Scherz & Monk, Chapter 8

Operational Amplifiers

A bare op amp is a useless light switch — the tiniest input difference slams its output to a power rail. The whole subject is one idea: wrap negative feedback around that absurd gain to trade it for a precise gain you choose with two resistors. Master two golden rules and you analyze almost any op-amp circuit by inspection.

Prerequisites: Ohm's law & KCL + voltage dividers + I = C·dV/dt (integrator only)
17
Chapters
5+
Simulations
0
Assumed Knowledge

Chapter 0: The Op Amp That Won't Amplify

You wire up a 741 — the most famous chip in electronics history. You connect a +15 V supply to one pin, −15 V to another. You apply a tiny test signal of +0.001 V to the (+) input, ground the (−) input, and probe the output, expecting a small amplified copy of your input. Maybe a few hundred millivolts.

Instead the meter reads +14 V. The output has slammed almost all the way to the positive supply rail. Puzzled, you flip the input to −0.001 V. The output instantly snaps to −14 V — pinned at the negative rail. There is no in-between. A whisker of input on either side of zero drives the output fully to one rail or the other.

This chip is not behaving like an amplifier at all. It behaves like a light switch: the slightest difference between its two inputs throws it hard one way or the other, with nothing graceful in the middle. So how does this twitchy, all-or-nothing device become the precise ×10 audio amplifier in your guitar pedal, the buffer in your sensor front-end, the integrator in an analog computer?

The answer is a single wire. The reason the bare op amp pins to a rail is that its open-loop gain A₀ is enormous — 100,000 or more. Output = A₀ × (V⁺ − V⁻). With A₀ = 200,000, an input difference of just 0.001 V "wants" to produce 200 V — impossible, so the output clips at the rail. The fix is negative feedback: run one wire from the output back to the (−) input. That feedback tames the wild gain into a precise, predictable closed-loop gain set entirely by external resistors. This whole chapter is the story of that wire.
The Bare Op Amp: A Light Switch, Not an Amplifier

Drag the input voltage on the (+) terminal of an open-loop 741 (gain A₀ = 200,000, rails ±14 V). Watch the output. Notice you can barely move the input off zero before the output is fully pinned. The "useful" region is microvolts wide.

V⁺ input (mV) +1.000
Open-loop gain A₀ 200000

Look at what the slider tells you. With A₀ = 200,000 and rails at ±14 V, the output only stays in its linear region while the input is between −14/200000 and +14/200000 volts — that is −70 µV to +70 µV. A window 140 microvolts wide. Outside it, the output is glued to a rail. No knob, no resistor, no skill makes a bare op amp into a clean amplifier. You must change the topology.

Why does a bare (open-loop) op amp with V⁺ = +1 mV and V⁻ = 0 slam its output to the positive rail?

Chapter 1: The Ideal Op-Amp Model

Before we tame the op amp, we need a clean mental model of what it does. Strip away the hundred transistors inside and an op amp reduces to one equation. It senses the voltage difference between its two inputs and multiplies it by a gigantic number:

Vout = A₀ · (V⁺ − V⁻)

V⁺ is the non-inverting input (marked +), V⁻ is the inverting input (marked −), and A₀ is the open-loop gain. The word "differential" is the key: the op amp ignores any voltage common to both inputs and amplifies only the difference. The ideal op amp is defined by three idealizations and one consequence.

1. Infinite open-loop gain. A₀ → ∞ (real: 10⁴ to 10⁶, i.e. 80–120 dB). This is what made the output pin to a rail in Chapter 0.

2. Infinite input impedance. Rin → ∞ (real: 10⁶ Ω for bipolar, 10¹² Ω for JFET). The inputs sip essentially no current — picoamps to nanoamps.

3. Zero output impedance. Rout → 0 (real: 10–1000 Ω). The output behaves like an ideal voltage source — it holds its voltage regardless of load.

Consequence. Because A₀ is so large, the only way Vout can be a finite, sensible number is if (V⁺ − V⁻) is almost exactly zero. Hold that thought — it becomes the virtual short.

Worked Example — How Tiny Is "Tiny"?

