The Op-Amp Bench

A gain so large it is useless on its own

13 minutes · the open loop, before any feedback exists

Recall From chapter 5: roughly what was the highest voltage gain a single bipolar stage could deliver? show answer

Chapter 5 found a ceiling near 2,900 for one bipolar stage, chapter 7 found something similar and messier for one FET stage. An op-amp is not one stage. It is several — a differential input pair, a gain stage, an output buffer — cascaded on one chip, and their gains multiply. A modest general-purpose part might quote an open-loop gain A₀ of 200,000. Better ones run into the millions.

The defining equation of an op-amp, before any wire connects its output back to its input, is almost insultingly simple:

V_out = A₀ (V₊ − V₋)

Two input terminals, their difference multiplied by an enormous number. That is the entire black box, for this lesson. No feedback, no resistors, nothing external at all yet.

Commit before you touch anything

A₀ = 200,000, and the supply rails limit the output to ±13.5 V. You slowly raise V₊−V₋ from 0. About how large a difference does it take to drive the output all the way to a rail?

Answer: B. 13.5 V ÷ 200,000 ≈ 67.5 µV. Drag the bench's input slider and watch the output hit a rail before the slider has moved any meaningful distance at all — the useful, unsaturated range of a bare op-amp is a sliver smaller than most op-amps' own input offset voltage.

A gain this large is a feature, used the wrong way

Left open-loop, an op-amp is not an amplifier in any practical sense — it is a comparator. Any input difference bigger than a few tens of microvolts, positive or negative, and the output is already pinned against a rail. That behaviour is not a flaw to be designed around; some circuits use it directly. A zero-crossing detector, a simple over-voltage alarm, a Schmitt trigger's core switching action — all of them want exactly this all-or-nothing response.

V₊ − V₋	V_out = A₀×difference	What actually happens
+10 µV	+2.0 V	within range — the only sliver where this formula still applies unclipped
+100 µV	+20 V	clipped at +13.5 V
−50 µV	−10 V	within range
−1 mV	−200 V	clipped at −13.5 V

“Infinite gain” is shorthand for “so large that the exact number stops mattering.” It is not a claim that A₀ is literally infinite — it is a working assumption that will turn out to be excellent once feedback is added in lesson 2, and this lesson is here so you know exactly what is being assumed away.

In the wild

Comparator ICs are this lesson, sold on purpose. A dedicated comparator chip is built and specified to behave exactly like this bench — fast, rail-to-rail, no attempt at linear operation — because some circuits want a decision, not an amplified copy.

Why op-amp datasheets bother quoting A₀ at all, if it's “basically infinite.” It sets exactly this threshold — how small an input imbalance is enough to saturate the part — which matters directly for anyone building a comparator or a precision zero-detector out of one.

Why touching the two input pins of a bare op-amp with your fingers makes the output flicker wildly. Skin resistance and stray pickup are enough, at these microvolt sensitivities, to swing the output between rails at whatever frequency the noise arrives.

Before you move on

With no feedback connected, what does an op-amp's output do for almost any nonzero input difference?

A₀ is so large that almost any input the device can distinguish from true zero is already enough to saturate the output. A describes what feedback will buy you starting next lesson — it is not what the bare device does.

Bench 01 · open loop, no feedback at all LINEAR

● Locked until you commit a prediction above.

Input difference V₊−V₋ 0 µV

−200 µV0+200 µV

Ideal output (A₀×V_id)—

Actual output—

The virtual short is a symptom, not a law

15 minutes · what feedback actually does to the two inputs

Recall From lesson 1: about how large an input difference was enough to saturate this op-amp's output? show answer

Every op-amp course eventually states the “golden rule”: the two inputs sit at the same voltage. Stated like that, it sounds like a property of the chip — something wired in, the way a bipolar transistor's base–emitter junction is wired in. It isn't. Watch what actually forces it.

Connect the output back to the inverting input through a wire — a unity-gain buffer, the simplest feedback loop there is. Now V₋ = V_out = A₀(V₊−V₋). Solve for V₋:

V₋ = V₊ · A₀/(1+A₀) — error = V₊−V₋ = V₊/(1+A₀)

With A₀ = 200,000, that error is V₊/200,001 — about five parts per million. Not zero. Astonishingly close to zero, because the loop keeps adjusting V_out until the difference driving it is tiny enough that A₀ times that tiny number lands back on V_out again, self-consistently.

Commit before you touch anything

You take the same unity-gain buffer and swap in a much cheaper op-amp, with A₀ = 2,000 instead of 200,000 — a hundred times smaller. What happens to how closely V₋ tracks V₊?

