Lecture 22: Op Amp Circuits — Subtractor, Integrator, Differentiator

What this lecture covers

Continuing the op amp unit, this lecture applies the ideal op amp model with no new theory, only the same four properties introduced previously (infinite gain, infinite input resistance, zero output resistance, and negative feedback) plus a new shortcut called the "v plus equals v minus method" (also called virtual ground) for analyzing op amp circuits under negative feedback quickly. Using these tools, the lecture builds up progressively more useful circuits: a subtractor circuit analyzed two ways (direct node analysis and superposition), then an integrator, then a differentiator, each designed step by step from first principles rather than presented as a finished formula.

After watching, you can apply the virtual-ground shortcut to analyze feedback op amp circuits quickly, recognize the inverting and noninverting connection patterns, and understand how routing a current through a capacitor via an op amp's virtual ground node produces integration or differentiation of a voltage signal.

Key ideas

• The v-plus-equals-v-minus (virtual ground) method: under negative feedback, when an op amp's output is not saturated, its two input terminals sit at essentially the same voltage because even a microvolt-scale difference, multiplied by the enormous gain, is enough to produce a normal output.
• Subtractor circuit: an op amp with v1 fed to the noninverting input through a resistor divider and v2 fed to the inverting input through a resistor with feedback produces vOUT = (R2/R1) * (v1 - v2).
• Two valid analysis routes: the same subtractor circuit can be solved either by writing node equations directly using the virtual ground shortcut, or by superposition (zeroing one source at a time and adding the resulting outputs).
• Inverting and noninverting connection patterns: two recurring building-block circuits, worth memorizing, where the output is either -(R2/R1)*vIN (inverting) or ((R1+R2)/R1)*vIN (noninverting).
• Sign convention warning: because the inverting input is a virtual ground, a positive voltage drop across the feedback resistor corresponds to a negative contribution to the output; forgetting this sign is a common mistake.
• Integrator design by reflection: routing an input voltage through a resistor into the op amp's virtual ground node converts it into a current (i = vI/R), which is then passed through a feedback capacitor so the output voltage becomes proportional to the integral of the input.
• Differentiator design by reflection: applying the input voltage directly across a capacitor produces a current proportional to dvI/dt; passing that current through a feedback resistor converts it back into a voltage, giving an output proportional to the derivative of the input.
• Design pattern reuse: both circuits are built by the same two-step trick: create the right kind of current using a capacitor or resistor, then use the op amp's virtual ground to convert that current into the desired output voltage.

Walkthrough

Review of the ideal op amp and the virtual ground method (0:17)

The instructor reviews the op amp's ideal properties (huge gain, infinite input resistance, zero output resistance) and introduces the "v plus equals v minus" shortcut: under negative feedback with the output not saturated, the two input terminals are approximately equal because the gain is so large that only a microvolt-scale difference is needed to sustain a normal output.

Building and analyzing a subtractor circuit (5:33)

Using two voltage sources v1 and v2 connected through resistor networks to the noninverting and inverting inputs, the instructor applies the virtual ground method and KCL to derive vOUT = (R2/R1)*(v1 - v2). He explicitly warns that the drop across the feedback resistor enters the output with a negative sign, a frequent source of errors.

Re-deriving the subtractor with superposition (17:52)

The instructor re-solves the same circuit by superposition, building two subcircuits (one with v1 zeroed, one with v2 zeroed) and identifying each as an instance of the inverting or noninverting connection pattern from earlier lectures. Adding the two partial outputs reproduces the same subtractor result, reinforcing that these two connection patterns are worth memorizing.

Designing the integrator (26:07)

Starting from the goal of integrating a voltage, the instructor first tries feeding the input through a plain resistor into a capacitor, showing this only approximates an integral when the RC time constant is very large, an unsatisfying result. He then uses the op amp's virtual ground property: since the inverting input sits at 0 V, a resistor from the input voltage to that node produces a current exactly proportional to the input, i = vI/R, which can then be routed through a feedback capacitor to produce a true integral, vO = -(1/RC) * integral(vI dt).

Designing the differentiator (39:27)

Reversing the integrator's logic, the instructor applies the input voltage directly across a capacitor connected to the virtual ground node, producing a current i = C*dvI/dt. Routing this current through a feedback resistor converts it back into a voltage, giving vO = -RC * dvI/dt. A demo confirms the results: a square-wave input to the integrator produces a triangular output, and the same input to the differentiator produces sharp spikes at each transition.

Before you watch

• Watch the previous lecture on the operational amplifier abstraction, since this lecture assumes familiarity with the ideal op amp model and negative feedback.
• Review superposition and the node method from earlier in the course, both used to analyze the subtractor circuit.
• Recall the element relationships for capacitors (i = C*dv/dt) from the RC and RLC lectures, since they are reused to build the integrator and differentiator.

Check your understanding

1. Why does negative feedback make v+ approximately equal to v- in an op amp circuit that is not saturated, and why does this shortcut give the same answer as full analysis with the dependent-source model?
2. In the inverting connection pattern, why does the output carry a negative sign relative to the voltage drop across the feedback resistor?
3. How does connecting a resistor from the input voltage to the op amp's virtual ground node convert that voltage into a current proportional to it?
4. Explain the two-step design logic that turns a capacitor and a resistor into either an integrator or a differentiator, depending on which element the input drives and which element provides feedback.