Question

In the circuit shown, the open loop gain of the operational amplifier is \(A_0=105\), with \(R_{in}=\infty\ \Omega\) and \(R_{out}=0\ \Omega\).

The circuit is a standard inverting amplifier: \(V_{in}=100\) mV drives the inverting input through a \(5\ k\Omega\) input resistor, and a \(100\ k\Omega\) resistor feeds back from \(V_{out}\) to the inverting input, with the non-inverting input grounded. What is the voltage gain of the circuit? (Round off to two decimal places)

• A. \(-16.67\)
• B. \(-20.00\)
• C. \(-21.00\)
• D. \(-12.67\)

Hint:

Use the finite-gain inverting amplifier formula: gain = -(Rf/Rin) divided by [1 + (1+Rf/Rin)/A0].

Answer 1

The Correct Option is A

Step 1: Set up the finite-gain inverting amplifier formula.
For an ideal op-amp with infinite gain, the inverting amplifier gain is simply \(-R_f/R_{in}\). With a finite open loop gain \(A_0\), a small error term must be included. The exact closed loop gain is
\[ \frac{V_{out}}{V_{in}}=\frac{-R_f/R_{in}}{1+\dfrac{1+R_f/R_{in}}{A_0}} \]

Step 2: Plug in the resistor values.
\[ \frac{R_f}{R_{in}}=\frac{100\text{ k}\Omega}{5\text{ k}\Omega}=20 \]

Step 3: Compute the correction term.
\[ \frac{1+R_f/R_{in}}{A_0}=\frac{1+20}{105}=\frac{21}{105}=0.2 \]

Step 4: Compute the denominator.
\[ 1+0.2=1.2 \]

Step 5: Compute the gain.
\[ \frac{V_{out}}{V_{in}}=\frac{-20}{1.2}=-16.67 \]

Step 6: Analyze the options.

(A) -16.67: Matches the exact finite-gain result. Correct.

(B) -20.00: This is just \(-R_f/R_{in}\), the ideal infinite-gain answer, which ignores the finite \(A_0\) correction. Incorrect.

(C) -21.00: Comes from wrongly adding \(1+R_f/R_{in}\) directly to the ratio instead of dividing it by \(A_0\) first. Incorrect.

(D) -12.67: Comes from a setup error in the correction term. Incorrect.

Final Answer:
\[ \boxed{-16.67} \]