← Analog & Mixed-Signal Interviews
Interview Question · Concept
kT/C Noise on a Sampling Capacitor
The Question
The interviewer asks
A resistor charges a capacitor. What is the total thermal noise voltage on the capacitor, and why does the resistor value not matter?
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Reveal the Model Answer
The total noise is √(kT/C), about 64.4 µV rms for 1 pF at 27 °C. A bigger resistor makes more noise per hertz, 4kTR. But it also narrows the bandwidth, 1/(4RC) in noise terms, by the same factor. The two cancel, so only C sets the total. To halve the noise you need four times the capacitance.
- R = 100 kΩ
- R = 10 kΩ
- R = 1 kΩ
- Frequency 100 kHzR = 100 kΩ = 12.87 µV · R = 10 kΩ = 4.071 µV · R = 1 kΩ = 1.287 µV
At 100 kHz the 100 kΩ circuit has gathered the most noise: its resistor is the noisiest per hertz.
- Frequency 10 MHzR = 100 kΩ = 61.06 µV · R = 10 kΩ = 38.47 µV · R = 1 kΩ = 12.87 µV
By 10 MHz the 100 kΩ curve has stopped rising. The 10 kΩ curve is still climbing and the 1 kΩ curve has barely started.
- Frequency 10 GHzR = 100 kΩ = 64.37 µV · R = 10 kΩ = 64.34 µV · R = 1 kΩ = 64.05 µV
All three end at the same total, 64.37 µV: 1 kΩ, 10 kΩ and 100 kΩ alike.
Read the plot as a table
| Frequency | R = 100 kΩ | R = 10 kΩ | R = 1 kΩ |
|---|---|---|---|
| 1 Hz | 0 V | 0 V | 0 V |
| 10 Hz | 122.1 nV | 38.62 nV | 12.21 nV |
| 100 Hz | 405.1 nV | 128.1 nV | 40.51 nV |
| 1 kHz | 1.287 µV | 406.9 nV | 128.7 nV |
| 10 kHz | 4.071 µV | 1.287 µV | 407.1 nV |
| 100 kHz | 12.87 µV | 4.071 µV | 1.287 µV |
| 1 MHz | 38.47 µV | 12.87 µV | 4.071 µV |
| 10 MHz | 61.06 µV | 38.47 µV | 12.87 µV |
| 100 MHz | 64.05 µV | 61.06 µV | 38.47 µV |
| 1 GHz | 64.34 µV | 64.05 µV | 61.06 µV |
| 10 GHz | 64.37 µV | 64.34 µV | 64.05 µV |
| 100 GHz | 64.38 µV | 64.37 µV | 64.34 µV |
| 1 THz | 64.38 µV | 64.38 µV | 64.37 µV |
Running total of the noise on 1 pF behind three resistors. All three end at the same 64.37 µV.
The RC bench from lesson E7.2: 10 kΩ into 1 pF, noise integrated far past the pole.
Run the netlist to draw its plots here.
vn = 6.437300e-05 ktc = 6.437404e-05Self-Check
Tick each key point you covered in your answer.
What Interviewers Listen For (Fresher)
A good fresher answer gives √(kT/C) and its value for 1 pF. It explains why the resistor cancels out of the total.
What Interviewers Listen For (Mid-Level)
A good mid-level answer links kT/C to ADC resolution: each extra bit costs four times the capacitance and the drive current. It integrates the noise far enough past the pole.
Cross-Questions
Where the interviewer goes next. Open a node to practise that follow-up.
WhyProve ItCompareScale It
- kT/C Noise on a Sampling Capacitor
-
WhyWhy the Resistor Drops Out of kT/CL1 JuniorL2 Mid-Level
Show in words why the resistor cancels out of the kT/C result.
2:30You have two and a half minutes. Speak or type, then press Get Feedback under your answer. No credits needed.
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Reveal the Model Answer
The noise density is 4kTR, so ten times R gives ten times the power per hertz. The RC filter's noise bandwidth is 1/(4RC), so ten times R gives a tenth of the bandwidth. The total is density times bandwidth, 4kTR · 1/(4RC) = kT/C. The R cancels.
AC analysisDensity and Running Total, 10 kΩ and 100 kΩ on 1 pFAbove: the noise density on the capacitor, in V/√Hz. Below: the running total, in V rms. Whole plot- 100 kΩ density
- 10 kΩ density
- 100 kΩ total
- 10 kΩ total
- Frequency 1 kHz100 kΩ density = 40.71 nV/rtHz · 10 kΩ density = 12.87 nV/rtHz · 100 kΩ total = 1.287 µV · 10 kΩ total = 406.9 nV
Far below both corners each density is its resistor's own: 40.71 nV/√Hz for 100 kΩ, 12.87 nV/√Hz for 10 kΩ. Ten times the resistance gives √10 times the density.
