Engineering & Technology · Electronics, Devices & Signal Processing
Noise, distortion, bandwidth and electronic performance limits
A larger gain makes a tiny signal visible, but it also enlarges noise already present at the input. A wider bandwidth follows faster events while admitting more noise. Electronic performance is a set of linked limits, so one impressive number rarely defines the whole useful system.
Electrical Circuits supplies charge/current balance, stored energy, sinusoidal phase, rms and the unloaded RC response. Material Structure & Properties supplies the distinction between a material description and a complete product. This chapter explains bias and feedback before using gain. It introduces spectra through an explicit two-tone sum and sampling through points on known curves. All parameters, code intervals, timing values and noise examples are original teaching models; no physical construction or performance qualification is implied.
Back to the chapterAfter reading, you can
Compare noise, deterministic distortion and bandwidth limits with explicit conditions and consistent performance metrics.
Name the contribution before combining it
Noise can be described statistically, while distortion is a systematic change related to the signal, such as harmonics from a nonlinear transfer curve. An unwanted external periodic pickup is an interference signal; calling everything noise can hide whether averaging or a different model would help. Offset is another distinct contribution: a stable added voltage is not random scatter.
Input-referred noise expresses an output contribution divided by the relevant gain so stages can be compared on a common basis. Voltage-noise density has units V/√Hz. For a flat density e and an ideal rectangular equivalent noise band B, rms noise is e√B. A real filter requires its equivalent noise bandwidth, which need not equal its labelled −3 dB bandwidth.
Uncorrelated zero-mean rms noise contributions combine by the square root of summed squares. Correlation changes that account. Amplifier input-current noise flowing through source impedance also generates voltage noise, so a low voltage-noise specification alone does not establish the best whole circuit.
Turn a ratio into a logarithmic level
A decibel, written dB, expresses a logarithmic ratio, not an amount of noise by itself. The base-ten logarithm log10(r) answers: “Which exponent of 10 gives the positive number r?” Thus log10(10)=1, log10(100)=2 and log10(1000)=3, because 1000=10³. Ratios must compare compatible quantities so their units cancel. A statement such as 60 dB needs the numerator, denominator and reference conditions to be meaningful.
For a positive power ratio P1/P0, its decibel value is 10 log10(P1/P0). For a same-kind amplitude ratio A1/A0, the amplitude-level convention is 20 log10(A1/A0). The two describe the same physical level difference when the power ratio is the square of the amplitude ratio: 10 log10[(A1/A0)²]=20 log10(A1/A0). For example, rms voltages across the same resistance obey P=V²/R, so the resistance cancels in their power ratio. With different resistances, a voltage ratio alone does not determine the power ratio.
For this lesson’s signal and noise rms voltages at the same reference and in the stated band, 1 mV divided by 1 μV is 1000=10³. Therefore 20 log10(1000)=20×3=60 dB. The corresponding equivalent power ratio across one common resistance is 1000²=10⁶, and 10 log10(10⁶)=60 dB too. Applying 10 to the voltage ratio would instead give 30 dB and mix the two conventions. The reference here is the noise rms itself, not an unstated absolute voltage.
Multiplying an amplitude ratio by ten adds 20 dB; multiplying a power ratio by ten adds 10 dB. For the same 1 mV signal, raising noise from 1 to 10 μV reduces the amplitude ratio from 1000 to 100 and SNR from 60 to 40 dB. A ratio below one has a negative level, without implying negative power. Zero cannot be inserted into a finite logarithm. A negative voltage gain also needs its magnitude for an amplitude level; its inversion or phase is separate information.
Quantify the chosen noise-band model
The original plot assumes flat 10 nV/√Hz input noise and an ideal rectangular noise bandwidth. At 100 Hz, rms noise is 100 nV; at 10,000 Hz it is 1000 nV. A hundredfold bandwidth increase produces tenfold rms noise under these assumptions. It does not mean every wider-band real amplifier follows this same curve.
Reducing the band is useful only if the intended signal remains represented. Very slow measurements may also encounter low-frequency excess noise, drift or environmental changes; indefinite averaging is not a promise of arbitrarily accurate results. Record filtering, time window and reference point when comparing quoted noise levels.
Separate small-signal speed from large-signal demands
Bandwidth describes a frequency response under stated conditions. Slew rate limits output rate of change, often expressed in V/μs. A sine of peak amplitude Vp and frequency f requires maximum slope 2πfVp. A device may have sufficient small-signal bandwidth while failing this large-signal demand. Settling additionally concerns reaching and staying within a specified error band after a change.
