Question

For the inverting operational amplifier circuit shown below, determine the closed-loop voltage gain (\(A_{cl} = V_{out}/V_{in}\)). The op-amp has an open-loop gain \(A_{OL} = 10^5\).

• A. -20
• B. 20
• C. -19.996
• D. -20.042

Hint:

When an op-amp's open-loop gain (\(A_{OL}\)) is finite, the actual closed-loop gain will always have a smaller magnitude than the ideal gain. For an inverting amplifier, this means the gain will be slightly closer to zero (e.g., -19.996 instead of -20).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the closed-loop voltage gain of an inverting op-amp. We are given the values of the input resistor, feedback resistor, and a finite open-loop gain. This means we should calculate the precise gain, not just the ideal one.
Step 2: Key Formula or Approach:
For an ideal op-amp (\(A_{OL} \to \infty\)), the gain of an inverting amplifier is: \[ A_{cl, ideal} = -\frac{R_f}{R_{in}} \] For a non-ideal op-amp with finite open-loop gain \(A_{OL}\), the formula for the closed-loop gain is: \[ A_{cl} = \frac{A_{cl, ideal}}{1 - \frac{\beta}{A_{OL}}} \quad \text{or more commonly written as} \quad A_{cl} = \frac{-R_f/R_{in}}{1 + \frac{1 + R_f/R_{in}}{A_{OL}}} \] Where the term \((1 + R_f/R_{in})\) is the non-inverting gain or "noise gain".
Step 3: Detailed Explanation:
1. Calculate the ideal closed-loop gain:
\[ A_{cl, ideal} = -\frac{R_f}{R_{in}} = -\frac{100 \, k\Omega}{5 \, k\Omega} = -20 \] 2. Calculate the non-ideal closed-loop gain:
Now, we use the formula for the non-ideal case with the given values:

• \(R_f = 100 \, k\Omega\)
• \(R_{in} = 5 \, k\Omega\)
• \(A_{OL} = 10^5\)
• \(R_f/R_{in} = 20\)

\[ A_{cl} = \frac{-20}{1 + \frac{1 + 20}{10^5}} \] \[ A_{cl} = \frac{-20}{1 + \frac{21}{100000}} \] \[ A_{cl} = \frac{-20}{1 + 0.00021} \] \[ A_{cl} = \frac{-20}{1.00021} \approx -19.995800... \] This value can be rounded to -19.996. The value -19.99 shown in the image is a close approximation.
Step 4: Final Answer:
The calculated closed-loop gain, accounting for the finite open-loop gain, is approximately -19.996. This is very close to the ideal value of -20, showing that for a high \(A_{OL}\), the ideal formula is a very good approximation.