Suppose A₀ = 100,000 and you want a sensible output of Vout = 5 V. Then the required input difference is:

V⁺ − V⁻ = Vout / A₀ = 5 / 100000 = 50 µV

Fifty microvolts. To get 5 V out, the two inputs need only differ by 50 millionths of a volt. This is why, in any working feedback circuit, we can treat the two inputs as being at the same voltage — the error is microscopic. The infinite gain is not a bug; it is the lever that forces precision.

Infinite gain is a feature, not a problem. A huge A₀ means even a microvolt of error between the inputs produces a large correction at the output. When you wrap feedback around it, that correction relentlessly drives the input difference toward zero. The bigger A₀ is, the tinier the residual error — so the closed-loop gain depends almost entirely on your resistors and almost not at all on the op amp itself. Stability through overwhelming force.
The Differential Transfer Curve

Vout vs the input difference (V⁺−V⁻), in microvolts. The slope is A₀. Crank A₀ up and the linear ramp gets so steep it becomes a near-vertical line through the origin — the seed of the "light switch" behavior. The shaded band is the linear region.

Open-loop gain A₀ 100000
Rail voltage (±V) 14
An ideal op amp has A₀ = 100,000. To produce a clean output of Vout = 8 V, what input difference (V⁺−V⁻) is required?

Chapter 2: The Golden Rules

Here is the good news that makes op-amp circuits tractable: you almost never need that messy Vout = A₀(V⁺−V⁻) equation. As long as the op amp is operating with negative feedback and is not pinned to a rail, two simple rules collapse nearly every op-amp circuit into one-line algebra.

Golden Rule 1 — No input current
The inputs draw no current: I⁺ = I⁻ = 0. This follows from the (near-)infinite input impedance. Whatever current flows toward an input pin, none of it goes into the op amp — it all flows through the external components. This is what lets us apply KCL at the input node and ignore the chip entirely.
Golden Rule 2 — The inputs are equal
With negative feedback, the op amp drives its output to whatever value makes V⁺ = V⁻. Not because they are wired together — they are not — but because the huge gain corrects any difference to near zero. This is the virtual short. If V⁺ is grounded, then V⁻ sits at 0 V too: a virtual ground.

That's it. Rule 1 says "the chip is invisible to current." Rule 2 says "the chip forces its two inputs to match." Combine them and you can analyze inverting amps, buffers, summers, and integrators with nothing more than Ohm's law and KCL.

Worked Example — Rules in Action

Consider an inverting amplifier (next chapter) with input resistor Rin = 10 kΩ from a source Vin = +0.5 V into the (−) node, the (+) input grounded, and feedback resistor Rf = 100 kΩ from output back to (−).

Gain = −5 V / 0.5 V = −10, entirely from the resistor ratio 100k/10k. We never touched A₀. That is the power of the golden rules.

When the rules do NOT apply. The golden rules assume the op amp is in its linear region with negative feedback active. They fail in two cases: (1) the output is saturated against a rail (then V⁺ ≠ V⁻ — it can't correct anymore), and (2) the feedback is positive (routed to the + input), as in a comparator with hysteresis, where the op amp deliberately races to a rail instead of balancing. Recognizing which regime you're in is half the battle.
Golden Rules Visualizer — Current Flows Around, Not Through

An inverting stage. Drag Vin. Animated dots show current entering through Rin, refusing to enter the op amp (Rule 1), and continuing through Rf. The (−) node label confirms it sits at 0 V (Rule 2, virtual ground).

Vin (V) +0.50
An op amp's (+) input is grounded and it has negative feedback to the (−) input. What is the voltage at the (−) input?

Chapter 3: Negative Feedback & the Virtual Short

Golden Rule 2 sounds like magic: how can two unconnected wires hold the same voltage? The mechanism is a feedback loop that is constantly self-correcting, faster than you can perceive. It is worth seeing the loop turn, step by step, because once you understand it you'll trust the virtual short in every circuit.