Performance ratios also have different denominators. Here SNR compares signal rms with noise rms, excluding separately identified distortion; THD compares combined harmonic rms with fundamental rms. SINAD includes both noise and distortion. State bandwidth, amplitude and exclusions because published terminology can vary. A narrow spectral bin floor is not total integrated noise.
Compare a noise account with a slope requirement
In an original input-referred example, two uncorrelated zero-mean contributions are 0.6 μV rms and 0.8 μV rms in the same band. Total noise is √(0.6²+0.8²)=1.0 μV rms. For a 1 mV rms signal, amplitude ratio is 1000 and SNR=20 log10(1000)=60 dB. Multiplying both signal and this noise by gain 100 leaves that ratio unchanged.
If a separate harmonic contribution is 10 μV rms, THD=10/1000=1% under our definition. Combined noise-plus-distortion rms is √(1²+10²)≈10.05 μV, giving SINAD≈39.96 dB. A low random-noise number alone would conceal that distortion.
Separately, a requested 3 V-peak sine at 10 kHz requires about 188,496 V/s, or 0.1885 V/μs. A hypothetical 0.1 V/μs slew limit cannot reproduce that requested slope. This comparison is a model constraint, not a complete bandwidth or product-selection test.
For a fresh same-port example, take signal 2 mV rms and total noise 5 μV rms in one declared band. The amplitude ratio is 400. Write 400=4×100; log10(400)=log10(4)+2≈0.60206+2, so SNR≈20×2.60206=52.041 dB. The logarithm of a product adds the exponents of its factors, which is why successive ratio changes become additions of decibel changes. If the same signal is compared with twice the noise, 10 μV, its ratio is 200 and SNR≈46.021 dB: a reduction of about 6.021 dB, not a halving of the decibel number.
As a separate reference-condition check, 10 V rms across 100 Ω dissipates 1 W, whereas 1 V rms across 10 Ω dissipates 0.1 W. Their voltage-amplitude ratio is ten, conventionally 20 dB, but their actual power ratio is also ten, or 10 dB, because resistance did not cancel. Always name the ratio being reported; the 20-times voltage shortcut cannot replace a power account across unlike loads.
Try the reasoning
Do two 1 μV rms noises always combine to √2 μV?
Show the answer and reasoning
Only under appropriate zero-mean uncorrelated assumptions; correlation or different reference bands changes the calculation.
Does gain automatically improve the SNR of existing input noise?
Show the answer and reasoning
No. Equal linear amplification of signal and that noise leaves their ratio unchanged; additional stage noise may worsen it.
Why report SINAD as well as a noise-only ratio when distortion matters?
Show the answer and reasoning
SINAD includes deterministic distortion contributions that a noise-only SNR can exclude.
A gain is −10 V/V. Is its voltage-amplitude level −20 dB, and does that level describe the inversion?
Show the answer and reasoning
Its magnitude is ten, so its amplitude level is +20 dB. The minus sign indicates voltage inversion, which must be reported separately. A negative decibel value would instead indicate a magnitude ratio below one.
At fixed signal and band, noise rms grows from 2 to 20 μV. How does the SNR level change, and why?
Show the answer and reasoning
The signal-to-noise amplitude ratio falls by a factor of ten. Its base-ten logarithm therefore decreases by one, giving a 20 dB decrease under the stated amplitude-level convention. This is not a tenfold decrease in the numerical dB value.
An rms voltage ratio is two. What are its amplitude level and its equivalent power ratio across the same resistance?
Show the answer and reasoning
The amplitude level is 20 log10(2)≈6.021 dB. Equal resistance makes the power ratio 2²=4, so 10 log10(4) gives the same 6.021 dB. Across different impedances, the voltage ratio alone does not establish that power ratio.
What to carry forward
Attach every performance number to its band, amplitude, model and exclusions. Noise, distortion, slew and settling answer different questions.
References
Teaching resources and research supporting this lesson’s explanations and examples.
- Understanding Operational Amplifier Specifications, SLOA011BJim Karki; Texas InstrumentsTeaching resource · PDF
- MT-047 — Op Amp NoiseAnalog Devices tutorial teamTeaching resource · PDF
- MT-003 — Understand SINAD, ENOB, SNR, THD, THD + N, and SFDRWalt Kester; Analog DevicesTeaching resource · PDF
- Guide to the SI, Special Publication 811 — section 8.7 Logarithmic quantities and units, selected mathematical definitionsNational Institute of Standards and TechnologyTeaching resource