- Frequency 1.592 MHz100 kΩ density = 28.79 nV/rtHz · 10 kΩ density = 12.81 nV/rtHz · 100 kΩ total = 45.52 µV · 10 kΩ total = 16.22 µV
The 100 kΩ corner, 1.592 MHz. Its running total is 45.52 µV, half of its final noise power. The 10 kΩ total is only 16.22 µV.
- Frequency 15.92 MHz100 kΩ density = 4.052 nV/rtHz · 10 kΩ density = 9.103 nV/rtHz · 100 kΩ total = 62.3 µV · 10 kΩ total = 45.52 µV
The 10 kΩ corner, ten times higher at 15.92 MHz. Its total has now reached 45.52 µV too: half of the same final power.
- Frequency 1 GHz100 kΩ density = 64.8 pV/rtHz · 10 kΩ density = 204.9 pV/rtHz · 100 kΩ total = 64.34 µV · 10 kΩ total = 64.05 µV
Both totals have settled at 64.37 µV. The taller curve is narrower by exactly as much as it is taller.
Read the plot as a table
Frequency 100 kΩ density 10 kΩ density 100 kΩ total 10 kΩ total 1 Hz 40.71 nV/rtHz 12.87 nV/rtHz 0 V 0 V 10 Hz 40.71 nV/rtHz 12.87 nV/rtHz 122.1 nV 38.62 nV 100 Hz 40.71 nV/rtHz 12.87 nV/rtHz 405.1 nV 128.1 nV 1 kHz 40.71 nV/rtHz 12.87 nV/rtHz 1.287 µV 406.9 nV 10 kHz 40.71 nV/rtHz 12.87 nV/rtHz 4.071 µV 1.287 µV 100 kHz 40.63 nV/rtHz 12.87 nV/rtHz 12.87 µV 4.071 µV 1 MHz 34.47 nV/rtHz 12.85 nV/rtHz 38.47 µV 12.87 µV 1.592 MHz 28.79 nV/rtHz 12.81 nV/rtHz 45.52 µV 16.22 µV 10 MHz 6.399 nV/rtHz 10.9 nV/rtHz 61.06 µV 38.47 µV 15.92 MHz 4.052 nV/rtHz 9.103 nV/rtHz 62.3 µV 45.52 µV 100 MHz 647.9 pV/rtHz 2.024 nV/rtHz 64.05 µV 61.06 µV 1 GHz 64.8 pV/rtHz 204.9 pV/rtHz 64.34 µV 64.05 µV 10 GHz 6.48 pV/rtHz 20.49 pV/rtHz 64.37 µV 64.34 µV 100 GHz 648 fV/rtHz 2.049 pV/rtHz 64.38 µV 64.37 µV 1 THz 64.8 fV/rtHz 204.9 fV/rtHz 64.38 µV 64.38 µV Ten times the resistance gives a taller density but a corner ten times lower. Both totals settle at 64.37 µV.
Self-Check
Tick each key point you covered in your answer.
-
Prove ItSolve E7.2.1: A Hold Capacitor for 12 BitsSimulator-graded
Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
-
WhyWhy Noise Bandwidth Is Wider Than the CornerL1 JuniorL2 Mid-Level
Why is the noise bandwidth π/2 times the -3 dB bandwidth, and not equal to it?
2:30You have two and a half minutes. Speak or type, then press Get Feedback under your answer. No credits needed.
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Reveal the Model Answer
A first-order filter does not cut off sharply. Its response falls slowly past the corner, so noise above f_-3dB still gets through. Integrating the whole curve gives the same total as a sharp filter π/2 times wider. That width is 1/(4RC), against 1/(2πRC) for the corner.
Self-Check
Tick each key point you covered in your answer.
Missed Corner
"Integrate the noise up to the -3 dB frequency."
The skirt above the corner carries real noise. Stopping at f_-3dB keeps only half of the total power.
Source Lesson E7.2 key points
-
Scale ItkT/C and ADC ResolutionL2 Mid-LevelL3 Senior
How does kT/C limit the resolution of a sampling ADC?
2:30You have two and a half minutes. Speak or type, then press Get Feedback under your answer. No credits needed.
Speech-to-text runs in your browser. Chrome and Edge may send audio to their own speech service.
You can edit your answer before you send it. Feedback uses AI and checks your answer only against our written answer. No credits are used and your answer is not saved.
Reveal the Model Answer
Each extra bit halves the quantization step, so the sampled noise must halve too. Noise goes as one over the square root of C, so each bit needs four times the capacitance. A bigger sampling capacitor also needs more drive current to charge in the same time. That makes kT/C a power limit for high-resolution converters.
