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Interview Question · Project Defence

Defending an Analog Block You Designed

L2 Mid-LevelL3 Seniorproject defencedesign review

The Question

The interviewer asks

Walk me through an analog block you designed. Cover the specification, how you sized it, what went wrong, and how you knew it was ready.

Reveal the Model Answer

A strong answer starts from the specification lines and the margins you set. It explains the sizing method with one or two real numbers. An example is g_m/I_D charts, which plot transconductance per unit current against current density. It tells one problem in detail, such as poor phase margin at the heaviest load, with what you tried and what worked. It ends with the evidence: corners, Monte Carlo, start-up and post-layout results.

From Brief to Sized OTAArrows show which number feeds the next decision.
From brief to sized OTAArrows show which number feeds the next decision. component fu and CL 10 MHz into 2 pF component gm of M1 and M2 ≥ 125.7 µS; aim 157µS component Current budget 2 µA bias + 16 µA tail component gm/ID ≥ 19.6 /V; pick 22 component Width from chart 0.181 µA/µm: W = 40µm component Length for gain L = 1 µm for M1-M4 component Input range tail 8 × 8/2, load 4/1 component Simulate andcompare every line at 0.7, 0.9,1.1 V 2π·fu·CLI_D ≈ 8 µA
  1. Speed sets gm

    Into a capacitive load, fu = gm1/(2πCL), so gm1 ≥ 2π × 10 MHz × 2 pF = 125.7 µS. The output node carries more than CL, and the low end of the input range costs current, so aim 25 % higher: 157 µS.

  2. Current sets gm/ID

    Keep 2 µA for the bias diode and copy it eight times for a 16 µA tail: about 8 µA per input device. Then gm/ID ≥ 157/8 = 19.6 V−1. Pick 22, in moderate inversion.

  3. The chart sets W; gain sets L

    The chart below gives 0.181 µA/µm at gm/ID = 22, so W = 8 µA / 0.181 µA/µm = 44 µm: take 40. Gain is gm/(gds2 + gds4); at a fixed current only L moves the gds terms, and L = 1 µm clears 40 dB in E5.3's OTA.

  4. Range, then simulate

    The lower edge is VGS1 + VDsat5: a wide pair keeps VGS1 low and a wide tail (8 × 8/2) keeps VDsat5 low. The upper edge is VDD − VSG3 + Vth1: a 4/1 load keeps VSG3 small. Then run every line of the brief.

Read the diagram as text
  • fu and CL: 10 MHz into 2 pF
  • gm of M1 and M2: ≥ 125.7 µS; aim 157 µS
  • Current budget: 2 µA bias + 16 µA tail
  • gm/ID: ≥ 19.6 /V; pick 22
  • Width from chart: 0.181 µA/µm: W = 40 µm
  • Length for gain: L = 1 µm for M1-M4
  • Input range: tail 8 × 8/2, load 4/1
  • Simulate and compare: every line at 0.7, 0.9, 1.1 V
  • fu and CL → gm of M1 and M2: 2π·fu·CL
  • gm of M1 and M2 → gm/ID
  • Current budget → gm/ID: I_D ≈ 8 µA
  • gm/ID → Width from chart
  • Width from chart → Length for gain
  • Length for gain → Input range
  • Input range → Simulate and compare

One way to tell the sizing story: each decision fed by a number, then checked by simulation.

TransientThe Worked Answer, Stepped at Three Loads10 mV steps into the buffer, each output as a fraction of its own final step.
  • CL 1 pF
  • CL 2 pF
  • CL 5 pF
00.20.40.60.81.01.2Fraction of the step00.2 µs0.4 µs0.6 µs0.8 µs1.0 µs1.2 µs1.4 µsTimefinal value00.20.40.60.81.01.2Fraction of the step00.5 µs1.0 µs1.5 µsTimefinal value
  1. Time 144.6 nsCL 1 pF = 0.9525 · CL 2 pF = 0.9633 · CL 5 pF = 1.021

    At 5 pF the output overshoots by 2.1 %. At 1 and 2 pF it never passes its final value.

  2. Time 173.1 nsCL 1 pF = 0.99 · CL 2 pF = 0.9935 · CL 5 pF = 1.004

    Within 1 % of the final value: 62.3 ns after the step at 5 pF, 66.0 ns at 2 pF and 73.1 ns at 1 pF. The light loads approach slowly from below.

Read the plot as a table
TimeCL 1 pFCL 2 pFCL 5 pF
0 s1.672 µ1.672 µ1.672 µ
144.6 ns0.95250.96331.021
173.1 ns0.990.99351.004
200 ns0.99770.99870.9998
400 ns111
600 ns111
800 ns111
1 µs111
1.2 µs111
1.4 µs111

The kind of proof a defence needs: one compensation stepped at three loads, with the measured overshoot.

