123-ME · Operational Amplifiers

Operational Amplifiers

Slides 10.1 & 13.2 — ideal op-amp properties · negative feedback · virtual ground · voltage follower · the inverting amplifier

Every circuit up to this point has been passive: resistors, capacitors and inductors can divide, delay and phase-shift a signal, but none of them can make it bigger. A voltage divider always hands you back less than you put in. The operational amplifier is the first active device in this course — it draws energy from a power supply and uses it to produce an output larger than its input. Two ideas do almost all the work here, and if you leave with only these two you can still answer the exam question: negative feedback forces the two input terminals to the same voltage, and no current flows into either input. Everything else, including the famous $A_v = -R_f/R_{in}$ , falls out of applying KCL once with those two facts in hand.

Pacing

Budget roughly 35–45 minutes. Ten of those go to the ideal-property list and why virtual ground happens (the conceptual half), and the rest to the inverting-amp derivation and practice. This is a short tier by exam weight, but it is also the tier where the analysis method is completely different from everything before it — you do not write node equations for the whole circuit, you write one KCL equation at one node.

The device and its symbol

An op-amp is a high-gain differential amplifier: it looks at the difference between its two input terminals and drives its output to an enormously amplified version of that difference. The symbol is a triangle pointing in the direction of signal flow, with an inverting input (marked $-$ ), a non-inverting input (marked $+$ ), and a single output. It also has two power-supply terminals — usually drawn only when they matter, but always physically present, because a device that produces more power at its output than arrives at its input must be getting the difference from somewhere.

The op-amp symbol. The supply rails are the only reason the device can amplify — and, as we will see, the only reason the ideal model eventually fails.

The four ideal properties

Real op-amps are complicated. The course model is not. Four idealizations turn the device into something you can analyze with the KCL you already know:

Infinite input impedance. $Z_{in} \to \infty$ , so no current flows into either input terminal: $i_+ = i_- = 0$ .

This is the single most useful fact for analysis. Any current arriving at an input node has to leave through some other branch — it cannot disappear into the chip.

Zero output impedance. $Z_{out} = 0$ , so the output behaves as an ideal voltage source — its voltage does not sag no matter what you connect to it.

This is what lets you cascade stages and analyze each one separately: the next stage cannot load the previous one.

Infinite open-loop gain. $A \to \infty$ . Real devices are around $10^5$ or more, which for our purposes is the same as infinite.

"Open-loop" means the bare device with no feedback wired around it. This is the property that manufactures virtual ground.

Very large bandwidth. The gain does not fall off with frequency over the range we care about.

Lets us treat the same DC gain formula as valid for the audio-frequency sinusoids in the examples.

Negative feedback — and why $v_+ = v_-$

Take property 3 seriously for a moment. With $A = 10^5$ and supply rails at $\pm 15$ V, the largest input difference the op-amp can amplify without slamming into a rail is

how small the input difference must be

|v_+ - v_-| \le \frac{|v_{out}|_{\max}}{A} = \frac{15\ \text{V}}{10^5} = 150\ \mu\text{V}

A hundred and fifty microvolts. If the two inputs differ by any more than that, the output is pinned against a supply rail and the device is no longer amplifying anything. Used open-loop, an op-amp is therefore useless as an amplifier — it is a comparator, an all-or-nothing switch.

Negative feedback fixes this. Wire a path from the output back to the inverting ( $-$ ) input. Now the loop is self-correcting: if $v_-$ drifts below $v_+$ , the difference $v_+ - v_-$ goes positive, the output swings up, and because the feedback path carries that rise back to $v_-$ , the node is pushed back up. The correction always opposes the error — hence negative feedback. The loop settles at exactly the point where the error is small enough that the huge gain produces the finite output the circuit needs, and since the gain is enormous, that error is essentially zero:

the feedback constraint (course notation)

v_+ = v_-

valid only when negative feedback is present and the output is not saturated

Read this carefully — it is the most misquoted line in the chapter

Negative feedback forces $v_- \to v_+$ . It does not force either one to zero. The node is at 0 V only when $v_+$ happens to be tied to real ground — which is exactly the case in the inverting amplifier, and that is where the name virtual ground comes from: the inverting node acts like ground (0 V) without being wired to it. In the voltage follower below, the same feedback rule gives $v_- = v_+ = v_{in} \ne 0$ . Same rule, different value. Ask "what is $v_+$ ?" first, every time.

