The Differential Amplifier Trap: A Worked Problem

When studying examples of op amp circuits for exams or bench design, the differential amplifier (subtractor) is a staple. It looks simple, but it hides a massive trap for students who memorize formulas without understanding the underlying nodal analysis. Below is a classic exam-style problem featuring an unbalanced resistor bridge. We will use a precision FET-input op amp like the Texas Instruments OPA2134, powered by standard ±15V rails.

Problem Statement

Given: An ideal op amp configured as a differential amplifier.

  • Inverting input resistor ($R_1$) = $10 k\Omega$ connected to $V_1 = 1.0V$
  • Feedback resistor ($R_f$) = $100 k\Omega$ connected to $V_{out}$
  • Non-inverting input resistor ($R_2$) = $12 k\Omega$ connected to Ground
  • Non-inverting source resistor ($R_3$) = $100 k\Omega$ connected to $V_2 = 1.2V$
  • Op amp supply rails: $\pm 15V$

Find: The exact output voltage ($V_{out}$).

The Trap: Many students immediately apply the shortcut formula $V_{out} = \frac{R_f}{R_1}(V_2 - V_1)$. If you do that here, you get $10 \times (1.2 - 1.0) = +2.0V$. This is completely wrong. That formula only applies when the resistor ratios are perfectly matched ($\frac{R_f}{R_1} = \frac{R_3}{R_2}$). Here, $\frac{100k}{10k} = 10$, but $\frac{100k}{12k} = 8.33$. The bridge is unbalanced, and the common-mode rejection is compromised.

Step-by-Step Algebraic Solution

Which method applies and why? We use Nodal Analysis combined with the Virtual Short principle. Superposition works, but nodal analysis is less prone to sign errors when dealing with unbalanced non-inverting voltage dividers. The virtual short dictates that for an ideal op amp in a negative feedback loop, the voltage at the inverting terminal ($V_-$) equals the voltage at the non-inverting terminal ($V_+$).

Step 1: Solve for the Non-Inverting Node ($V_+$)

The non-inverting input draws zero current (ideal op amp infinite input impedance). Therefore, $R_3$ and $R_2$ form a simple unloaded voltage divider between $V_2$ and Ground.

  1. $V_+ = V_2 \times \left( \frac{R_2}{R_2 + R_3} \right)$
  2. $V_+ = 1.2V \times \left( \frac{12k\Omega}{12k\Omega + 100k\Omega} \right)$
  3. $V_+ = 1.2V \times \left( \frac{12}{112} \right)$
  4. $V_+ = 1.2V \times 0.107143$
  5. $V_+ = 0.12857V$

Step 2: Apply the Virtual Short

Because of the negative feedback through $R_f$, the op amp adjusts its output to force the inverting terminal to match the non-inverting terminal.

  1. $V_- = V_+$
  2. $V_- = 0.12857V$

Step 3: Write Kirchhoff's Current Law (KCL) at $V_-$

The sum of currents entering the inverting node must equal zero. Current flows from $V_1$ through $R_1$, and from $V_{out}$ through $R_f$.

  1. $\frac{V_1 - V_-}{R_1} + \frac{V_{out} - V_-}{R_f} = 0$
  2. $\frac{1.0V - 0.12857V}{10k\Omega} + \frac{V_{out} - 0.12857V}{100k\Omega} = 0$
  3. $\frac{0.87143V}{10k\Omega} = - \frac{V_{out} - 0.12857V}{100k\Omega}$

Step 4: Isolate and Solve for $V_{out}$

Multiply both sides by $100k\Omega$ to clear the denominator on the right.

  1. $0.87143V \times \left(\frac{100k\Omega}{10k\Omega}\right) = - (V_{out} - 0.12857V)$
  2. $0.87143V \times 10 = -V_{out} + 0.12857V$
  3. $8.7143V = -V_{out} + 0.12857V$
  4. $V_{out} = 0.12857V - 8.7143V$
  5. $V_{out} = -8.5857V$

Rounding to standard engineering precision, $V_{out} = -8.59V$.

Sanity Checks and Real-World Verification

Before moving to the next exam question or soldering your prototype, you must verify the result. Here is how to independently check your work.

1. Order of Magnitude and Saturation Check

Is $-8.59V$ physically possible? The OPA2134 is powered by $\pm 15V$ rails. Its output voltage swing specification guarantees it can swing to within 1V of the rails (i.e., $\pm 14V$). Since $-8.59V$ is well within the $-14V$ to $+14V$ linear range, the op amp is not saturated. The units are correct (Volts), and the magnitude makes sense given the $10\times$ gain applied to the $\sim 0.87V$ potential difference across $R_1$.

2. Independent Verification via Bench or Simulation

If you build this on a breadboard, do not just trust the math. Texas Instruments Precision Labs emphasizes accounting for non-ideal DC parameters. The OPA2134 has a typical input offset voltage ($V_{os}$) of $1mV$. In this circuit, the noise gain is $1 + \frac{R_f}{R_1} = 11$. Therefore, the offset voltage is amplified by 11, introducing up to $\pm 11mV$ of error at the output. When measuring with a Fluke 87V multimeter, an actual reading between $-8.57V$ and $-8.61V$ confirms your theoretical math is correct.

Pro-Tip: If you simulate this in LTspice to verify, ensure you select a specific op amp model (like the OPA2134 or LT1001) rather than the default ideal 'opamp' symbol, which lacks realistic rail limits and phase margins.

FAQ: Common Questions on Examples of Op Amp Circuits

What are the most common examples of op amp circuits in exams?

University and certification exams heavily favor circuits that test your understanding of feedback and nodal analysis. The top four are:

  • Inverting Amplifier: Tests basic virtual ground concepts and gain inversion.
  • Non-Inverting Amplifier: Tests the voltage divider at the feedback node and the $1 + \frac{R_f}{R_1}$ gain formula.
  • Transimpedance Amplifier (TIA): Tests current-to-voltage conversion, often using a photodiode model.
  • Differential/Instrumentation Amplifiers: Tests superposition, common-mode rejection ratio (CMRR), and bridge balancing (as demonstrated above).

For a deeper dive into the semiconductor physics behind these, All About Circuits provides excellent foundational reading on differential configurations.

How do non-ideal parameters affect examples of op amp circuits?

In textbook examples, we assume infinite open-loop gain, infinite input impedance, and zero output impedance. On the bench, three parameters ruin your day:

  1. Input Offset Voltage ($V_{os}$): Acts like a tiny DC battery in series with the non-inverting input. It gets multiplied by the noise gain, causing DC errors at the output.
  2. Input Bias Current ($I_b$): Real op amps (especially bipolar ones like the LM741) draw nanoamps or microamps into their inputs. If the Thevenin equivalent resistance seen by both inputs isn't matched, this current creates an unwanted voltage drop that mimics an offset voltage.
  3. Slew Rate: The maximum rate of change of the output voltage (e.g., $20V/\mu s$ for the OPA2134). If your AC signal demands a faster transition than the slew rate allows, the output distorts into a triangle wave.

Why does my physical op amp circuit clip when the math says it shouldn't?

If your algebra says $V_{out}$ should be $12V$, but your oscilloscope shows a flat line at $10.5V$, you are hitting the output voltage swing limit. Standard op amps are not 'rail-to-rail' output. A classic LM741 powered by $\pm 15V$ will typically clip around $\pm 13V$ under light loads, and even lower (e.g., $\pm 11V$) if driving a heavy load like a $50\Omega$ resistor. Always check the 'Output Voltage Swing' table in the datasheet under your specific load resistance ($R_L$) and supply voltage conditions.