Answer: B. Error = V₊/(1+A₀), so a hundred-fold smaller A₀ gives a roughly hundred-fold larger error. Drag the bench's A₀ slider down and watch the error readout grow in direct proportion. The “virtual short” was never exact — it was always A₀-dependent, and it only looks like a law because real op-amps make A₀ so large the dependence disappears from view.

Feedback is doing the same job chapter 3 already showed you

This is the same negative-feedback story as chapter 3's emitter follower, run with a much bigger loop gain. There, the transistor's own g_m and the emitter resistor set how tightly the output tracked the input. Here, A₀ plays that role directly, undiluted by any resistor — which is exactly why an op-amp's virtual short is so much tighter than a single transistor stage's approximate tracking ever was.

A₀	Tracking error, V₊ = 1 V	Parts per million
2,000	0.4995 mV	~500 ppm
20,000	49.998 µV	~50 ppm
200,000	5.0000 µV	~5 ppm
2,000,000	0.50000 µV	~0.5 ppm

“V₊ = V₋” is the A₀→∞ limit of a real feedback loop, not a wire that appears when you add feedback. Treating it as exact is an excellent, load-bearing approximation — the rest of this chapter uses it freely — but lesson 3 onward will show you precisely what it costs to make that approximation, in numbers you can put on a spec sheet.

In the wild

Why precision instrumentation amplifiers advertise their open-loop gain in the millions. A higher A₀ buys a tighter virtual short directly — this is the entire reason that number is on the datasheet, not a marketing flourish.

Chopper-stabilised op-amps. Some precision parts periodically re-null their own input offset using an internal switching trick, effectively behaving as if A₀ were even larger than the raw silicon provides — because for many precision applications, closing this exact gap is the entire product.

Why an op-amp buffer is a better voltage reference driver than a plain wire. A wire has zero drive strength; this buffer holds V_out equal to V₊ to within microvolts while supplying whatever current the load demands — chapter 3's follower idea, with a vastly smaller error term.

Before you move on

In a closed feedback loop, what actually determines how closely V₋ tracks V₊?

B is the idealisation this lesson is built to correct. C matters starting in lesson 3, but even with resistors fixed, this lesson's bench shows the tracking error still moves when A₀ alone is changed — so the resistors cannot be the whole story.

Bench 02 · the virtual short, and its error term TRACKING

● Locked until you commit a prediction above.

Open-loop gain A₀ 200,000

20020,0002,000,000

V₊1.0000 V

V₋—

Error, V₊−V₋—

Two configurations, and the error each one carries

15 minutes · inverting and non-inverting gain, exact and approximate

Recall From lesson 2: what is the tracking error between V₊ and V₋, in terms of A₀? show answer

The two standard op-amp gain circuits both build on lesson 2's near-equal inputs. In the inverting configuration, V_in reaches V₋ through R_in, and R_f feeds V_out back to that same node, with V₊ grounded. In the non-inverting configuration, V_in drives V₊ directly, and R_in, R_f form a divider from V_out back down to ground at V₋. Solve each circuit exactly, keeping A₀ finite instead of assuming V₊=V₋ outright, and both reduce to the same shape:

A_cl = A_ideal / (1 + NG/A₀) NG = 1 + R_f/R_in

A_ideal is −R_f/R_in for the inverting case and 1+R_f/R_in for the non-inverting case — the two familiar textbook formulas. NG, the noise gain, is the same expression, 1+R_f/R_in, in both circuits regardless of which one you're using. It is the gain lesson 2's error term gets multiplied through, and you will meet it again by name in lesson 4.

Commit before you touch anything

Two inverting amplifiers, same A₀ = 200,000: one built for a gain of −10 (R_f=100 kΩ, R_in=10 kΩ), one for a gain of −1,000 (R_f=1 MΩ, R_in=1 kΩ). Which one sits closer to its own ideal value, percentage-wise?

Answer: B. Percent error ≈ NG/A₀. NG = 11 for the first design, 1,001 for the second — roughly a hundredfold difference, and the error tracks it directly: about 0.0055% versus about 0.5%. Asking for more closed-loop gain always costs you some of this accuracy; there is no way to have both a huge A_ideal and a vanishing error from the same fixed A₀.

