← Analog & Mixed-Signal Interviews
Interview Question · Code Writing
Monte Carlo on an Amplifier's Offset
- Likely at Infineon
- Likely at NVIDIA
- Likely at AMD
The Question
The interviewer asks
Set up a Monte Carlo run for an OTA's offset. What do you report from it, and what number do you sign off?
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Reveal the Model Answer
Give every transistor in every copy its own random draw. In sky130, turn mismatch on with MC_MM_SWITCH=1 and fix the seed, so the run repeats. Report the mean and σ. The mean is the systematic offset plus a small sampling error. Sign off at the mean plus or minus 3σ, not at the worst die you happened to sample. In the lesson 30 dies gave a mean of -2.62 mV and σ of 1.83 mV.
- The input pair
A threshold shift of M2 against M1 appears at the input one to one.
- The load mirror
A shift of M4 against M3 changes M4's current by gm3ΔV. M2 must absorb it, so the input sees ΔV·gm3/gm1, 0.57 times the shift.
- The bias
MB and M5 change the tail current, which both halves share: what is left after the pair and load parts spreads by 24 µV.
Read the circuit as text
- M3 PMOS transistor (load 8/1): gate →
n1, source →VDD, drain →n1 - M4 PMOS transistor (load 8/1): gate →
n1, source →VDD, drain →n2 - M1 NMOS transistor (pair 10/1): gate →
n3, drain →n1, source →n4 - M2 NMOS transistor (pair 10/1): gate →
n2, drain →n2, source →n4 - IB current source (20 µA): + →
VDD, − →n5 - MB NMOS transistor (8/1): gate →
n5, drain →n5, source →0 - M5 NMOS transistor (8/1): gate →
n5, drain →n4, source →0 - out port: terminal →
n2 - in port: terminal →
n3
OTA-A of E5.3 (W/L in µm). In SKY130 the mismatch model shifts the threshold of each transistor only.
Read the histogram as a table
| Offset from | to | Dies |
|---|---|---|
| -9 mV | -8 mV | 0 |
| -8 mV | -7 mV | 0 |
| -7 mV | -6 mV | 0 |
| -6 mV | -5 mV | 3 |
| -5 mV | -4 mV | 5 |
| -4 mV | -3 mV | 5 |
| -3 mV | -2 mV | 6 |
| -2 mV | -1 mV | 7 |
| -1 mV | 0 V | 2 |
| 0 V | 1 mV | 1 |
| 1 mV | 2 mV | 0 |
| 2 mV | 3 mV | 0 |
| 3 mV | 4 mV | 1 |
30 values in 13 bins.
Offset of 30 Monte Carlo dies, seed 5: mean -2.62 mV, σ 1.83 mV.
Self-Check
Tick each key point you covered in your answer.
What Interviewers Listen For (Junior)
A good junior answer turns mismatch on with a fixed seed, so the run repeats. It reports the mean and σ of the offset.
What Interviewers Listen For (Senior)
A good senior answer splits σ into the share of each device pair, and makes only the dominant device bigger. It states how well N samples know σ.
Cross-Questions
Where the interviewer goes next. Open a node to practise that follow-up.
WhyWhat IfProve ItCompare
- Monte Carlo on an Amplifier's Offset
-
WhyPelgrom's LawL1 JuniorL2 Mid-Level
Why does a bigger device match better, and by how much?
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Reveal the Model Answer
Random variations in doping and edges average out over a larger gate. For a pair, σ of the threshold difference is A_VT divided by √(W·L). Four times the area halves it. For sky130 the lesson found A_VT of about 4.75 mV·µm for the NMOS and 8.28 mV·µm for the PMOS.
DC sweepThreshold Mismatch of a Pair Against 1/√(W·L)The straight line is Pelgrom's law with AVT = 4.746 mV·µm. The curve adds the model's size factor for square devices. The points are 20 measured pairs at each size. Whole plot- Pelgrom, 4.746 mV·µm
- model, square devices
- 1/sqrt(W·L) (1/µm) 0.25Pelgrom, 4.746 mV·µm = 1.187 mV · model, square devices = 1.23 mV
The 4/4 pairs, 16 µm2: 1.131 mV measured, 1.230 mV on the model curve.
- 1/sqrt(W·L) (1/µm) 0.5Pelgrom, 4.746 mV·µm = 2.373 mV · model, square devices = 2.677 mV
The 2/2 pairs, 4 µm2: 2.895 mV measured against 2.677 mV on the curve. A quarter of the area, about twice the mismatch.
