If you are asking what is an example of a negative feedback loop in practical electronics, the most foundational and universally tested answer is the non-inverting operational amplifier (op-amp) circuit. In a negative feedback loop, a portion of the output signal is routed back to the input in a way that opposes changes, forcing the system to stabilize at a precise, predictable state. In an op-amp, this means feeding a voltage divider from the output back into the inverting (-) input, which forces the op-amp to adjust its output until the inverting and non-inverting inputs are at the exact same voltage.

This walkthrough breaks down a classic exam problem on this exact topology, showing every algebraic step, identifying the common traps, and providing a concrete decision framework for selecting real-world components.

Practice Problem: Designing a Non-Inverting Amplifier

📝 EXAM PROBLEM STATEMENT
Design a non-inverting amplifier using an LM358 op-amp powered by a single 12V DC supply. The circuit must amplify a 0.5V peak sine wave to exactly 6.0V peak (a voltage gain of 12 V/V). The ground reference resistor ($R_g$) is fixed at 10 kΩ to minimize thermal noise. Calculate the exact required value for the feedback resistor ($R_f$) to achieve this gain. Verify if the output will clip given the power supply constraints.

Which Theorem/Method Applies and Why?

To solve this, we apply the Ideal Op-Amp Golden Rules combined with Kirchhoff’s Current Law (KCL). We use the Golden Rules because, in a negative feedback configuration, the op-amp's massive open-loop gain forces the differential input voltage to virtually zero. KCL allows us to map the current flow through the feedback network to derive the closed-loop gain equation. All About Circuits provides an excellent primer on these foundational rules.

⚠️ THE TRAP: The most common exam mistake is blindly applying the inverting amplifier gain formula ($A_v = -R_f / R_g$) or forgetting the "+1" in the non-inverting formula. If you calculate $R_f = 120 k\Omega$ by multiplying 12 by 10k, you have fallen into the trap of ignoring the baseline unity gain inherent to the non-inverting topology.

Step-by-Step Algebraic Solution

  1. Define the Nodes: Let the non-inverting input be $V_+$, the inverting input be $V_-$, the input signal be $V_{in}$, and the output be $V_{out}$.
  2. Apply Golden Rule 1 (Virtual Short): With negative feedback active, the op-amp drives its output to make $V_- = V_+$. Since $V_+$ is connected directly to $V_{in}$, we establish that $V_- = V_{in}$.
  3. Apply Golden Rule 2 (Infinite Input Impedance): No current flows into the op-amp's input pins. Therefore, the current flowing through $R_g$ to ground must exactly equal the current flowing through $R_f$ from the output.
  4. Write the KCL Equation at the $V_-$ Node:
    $I_{Rg} = I_{Rf}$
    $(V_- - 0) / R_g = (V_{out} - V_-) / R_f$
  5. Substitute $V_-$ with $V_{in}$:
    $V_{in} / R_g = (V_{out} - V_{in}) / R_f$
  6. Rearrange to solve for the Gain ($A_v = V_{out} / V_{in}$):
    Multiply both sides by $R_f$: $V_{in} \times (R_f / R_g) = V_{out} - V_{in}$
    Add $V_{in}$ to both sides: $V_{out} = V_{in} + V_{in} \times (R_f / R_g)$
    Factor out $V_{in}$: $V_{out} = V_{in} \times [1 + (R_f / R_g)]$
    Divide by $V_{in}$: $A_v = 1 + (R_f / R_g)$
  7. Plug in the Known Values:
    We need a gain ($A_v$) of 12, and $R_g$ is 10,000 Ω.
    $12 = 1 + (R_f / 10,000)$
  8. Isolate $R_f$:
    Subtract 1 from both sides: $11 = R_f / 10,000$
    Multiply by 10,000: $R_f = 110,000 \Omega$

Final Answer: The required feedback resistor is 110 kΩ.

Answer Sanity Check

Does this make physical sense? A gain of 12 V/V means the output should be 12 times the input. $0.5V \times 12 = 6.0V$. The Texas Instruments LM358 datasheet specifies that the maximum high-level output voltage ($V_{OH}$) on a 12V supply is typically $V_{CC} - 1.5V$, or 10.5V. Since our required peak output of 6.0V is well below the 10.5V saturation limit, the signal will not clip. Furthermore, 110 kΩ is a standard E24 series resistor value, meaning this is practically buildable without combining series/parallel resistors.

Decision Tree: Selecting Real-World Feedback Resistor Values

In an exam, $R_g = 10 k\Omega$ is given. On the workbench, you must choose both $R_g$ and $R_f$. Use this decision matrix to terminate your design with a concrete pick based on your application constraints.

Application Condition Engineering Constraint Concrete Pick ($R_g$ / $R_f$)
Battery-Powered / Low Quiescent Current Must minimize current draw through the feedback divider to extend battery life. 100 kΩ / 1.1 MΩ
High-Speed / High-Frequency RF Must minimize parasitic capacitance effects and Johnson-Nyquist thermal noise. 1 kΩ / 11 kΩ
General-Purpose DC / Audio (Default) Balances power consumption, noise, and avoids loading the op-amp output stage. 10 kΩ / 110 kΩ
✅ DEFAULT RECOMMENDATION: Unless you have a specific high-speed or ultra-low-power constraint, always default to the 10 kΩ / 110 kΩ pair. It keeps the feedback network current at a safe ~0.5 mA, easily sourced by the LM358's output stage, while keeping thermal noise negligible for audio and DC sensor applications.

Independent Verification & Common Exam Traps

Never trust a single calculation on a critical design. Here is how you verify the answer independently:

  1. Simulation (LTspice): Drop an LT1001 or generic op-amp model into LTspice. Apply a 0.5V DC source to the non-inverting pin. Place 10k to ground and 110k from output to the inverting pin. Run an operating point (.op) simulation. The output node must read exactly 6.000V.
  2. Physical Bench Test: Build the circuit on a breadboard. Power the LM358 with a bench supply set to 12.0V. Inject a 1 kHz, 0.5 Vpp sine wave from a function generator. Probe the output with an oscilloscope. You should measure exactly 6.0 Vpp. If you measure slight clipping on the positive peaks, your function generator is actually outputting >0.5V, or your 12V rail is sagging under load.

Frequently Asked Questions

Q: What happens if I accidentally swap $R_f$ and $R_g$?

If you put 10 kΩ in the feedback path and 110 kΩ to ground, your gain equation becomes $1 + (10,000 / 110,000) = 1.09 V/V$. Your 0.5V input will only amplify to 0.545V. The negative feedback loop is still stable, but the attenuation of the voltage divider heavily reduces the closed-loop gain.

Q: Why not just use a single resistor instead of a divider for feedback?

A single resistor from output to inverting input (with the inverting input tied directly to ground via a wire) creates a voltage follower (unity gain buffer). The gain is exactly 1 V/V. To achieve any gain greater than 1, you must use a voltage divider to feed back only a fraction of the output, forcing the op-amp to drive its output higher to satisfy the virtual short condition.

Q: Does the negative feedback loop guarantee stability?

At DC and low frequencies, yes. However, at high frequencies, internal op-amp pole delays introduce phase shift. If the phase shift reaches 180 degrees before the loop gain drops below 1, the negative feedback turns into positive feedback, and the circuit will oscillate. This is why high-speed designs require a small compensation capacitor in parallel with $R_f$ to roll off high-frequency gain.