Question

Consider an ideal OP-AMP circuit as shown in the figure.

The resistances \(R_1=R_2=R_3=R_4=50\text{ k}\Omega\).
The magnitude of the closed loop gain is (rounded off to two decimal places).

Hint:

Redraw the R2-R3-R4 T-network as one equivalent feedback resistor, then use the usual inverting-amplifier gain formula.

Answer 1

Correct Answer: 3

Step 1: Look at the circuit topology.
The signal \(v_i\) drives the inverting input of the op-amp through \(R_1\). The non-inverting input is tied to ground, so by the virtual short property of an ideal op-amp in negative feedback, the inverting input also sits at \(0\) V.
The feedback path is not a single resistor. It is a T-network: \(R_2\) runs from the inverting node to an internal node, call it \(X\), then \(R_4\) runs from \(X\) down to ground, and \(R_3\) runs from \(X\) to the output \(v_o\). This T-network lets the circuit reach a gain larger than \(1\) using only equal valued resistors.

Step 2: Apply KCL at the inverting node.
No current enters the ideal op-amp input, so whatever current comes in through \(R_1\) must leave through \(R_2\) toward node \(X\).
\[ \frac{v_i-0}{R_1}=\frac{0-V_X}{R_2} \]
This gives
\[ V_X=-\frac{R_2}{R_1}\,v_i \]

Step 3: Apply KCL at node X.
At node \(X\), the current that arrived from \(R_2\) splits into two paths: one through \(R_4\) to ground, and one through \(R_3\) to the output.
\[ \frac{0-V_X}{R_2}=\frac{V_X-0}{R_4}+\frac{V_X-v_o}{R_3} \]

Step 4: Put in the equal resistor values.
Here \(R_1=R_2=R_3=R_4=R=50\text{ k}\Omega\). From Step 2, \(V_X=-v_i\). Substitute this into the Step 3 equation and cancel the common \(R\):
\[ -V_X=V_X+V_X-v_o \]
\[ v_o=3V_X \]

Step 5: Solve for the gain.
\[ v_o=3(-v_i)=-3v_i \]
So the closed loop gain is
\[ \frac{v_o}{v_i}=-3 \]
The negative sign only shows that the circuit is inverting, so the magnitude asked in the question is
\[ \boxed{3.00} \]