Self-Check

Tick each key point you covered in your answer.

What Interviewers Listen For (Mid-Level)

A good mid-level answer tells a structured story with measured numbers. It describes one real problem and how it was solved.

Cross-Questions

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

WhyWhat IfProve ItBreak ItCompareScale It

  1. Defending an Analog Block You Designed
  2. Break ItA Post-Layout SurpriseL2 Mid-LevelL3 Senior

    Tell me about a time your block behaved differently after layout. What changed, and what did you learn?

    Reveal the Model Answer

    A good story names the effect, such as extra parasitic capacitance that cut phase margin or a gradient that added offset. It shows how it was found, by post-layout simulation or measurement, and how it was fixed. It ends with what changed in the method, such as margins kept for parasitics or common-centroid placement planned from the start.

    Self-Check

    Tick each key point you covered in your answer.

  3. CompareHigh vs Low gm/IDL1 JuniorL2 Mid-Level

    Compare biasing a transistor at high g_m/I_D with biasing it at low g_m/I_D.

    Reveal the Model Answer

    High g_m/I_D, near weak inversion, gives the most transconductance per microamp. But the device is wide and slow, and its threshold mismatch becomes a larger current error. Low g_m/I_D, strong inversion, costs current but gives small, fast devices and better current matching. Input pairs often sit high for noise and offset; mirrors and fast stages sit lower.

    Self-Check

    Tick each key point you covered in your answer.

  4. Scale ItReviewing an Analog Block Before Tape-OutL3 SeniorL4 Staff

    You review a colleague's op-amp and bias block a month before tape-out. What evidence do you ask for before you accept it?

    Reveal the Model Answer

    Ask for every specification line with its measured margin at every corner, crossed with supply and temperature. Ask for Monte Carlo mean plus or minus 3σ on offset and matching. Check stability at the worst load and feedback factor, with a step as well as a margin. Check start-up with supply ramps, fast and cold-slow. Finally, ask which known faults the benches were shown to catch.

    One Bench Against Five BuffersArrows carry the same stimulus to each copy and the measured lines back to the verdict.
    One bench against five buffersArrows carry the same stimulus to each copy and the measured lines back to the verdict. component Your bench stimulus and measlines component Final design component No Rz phase margin component Bias 5 µA slew rate component Load mirror 4/2 input range component M2 22/2 offset component Verdicts pass, fail × 4
    1. Dynamic faults

      No Rz and a starved bias leave the settled DC levels right, so only the settling lines catch them for sure.

    Read the diagram as text
    • Your bench: stimulus and meas lines
    • Final design
    • No Rz: phase margin
    • Bias 5 µA: slew rate
    • Load mirror 4/2: input range
    • M2 22/2: offset
    • Verdicts: pass, fail × 4
    • Your bench → Final design
    • Your bench → No Rz
    • Your bench → Bias 5 µA
    • Your bench → Load mirror 4/2
    • Your bench → M2 22/2
    • Final design → Verdicts
    • No Rz → Verdicts
    • Bias 5 µA → Verdicts
    • Load mirror 4/2 → Verdicts
    • M2 22/2 → Verdicts
    AC analysisLoop Gain of OPA-B in the Sf Corner, Nominal and Cold at Low SupplyThe same op-amp, the same corner. Only the supply (1.8 → 1.62 V) and the temperature (27 → −40 °C) change.
    • sf, 1.8 V, 27 °C
    • sf, 1.62 V, −40 °C
    -60 dB-40 dB-20 dB020 dB40 dB60 dB80 dB100 dBLoop gain1 Hz100 Hz10 kHz1 MHz100 MHzFrequencyA0 = 79.8 dBA0 = 60.1 dB-100 dB-50 dB050 dB100 dBLoop gain1 Hz1 kHz1 MHz1 GHzFrequencyA0 = 79.8 dBA0 = 60.1 dB
    Read the plot as a table
    Frequencysf, 1.8 V, 27 °Csf, 1.62 V, −40 °C
    1 Hz79.77 dB60.13 dB
    10 Hz79.77 dB60.13 dB
    100 Hz79.76 dB60.13 dB
    1 kHz79.17 dB60.12 dB
    10 kHz67.83 dB59.42 dB
    100 kHz48.11 dB47.38 dB
    1 MHz28.11 dB27.61 dB
    10 MHz7.725 dB7.411 dB
    100 MHz-18.93 dB-17 dB
    1 GHz-49.24 dB-45.32 dB

    One corner, two conditions. Only the supply and the temperature change, and the loop gain changes with them.

