frac{V_{\mathrm{in}}-V_{-}}{R_1}=\frac{V_{-}-V_{\mathrm{out}}}{R_f}$

Substituting $V_{-}=-\frac{V_{\mathrm{out}}}{A_0}$ into the KCL equation: $\frac{V_{\mathrm{in}}+\frac{V_{\mathrm{out}}}{A_0}}{R_1}=\frac{-\frac{V_{\mathrm{out}}}{A_0}-V_{\mathrm{out}}}{R_f}$

Separating the terms for $V_{\mathrm{in}}$ and $V_{\mathrm{out}}$: $\frac{V_{\mathrm{in}}}{R_1}=-V_{\mathrm{out}}\left(\frac{1}{R_f}+\frac{1}{A_0{R}_f}+\frac{1}{A_0{R}_1}\right)$

Rearranging the equation to solve for the closed-loop voltage gain $A_v=\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}}$: $A_v=\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}}=-\frac{R_f}{R_1+\frac{R_1+R_f}{A_0}}$

Substituting the Numerical Values:
Let's substitute $R_1=5\mathrm{k\Omega }$, $R_f=100\mathrm{k\Omega }$, and $A_0={10}^5$: $A_v=-\frac{100}{5+\frac{5+100}{{10}^5}}$

Simplifying the denominator: $A_v=-\frac{100}{5+\frac{105}{100000}}=-\frac{100}{5+0.00105}=-\frac{100}{5.00105}\approx -19.9958$

Rounding to two decimal places, the closed-loop voltage gain of the circuit is –20.00.