Op Amp Circuits Examples: Solving the Mismatched Differential Amplifier Trap

When analyzing op amp circuits examples for exams or bench debugging, the differential amplifier is the most common trap. For the mismatched TL072 circuit detailed below ($V_1=2V$, $V_2=1V$, $R_f=100k\Omega$, $R_1=10k\Omega$, $R_3=10k\Omega$, $R_4=90k\Omega$), the exact output voltage is -10.1V. Most students blindly apply the standard difference formula and guess -10V, failing the exam. Here is the complete algebraic breakdown using superposition, the exact trap that ruins your common-mode rejection, and how to verify it on the bench.

The Problem Statement: A Differential Amplifier with a Twist

Step-by-Step Algebraic Solution

Which method applies and why? We must use the Superposition Theorem. Because the op-amp is operating in its linear region with negative feedback, the circuit behaves linearly. Superposition allows us to isolate the effect of each independent voltage source ($V_1$ and $V_2$) by grounding the other, solving two simpler circuits, and summing the results. Nodal analysis works too, but superposition maps directly to standard inverting/non-inverting gain blocks.

Step 1: The Inverting Contribution (Ground $V_2$)

Set $V_2 = 0V$. The non-inverting input is now tied to ground through $R_3$ and $R_4$. Because the op-amp is ideal, no current flows into the input pins. The voltage at the non-inverting pin ($V_+$) is $0V$. Due to the virtual short, the inverting pin ($V_-$) is also at $0V$ (virtual ground).

The circuit is now a standard inverting amplifier. The gain equation is:

$$A_{inv} = -\frac{R_f}{R_1}$$

Substituting our values:

$$A_{inv} = -\frac{100k\Omega}{10k\Omega} = -10$$

The output contribution from $V_1$ is:

$$V_{out1} = A_{inv} \times V_1 = -10 \times 2.0V = -20.0V$$

Step 2: The Non-Inverting Contribution (Ground $V_1$)

Set $V_1 = 0V$. The inverting input is tied to ground via $R_1$. The circuit is now a standard non-inverting amplifier, but the input signal ($V_2$) is attenuated by a voltage divider before reaching the op-amp pin.

First, calculate the voltage at the non-inverting pin ($V_+$) using the voltage divider formed by $R_3$ and $R_4$:

$$V_+ = V_2 \times \left( \frac{R_4}{R_3 + R_4} \right)$$

$$V_+ = 1.0V \times \left( \frac{90k\Omega}{10k\Omega + 90k\Omega} \right) = 1.0V \times \left( \frac{90}{100} \right) = 0.9V$$

Next, apply the non-inverting gain formula. The gain is determined by the feedback network ($R_f$ and $R_1$) relative to the virtual short at $V_-$:

$$A_{non} = 1 + \frac{R_f}{R_1}$$

$$A_{non} = 1 + \frac{100k\Omega}{10k\Omega} = 1 + 10 = 11$$

The output contribution from $V_2$ is:

$$V_{out2} = A_{non} \times V_+ = 11 \times 0.9V = 9.9V$$

Step 3: Summation

By superposition, the total output voltage is the sum of the individual contributions:

$$V_{out} = V_{out1} + V_{out2}$$

$$V_{out} = -20.0V + 9.9V = \mathbf{-10.1V}$$

The Trap, Sanity Check, and Verification

Answer Sanity Check

1. Order of Magnitude & Units: The result is in volts, and the magnitude (~10V) aligns with the input scale (1-2V) multiplied by the gain factor (~10).
2. Rail Limits: The TL072 on $\pm 15V$ rails has a typical maximum output swing of $\pm 13.5V$ (and absolute maximum of $\pm 15V$). Our calculated $-10.1V$ is well within the linear operating region, confirming our assumption that the op-amp did not saturate.
3. Directionality: The inverting input (2V) is higher than the non-inverting input (1V). With a dominant inverting gain path, a negative output is logically sound.

How to Verify Independently

Never trust hand-calculated op amp circuits examples on the first pass. Verify using:

• SPICE Simulation: Draft the schematic in LTspice. Use the generic UniversalOpAmp2 or import the TL072 model. Run an .op (operating point) simulation. The netlist will report V(out) = -10.1V.
• Bench Verification: Build the circuit on a breadboard. Use 1% tolerance metal film resistors (a standard 90k\Omega 1% resistor is available, or series a 80k and 10k). Power the TL072 with a dual-track bench supply set to $\pm 15V$. Measure $V_{out}$ with a 4.5-digit digital multimeter (DMM). You should read between -10.05V and -10.15V, accounting for the op-amp's input offset voltage ($V_{os}$) and resistor tolerances.

Frequently Asked Questions (FAQ)

Why do my op amp circuits examples always assume ideal infinite gain?

In introductory exams, assuming infinite open-loop gain ($A_{OL} = \infty$) simplifies the math by enforcing the "virtual short" rule ($V_+ = V_-$). In reality, a TL072 has an $A_{OL}$ of about 200 V/mV (106 dB). For a closed-loop gain of 10, the loop gain is still massive, meaning the virtual short error is in the microvolt range—far below the noise floor of standard bench equipment. We only abandon the ideal assumption when designing high-frequency circuits (where gain rolls off) or ultra-high-precision DC circuits (where finite gain and $V_{os}$ dominate the error budget).

How do I choose the right op-amp for high-precision differential circuits?

If you are building a difference amplifier where resistor matching is a pain, do not use a generic TL072 or LM741. Instead, use a dedicated instrumentation amplifier like the INA128 or AD620, which has laser-trimmed internal resistors guaranteeing a CMRR of 100dB+. If you must use discrete op-amps, select a precision part like the OPA2188 (zero-drift, max $V_{os}$ of 25\mu V) and use a matched resistor network IC (like the LT5400) rather than discrete 1% resistors to ensure the ratios track perfectly across temperature changes.

What happens if the op-amp output saturates in a difference amplifier?

If your calculated $V_{out}$ exceeds the op-amp's output swing limits (e.g., calculating -16V on a $\pm 15V$ TL072 supply), the output transistor hits the rail and clips. The virtual short breaks down immediately; $V_-$ no longer equals $V_+$. The op-amp essentially acts as a comparator, and the feedback network ceases to control the gain. To fix this, you must either increase the supply rails (if the op-amp's absolute maximum ratings allow), reduce the closed-loop gain, or attenuate the input signals before they reach the summing junctions.