    Self-Check

    Tick each key point you covered in your answer.

    What Interviewers Listen For (Senior)

    A good senior answer asks for PVT corners crossed with Monte Carlo, and for stability in the worst case. It also asks for start-up transients and proof that the benches work.

    Missed Corner

    "It meets spec at typical with an ideal bias source."

    An ideal bias source hides the biggest process effect. The real reference current moves with process, supply and temperature, so the corners must use it.

    Source ASIC Point lesson E7.6

    Half-Truth

    "All the DC checks pass, so the benches are good."

    DC-level checks never see a missing R_z or a starved bias. A bench is trusted only after it fails known faults.

    Source ASIC Point lesson E8.4

    Wrong Altitude

    Reviews only the schematic numbers.

    Layout adds gradients and parasitics, and system use adds loads and supply noise. A senior review asks how those were covered.

  5. WhyWhy a Bench Must Fail Known FaultsL2 Mid-LevelL3 Senior

    Why should a test bench be shown to fail on purpose before you trust it?

    Reveal the Model Answer

    A bench that passes everything may not be measuring what you think. Insert known faults, such as a missing R_z, a starved bias or a swapped input. A good bench fails each one with a clear reason. In the lesson, DC-level checks missed both a missing R_z and a starved bias.

    TransientThe Final Design and the Two Dynamic Faults
    • final design
    • no R_z
    • bias 5 µA
    -20 mV-15 mV-10 mV-5 mV05 mV10 mV15 mV20 mVv(out) − v(dac)0100 ns200 ns300 ns400 ns500 ns600 ns700 ns800 ns900 nsTime+½ LSB−½ LSB-20 mV-10 mV010 mV20 mVv(out) − v(dac)0200 ns400 ns600 ns800 nsTime+½ LSB−½ LSB
    1. Time 370 nsfinal design = 61.54 µV · no R_z = 17.13 mV · bias 5 µA = 164.3 mV

      70 ns after the fall only the final design has settled for good. Without Rz the output rings until 112.3 ns; the starved copy slews until 176.1 ns.

    Read the plot as a table
    Timefinal designno R_zbias 5 µA
    0 s50.28 µV50.28 µV93.29 µV
    100 ns50.28 µV50.28 µV93.29 µV
    200 ns215.5 µV-3.915 mV-29.01 mV
    300 ns789.4 µV819.1 µV-6.087 mV
    370 ns61.54 µV17.13 mV164.3 mV
    400 ns51.91 µV2.84 mV34.21 mV
    500 ns-1.362 mV-1.29 mV5.276 mV
    600 ns96.08 µV-1.484 mV22.49 mV
    700 ns75.11 µV110.3 µV902.4 µV
    800 ns96.97 µV155.3 µV108.6 µV
    900 ns96.97 µV96.41 µV129.2 µV

    Measured on ngspice 42; clipped at ±20 mV.

    Self-Check

    Tick each key point you covered in your answer.

  6. Prove ItSolve E8.4.1: Break the DAC BufferSimulator-graded

    Show it, don't just say it: this Silicon Lab problem is graded by the simulator.

    Open Problem E8.4.1

  7. What IfChoosing the Faults to Plant in a BenchL2 Mid-LevelL3 Senior

    Which faults would you plant in a DAC output buffer bench, and which check should catch each one?

    Reveal the Model Answer

    Plant faults that a weak bench would miss. Remove R_z: the step settling check at speed must show ringing. Starve the bias: the input-range check at speed must fail near the top code. Swap the inputs: the open-loop sign check must fail. Each check that never fails is measuring nothing useful.

    Self-Check

    Tick each key point you covered in your answer.

    Tool as Truth

    "The DC sweeps pass with every fault, so the faults are harmless."

    DC checks cannot see stability or slew. In the lesson they missed both a missing R_z and a starved bias.

    Source Lesson E8.4 key points

  8. CompareProcess Corners vs Monte CarloL1 JuniorL2 Mid-Level

    Compare process corners with a mismatch Monte Carlo run. Do you need both?

    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.

  9. 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.

    Open Problem E7.6.1

  10. Scale ItWriting a Bias Block SpecificationL3 Senior

    You must hand over a bias block to other designers. How do you write its specification and prove it?

    Reveal the Model Answer

    Write each requirement as one line with a limit and a measurement, set after measuring a reference design at every corner. Use one bench per check. Run a supply sweep and a temperature and supply sweep for the worst case. Add an output sweep for compliance and power-up transients, fast and slow. A constant-g_m reference moves its current on purpose, so specify the tail current as a band.