The voltage follower

The simplest possible feedback circuit: connect the output directly to the inverting input — 100% of the output fed back, no resistors at all. Apply the input to the non-inverting terminal.

Voltage follower (unity-gain buffer). Gain of exactly 1 — which sounds pointless until you ask what it does to impedance.

The analysis is one line. The feedback wire makes $v_- = v_{out}$ by inspection. The feedback constraint gives $v_- = v_+$ . And $v_+ = v_{in}$ because that is where the source is connected. Chain them:

voltage follower

v_{out} = v_- = v_+ = v_{in} \qquad\Longrightarrow\qquad A_v = 1

Gain of one. Why build it? Because of properties 1 and 2. The source sees the op-amp's infinite input impedance, so it delivers no current and is not loaded at all; the load sees the op-amp's zero output impedance, so it can draw whatever current it likes without the voltage sagging. A follower is an impedance transformer: it copies a voltage from a weak, high-impedance source onto a strong, low-impedance one. Put one between a sensor and a resistive network and the divider ratio you calculated on paper is the ratio you actually measure.

The inverting amplifier

Now the configuration that actually produces gain. Ground the non-inverting input. Feed the signal into the inverting input through a resistor $R_{in}$ , and close the feedback loop with a resistor $R_f$ from the output back to that same node.

The inverting amplifier.

R_{in}

sets how much current the input produces;

R_f

sets how much voltage that current develops at the output. The marked node is the virtual ground.

Deriving $A_v = -R_f/R_{in}$

Everything now follows from two facts and one KCL equation. Note how little of the circuit we actually touch — we never write an equation for the output node, and we never need to know anything about what is inside the triangle.

Find $v_+$ . The non-inverting input is wired straight to ground, so $v_+ = 0$ .

Apply the feedback constraint. $R_f$ runs from the output back to the inverting input, so this is negative feedback and $v_- = v_+ = 0$ .

This is the virtual ground. The node sits at 0 V, but no wire connects it to ground — no current drains away there.

Find the input current. $R_{in}$ has $v_{in}$ on its left end and the virtual ground (0 V) on its right, so by Ohm's law

i_{in} = \frac{v_{in} - 0}{R_{in}} = \frac{v_{in}}{R_{in}}

Note the side benefit: the source always sees exactly

R_{in}

, no matter what

R_f

is. The input impedance of an inverting amplifier is

R_{in}

KCL at the inverting node. Three branches meet there: $R_{in}$ , $R_f$ , and the op-amp input terminal. The terminal carries zero current (property 1), so all of $i_{in}$ must continue into $R_f$ :

i_f = i_{in} = \frac{v_{in}}{R_{in}}

This is the whole trick. The op-amp input is a dead end, so the feedback resistor has no choice but to swallow the entire input current.

Walk across $R_f$ to the output. Starting at the virtual ground (0 V) and travelling in the direction of $i_f$ , we drop $i_f R_f$ :

v_{out} = 0 - i_f R_f = -\frac{v_{in}}{R_{in}} R_f

Divide out $v_{in}$ .

A_v = \frac{v_{out}}{v_{in}} = -\frac{R_f}{R_{in}}

inverting amplifier — closed-loop gain

A_v = -\frac{R_f}{R_{in}} \qquad\qquad v_{out} = -\frac{R_f}{R_{in}}\, v_{in} \qquad\qquad Z_{in} = R_{in}

What the minus sign means

It is a $180^\circ$ phase inversion, not an error. Positive in, negative out. Feed a sinusoid and the output is the same sinusoid flipped upside down. Feed $+1$ V into a circuit with $A_v = -10$ and the output is $-10$ V, which is only possible if the negative supply rail is at least $-10$ V — a point we return to below.

