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: Recall the T-network to single resistor trick.
When a feedback path made of three resistors is arranged as $R_2$ in series from the inverting node to a middle point, $R_4$ from that middle point to ground, and $R_3$ from the middle point to the output, the whole T-network behaves like one plain feedback resistor of value
\[ R_{f,eq}=R_2+R_3+\frac{R_2 R_3}{R_4} \]
This comes from a Y to delta type transform of the T-network as seen between the inverting node and the output.

Step 2: Plug in the resistor values.
Here $R_2=R_3=R_4=50$ k$\Omega$, so
\[ R_{f,eq}=50+50+\frac{50\times50}{50}=50+50+50=150\text{ k}\Omega \]

Step 3: Use the plain inverting amplifier gain formula.
For a simple inverting amplifier with input resistor $R_1$ and feedback resistor $R_f$, the gain is
\[ \frac{v_o}{v_i}=-\frac{R_f}{R_1} \]
Here $R_1=50$ k$\Omega$ is still the input resistor, so
\[ \frac{v_o}{v_i}=-\frac{150}{50}=-3 \]

Step 4: State the magnitude.
The T-network lets three $50$ k$\Omega$ resistors act like a single $150$ k$\Omega$ feedback resistor, giving a gain of $3$ with resistors far smaller than $150$ k$\Omega$ each. This is exactly why designers use a T-network when a huge feedback resistor is hard to build.
\[ \boxed{3.00} \]