Interview Question · Code Reading

Input-Referred Noise of an OTA

L1 JuniorL2 Mid-Levelnoiseota

Likely at Analog Devices
Likely at Intel

The interviewer asks

This bench runs a noise analysis on a five-transistor OTA. Why does it read the input-referred spectrum, and which devices dominate it?

spice

XM5 tail nb 0 0 sky130_fd_pr__nfet_01v8 W=8 L=1
XM1 x inp tail 0 sky130_fd_pr__nfet_01v8 W=10 L=1
XM2 out inn tail 0 sky130_fd_pr__nfet_01v8 W=10 L=1
XM3 x x vdd vdd sky130_fd_pr__pfet_01v8 W=8 L=1
XM4 out x vdd vdd sky130_fd_pr__pfet_01v8 W=8 L=1
CL out 0 1p
VIP inp 0 DC 0.9 AC 1
noise v(out) VIP dec 10 1 1g 1
noise v(out) VIP dec 10 10 100k

Reveal the Model Answer

Input-referred noise is output noise divided by gain, so amplifiers with different gains can be compared fairly. In a 5T OTA the input pair dominates. The load mirror adds its noise scaled by g_m3/g_m1, and the tail cancels as common mode. Here flicker noise dominates up to about 939 kHz. Above that, the thermal floor needs the model's γ, about 2 for the NMOS pair, not the textbook 2/3.

OTA-A and Where Its Noise Comes FromThe same OTA-A as E5.3. Each transistor is a noise current between its drain and source.

The input pair
A noise current in M1 flows through M3 and M4 copies it to the output. A noise current in M2 reaches the output directly. Both act exactly like an input signal, so they refer to the input divided by g_m1.
i_n1 · i_n2
The load mirror
A noise current in M3 or M4 also lands on the output. To make the same current, the input would need i_n3/g_m1, so a load device's gate noise reaches the input multiplied by g_m3/g_m1.
i_n4
The tail
A noise current in M5 splits equally between the two halves. M4 copies M1's half to the output while M2 takes the same half away, so it cancels there: it is common mode.
i_n5

Read the circuit as text

M3 PMOS transistor (8/1): gate → x, source → VDD, drain → x
M4 PMOS transistor (8/1): gate → x, source → VDD, drain → n1
M1 NMOS transistor (10/1): gate → n2, drain → x, source → tail
M2 NMOS transistor (10/1): gate → n3, drain → n1, source → tail
IB current source (20 µA): + → VDD, − → n4
MB NMOS transistor (8/1): gate → n4, drain → n4, source → 0
M5 NMOS transistor (8/1): gate → n4, drain → tail, source → 0
C_L capacitor (1 pF): pin 1 → n1, pin 2 → 0
out port: terminal → n1
+ port: terminal → n2
− port: terminal → n3

NMOS bodies at ground, PMOS bodies at VDD. W/L in µm.

AC analysisOTA-A Input-Referred Noise: Flicker and Thermal PartsThe total, the flicker part of M1 to M4, and their channel thermal part, all referred to the + input.

total
flicker part
thermal part

Frequency 1 kHztotal = 412.6 nV/rtHz · flicker part = 411.9 nV/rtHz · thermal part = 22.65 nV/rtHz
At 1 kHz the total is 412.6 nV/√Hz, almost all flicker noise. The thermal part is only 22.65 nV/√Hz.
Frequency 938.9 kHztotal = 32.01 nV/rtHz · flicker part = 22.6 nV/rtHz · thermal part = 22.65 nV/rtHz
At 938.9 kHz the flicker and thermal parts are equal: the amplifier's flicker corner.
Frequency 10 MHztotal = 24.15 nV/rtHz · flicker part = 8.318 nV/rtHz · thermal part = 22.65 nV/rtHz
At 10 MHz the total is 24.15 nV/√Hz: thermal 22.65 nV/√Hz plus a flicker part that is still there.

Read the plot as a table

Frequency	total	flicker part	thermal part
1 Hz	8.117 µV/rtHz	8.117 µV/rtHz	22.65 nV/rtHz
10 Hz	2.98 µV/rtHz	2.979 µV/rtHz	22.65 nV/rtHz
100 Hz	1.104 µV/rtHz	1.104 µV/rtHz	22.65 nV/rtHz
1 kHz	412.6 nV/rtHz	411.9 nV/rtHz	22.65 nV/rtHz
10 kHz	156.2 nV/rtHz	154.5 nV/rtHz	22.65 nV/rtHz
100 kHz	62.46 nV/rtHz	58.2 nV/rtHz	22.65 nV/rtHz
938.9 kHz	32.01 nV/rtHz	22.6 nV/rtHz	22.65 nV/rtHz
1 MHz	31.57 nV/rtHz	21.98 nV/rtHz	22.65 nV/rtHz
10 MHz	24.15 nV/rtHz	8.318 nV/rtHz	22.65 nV/rtHz
100 MHz	24.07 nV/rtHz	3.141 nV/rtHz	22.78 nV/rtHz
1 GHz	43.61 nV/rtHz	2.122 nV/rtHz	39.4 nV/rtHz

OTA-A's input-referred noise. Flicker noise rules at low frequency; the two parts meet at about 939 kHz.