Design	R_f	R_in	Ideal gain	NG	Actual gain	Error
Modest	100 kΩ	10 kΩ	−10	11	−9.99945	0.0055%
Aggressive	1 MΩ	1 kΩ	−1,000	1,001	−995.02	0.498%

The noise gain, not the signal gain, is what sets the error. This is the first hint of an idea that will pay off directly in lesson 4: it is NG = 1+R_f/R_in that appears in every denominator this chapter derives, in both configurations, for both gain error and — next lesson — bandwidth. Whatever number you actually plan to multiply your signal by, the resistor ratio's cost is billed in this other currency.

In the wild

Why a precision instrumentation amplifier is usually built from three op-amps at modest individual gain, not one at high gain. Splitting a needed gain of 1,000 across two stages of about 32 each keeps every individual NG small, and the compounded error stays far below what one aggressive stage would carry alone.

Why datasheets specify “gain error” or “gain accuracy” only alongside a stated closed-loop gain. The number is meaningless without NG attached — exactly this lesson's finding, restated as a spec.

Audio mixing consoles, and why channel trim pots rarely run at extreme boost. Pushing one gain stage to its limit compounds this error with the noise-gain-dependent bandwidth loss of lesson 4 — two separate costs for the same design decision.

Before you move on

What quantity determines the percentage gain error of a closed-loop op-amp stage, for a given A₀?

A comes close for a high-gain inverting stage, where NG≈Rf/Rin, but the two diverge sharply at low gain — a unity-gain inverter has Rf/Rin=1 but NG=2, and the bench's error readout follows NG, not Rf/Rin, at every setting.

Bench 03 · closed-loop gain, exact vs ideal INVERTING

● Locked until you commit a prediction above.

Target ideal gain magnitude 10×

1×10×1,000×

Ideal gain—

Actual gain—

Noise gain NG—

Error—

The bandwidth you traded for gain

15 minutes · gain-bandwidth product, and where it comes from

Recall From lesson 3: what did the noise gain NG turn out to equal, in both configurations? show answer

A₀ = 200,000 was never a number that held at every frequency — it is the open-loop gain at DC. Internally, an op-amp is dominated by one deliberately-placed capacitor that rolls its gain off at a single, gentle 6 dB-per-octave slope, starting from a corner frequency f₀ that can sit as low as a few hertz. The product of gain and bandwidth stays constant along that slope, all the way up to a frequency called the gain-bandwidth product, GBW = A₀ f₀. A typical general-purpose part specifies GBW directly — 1 MHz is a common, unglamorous figure.

Feed that frequency-dependent A(f) through lesson 3's exact gain formula and the closed-loop response inherits a −3 dB point of its own:

f_cl = GBW / NG

Commit before you touch anything

GBW = 1 MHz. One stage is set for a gain of −10 (NG=11); another, for −1,000 (NG=1,001). Which has the wider closed-loop bandwidth, and by roughly what factor?

Answer: B. f_cl = GBW/NG: about 91 kHz for the gain-of-−10 stage, about 1 kHz for the gain-of-−1,000 stage — roughly the same hundredfold ratio lesson 3's error carried, because the same NG is doing both jobs. Drag the bench's gain slider up and watch the curve's knee slide left in lock-step.

One knob, two different bills

Lessons 3 and 4 are not two separate costs of using feedback — they are the same NG, charged twice, for two different things. Ask for more closed-loop gain from a fixed A₀ and GBW, and you simultaneously buy a larger gain error and a narrower bandwidth, in the same proportion, because both bills are addressed to the identical number.

NG	f_cl = GBW/NG	Gain error, from lesson 3
11 (gain −10)	90.9 kHz	0.0055%
101 (gain −100)	9.9 kHz	0.050%
1,001 (gain −1,000)	999 Hz	0.498%

This is chapter 5's ceiling, reappearing in a new shape. A single bipolar or FET stage traded away gain against a fixed physical limit — V_A/V_T, or the FET's moving ceiling. An op-amp trades gain against bandwidth instead, at a rate fixed by GBW, which is itself set by the internal compensation capacitor. Different currency, same lesson: nothing in this course hands you gain for free.

In the wild

Why an audio op-amp used as a 60 dB (×1000) preamp needs a much higher-GBW part than one used as a ×2 buffer. This lesson's formula, read directly off a datasheet's GBW spec before any circuit is even built.

Decompensated op-amps. Some parts deliberately remove internal compensation to push GBW higher, at the cost of only being stable above some minimum closed-loop gain — a trade this lesson explains, chapter 12 will explain why it's dangerous to ignore.

Why a op-amp-based active filter's cutoff frequency creeps as you increase its passband gain. If GBW isn't comfortably above the filter's own corner frequency, f_cl starts competing with the filter's intended response instead of staying safely out of its way.