Why the answer depends only on the ratio

The gain contains no op-amp parameter at all — not $A$ , not the input impedance, nothing. That is the deep reason feedback is used everywhere in electronics: it trades away gain you have in absurd excess for gain you can predict. Two resistors you can buy to 1% tolerance now set the gain, instead of a transistor parameter that varies by a factor of three between chips off the same wafer.

Worked example — sizing an inverting stage

A sensor produces $v_{in} = 1$ V and must be amplified to $-10$ V, and the sensor may not be loaded by more than 1 mA. Choose $R_{in}$ and $R_f$ .

Step 1 — gain required. $A_v = v_{out}/v_{in} = -10/1 = -10$ , so $R_f/R_{in} = 10$ .

Step 2 — loading sets $R_{in}$ . The source sees $Z_{in} = R_{in}$ , so $i_{in} = 1\ \text{V}/R_{in} \le 1$ mA requires $R_{in} \ge 1$ kΩ. Take $R_{in} = 1$ kΩ.

Step 3 — ratio sets $R_f$ . $R_f = 10 R_{in} = 10$ kΩ.

Step 4 — check. $i_{in} = 1\ \text{V}/1\ \text{k}\Omega = 1$ mA, all of it through $R_f$ , so $v_{out} = 0 - (1\ \text{mA})(10\ \text{k}\Omega) = -10$ V. ✓

R_{in} = 1

kΩ,

R_f = 10

kΩ, giving

A_v = -10

V/V.

Where the ideal model breaks: saturation

The gain formula has no upper limit built into it — nothing in $-R_f/R_{in}$ knows about the power supply. But the op-amp cannot produce an output voltage larger than the rails feeding it. Once the ideal prediction $-\tfrac{R_f}{R_{in}} v_{in}$ exceeds $+V_{sat}$ or falls below $-V_{sat}$ , the output simply stops there and the feedback loop loses control — with the output pinned, it can no longer correct the input error, so $v_- \ne v_+$ and the virtual ground assumption itself fails. This is clipping: a sinusoid comes out with its peaks flattened.

Transfer characteristic. The straight middle section — negative slope, because the amplifier inverts — is where every formula in this chapter applies. Outside it, only "the output equals the rail" is true.

the linear-region condition — check this on every problem

\left|\frac{R_f}{R_{in}}\, v_{in}\right| \le V_{sat} \qquad\Longleftrightarrow\qquad |v_{in}| \le \frac{R_{in}}{R_f} V_{sat}

Widget — gain, rails and clipping

Drag $R_f$ and $R_{in}$ and watch the gain change. Then push the input amplitude up, or the rails down, until the dashed ideal-model trace leaves the shaded band — that is the moment the virtual-ground model stops describing the real circuit, and the solid trace (what the op-amp actually does) flattens against the rail.

Inverting amp — gain & saturation linear

R_f = 10.0 kΩ R_in = 2.0 kΩ input amplitude = 0.50 V supply rails ±V_sat = 15.0 V

A_v = −R_f/R_in = −5.00

ideal peak |A_v·v_in| = 2.50 V

actual output peak = 2.50 V

Z_in = R_in = 2.0 kΩ

clipped fraction of cycle = 0.0 %

Practice problems

Verification note — read once

No tutorial, exercise or exam in this course's source material touches op-amps, so there is no official answer key to check these against. Every problem below was therefore solved two structurally independent ways, and each solution states which two. Problems whose answer could not be confirmed twice were not written. If you get a different answer, work the second method listed and see which one your result matches.

P1 — Basic inverting stage warm-up

An inverting amplifier is built with $R_{in} = 2.5$ kΩ, $R_f = 20$ kΩ, and supply rails at $\pm 15$ V. A DC input of $v_{in} = 0.6$ V is applied.

(a) Find the closed-loop gain $A_v$ . (b) Find $v_{out}$ . (c) Find the current delivered by the source. (d) Confirm the amplifier is operating in its linear region.

Method A — virtual ground and KCL (the derivation route).