The Self-Correcting Loop

Imagine a non-inverting amplifier. You feed Vin into the (+) input, and a fraction of the output is fed back to the (−) input through a divider. Walk the loop:

  1. Say V⁺ momentarily exceeds V⁻ by a hair. The op amp sees a positive difference and its output rises (output = A₀ × difference).
  2. The rising output is divided down and fed back to V⁻, which therefore also rises.
  3. V⁻ rising shrinks the difference (V⁺−V⁻), which slows the output's rise.
  4. The loop settles at the exact point where the difference is essentially zero — i.e. V⁻ = V⁺. Equilibrium.

This is a stabilizing, negative feedback loop: any deviation produces a correction that opposes the deviation. The op amp behaves like a tireless servo whose only job is to make its two inputs equal. We call the resulting V⁺ = V⁻ condition a virtual short — "short" because the voltages match as if wired together, "virtual" because no current crosses (Rule 1).

Negative feedback ⇒ V⁺ = V⁻      (+) grounded ⇒ V⁻ ≈ 0 = virtual ground

Why "Virtual" Matters

A real short (a wire) between the inputs would let current flow and would short out your signal. The virtual short gives you the voltage equality without the current path. The (−) node is held at the (+) node's voltage by the op amp's output muscle, not by a copper connection — so you can still run resistors into that node and do useful arithmetic with the currents.

The op amp watches one thing and controls one thing. It watches (V⁺−V⁻) and controls Vout. With negative feedback in place, you can predict its behavior backwards: the output will go wherever it must to satisfy V⁻ = V⁺. Stop thinking "what does the chip output for this input?" and start thinking "what output makes the inputs equal?" That inversion is the single most useful habit in analog design.
Virtual-Short Visualizer — V⁻ Tracks V⁺ Instantly

Drag the (+) input voltage. The (−) input (fed back from the output) snaps to follow it, and the error meter (V⁺−V⁻) stays pinned at essentially zero. Increase A₀ and the residual error shrinks toward nothing.

V⁺ input (V) +2.0
Open-loop gain A₀ 100000
What physically holds the (−) input at the same voltage as the (+) input in a negative-feedback circuit?

Chapter 4: The Inverting Amplifier

Now we cash in. The inverting amplifier is the workhorse configuration, and with the golden rules its gain falls out in three lines. The circuit: source Vin through input resistor Rin into the (−) node; feedback resistor Rf from the output back to that same (−) node; the (+) input grounded.

Deriving the Gain

Because (+) is grounded, Rule 2 makes the (−) node a virtual ground at 0 V. The current arriving through Rin is:

Iin = (Vin − 0) / Rin = Vin / Rin

By Rule 1 none of it enters the op amp, so it all continues through Rf toward the output. The voltage at the output end of Rf is the virtual-ground voltage (0) minus the drop across Rf:

Vout = 0 − Iin·Rf = −(Vin/Rin)·Rf  ⇒  Gain = Vout/Vin = −Rf/Rin

The minus sign means the output is inverted — a positive input gives a negative output, and the waveform flips upside down. The magnitude is set purely by the ratio Rf/Rin. The op amp's own A₀ never appears.

Worked design — build a ×(−10) amplifier. You want a gain of −10. Pick Rin = 10 kΩ (a sane value — big enough not to load the source, small enough to avoid noise). Then Rf = 10 × Rin = 100 kΩ. Check: with Vin = +0.5 V, Vout = −(100k/10k)×0.5 = −5 V. For best DC accuracy add a bias-compensation resistor in series with the (+) input equal to Rin∥Rf = (10k×100k)/(110k) ≈ 9.1 kΩ, so both inputs see the same source resistance and the bias currents cancel.

Why the Virtual Ground Sets the Input Impedance

A subtle bonus: because the (−) node is held at 0 V, the source "sees" exactly Rin to ground — the input impedance of the whole stage is simply Rin. That makes it easy to set, but it also means a low Rin loads your source. It is the classic trade-off of the inverting topology, and the reason the non-inverting amp (next chapter) exists.

SHOWCASE — Inverting / Non-Inverting Gain Explorer

The big one. A sine wave enters on the left. Choose the mode, then set Rf and Rin. Watch the output sine on the right scale by the gain — and, in inverting mode, flip. The live gain readout shows −Rf/Rin or 1+Rf/Rin. If the scaled output would exceed the ±12 V rails, it clips.