Before you move on

An op-amp has GBW = 4 MHz. A stage is built with NG = 40. Roughly what is its closed-loop −3 dB bandwidth?

f_cl = GBW/NG = 4,000,000/40 = 100,000 Hz. B inverts the division; C is the open-loop bandwidth, not the closed-loop one — they are equal only in the limiting case NG=1, a unity-gain buffer.

Bench 04 · the closed-loop response, against frequency FLAT

● Locked until you commit a prediction above.

Noise gain NG 11

1111,001

Closed-loop bandwidth—

GBW (fixed)1.00 MHz

Slew rate: a limit small signals never meet

14 minutes · a completely different kind of bandwidth

Recall From lesson 4: does the closed-loop −3 dB bandwidth depend on how large the input signal is? show answer

Lesson 4's bandwidth came from a linear model — one pole, one time constant, completely indifferent to signal size. It is genuinely true that a 1 millivolt sine and a 1 volt sine see the identical −3 dB corner, provided both stay small enough for the internal stages to behave linearly. Slew rate is the limit that shows up only once they don't.

Internally, the op-amp's compensation capacitor has to be charged and discharged by a limited internal current, which puts a hard ceiling on how fast V_out can move — volts per microsecond, quoted directly on every datasheet as SR. Ask the output to move faster than that, for any reason, and it simply can't — it ramps at exactly SR until it catches up.

Commit before you touch anything

SR = 0.5 V/µs. You drive a unity-gain buffer with a fast, clean 5 V step. About how long does the output take to get there?

Answer: B. 5 V ÷ 0.5 V/µs = 10 µs, and watch the bench trace it as a straight ramp, not a curve. A full step this large is not a small signal in lesson 4's sense at all — the internal stages are running flat out, not settling toward a target linearly.

The two limits, on the same signal

For a sine wave of peak amplitude V_pk at frequency f, the fastest the ideal waveform ever needs to move is its peak slope, 2πf V_pk. Set that equal to SR and solve for f, and you get the full-power bandwidth:

FPBW = SR / (2πV_pk)

This is a completely different number from lesson 4's f_cl, and for a large swing it is very often the smaller, binding one.

Output swing V_pk	FPBW = SR/(2πV_pk)	Compare: this bench's f_cl at NG=11
1 V	79.6 kHz	90.9 kHz
2 V	39.8 kHz
5 V	15.9 kHz
10 V	7.96 kHz

A datasheet's bandwidth number and its slew-rate number answer two different questions, and only the smaller one protects you. A one-volt swing at 50 kHz sails under both limits here. A ten-volt swing at that same 50 kHz is comfortably inside lesson 4's small-signal bandwidth, and yet the output cannot possibly keep up — FPBW at 10 V is only 7.96 kHz. The waveform doesn't distort gently the way lesson 1–7's clipping did; it turns into a triangle wave, ramping at a fixed rate regardless of what the input asked for.

In the wild

Why a headphone amp datasheet lists slew rate separately from bandwidth. A 20 kHz audio signal at full swing needs real slew rate, not just a high small-signal bandwidth number — the two specs are protecting against genuinely different failure modes.

Slew-induced distortion, and why it sounds different from ordinary clipping. Rail clipping (lesson 6) flattens only the very top and bottom of a wave. Slew limiting reshapes the whole edge into a straight ramp, which sounds harsher for the same peak signal level.

Why video and RF op-amps quote slew rates in thousands of volts per microsecond. A fast-changing video signal needs to swing several volts in nanoseconds — ordinary general-purpose parts, with SR in the fractions of a volt per microsecond, are nowhere close.

Before you move on

An op-amp's small-signal bandwidth at some gain setting is comfortably above the frequency you're using. Is the output guaranteed to be a clean, undistorted copy of the input?

B ignores that lesson 4's bandwidth is a small-signal, linear-model result — it says nothing about large swings. C names a real, separate limit (lesson 6), but it isn't the one this question describes; slew limiting can bite well before either input reaches a rail.

Bench 05 · slew-limited response, in time LINEAR

● Locked until you commit a prediction above.