- 1/sqrt(W·L) (1/µm) 1Pelgrom, 4.746 mV·µm = 4.746 mV · model, square devices = 6.398 mV
The 1/1 pairs, 1 µm2: the model curve is 6.398 mV, 35 % above the straight line's 4.746 mV. That is the model's size factor. The 20 pairs measure 5.103 mV.
Read the plot as a table
1/sqrt(W·L) (1/µm) Pelgrom, 4.746 mV·µm model, square devices 0.2 949.2 µV 968.3 µV 0.25 1.187 mV 1.23 mV 0.3 1.424 mV 1.501 mV 0.4 1.898 mV 2.07 mV 0.5 2.373 mV 2.677 mV 0.6 2.848 mV 3.325 mV 0.7 3.322 mV 4.018 mV 0.8 3.797 mV 4.759 mV 0.9 4.271 mV 5.551 mV 1 4.746 mV 6.398 mV Measured on ngspice 42 with 20 pairs per size and seed 5. The 10/1 device is not square, so it sits near the line and off the curve.
Self-Check
Tick each key point you covered in your answer.
-
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.
-
WhyWhy Mismatch Falls as the Square Root of AreaL2 Mid-Level
Why does mismatch fall with 1/√(W·L), and not with 1/(W·L)?
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 threshold error comes from many small random events, such as single dopant atoms, spread over the gate. Their number grows with the area. The spread of a sum of random events grows only as its square root. Averaged over the area, the error therefore falls as 1/√(W·L).
Self-Check
Tick each key point you covered in your answer.
Misconception
"Double the width to halve the offset."
Doubling W doubles the area, which cuts σ only by √2. Halving σ needs four times the area.
Source Pelgrom, Duinmaijer and Welbers, Matching Properties of MOS Transistors, IEEE JSSC 1989
-
What IfWhich Device to Make BiggerL2 Mid-LevelL3 Senior
Asked if they say to make every device bigger
What if you can afford more area for only one device pair? Which do you grow?
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
Grow the pair whose share dominates the offset. Each pair's part adds as a power: the input pair one to one, the load mirror scaled by g_m3/g_m1. Four times the area of every device halves σ. Sixteen times the area of one pair leaves the other pair's part untouched, so it helps only if that pair dominated.
DC sweepPair Part and Load Part of Each Die, Pair 10/1The lab's 30 dies. Neither part dominates. Whole plot- offset − mean
- pair part
- load part
- Die 8offset − mean = -2.568 mV · pair part = -3.491 mV · load part = 869.4 µV
Die 8: the pair part is −3.49 mV and the load part +0.87 mV, together −2.62 mV. The die sits 2.57 mV below the mean: the two parts leave 0.05 mV over.
- Die 17offset − mean = 5.798 mV · pair part = 2.062 mV · load part = 3.728 mV
Die 17: +2.06 mV from the pair and +3.73 mV from the load, so the die sits 5.80 mV above the mean.
Read the plot as a table
Die offset − mean pair part load part 5 -1.543 mV -1.144 mV -434.1 µV 8 -2.568 mV -3.491 mV 869.4 µV 10 1.395 mV 146.1 µV 1.228 mV 15 -1.191 mV -888.5 µV -319.6 µV 17 5.798 mV 2.062 mV 3.728 mV 20 1.413 mV 778.9 µV 682.6 µV 25 -392.1 µV 1.437 mV -1.81 mV 30 -636.4 µV -559 µV -94.61 µV Each die's offset split into the pair part and the load part. Here neither dominates, so growing one pair alone helps little.
Self-Check
Tick each key point you covered in your answer.
-
CompareProcess Corners vs Monte CarloL1 JuniorL2 Mid-Level
Compare process corners with a mismatch Monte Carlo run. Do you need both?
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.
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Reveal the Model Answer
Yes. A process corner moves every transistor of a type together, the global spread between wafers. Mismatch moves each transistor alone, the local spread on one die. Offset comes mostly from mismatch, while bias currents and speed move mostly with corners. Sign-off runs corners crossed with supply and temperature, plus Monte Carlo.
Self-Check
Tick each key point you covered in your answer.
-
Prove ItSolve E7.6.1: Nine Settings, One VerdictSimulator-graded
Show it, don't just say it: this Silicon Lab problem is graded by the simulator.
-
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.
What People Get Wrong
Common wrong answers and why they fail.
Misconception
"Sign off at the worst die in the Monte Carlo run."
The worst of 30 samples depends on luck. The lesson's worst sampled die was -5.19 mV, while mean minus 3σ is about -8.1 mV. Use the statistics, not the sample.
Source ASIC Point lesson E7.5 (measured)
Half-Truth
"Thirty samples give an exact σ."
Thirty samples know σ only to about ±13 %. Add margin, or run more samples, before trusting a tight budget.
Source ASIC Point lesson E7.4