$v_+ = 0$ (grounded), and negative feedback via $R_f$ gives $v_- = v_+ = 0$ : virtual ground. The current through $R_{in}$ is

i_{in} = \frac{v_{in} - 0}{R_{in}} = \frac{0.6\ \text{V}}{2500\ \Omega} = 240\ \mu\text{A}

No current enters the op-amp input, so KCL at the inverting node sends all $240\ \mu$ A through $R_f$ . Walking from the node to the output across $R_f$ in the direction of that current:

v_{out} = 0 - i_f R_f = -(240\times10^{-6})(20\,000) = -4.8\ \text{V}

The source current is exactly $i_{in} = 240\ \mu$ A, since the source sees $Z_{in} = R_{in} = 2.5$ kΩ.

Method B — finite-gain analysis, taking the limit (never assumes virtual ground).

Drop the ideal-gain assumption entirely. Let the open-loop gain be a finite $A$ , and let the inverting node sit at an unknown $v$ . The device relation is $v_{out} = A(v_+ - v_-) = A(0 - v) = -Av$ , hence $v = -v_{out}/A$ . Writing KCL at the inverting node with only "no input current" assumed:

\frac{v_{in} - v}{R_{in}} + \frac{v_{out} - v}{R_f} = 0

Substituting $v = -v_{out}/A$ and multiplying through by $R_{in}R_f$ :

R_f\!\left(v_{in} + \frac{v_{out}}{A}\right) + R_{in} v_{out}\!\left(1 + \frac{1}{A}\right) = 0

v_{out} = -\frac{(R_f/R_{in})\, v_{in}}{1 + \dfrac{1 + R_f/R_{in}}{A}}

With $R_f/R_{in} = 8$ , $v_{in} = 0.6$ V and a realistic $A = 10^5$ :

v_{out} = -\frac{(8)(0.6)}{1 + 9/10^{5}} = -\frac{4.8}{1.00009} = -4.79957\ \text{V}

and as $A \to \infty$ this tends to exactly $-4.8$ V. Agreement with Method A to five significant figures. As a bonus this method produces the virtual ground rather than assuming it: $v = -v_{out}/A = 4.8/10^5 = 48\ \mu$ V, i.e. 0 V to any precision that matters.

Consistency back-check. Plug $v_{out} = -4.8$ V back into the original circuit: current into the node through $R_{in}$ is $(0.6-0)/2500 = 240\ \mu$ A; current out of the node through $R_f$ is $(0-(-4.8))/20\,000 = 240\ \mu$ A. In equals out, with zero into the op-amp. KCL holds. ✓

(d) Linear region. $|v_{out}| = 4.8\ \text{V} \le V_{sat} = 15$ V, so the output is well inside the rails and the linear model is valid. ✓

(a)

A_v = -8

V/V · (b)

v_{out} = -4.8

V · (c)

i_{in} = 240\ \mu

A · (d)

4.8\ \text{V} \le 15

V — linear.
Verified by: virtual-ground KCL and finite-open-loop-gain limit analysis.

P2 — Two inverting stages in cascade core

Two inverting amplifiers are chained: the output of stage 1 drives the input resistor of stage 2. Stage 1 has $R_{in1} = 1$ kΩ, $R_{f1} = 5$ kΩ. Stage 2 has $R_{in2} = 2$ kΩ, $R_{f2} = 6$ kΩ. Both op-amps run on $\pm 12$ V rails. The input is $v_{in} = 0.2$ V DC.

(a) Find the intermediate voltage $v_1$ at the output of stage 1. (b) Find $v_{out}$ . (c) What is the overall gain, and why is it positive? (d) Confirm neither op-amp saturates. (e) Why does stage 2's input resistor not change stage 1's answer?

Method A — stage-by-stage closed-loop gain formula.

A_{v1} = -\frac{R_{f1}}{R_{in1}} = -\frac{5}{1} = -5 \qquad A_{v2} = -\frac{R_{f2}}{R_{in2}} = -\frac{6}{2} = -3

v_1 = A_{v1} v_{in} = (-5)(0.2) = -1.0\ \text{V}

v_{out} = A_{v2} v_1 = (-3)(-1.0) = +3.0\ \text{V}

Overall $A_v = A_{v1}A_{v2} = (-5)(-3) = +15$ .