    The Bias Block, Piece by PieceArrows show where a current or a gate voltage goes. W/L in µm.
    The bias block, piece by pieceArrows show where a current or a gate voltage goes. W/L in µm. component Start-up XS1-XS3: kick atpower-up, then off component Reference Beta multiplier, K = 4,RS a res_high_po: 10µA component PMOS copies XP3 4/1: 10 µA; XP41/1: 2.5 µA component Cascode bias XBC 0.5/1.5 diodesets node c component Reference stack XC1 over XB1, 16/1units: node g output Tail output XC2 over XB2, twounits: 20 µA output PMOS bias output XB3 sinks 10 µA fromXPB 8/1 kickgate b2.5 µA10 µAgate cgate ggate g
    1. Reference and start-up

      The beta multiplier of E4.4 sets 10 µA with a res_high_po resistor (W = 0.35 µm, L = 8.54 µm). The start-up of E4.5 kicks it on at power-up and then switches off.

    2. Send currents

      PMOS devices on the reference's gate line b copy its current: 10 µA for the NMOS reference stack and 2.5 µA for the cascode bias. Currents travel; gate voltages stay local (E4.2).

    3. Wide-swing cascode

      The 2.5 µA diode XBC sets the cascode gate c. The reference stack sets the bottom gate g with its own cascode, so both bottom devices see the same drain voltage (E4.3). The tail is two units of the stack: 20 µA.

    4. PMOS bias output

      One more unit on gate g sinks 10 µA from a diode-connected 8/1 PMOS. Its gate is pbias: any 8/1 PMOS with its source at VDD and this gate voltage sources about 10 µA.

    Read the diagram as text
    • Start-up: XS1-XS3: kick at power-up, then off
    • Reference: Beta multiplier, K = 4, RS a res_high_po: 10 µA
    • PMOS copies: XP3 4/1: 10 µA; XP4 1/1: 2.5 µA
    • Cascode bias: XBC 0.5/1.5 diode sets node c
    • Reference stack: XC1 over XB1, 16/1 units: node g
    • Tail output: XC2 over XB2, two units: 20 µA
    • PMOS bias output: XB3 sinks 10 µA from XPB 8/1
    • Start-up → Reference: kick
    • Reference → PMOS copies: gate b
    • PMOS copies → Cascode bias: 2.5 µA
    • PMOS copies → Reference stack: 10 µA
    • Cascode bias → Tail output: gate c
    • Reference stack → Tail output: gate g
    • Reference stack → PMOS bias output: gate g

    The full netlist is the Try it below. Units are whole devices (m=2 for the tail), so the 2:1 ratio does not depend on drawn widths matching.

    Self-Check

    Tick each key point you covered in your answer.

  11. What IfA DAC Buffer's Input Range at the CornersL2 Mid-LevelL3 Senior

    Asked if the block drives a DAC or ADC

    What if a DAC output buffer passes at typical but a corner moves its input range?

    Reveal the Model Answer

    A corner can push an edge of the input range past the top or bottom code. Then the buffer's tail leaves saturation at that code and the output error jumps. Choose the DAC references inside the buffer's range at speed, not just where the DC error is small. Check them at every corner.

    DC sweepDC Error at Sf, −40 °C: Versions 2 and 3
    • version 2
    • version 3
    -5 mV05 mV10 mV15 mVv(out) − v(in)0.7 V0.8 V0.9 V1.0 V1.1 V1.2 V1.3 VInput+½ LSBcode 63-5 mV05 mV10 mV15 mVv(out) − v(in)0.8 V1.0 V1.2 VInput+½ LSBcode 63
    Read the plot as a table
    Inputversion 2version 3
    700 mV-149.8 µV-88.19 µV
    800 mV13.66 µV37.73 µV
    900 mV48.87 µV64.35 µV
    1 V93.74 µV89.88 µV
    1.1 V630.6 µV211.5 µV
    1.2 V10.28 mV2.951 mV
    1.3 V47.59 mV33.11 mV

    DC error at the sf corner and -40 °C. At the top code, version 2 errs by 9.02 mV and version 3 by 2.42 mV.

    Self-Check

    Tick each key point you covered in your answer.

What People Get Wrong

Common wrong answers and why they fail.

Delivery

Lists simulator commands and tool names instead of decisions.

The interviewer wants the design decisions and the evidence behind them. Tools are the smallest part of the story.

Buzzword Without Mechanism

Says "it met all specs" without saying at which corners.

A spec met at typical only is not met. Name the corners, supply and temperature range, and the Monte Carlo limits.

Wrong Altitude

A senior candidate who cannot name a trade-off they made.

Every analog block trades power, area, speed, noise and accuracy. Not naming one suggests the decisions were made by someone else.

Learn It, Then Prove It