Rf (kΩ) 100
Rin (kΩ) 10
Input amplitude (V) 1.0
An inverting amplifier has Rin = 2 kΩ and Rf = 20 kΩ. What is its voltage gain?

Chapter 5: Non-Inverting Amplifier & the Buffer

What if you want gain without the inversion, and you don't want to load your source through a low Rin? Move the signal to the (+) input instead. Now the source connects to the (+) input directly — which, by Rule 1, draws no current — so the input impedance is gigantic. The feedback network (Rf from output to (−), and Rin from (−) to ground) forms a divider.

Deriving the Gain

By Rule 2, V⁻ = V⁺ = Vin. The (−) node sits at the midpoint of a voltage divider formed by Rf (top) and Rin (bottom) between Vout and ground:

V⁻ = Vout · Rin/(Rin+Rf) = Vin

Solving for Vout:

Vout = Vin · (Rin+Rf)/Rin  ⇒  Gain = 1 + Rf/Rin

Notice three things. The gain is positive (no flip). It can never be less than 1 — the "1+" floor is unavoidable here. And the input impedance is essentially infinite, because the source feeds the bare (+) pin.

Worked design — build a ×5 non-inverting amp. You need gain = 5, so 1 + Rf/Rin = 5, meaning Rf/Rin = 4. Choose Rin = 1 kΩ → Rf = 4 kΩ. Check: Vin = +1 V gives Vout = (1 + 4k/1k)×1 = +5 V, same polarity. Compare with the inverting amp: to get ×5 there you'd need Rf/Rin = 5 and you'd flip the sign.

The Voltage Follower (Buffer): Gain = +1

Take the non-inverting amp to its limit: set Rf = 0 (wire the output straight to (−)) and remove Rin (open it). Then gain = 1 + 0 = +1. The output simply copies the input. Useless as an amplifier — but priceless as an impedance buffer. It presents near-infinite impedance to the source (drawing no current from it) and near-zero impedance to the load (driving it stiffly).

The follower is the great isolator. A weak, high-impedance source — a pH probe, a photodiode, a long potentiometer — collapses the moment you connect a real load to it. Slip a voltage follower between source and load: the source sees an open circuit (no loading, no droop), and the load gets driven by the op amp's stiff output. Same voltage, transformed impedance. This is why nearly every sensor front-end starts with a buffer.
Buffer as Impedance Isolator

A high source resistance Rsrc feeds a load Rload. Toggle the buffer in and out. Without the buffer the divider droops the voltage badly; with it, the load voltage equals the source voltage no matter how heavy the load.

Source resistance (kΩ) 50
Load resistance (kΩ) 10
A non-inverting amplifier has Rf = 9 kΩ and Rin = 1 kΩ. What is its gain?

Chapter 6: Summing & Difference Amplifiers

The virtual ground of the inverting amp is a gift that keeps giving. Because the (−) node is pinned at 0 V, you can run several input resistors into it at once, and each one delivers its own current independently — the node doesn't "see" the others. By KCL all those currents add and flow through Rf. You have built an analog adder.

The Summing Amplifier

With inputs V₁, V₂, V₃ through resistors R₁, R₂, R₃ into the virtual ground, each contributes current Vk/Rk. KCL at the node: the sum of input currents equals the feedback current, so:

Vout = −Rf·(V₁/R₁ + V₂/R₂ + V₃/R₃)

If all resistors are equal (R₁ = R₂ = R₃ = Rf = R), this simplifies to a clean weighted-free sum:

Vout = −(V₁ + V₂ + V₃)
Worked example — the analog adder. Let R₁ = R₂ = R₃ = Rf = 10 kΩ, with V₁ = +1 V and V₂ = +2 V (V₃ grounded). Then Vout = −(1 + 2 + 0) = −3 V. The circuit literally added two voltages. Want a weighted mixer instead? Make Rf/R₁ = 2 and Rf/R₂ = 1 and you get Vout = −(2V₁ + V₂). This is exactly how an analog audio mixer blends channels, and how the resistor-ladder DAC builds an output from binary-weighted bits.