The noise bench from lesson E7.3: OTA-A open loop.

SPICEngspice 42 · simulator service

Edit the netlist, then run it

Expected: vin_10 = 2.979765e-06 vin_1k = 4.125976e-07 vin_10meg = 2.414615e-08 vin_band = 4.192227e-05

* E7.3 lab: input-referred noise of OTA-A, open loop, C_L = 1 pF, inputs at 0.9 V
.lib sky130 tt
VDD vdd 0 1.8
IB vdd nb 20u
XMB nb nb 0 0 sky130_fd_pr__nfet_01v8 W=8 L=1
XM5 tail nb 0 0 sky130_fd_pr__nfet_01v8 W=8 L=1
XM1 x inp tail 0 sky130_fd_pr__nfet_01v8 W=10 L=1
XM2 out inn tail 0 sky130_fd_pr__nfet_01v8 W=10 L=1
XM3 x x vdd vdd sky130_fd_pr__pfet_01v8 W=8 L=1
XM4 out x vdd vdd sky130_fd_pr__pfet_01v8 W=8 L=1
CL out 0 1p
VIP inp 0 DC 0.9 AC 1
LF out inn 1T
CF inn 0 1
.control
noise v(out) VIP dec 10 1 1g 1
noise v(out) VIP dec 10 10 100k
ac dec 10 1 1g
let vin = noise1.inoise_spectrum
let vout = noise1.onoise_spectrum
let gain = mag(v(out))
let one = 1+0*gain
let vin_band = noise4.inoise_total*one
meas ac a_1k FIND gain AT=1k
meas ac vout_1k FIND vout AT=1k
meas ac vin_10 FIND vin AT=10
meas ac vin_1k FIND vin AT=1k
meas ac vin_100k FIND vin AT=100k
meas ac vin_1meg FIND vin AT=1meg
meas ac vin_10meg FIND vin AT=10meg
meas ac vin_band FIND vin_band AT=1k
.endc
.end

Self-Check

Tick each key point you covered in your answer.

Key point 1 Input-referred noise is output noise divided by gain, so amplifiers with different gains can be compared fairly. Key point 2 In a 5T OTA the input pair dominates. Key point 3 The load mirror adds its noise scaled by g_m3/g_m1, and the tail cancels as common mode. Key point 4 Here flicker noise dominates up to about 939 kHz. Key point 5 Above that, the thermal floor needs the model's γ, about 2 for the NMOS pair, not the textbook 2/3.

What Interviewers Listen For (Junior)

A good junior answer divides the output noise by the gain, and names the input pair as the main source. It keeps flicker noise apart from thermal noise.

Cross-Questions

Where the interviewer goes next. Open a node to practise that follow-up.

WhyProve ItCompareScale It

Input-Referred Noise of an OTA
WhyWhy the Tail Current Source Adds No NoiseL2 Mid-Level
Why does noise from the tail current source not appear at the OTA's input?

Reveal the Model Answer

Tail noise changes the current in both halves of the pair equally. That is a common-mode change. The OTA rejects common mode, so the tail noise barely reaches the differential output. It returns if the halves are badly mismatched.

Self-Check
Tick each key point you covered in your answer.
Key point 1 Tail noise changes the current in both halves of the pair equally. Key point 2 That is a common-mode change. Key point 3 The OTA rejects common mode, so the tail noise barely reaches the differential output. Key point 4 It returns if the halves are badly mismatched.
- Lesson E7.3Input-Referred Noise of an Amplifier
CompareWhy the Load Mirror Noise Does Not CancelL2 Mid-Level
Compare the load mirror with the tail. Why does the load's noise reach the input when the tail's does not?

Reveal the Model Answer

Each load device's noise current flows into one side of the OTA only. So it is a differential signal, and the OTA does not reject it. Referred to the input, the load's noise appears scaled by g_m3/g_m1. A smaller g_m3/g_m1 lowers its share.

Self-Check
Tick each key point you covered in your answer.
Key point 1 Each load device's noise current flows into one side of the OTA only. Key point 2 So it is a differential signal, and the OTA does not reject it. Key point 3 Referred to the input, the load's noise appears scaled by g_m3/g_m1. Key point 4 A smaller g_m3/g_m1 lowers its share.

Misconception

"The load is a current source, so it adds no noise."

Its devices make thermal and flicker noise like any transistor. That noise reaches the output with the signal.

Source Lesson E7.3 key points
- Lesson E7.3Input-Referred Noise of an Amplifier

CompareFlicker Noise vs Thermal NoiseL1 JuniorL2 Mid-Level

Compare flicker noise with thermal noise in a MOSFET. How do you reduce each one?

Reveal the Model Answer

Thermal noise is white, 4kTγg_m in the channel, and falls when g_m rises. Flicker noise falls with frequency and with gate area. The flicker corner is where the two meet. In the lesson, 16 times the area moved the corner from 1.29 MHz to 48 kHz. Read a device's noise over the band you use, because a 100 kHz number says nothing about 10 Hz.