Method B — node-by-node KCL from scratch, no gain formula used.

Node a (stage 1's inverting input): grounded $v_+$ plus negative feedback gives $v_a = 0$ . KCL with zero op-amp input current:

\frac{v_{in} - 0}{R_{in1}} + \frac{v_1 - 0}{R_{f1}} = 0 \;\Longrightarrow\; \frac{0.2}{1000} + \frac{v_1}{5000} = 0 \;\Longrightarrow\; v_1 = -5000\!\left(\frac{0.2}{1000}\right) = -1.0\ \text{V}

Node b (stage 2's inverting input): same reasoning gives $v_b = 0$ , and

\frac{v_1 - 0}{R_{in2}} + \frac{v_{out} - 0}{R_{f2}} = 0 \;\Longrightarrow\; \frac{-1.0}{2000} + \frac{v_{out}}{6000} = 0 \;\Longrightarrow\; v_{out} = \frac{6000}{2000} = +3.0\ \text{V}

Both methods give $v_1 = -1.0$ V and $v_{out} = +3.0$ V. Sanity-check the currents too: stage 1 passes $0.2/1\text{k} = 200\ \mu$ A through $R_{f1}$ , giving a 1.0 V drop ✓; stage 2 passes $1.0/2\text{k} = 500\ \mu$ A through $R_{f2}$ , giving a 3.0 V drop ✓.

(c) The gain is positive because each stage inverts, and two inversions cancel — $180^\circ + 180^\circ = 360^\circ$ . Cascading inverting stages is the standard way to get a large non-inverting gain with predictable resistor ratios.

(d) Stage 1 output: $|-1.0| \le 12$ V ✓. Stage 2 output: $|3.0| \le 12$ V ✓. Both linear. (Always check the intermediate node too — a cascade can saturate in the middle while its final output looks harmless.)

(e) Because an ideal op-amp has zero output impedance (property 2). Stage 1's output is an ideal voltage source, so the 2 kΩ that stage 2 hangs on it draws current without pulling the voltage down. Two passive networks chained like this would load each other and you could not multiply their ratios; op-amp stages you can.

(a)

v_1 = -1.0

V · (b)

v_{out} = +3.0

V · (c)

A_v = +15

, two inversions cancel · (d) both within

\pm 12

V — linear.
Verified by: cascaded closed-loop gain formula and independent node-by-node KCL, cross-checked against branch currents.

P3 — Designing against the rails challenge — where the ideal model fails

An inverting amplifier has $R_{in} = 4$ kΩ and runs on $\pm 10$ V rails. Its input is a sinusoid of amplitude $0.5$ V (peak).

(a) What is the largest $R_f$ for which the output peak does not exceed the rails? (b) A technician instead fits $R_f = 100$ kΩ. What does the output look like? (c) For what fraction of each cycle is the output clipped? (d) During clipping, what is the actual voltage at the inverting input — and what does that tell you about virtual ground?

(a) Method A — gain formula against the rail. The output peak is $|A_v|\,(0.5\ \text{V})$ and must not exceed 10 V:

|A_v| \le \frac{10\ \text{V}}{0.5\ \text{V}} = 20 \qquad\Longrightarrow\qquad \frac{R_f}{4\ \text{k}\Omega} \le 20 \qquad\Longrightarrow\qquad R_f \le 80\ \text{k}\Omega

(a) Method B — current route, independent of the gain formula. At the input peak the virtual ground makes the current through $R_{in}$ equal to $0.5/4000 = 125\ \mu$ A, and all of it flows through $R_f$ . The voltage developed across $R_f$ is the output magnitude, so we need $(125\ \mu\text{A})R_f \le 10$ V, giving $R_f \le 10/(125\times10^{-6}) = 80$ kΩ. Both give exactly 80 kΩ.