Sine amplitude V_pk 1.0 V

0.1 V5 V12 V

Frequency 20 kHz

1 kHz20 kHz200 kHz

Required slope—

Device SR0.50 V/µs

The rails are real, no matter how ideal the rest is

14 minutes · the last edge of the black box

Recall From chapter 4: what happened to a bipolar stage's output once the signal pushed it into its supply rail? show answer

Lessons 2 through 5 all quietly assumed V_out stays inside the supply rails. It is worth closing this chapter by lifting that assumption, because it is the one non-ideality with absolutely nothing subtle about it: an op-amp cannot output a voltage its own power pins don't span, full stop, and most parts fall a volt or two short of the rails themselves — the datasheet parameter is usually called output swing, not rail-to-rail, unless the part is specifically designed and marketed for it.

This bench uses V_OH = +13.5 V and V_OL = −13.5 V on ±15 V supplies — 1.5 V of headroom lost on each side, typical for an older general-purpose part.

Commit before you touch anything

An inverting stage is set for a gain of −10. How large can V_in get, peak, before V_out starts clipping against ±13.5 V?

Answer: A. 13.5 V ÷ 10 = 1.35 V. Drag the input amplitude past that point on the bench and watch the top and bottom of the sine flatten while the middle keeps tracking the input — exactly chapter 4's clipping shape, on a completely different device.

The same shape, a different cause

Chapter 4's clipping came from a transistor running out of headroom against its own collector resistor and supply. This clipping comes from the op-amp's own output stage running out of room against its supply pins — internally, in fact, built from exactly the kind of push-pull output stage chapter 6 covered. The waveform looks identical on a scope: flat top, flat bottom, a straight-line edge where a curve used to be. The two chapters gave you two different reasons the same picture can occur, and a real circuit can hit either one, or both, depending on where the limited headroom actually is.

Closed-loop gain	Onset of clipping, V_in peak
−1	13.5 V
−10	1.35 V
−100	0.135 V

Every non-ideality this chapter covered fails differently, and knowing which one is biting matters. Rail clipping (this lesson) flattens the peaks while the zero crossings stay clean. Slew limiting (lesson 5) turns a sine into a triangle, with the distortion often worst through the zero crossing, where the ideal slope is steepest. Gain error (lesson 3) doesn't distort the shape at all — it just scales the whole thing slightly wrong. Three different signatures on an oscilloscope, three different fixes.

In the wild

Why professional audio gear runs op-amps on ±15 to ±18 V rails instead of ±5 V. More rail headroom directly buys more undistorted swing before this lesson's clipping sets in — one of the plainest reasons pro gear sounds cleaner at high levels than consumer gear built around the same op-amp.

Rail-to-rail output op-amps, and why they still aren't quite rail-to-rail. Modern parts get V_OH/V_OL to within millivolts of the supply rather than volts — a real improvement, not a change in the underlying limit this lesson describes.

Why a single-supply (0 V to +5 V) op-amp circuit needs a deliberate mid-supply bias point. Without one, half of any AC signal is asking the output to go negative, which a single supply cannot do at all — the most common beginner mistake this exact non-ideality causes.

Before you move on

Reaching back to chapter 6: what internal circuit stage inside a typical op-amp is directly responsible for delivering its output current and setting its rail-headroom limit?

The input pair (B) sets offset and bias-current behaviour, and the compensation capacitor (C) sets GBW — neither one drives the load or defines how close the output can approach the rails. That is the output stage's job, and it is built the same way chapter 6 taught, just inside the same package as everything else in this chapter.

Bench 06 · output swing against the rails LINEAR

● Locked until you commit a prediction above.

Input amplitude V_in 0.50 V pk

0.05 V1.35 V3.0 V

Ideal output peak—

Actual output peak—

✓

Chapter checkpoint

8 minutes · six questions, and two of them reach back

Question 1

With no feedback, an op-amp with A₀ = 500,000 and ±13 V rails receives a 200 µV input difference. What does the output do?

The arithmetic in A is exactly right and exactly why the answer isn't A: an ideal, unclamped op-amp would demand 100 V, but the physical device cannot exceed its own supply rails, so it clips at 13 V long before the open-loop formula's answer is reached.

Question 2

Why does the “virtual short” (V₊=V₋) get more accurate as A₀ increases?

B is the idealisation lesson 2 corrected. C mixes up an unrelated non-ideality (input bias current, briefly mentioned in chapter 7's closing notes) with the one actually at work here.

Question 3

Reaching back to chapter 5: what did a single bipolar stage's intrinsic gain ceiling, g_mr_o, turn out to equal?

This is exactly why an op-amp needs several cascaded gain stages to reach A₀ figures in the hundreds of thousands — one stage alone, bipolar or FET, tops out nowhere near that high.

Question 4

An op-amp has GBW = 2 MHz. A non-inverting stage is built with R_f = 9 kΩ, R_in = 1 kΩ. Roughly what closed-loop bandwidth results?