AC analysisDrain-Current Noise at 10 µA: W/L = 10/1 and 40/4The total density of each device, and the channel thermal part of the 10/1 device alone.

10/1 total
40/4 total
10/1 thermal part

Frequency 10 Hz10/1 total = 334.5 pA/rtHz · 40/4 total = 86.14 pA/rtHz · 10/1 thermal part = 2.389 pA/rtHz
At 10 Hz flicker noise dominates: 334.5 pA/√Hz for 10/1 and 86.11 pA/√Hz for 40/4. Sixteen times the area gives 15.09 times less noise power.
Frequency 48.08 kHz10/1 total = 9.808 pA/rtHz · 40/4 total = 3.467 pA/rtHz · 10/1 thermal part = 2.389 pA/rtHz
The 40/4 device's flicker corner, 48.08 kHz: here its flicker and thermal parts are equal.
Frequency 1.292 MHz10/1 total = 3.378 pA/rtHz · 40/4 total = 2.529 pA/rtHz · 10/1 thermal part = 2.389 pA/rtHz
The 10/1 device's corner is 27 times higher, at 1.292 MHz.
Frequency 100 MHz10/1 total = 2.42 pA/rtHz · 40/4 total = 2.455 pA/rtHz · 10/1 thermal part = 2.389 pA/rtHz
At 100 MHz only thermal noise is left: 2.389 pA/√Hz for 10/1 and 2.453 pA/√Hz for 40/4, almost the same, because g_m is almost the same.

Read the plot as a table

Frequency	10/1 total	40/4 total	10/1 thermal part
1 Hz	879.8 pA/rtHz	226.5 pA/rtHz	2.389 pA/rtHz
10 Hz	334.5 pA/rtHz	86.14 pA/rtHz	2.389 pA/rtHz
100 Hz	127.2 pA/rtHz	32.83 pA/rtHz	2.389 pA/rtHz
1 kHz	48.41 pA/rtHz	12.69 pA/rtHz	2.389 pA/rtHz
10 kHz	18.54 pA/rtHz	5.33 pA/rtHz	2.389 pA/rtHz
48.08 kHz	9.808 pA/rtHz	3.467 pA/rtHz	2.389 pA/rtHz
100 kHz	7.386 pA/rtHz	3.042 pA/rtHz	2.389 pA/rtHz
1 MHz	3.574 pA/rtHz	2.547 pA/rtHz	2.389 pA/rtHz
1.292 MHz	3.378 pA/rtHz	2.529 pA/rtHz	2.389 pA/rtHz
10 MHz	2.594 pA/rtHz	2.467 pA/rtHz	2.389 pA/rtHz
100 MHz	2.42 pA/rtHz	2.455 pA/rtHz	2.389 pA/rtHz
1 GHz	2.394 pA/rtHz	2.473 pA/rtHz	2.39 pA/rtHz

Two devices at 10 µA. Sixteen times the gate area moves the flicker corner from 1.29 MHz to 48 kHz.

Self-Check

Tick each key point you covered in your answer.

Key point 1 Thermal noise is white, 4kTγg_m in the channel, and falls when g_m rises. Key point 2 Flicker noise falls with frequency and with gate area. Key point 3 The flicker corner is where the two meet. Key point 4 In the lesson, 16 times the area moved the corner from 1.29 MHz to 48 kHz. Key point 5 Read a device's noise over the band you use, because a 100 kHz number says nothing about 10 Hz.

Prove ItSolve E7.5.1: Offset by the Square MicrometreSimulator-graded

Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
Open Problem E7.5.1
Scale ItSplitting a Noise Budget Across a Signal ChainL3 Senior
You own the front end of a sensor chain. How do you split the noise budget between the amplifier, the sampling capacitor and later stages?

Reveal the Model Answer

Independent noise sources add as powers, so write the budget in squared volts. Refer every stage to the input by dividing by the gain in front of it. The first stage usually dominates, so give it most of the budget and enough gain. Size the sampling capacitor from kT/C. Check the total over the real signal band, flicker included.

Self-Check
Tick each key point you covered in your answer.
Key point 1 Independent noise sources add as powers, so write the budget in squared volts. Key point 2 Refer every stage to the input by dividing by the gain in front of it. Key point 3 The first stage usually dominates, so give it most of the budget and enough gain. Key point 4 Size the sampling capacitor from kT/C. Key point 5 Check the total over the real signal band, flicker included.

What People Get Wrong

Common wrong answers and why they fail.

Misconception

"Compare the two amplifiers by their output noise."

A higher-gain amplifier shows more output noise even if it is quieter. Refer the noise to the input before comparing.

Source ASIC Point lesson E7.3

Spec Misquote

"Use γ = 2/3 for the MOSFET thermal noise."

That is the long-channel value. The sky130 models give about 2 for the NMOS and about 1 for the PMOS load, so a 2/3 estimate is far too low.

Source ASIC Point lessons E7.1 and E7.3 (measured)

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