04 · Advanced

Loop gain & closed-loop gain

Quantify the correction mechanism introduced in chapter 03. Start from the subtractive loop e=xs−βxo and forward relation xo=Ae, define the loop gain L=Aβ under an explicit sign convention, derive the exact closed-loop gain A/(1+Aβ), then connect large loop gain to finite-gain error, desensitization, noise gain, input/output resistance changes and the frequency range over which ideal op-amp formulas remain accurate. Keep stability margins for chapter 05.

220–290 minAdvanced
In this lesson
  • Write the canonical subtractive-loop equations e=xs−βxo and xo=Ae before using a memorized feedback formula.
  • Define L=Aβ as the loop-gain magnitude for the stated negative-feedback sign convention and distinguish A, β, L and closed-loop gain.
  • Derive the exact closed-loop relation xo/xs=A/(1+Aβ)=A/(1+L).
  • Rewrite the closed-loop gain as (1/β)L/(1+L) and quantify the finite-loop-gain error relative to 1/β.
  • Derive e/xs=1/(1+L) and interpret 1+L as the error-reduction/desensitization factor.
  • Derive the sensitivity of closed-loop gain to forward gain, S_A=1/(1+L), while recognizing that β accuracy becomes important at large L.
  • Relate the inverting amplifier loop gain to its noise gain, with β=1/Gnoise when the signal source is zeroed.
  • Quantify input- and output-resistance changes for the voltage-series feedback topology, with explicit topology/model caveats.
  • Use frequency-dependent A(s), β(s) and L(s) to explain why closed-loop accuracy degrades as loop gain falls with frequency without yet performing a phase-margin analysis.
  • Explain why practical loop-gain measurement or simulation must preserve the operating point and relevant loading when the loop is opened or injected into.
Starting question

Chapter 03 showed that feedback corrects error. This chapter asks how much correction is available. The answer is not the closed-loop gain itself: it is the loop gain, the round-trip transmission that determines how strongly the circuit can suppress error and uncertainty.

04.1Start from two equations, not from a slogan

Figure 4.1 — Finite-gain non-inverting loop

The one question is: which physical quantities correspond to A, β and L=Aβ?

+−Axsxoxf=βxoR1R2+VS−VS

xs fixes v+. The R1–R2 network samples xf=βxo from the output and returns it to v−. The op-amp converts the error e=xs−xf to output with forward gain A. L=Aβ therefore measures the round-loop transmission before the subtractive comparison sign.

For the canonical scalar negative-feedback loop we choose the subtractive convention

e=xs−xf
xf=βxo
xo=Ae

A is the forward gain from error to output. β is the return factor from output to the comparison input. In this chapter we define the positive loop-gain magnitude L=Aβ for a loop whose subtraction has already been written explicitly in e=xs−xf. Other texts sometimes absorb the minus sign into the return ratio; sign conventions must therefore be read before comparing formulas.

04.2Exact closed-loop gain

Substituting xf=βxo into e and xo=Ae gives

xo=A(xs−βxo)
xo(1+Aβ)=Axs
ACL≡xo/xs=A/(1+Aβ)=A/(1+L)

This relation is exact for the stated linear single-loop scalar model. No infinite-gain approximation has been used.

04.3The ideal result is the large-loop-gain limit

ACL=(1/β)·L/(1+L)
ACL/(1/β)=L/(1+L)
[ACL−(1/β)]/(1/β)=−1/(1+L)

The familiar ACL≈1/β is therefore not a separate law. It is the asymptotic result |L|≫1 under stable negative feedback. A target accuracy can be converted into a minimum required loop gain instead of merely saying “A is large.”

Example: if L=99 at a given frequency, the closed-loop gain is 99/100=0.99 of its ideal 1/β value: a −1% finite-loop-gain error.

04.4Error suppression and desensitization

The comparison error follows directly from the same equations:

e/xs=1/(1+L)

Thus 1+L is the desensitization factor for this model. A large loop gain means that only a small error is needed to command the output.

The logarithmic sensitivity of closed-loop gain to forward gain A, with β fixed, is

SAACL=(A/ACL)·∂ACL/∂A=1/(1+L)

If A changes by a small fractional amount δA/A, the resulting fractional closed-loop gain change is approximately [1/(1+L)]δA/A. Feedback suppresses uncertainty in A; it does not make β irrelevant.

04.5At large L, accuracy migrates from A to β

SβACL=(β/ACL)·∂ACL/∂β=−L/(1+L)

When L≫1, SA→0 while Sβ→−1. The closed-loop gain becomes insensitive to uncertain amplifier gain but nearly inversely proportional to the feedback factor. This is why precision closed-loop gain requires a precise feedback network even when the op-amp has enormous open-loop gain.