The Difference Amplifier (Subtractor)

To subtract, feed one signal to the inverting side and one to the non-inverting side, with matched resistor pairs. With R₁ into (−) and R₂ as feedback, and a matching R₁/R₂ divider on the (+) input:

Vout = (R₂/R₁)·(V₂ − V₁)

This amplifies the difference between two inputs while rejecting any voltage they share in common — the seed of the instrumentation amplifier used to pull a tiny sensor signal out of a sea of common-mode noise (e.g. a strain-gauge bridge or an ECG electrode).

Why the virtual ground enables clean summing. In an ordinary resistor mixer, every input interacts with every other — change one and the others shift. The op-amp summer's virtual ground breaks that coupling: each input sees a fixed 0 V on the far side of its resistor, so its current depends only on its own voltage. The inputs are perfectly isolated even though they share a node. That independence is the whole reason this topology dominates analog computation.
Summing Amplifier — Watch the Currents Add

Three inputs into one virtual ground (all R = Rf = 10 kΩ). Set V₁, V₂, V₃ and watch the current bars sum, then invert into Vout = −(V₁+V₂+V₃). The output clips at ±12 V.

V₁ (V) +1.0
V₂ (V) +2.0
V₃ (V) +0.0
A summing amp has R₁=R₂=Rf=10 kΩ, V₁=+2 V, V₂=+3 V. What is Vout?

Chapter 7: Integrators & Differentiators

So far the feedback element has been a plain resistor. Replace it with a capacitor and the op amp starts doing calculus. The reason is the capacitor's defining law, I = C·dV/dt: a capacitor relates current to the rate of change of voltage. Put it in the feedback path and the math of the virtual ground turns into integration.

The Integrator

Take the inverting topology but swap Rf for a capacitor C. The input current is still I = Vin/R (virtual ground). That same current must flow into C, and for a capacitor the voltage builds as the integral of current. Working it through:

Vout = −(1/RC) · ∫ Vin dt

A constant input therefore produces a ramp output — the integral of a constant is a straight line. The slope of that ramp is Vin/(RC). Feed in a square wave and you get a triangle wave out; feed in a sine and you get a cosine (a 90° phase shift). The integrator is the heart of analog computers, ramp generators, and dual-slope ADCs.

Worked example — the ramp generator. Let R = 10 kΩ, C = 1 µF, so RC = (10×10³)(1×10⁻⁶) = 0.01 s. Apply a steady Vin = −1 V. Then dVout/dt = −Vin/(RC) = −(−1)/0.01 = +100 V/s. The output ramps upward at 100 volts per second — it would climb 1 V every 10 ms until it hits the rail. To stop a real integrator from drifting to a rail on DC offset, you bleed it off with a large resistor across C.

The Differentiator

Swap the components: capacitor on the input, resistor in the feedback. Now the input current is C·dVin/dt, and the output is:

Vout = −RC · dVin/dt

It outputs the slope of its input. A triangle wave in gives a square wave out; a sharp edge gives a spike. Differentiators are less common in practice because they amplify high-frequency noise (steep noise = big slope), so they're usually tamed with a small series resistor.

The capacitor is doing the calculus, not the op amp. All the integration happens because I = C·dV/dt is itself a differential equation. The op amp's only contribution is the virtual ground, which pins one plate of the capacitor at 0 V so the capacitor's current is cleanly determined by Vin/R alone. Swap R and C and you swap integration for differentiation — same op amp, opposite operation. Topology is everything.
Integrator Waveform Shaper — Square In, Triangle Out

A square wave drives the integrator. The output ramps up while the input is low and down while it's high — producing a triangle. Change R and C to change the ramp slope (1/RC). Switch the input to a sine to see the 90° phase-shifted (cosine) output.

R (kΩ) 10
C (µF) 1.0
An integrator has R = 10 kΩ, C = 1 µF (RC = 0.01 s) and a constant input Vin = −1 V. How does the output behave?