NG = 1+9k/1k = 10, so f_cl = 2,000,000/10 = 200,000 Hz. The formula uses noise gain, not signal gain — but for a non-inverting stage the two happen to be numerically identical, which is why this one comes out clean.

Question 5

A stage's small-signal bandwidth easily covers the frequency in use, yet the output still comes out visibly distorted into something close to a triangle wave at large amplitude. What is the most likely cause?

Gain error (B) scales the whole waveform slightly without reshaping it. A low noise gain (C) is not a problem at all — it means less error and more bandwidth, from lessons 3 and 4. A triangle-shaped large-signal distortion, with small-signal bandwidth ruled out, is lesson 5's signature.

Question 6

Reaching back to chapter 4: what shape does clipping take on an oscilloscope trace, whether it comes from a transistor stage or an op-amp's output stage?

B is lesson 5's slew-limiting signature, not clipping's. Clipping is a hard boundary — the output simply cannot go further — which shows up as a genuinely flat region, not a gradual rounding, whichever device is doing the clipping.

Checkpoint score

0 / 6

Answer all six. Four or more and the chapter has landed.

Every word this chapter introduced

Open-loop gain A₀ L1

V_out/(V₊−V₋) with no feedback connected. Enormous — often 10⁵ to 10⁷ — and by itself makes the device behave as a comparator, not an amplifier.

Comparator behaviour L1

The bare op-amp's tendency to saturate against a rail for almost any distinguishable nonzero input difference.

Virtual short L2

V₊≈V₋ under closed-loop feedback. An A₀→∞ approximation, not an exact equality.

Noise gain, NG L3

1+R_f/R_in. Governs both the closed-loop gain error and, in lesson 4, the closed-loop bandwidth — in both inverting and non-inverting configurations.

Closed-loop gain error L3

Approximately NG/A₀, the fractional shortfall of the actual gain below the ideal R_f/R_in-based formula.

Gain-bandwidth product, GBW L4

A₀·f₀, held constant along the open-loop roll-off. A single spec that sets bandwidth at any closed-loop gain.

Closed-loop bandwidth, f_cl L4

GBW/NG. The higher the gain you ask for, the narrower the bandwidth you get, at a rate fixed by GBW.

Slew rate, SR L5

The maximum rate, in volts per second, the output can change at all — a large-signal limit, completely separate from the small-signal f_cl.

Full-power bandwidth, FPBW L5

SR/(2πV_pk). The highest frequency a given output swing can reach before slew-rate limiting distorts it.

Output swing / rails L6

The actual voltage range the output can reach, typically falling a volt or more short of the supply rails on a non-rail-to-rail part.

Rail clipping L6

A flat-topped output when the demanded swing exceeds V_OH/V_OL — the same visible shape as chapter 4's transistor clipping, from a different physical cause.

Where this goes next

PART TWO

Sedra & Smith begins

Chapters 9–12 open up the black box this chapter deliberately kept shut, building the internal stages from transistors nobody can touch or replace individually.

CH 09

The Mirror Bench

Current mirrors and active loads — the actual circuit that gives a real op-amp's input stage its enormous, chapter-5-defying gain.

CH 10

The Difference Bench

The long-tailed pair and CMRR — what V₊ and V₋ are actually connected to, inside the package.

CH 12

The Loop Bench

Miller effect and phase margin — why that single internal compensation capacitor from lesson 4 is placed exactly where it is, and what happens to a feedback loop when it isn't enough.

About the simulations. This chapter's op-amp is modelled purely as a black box: a finite open-loop gain A₀, a single-pole frequency response set by GBW, a slew-rate limit SR, and hard output clamps at V_OH/V_OL — nothing about the transistors inside it is simulated, in keeping with the course plan's instruction to treat the op-amp as a sealed unit until chapters 9–12 open it up. Benches 2 and 3 solve the exact finite-gain nodal equations for each configuration directly, rather than assuming V₊=V₋ and only checking the assumption afterward. Bench 4's frequency response uses genuine complex arithmetic on the single-pole model, not a hand-drawn asymptote. Bench 5 numerically time-steps a first-order lag toward the target voltage, rate-limited to ±SR at every step, rather than sketching the slewed edge as a fixed shape. Two simplifications worth knowing: input bias current and input offset voltage — both genuine, nonzero, and frequently the dominant error source in a real precision circuit — are not modelled here, and every parameter is held constant with temperature and supply voltage, which real ones are not.