04.6Non-inverting and follower examples

For Figure 4.1, negligible input current gives

β=R1/(R1+R2)
L=Aβ
ACL=A/(1+Aβ)

For a voltage follower β=1, so L=A and ACL=A/(1+A). For R1=1 kΩ, R2=9 kΩ and A=10 000, β=0.1, L=1000 and ACL=10000/1001≈9.990 V/V rather than exactly 10 V/V.

04.7Inverting amplifier: loop gain follows noise gain

Figure 4.2 — Inverting loop with the source zeroed

The one question is: why is β the reciprocal of noise gain rather than signal gain?

+−AR1v−xoR2+VS−VS

To inspect the return path, an independent ideal voltage source is zeroed, so the left side of R1 becomes ground. Seen from xo, R2 and R1 then form the feedback divider to v−. Thus β=R1/(R1+R2)=1/Gnoise.

To determine the return factor, zero the independent signal source while preserving the feedback network. For an ideal zero-impedance voltage source, R1 then returns to ground. The fraction of output reaching the inverting input is

β=R1/(R1+R2)
Gnoise=1/β=1+R2/R1
L=Aβ=A/Gnoise

The signal gain remains approximately −R2/R1. Therefore signal gain and loop gain are not interchangeable; stability and finite-loop-gain accuracy of the inverting circuit track its noise gain.

04.8Impedance transformation is topology dependent

For the idealized voltage-series feedback topology of the non-inverting amplifier, the same desensitization factor modifies terminal resistances approximately as

Rin,f≈Rin(1+L)
Rout,f≈Rout/(1+L)

These formulas require the stated series-mixing / voltage-sampling topology and an appropriate unilateral feedback model. They are not universal rules for every feedback connection.

04.9Loop gain is frequency dependent

Real forward gain is A(s), and the return network can also be β(s). Therefore

L(s)=A(s)β(s)
ACL(s)=A(s)/[1+L(s)]

Even if |L| is enormous at DC, it falls as amplifier poles reduce |A|. The approximation ACL≈1/β consequently loses accuracy with frequency. For a single dominant pole and frequency-independent β, feedback broadens the first-order closed-loop bandwidth by approximately 1+L0 while reducing low-frequency gain, expressing the familiar gain–bandwidth trade.

This section deliberately stops before asking whether the complex denominator 1+L(s) can approach zero. That phase/stability question is chapter 05.

04.10Do not measure loop gain by simply deleting a wire

A practical loop-gain simulation introduces a test perturbation at a suitable point and measures the returned perturbation while preserving the DC operating point and the impedances seen by the surrounding circuit as closely as the chosen method requires. A naive open circuit can destroy bias, alter loading or even change which loop is being measured.

For the simple one-loop examples here, Aβ is visible algebraically. In transistor-level amplifiers and nested loops, loop-breaking and return-ratio methods must be applied with greater care.

04.11Mastery exercises

  1. A loop has A=2×105 and β=0.01. Find L, exact ACL, ideal 1/β and relative finite-loop-gain error.
  2. What minimum L is required to keep the magnitude of the finite-loop-gain error below 0.1% in the scalar model?
  3. If A changes by +10% while L=99, estimate the fractional change in ACL.
  4. For an inverting amplifier R1=10 kΩ and R2=90 kΩ, find signal gain, noise gain and β. If A=105, find L.
  5. For voltage-series feedback with L=999, estimate the factors by which Rin and Rout change.
  6. Explain why large DC loop gain alone is insufficient to guarantee accurate closed-loop gain at high frequency.
  7. Why can opening a feedback wire give a misleading loop-gain result in a biased transistor amplifier?
Solutions

1. L=2000; ACL=200000/(1+2000)=99.950 V/V, while 1/β=100 V/V. Relative error=−1/2001≈−0.0500%.

2. Require 1/(1+L)<0.001, hence L>999.

3. Sensitivity=1/(1+99)=0.01, so a +10% change in A produces approximately +0.1% change in ACL.

4. Signal gain≈−9; noise gain=10; β=0.1; L=10,000.

5. Rin rises by about 1000× and Rout falls by about 1000×, under the stated topology/model assumptions.

6. Because |A(s)| and therefore |L(s)| fall with frequency, increasing finite-loop-gain error; phase also becomes relevant for stability.

7. Opening the wire can change DC bias and impedances, so the measured path may no longer represent the loop transmission of the operating circuit.

Next: chapter 05 keeps L(s) complex and asks when the denominator 1+L(s) threatens stability.