Chapter 8: Comparators & Hysteresis

In Chapter 0 the bare op amp's "light switch" behavior was a bug. Now we make it a feature. Remove the negative feedback entirely and the op amp becomes a comparator: it asks one yes/no question — "is V⁺ above or below V⁻?" — and slams its output to the appropriate rail. Above the threshold → high; below → low. A 1-bit analog-to-digital decision.

The Open-Loop Comparator and Its Flaw

Tie a reference voltage Vref to (−) and your signal to (+). When the signal crosses Vref, the output flips. Perfect — until the signal is noisy. A real signal jitters as it crosses the threshold, and each tiny wiggle across the line produces another output flip. Near the crossing you get a burst of rapid, useless transitions called chatter. A thermostat built this way would click its relay dozens of times a second right at the set point.

Hysteresis: Positive Feedback to the Rescue

The cure is to add a little positive feedback — route a fraction of the output back to the (+) input through resistors R₁ (to the signal/reference) and R₂ (from the output). This creates two thresholds instead of one. When the output is high, the threshold sits higher; when low, it sits lower. The signal must travel all the way across a deadband to trigger the reverse flip:

±VT = ±Vout(sat) · R₁/R₂

This two-threshold behavior is hysteresis (the circuit is a Schmitt trigger). Once it flips high, noise can't flip it back until the signal drops well below the lower threshold — so the brief jitters that caused chatter are simply ignored. Clean, single transitions.

Worked example — sizing the deadband. Suppose the output saturates at ±12 V and you want thresholds at ±0.6 V (a 1.2 V-wide deadband to ride over your noise). Then R₁/R₂ = VT/Vout = 0.6/12 = 0.05. Choose R₂ = 100 kΩ → R₁ = 5 kΩ. Now a noise spike smaller than 1.2 V peak-to-peak right at the crossing cannot cause a false re-trigger. Make the deadband wider than your worst-case noise and chatter disappears.
Comparators are NOT linear amplifiers. A comparator deliberately lives in saturation — the golden rules (V⁺=V⁻) do not apply because there's no negative feedback to enforce them. Dedicated comparator chips (LM311, LM393) are optimized for fast switching and have open-collector outputs; using a general-purpose op amp as a comparator works but switches slowly. The hysteresis trick uses positive feedback, the mirror image of everything in Chapters 3–7.
Comparator with Hysteresis — Chatter vs Clean Transitions

A noisy signal sweeps across the threshold. With the hysteresis window at zero you see chatter — a burst of flips at every crossing. Widen the window and the two thresholds (dashed) open into a deadband; the output now makes clean single transitions. Add noise to stress-test it.

Hysteresis window (V) 0.00
Noise amplitude (V) 0.40
Why do practical comparators use hysteresis (positive feedback) rather than a single fixed threshold?

Chapter 9: Where Ideal Meets Real

The golden rules treat the op amp as perfect. Real chips are not. Three limitations bite hardest in practice, and knowing them is the difference between a circuit that works on paper and one that works on the bench.

Gain–Bandwidth Product (GBW)

An op amp's open-loop gain is huge at DC but rolls off with frequency. The product of gain and bandwidth is a constant called the gain–bandwidth product, equal to the unity-gain frequency fT:

A · f = fT  ⇒  BW = fT / (closed-loop gain)

For a 741, fT ≈ 1 MHz. So a gain of 10 gives a bandwidth of 1 MHz/10 = 100 kHz; a gain of 1000 leaves only 1 MHz/1000 = 1 kHz. You don't get high gain and high bandwidth from one stage — you trade one for the other. Need both? Cascade stages, or pick a faster op amp.

Slew Rate

Even within its bandwidth, the output can only change so fast. The maximum slope of the output is the slew rate, SR. For a 741, SR = 0.5 V/µs. A sine wave Vout = Vpeaksin(2πft) has a maximum slope of 2πf·Vpeak; demand a steeper slope than SR and the op amp can't keep up — the smooth sine degrades into a triangle. The largest undistorted frequency for a given amplitude is:

fmax = SR / (2π·Vpeak)
Worked example — the slew limit. A 741 (SR = 0.5 V/µs = 0.5×10⁶ V/s) drives a ±10 V (20 V peak-to-peak) sine. Maximum undistorted frequency: fmax = SR/(2πVpeak) = 0.5×10⁶/(2π×10) ≈ 8 kHz. Push past 8 kHz at full amplitude and your clean sine turns into a triangle — even though the 741's small-signal bandwidth extends much higher. Slew rate, not bandwidth, sets the limit for large signals. Want full-amplitude audio to 20 kHz? You need an op amp with SR > ~1.3 V/µs.