Self-Check
Tick each key point you covered in your answer.
-
ComparekT/C Noise vs Charge Injection in a Sample-and-HoldL1 JuniorL2 Mid-Level
Compare kT/C noise with charge injection as errors of a sample-and-hold switch.
2:30You have two and a half minutes. Speak or type, then press Get Feedback under your answer. No credits needed.
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You can edit your answer before you send it. Feedback uses AI and checks your answer only against our written answer. No credits are used and your answer is not saved.
Reveal the Model Answer
kT/C noise is random and set only by the hold capacitor, about 64 µV on 1 pF. Charge injection is a step left when the switch opens: channel charge and clock feedthrough land on the capacitor. It grows with switch width, falls with C_H, and depends on the input, so it is distortion, not just offset. A half-width dummy switch on the inverted clock cancels most of it.
TransientThe Hold Error When the Switch Opens, at Three InputsThe error v(hold) − v(in) of an NMOS switch (W = 1 µm) into 1 pF. Below: the clock on its gate. Whole plot- input 0.2 V
- input 0.5 V
- input 0.8 V
- clock
- Time 900 psinput 0.2 V = -201.2 pV · input 0.5 V = -741.1 pV · input 0.8 V = -4.223 nV · clock = 1.8 V
Tracking: each hold node equals its input, so every error is zero.
- Time 1.1 nsinput 0.2 V = -903.2 µV · input 0.5 V = -734.8 µV · input 0.8 V = -575.7 µV · clock = 0 V
The gate falls in 0.1 ns and the switch opens. Each hold node jumps down.
- Time 3 nsinput 0.2 V = -903.3 µV · input 0.5 V = -734.9 µV · input 0.8 V = -575.8 µV · clock = 0 V
Held: −0.903 mV at 0.2 V, −0.735 mV at 0.5 V, −0.576 mV at 0.8 V. The step shrinks as the input rises.
Read the plot as a table
Time input 0.2 V input 0.5 V input 0.8 V clock 0 s -201.2 pV -741.1 pV -4.311 nV 1.8 V 500 ps -201.2 pV -741.1 pV -4.26 nV 1.8 V 900 ps -201.2 pV -741.1 pV -4.223 nV 1.8 V 1 ns -201.2 pV -741.1 pV -4.215 nV 1.8 V 1.1 ns -903.2 µV -734.8 µV -575.7 µV 0 V 1.5 ns -903.3 µV -734.9 µV -575.8 µV 0 V 2 ns -903.3 µV -734.9 µV -575.8 µV 0 V 2.5 ns -903.3 µV -734.9 µV -575.8 µV 0 V 3 ns -903.3 µV -734.9 µV -575.8 µV 0 V Charge injection, measured: when the switch opens, each held voltage jumps by a step that depends on the input.
Self-Check
Tick each key point you covered in your answer.
-
Prove ItSolve E8.2.1: The Pedestal That Came with a Wider SwitchSimulator-graded
Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
-
Prove ItSolve E7.2.1: A Hold Capacitor for 12 BitsSimulator-graded
Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
What People Get Wrong
Common wrong answers and why they fail. Run It shows the wrong belief failing on a real simulator.
Misconception
"Use a bigger resistor and the capacitor sees more noise."
The density grows with R but the bandwidth shrinks with it. With 100 kΩ instead of 10 kΩ the total stays at 64.4 µV. Four times the capacitance halves it to 32.2 µV.
Run It
The same bench with R1 raised from 10 kΩ to 100 kΩ.
Run the netlist to draw its plots here.
vn = 6.437328e-05See the Fix
* E7.2 lab: the noise on C1 of an RC low-pass, R1 = 10 k, C1 = 1 pF
V1 in 0 DC 0 AC 1
R1 in out 10k
C1 out 0 4p
.control
* 100 points per decade from 1 Hz to 1 THz, far past the pole
noise v(out) V1 dec 100 1 1T
print onoise_total
* meas cannot read a noise plot: carry the total into a two-point dc run
dc V1 0 0.1 0.1
let one = 1+0*v(in)
let vn = noise2.onoise_total*one
* sqrt(kT/C) at 27 C; change 1p here when you change C1
let ktc = sqrt(boltz*(27+273.15)/4p)*one
meas dc vn FIND vn AT=0
meas dc ktc FIND ktc AT=0
.endc
.end
Prints: vn = 3.218662e-05 ktc = 3.218702e-05
Missed Corner
"Integrate the noise up to the -3 dB frequency."
The noise bandwidth of one pole is π/2 times the -3 dB frequency, and the tail above it counts. Run the sweep well past the corner with enough points per decade.
Source ASIC Point lesson E7.2