(b) With $R_f = 100$ kΩ, $A_v = -100/4 = -25$ , so the ideal model predicts a peak of $25 \times 0.5 = 12.5$ V. That exceeds the 10 V rails, so the output is a $180^\circ$ -inverted sinusoid whose peaks are flattened at exactly $\pm 10$ V — a clipped waveform. Around the zero-crossings, where the instantaneous output magnitude is below 10 V, the amplifier is still perfectly linear.

(c) Write $v_{out,ideal}(t) = -12.5\sin\theta$ . The output clips whenever $12.5\,|\sin\theta| > 10$ , i.e. $|\sin\theta| > 0.8$ . Solving $\sin\theta = 0.8$ gives $\theta = 53.13^\circ$ , so within the first half-cycle the clipped interval runs from $53.13^\circ$ to $180^\circ - 53.13^\circ = 126.87^\circ$ , a span of $73.74^\circ$ . By symmetry the negative half-cycle clips for the same span, so per full $360^\circ$ cycle:

\text{clipped fraction} = \frac{2(73.74^\circ)}{360^\circ} = \frac{147.48}{360} = 0.4097 \approx 41.0\%

Cross-check by a different route: the fraction of a cycle for which $|\sin\theta| \le 0.8$ is $\tfrac{4}{360}\arcsin(0.8) = \tfrac{4(53.13)}{360} = 0.5903$ , and $1 - 0.5903 = 0.4097$ . ✓ Same number from the complementary calculation.

(d) Method A — the assumption contradicts itself. Suppose virtual ground still held at the input peak. Then $i_f = 125\ \mu$ A and $v_{out} = -12.5$ V — impossible, since the op-amp cannot swing beyond $-10$ V. The assumption produces a contradiction, so it must be false during clipping.

(d) Method B — solve for the node voltage with the output forced to the rail. Let the inverting node be $v$ (unknown, no longer assumed zero) with $v_{out}$ pinned at $-10$ V. Only "no current into the op-amp input" is still true, so KCL at that node gives, at the input peak $v_{in} = 0.5$ V:

\frac{0.5 - v}{4000} + \frac{-10 - v}{100\,000} = 0

Multiplying by $100\,000$ : $\;25(0.5 - v) + (-10 - v) = 0 \Rightarrow 12.5 - 25v - 10 - v = 0 \Rightarrow 26v = 2.5$ , so

v = \frac{2.5}{26} = 0.0962\ \text{V} \approx 96\ \text{mV}

That is not 0 V — the virtual ground has failed. And it is self-consistent: the bare device would demand $v_{out} = A(v_+ - v_-) = 10^5(0 - 0.0962) = -9615$ V, hugely beyond the rail, which is exactly why the output stays pinned at $-10$ V. Both methods agree that virtual ground is a consequence of the feedback loop being in control, not an unconditional property of the wiring.

(a)

R_f \le 80

kΩ · (b) inverted sinusoid clipped flat at

\pm 10

V (ideal peak would be 12.5 V) · (c)

\approx 41.0\%

of each cycle · (d)

v_- \approx 96

mV, not 0 — virtual ground fails once the output saturates.
Verified by: (a) gain-formula-vs-rail and feedback-current-times-

R_f

; (c) direct clipped-span and complementary unclipped-span calculation; (d) proof-by-contradiction and explicit forced-rail KCL solve.

Exam checklist

The five moves

1. Identify $v_+$ — is the non-inverting input grounded, or driven? 2. Is there negative feedback (a path from output to the $-$ input)? If yes, set $v_- = v_+$ . 3. Write KCL at the inverting node only, remembering $i_- = 0$ . 4. Solve for $v_{out}$ . 5. Check $|v_{out}| \le V_{sat}$ — if it fails, the answer is the rail, not the formula.

Traps

Dropping the minus sign on $A_v$ . Forgetting that the input impedance is $R_{in}$ alone (not $R_{in} + R_f$ ) — the virtual ground terminates the input branch. Assuming $v_- = 0$ in a follower, where it equals $v_{in}$ . Forgetting to check the intermediate node for saturation in a cascade. And treating $v_- = v_+$ as always true: it requires negative feedback and an unsaturated output.