Offset, Bias, and Single Supply

Two more gotchas. Input offset voltage (741: ~2 mV) is a small built-in imbalance — the output isn't exactly zero when the inputs match; it gets multiplied by the gain, so at gain 1000 a 2 mV offset becomes 2 V of error. Input bias current (741: ~500 nA) flows into the inputs and drops voltage across your resistors — the reason for that bias-compensation resistor in Chapter 4. And a single-supply op amp (no negative rail) can't output negative voltages, so you must bias signals around a mid-supply reference.

Saturation & Slew-Rate Scope

A real-op-amp model (SR = 0.5 V/µs, rails ±12 V). Crank amplitude and frequency. The ideal output (dashed) is a clean sine; the real output (solid) clips at the rails when too big and triangulates when the demanded slope exceeds the slew rate. The readout flags which limit you've hit.

Amplitude (V) 8.0
Frequency (kHz) 5
A 741 has fT = 1 MHz. Configured for a closed-loop gain of 50, what is its approximate bandwidth?

Chapter 10: Connections & What Comes Next

One idea unifies this entire chapter: feedback trades raw gain for control. The op amp's absurd open-loop gain is not the point — it is the raw material. Wrap negative feedback around it and you get precise, resistor-defined behavior; wrap positive feedback and you get decisive switching. Everything else is choosing what to put in the feedback path.

ConfigurationKey equationWhat it does
Ideal modelVout = A₀(V⁺−V⁻)Huge gain on the input difference
Golden rulesI⁺=I⁻=0; V⁺=V⁻Collapse any feedback circuit to algebra
Virtual ground(+) grounded ⇒ V⁻≈0Pins the inverting node at 0 V
Inverting ampGain = −Rf/RinScaled, flipped output; Rin sets input Z
Non-inverting ampGain = 1 + Rf/RinScaled, same-polarity; huge input Z; gain ≥ 1
Voltage followerGain = +1Impedance buffer / isolator
Summing ampVout = −∑(Rf/Rk)VkAnalog adder / mixer / DAC
Difference ampVout = (R₂/R₁)(V₂−V₁)Subtractor / instrumentation front-end
IntegratorVout = −(1/RC)∫VindtRamp gen; square→triangle; analog calculus
DifferentiatorVout = −RC·dVin/dtSlope detector; edge→spike
Comparator + hyst.±VT = ±Vsat·R₁/R₂1-bit decision; Schmitt trigger
GBWBW = fT/gainGain–bandwidth trade-off
Slew ratefmax = SR/(2πVpeak)Large-signal speed limit

Where Op Amps Go From Here

Active filters (Ch 9) — put R and C networks in the feedback path of an op amp and you get filters with gain, sharp roll-off, and no bulky inductors. The integrator you just met is a single-pole low-pass filter in disguise.
Oscillators & timers (Ch 10) — combine the integrator and the Schmitt-trigger comparator and the output chases its own tail forever: a relaxation oscillator. The 555 timer is this idea hardened into a chip.
Power supplies (Ch 11) — a voltage regulator is an op amp comparing the output against a reference and driving a pass transistor to hold it steady. Negative feedback again, this time regulating volts and amps.
Audio (Ch 16) — preamps, mixers, and tone controls are inverting/summing/filter stages. The LM386 is a complete power op amp with gain selectable from 20 to 200.

Connections to Other Chapters

"The op amp's gain is so high that the only sensible thing the output can do is whatever makes the inputs equal."
— the virtual short, in one sentence

You now hold the master key to analog design. Two golden rules, one feedback wire, and a handful of resistor ratios let you read — and build — nearly every linear circuit in the book. Next, we put frequency-dependent components in that feedback path and shape signals across the spectrum.

What is the single unifying idea behind every op-amp configuration in this chapter?
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