Question

If the operational amplifier in the circuit below is ideal, the input impedance looking into terminal \(V_{in}\) is

• A. \(\infty\)
• B. \(-\dfrac{R \times R_1}{R_F}\)
• C. \(\dfrac{R \times R_F}{R_1}\)
• D. \(R\)

Hint:

No current enters the op-amp inputs; find I_in through R using the gain set by R1 and RF at the (-) node.

Answer 1

The Correct Option is B

Step 1: Label the two op-amp input nodes.
Let \(V_-\) be the voltage at the inverting input and \(V_+\) be the voltage at the non-inverting input. The non-inverting input connects directly, by a plain wire, to the terminal \(V_{in}\), and it also connects to the output \(V_{out}\) through resistor \(R\). The inverting input connects to ground through \(R_1\) and to the output through the feedback resistor \(R_F\).

Step 2: Apply the ideal op-amp rules.
For an ideal op-amp, no current flows into either input terminal, and the virtual short forces \(V_- = V_+\). Since \(V_{in}\) is wired straight to the \(+\) input with no resistor in between, \(V_+ = V_{in}\), so \(V_- = V_{in}\) too.

Step 3: Write the KCL equation at the inverting node.
At node \(V_-\), the current arriving from ground through \(R_1\) must equal the current leaving toward the output through \(R_F\) (no current is lost into the op-amp input):
\[ \frac{0 - V_-}{R_1} + \frac{V_{out} - V_-}{R_F} = 0 \]
Substituting \(V_- = V_{in}\):
\[ \frac{-V_{in}}{R_1} + \frac{V_{out}-V_{in}}{R_F} = 0 \]
Multiplying through by \(R_1 R_F\):
\[ -V_{in}R_F + R_1(V_{out}-V_{in}) = 0 \]
\[ R_1 V_{out} = V_{in}(R_F+R_1) \]
\[ V_{out} = V_{in}\left(1+\frac{R_F}{R_1}\right) \]

Step 4: Find the current drawn from the \(V_{in}\) source.
The \(V_{in}\) terminal connects only to the \(+\) input node, and that node's only other connection is resistor \(R\) to the output. Since the op-amp input draws no current, every bit of current supplied by the \(V_{in}\) source must flow out through \(R\) toward \(V_{out}\):
\[ I_{in} = \frac{V_+ - V_{out}}{R} = \frac{V_{in}-V_{out}}{R} \]
Substituting \(V_{out}\) from Step 3:
\[ I_{in} = \frac{V_{in} - V_{in}\left(1+\frac{R_F}{R_1}\right)}{R} = \frac{-V_{in}\frac{R_F}{R_1}}{R} = \frac{-V_{in}R_F}{R_1 R} \]

Step 5: Compute the input impedance.
\[ Z_{in} = \frac{V_{in}}{I_{in}} = \frac{V_{in}}{\dfrac{-V_{in}R_F}{R_1 R}} = \frac{-R_1 R}{R_F} = -\frac{R \times R_1}{R_F} \]
The negative sign is not a mistake; this circuit is a Negative Impedance Converter (NIC), a classic op-amp configuration that makes a source 'see' a negative resistance, because the positive feedback path through \(R\) pumps out more current as \(V_{in}\) rises rather than less.

Step 6: Why the other options are wrong.
Option (A) \(\infty\) would be true only if \(V_{in}\) fed a plain non-inverting amplifier with no feedback resistor tied back to the \(+\) node, drawing zero net current, but here \(R\) does carry current because of the feedback loop through \(R_F\) and \(R_1\).
Option (C) inverts the roles of \(R_F\) and \(R_1\) and drops the sign, which does not follow from the algebra above.
Option (D) ignores the gain set by \(R_F\) and \(R_1\) altogether.

Final Answer:
The input impedance looking into \(V_{in}\) is \(-\dfrac{R \times R_1}{R_F}\).
\[ \boxed{Z_{in} = -\frac{R \times R_1}{R_F}} \]