90+ free analog and mixed-signal questions on the MOSFET and gm/ID, single-stage amplifiers, current mirrors, differential pairs, feedback and stability, op-amps, noise, references and regulators, data converters, PLLs, mismatch and layout. Includes 18 worked answers, 8 solved numericals, Miller compensation explained pole by pole and a gm, ro and gain calculator.
580 questions with interview-ready answers and tips, 25 solved numericals, 20 SPICE circuit exercises and 30 debugging scenarios.
Analog interviews are conversations at a whiteboard. The interviewer draws a circuit and asks for its gain, its output resistance, its poles, its noise, and what you would change to meet a specification. There are fewer candidates than for digital roles, but the bar is high: you're expected to reason, estimate and explain trade-offs, not recite.
Strong answers draw the small-signal picture first, give a quick hand estimate with its approximation stated, and finish with what it means for the design. Start with the equations and the calculator, skim the 50 questions by topic, then try each worked answer and numerical before you open it. Device background is on the CMOS and semiconductor devices pages.
Small-signal models, the four basic stages, mirrors and the differential pair. Expect “what's the gain?”, “what's the output resistance?” and quick numericals.
Op-amps, compensation, noise, references and regulators: how you'd size a block to a specification and what you'd give up for speed, gain, noise or power.
Mismatch, layout, parasitics, PVT corners, Monte Carlo and debugging: “it passed every corner, so why does silicon ring?”
Sketch the small-signal picture and name which impedance each node sees before writing any formula.
Give the quick answer with its approximation stated (“assuming ro >> RD”), then add the correction if asked.
Every analog answer gives something up: gain for headroom, noise for power, speed for matching. Say which.
Hertz versus radians per second, and milli- versus micro-, cause most wrong numericals. Do a sanity check before you finish.
Say which corner is worst for the specification you're discussing: slow/hot for speed, fast/cold for stability, and mismatch for offset.
Mention matching, parasitics and coupling where they matter. It shows you've seen a block go to silicon.
Simple square-law and small-signal forms, good for hand estimates. Real designs check them against simulation and gm/ID lookup tables.
Square-law estimate for one transistor in saturation, plus what it means for a single-stage OTA driving CL.
“Which stage would you use here, and why?” comes up in almost every analog round. Learn each one's gain, impedances and catch.
| STAGE | VOLTAGE GAIN | INPUT RESISTANCE | OUTPUT RESISTANCE | TYPICAL USE | THE CATCH |
|---|---|---|---|---|---|
| Common source | −gmRout (high) | Very high (gate) | ro ∥ Rload (high) | Main gain stage | Miller multiplies Cgd |
| CS with degeneration | −Rload/(1/gm + RS) | Very high | Raised by (1 + gmRS) | Linear, predictable gain | Gain and headroom traded for linearity |
| Source follower | < 1, about gm/(gm + gmb) | Very high | ≈ 1/gm (low) | Buffer, level shift | Body effect and lost swing |
| Common gate | +gmRout | ≈ 1/gm (low) | High | Current input, wideband, cascode | Low input impedance loads the source |
| Cascode | −gm1(gm2ro2ro1 ∥ Rload) | Very high | ≈ gmro2 (very high) | High-gain single stage | Headroom and swing |
| Differential pair | gmRout (differential) | Very high | As the load | Input stage, rejects common mode | Needs a tail and, if fully differential, CMFB |
A sample of what analog interviewers ask most. The 580-question ebook goes much deeper on each topic and answers every question.
How analog designers see one transistor.
The four stages every circuit is built from.
Copying currents accurately across a chip.
The input stage of almost every amplifier.
Poles, zeros and why amplifiers ring.
Turning stages into a usable amplifier.
Where the floor of every analog spec comes from.
Clean, stable voltages for everything else.
Where analog meets digital.
The parts that make silicon match simulation.
580 questions in 27 chapters, all answered: 25 solved numericals, 20 SPICE exercises, 30 debugging scenarios, plus an annotated sign-off report and a 40-question rapid-fire round.
Sketch the circuit and say your answer aloud first, then open it. Each answer shows what a strong reply covers, what the interviewer is listening for, and the slip that costs candidates most.
A common-source stage with an ideal current-source load has gain A = gmro, because the only resistance at the output is the device's own ro. Any real load only adds parallel resistance and lowers it. So gmro, the intrinsic gain, is the most one transistor can give.
With the square law, gm = 2ID/Vov and ro ≈ 1/(λID), so the current cancels and gmro ≈ 2/(λVov). Short channels have a larger λ (channel-length modulation, DIBL), so intrinsic gain drops from around 100 in older nodes to around 10 in the most advanced ones.
Designers fight back with longer channels in analog devices, cascodes, which multiply output resistance by another gmro, and more gain stages.
gm/ID is transconductance efficiency: how much gm you get for each unit of current. It moves from about 25–30 S/A in weak inversion down to 5–10 S/A in strong inversion, and it maps one-to-one onto the inversion level.
The flow is: pick gm from the specification (for example GBW × 2πCL), pick gm/ID from the trade-off, which gives ID, then read the current density ID/W from simulated lookup tables, which gives W. The method works in every inversion region, where square-law hand sizing breaks down.
A cascode device stacked on top of a transistor holds the lower device's drain at an almost constant voltage. Any change at the output appears across the cascode device, which has its own gm working against it as local feedback. The result is Rout ≈ gm2ro2ro1, roughly gmro times higher than one device.
The cost:
The costs are about twice the area and current for the same noise, and, for fully differential outputs, a common-mode feedback loop. Nearly every precision input stage accepts those costs.
With open-loop gain A and feedback factor β, the closed-loop gain is A/(1 + Aβ) ≈ 1/β when the loop gain Aβ is large. You give up a factor of (1 + Aβ) in gain, and in return many properties improve by that same factor:
The price is the risk of instability, because the loop adds phase shift, so every feedback amplifier needs a phase-margin check.
Flicker (1/f) noise comes from carriers being trapped and released by defects at the gate oxide interface. Its power spectral density rises as 1/f and scales roughly as 1/(WLCox), so it dominates at low frequency, below the noise corner where it meets the thermal floor.
A single-stage OTA's GBW is gm1/(2πCL), so the required input gm comes first:
gm1 = 2π · GBW · CL = 2π × 50 MHz × 2 pF ≈ 0.63 mS pick gm/ID = 15 S/A (moderate inversion: good noise and offset) ID per side = 0.63 mS / 15 ≈ 42 µA -> tail ≈ 84 µA, call it 90 µA W = ID / (ID/W from the lookup table at your chosen L)
Then check the rest: slew rate Itail/CL = 45 V/µs; mirror poles well above 50 MHz, so give the load mirror a lower gm/ID; gain gm1(ro2 ∥ ro4) still meets the specification; headroom for the tail at the lowest input common-mode. Add margin, because GBW falls at the slow corner.
A common hand-design rule targets about 60° phase margin:
Combining the first two: gm2/CL ≥ 2.2 gm1/CC. With gm2 ≈ 10 gm1, that gives CC ≥ 0.22 CL.
A resistor RZ in series with CC moves the zero to 1/(CC(1/gm2 − RZ)). RZ = 1/gm2 pushes it to infinity; a slightly larger RZ moves it into the left half-plane, where it can cancel p2 and buy back phase. Make RZ from a triode MOSFET biased to track gm2 over corners, then confirm the margin with a loop-gain simulation at every corner and load.
Check it with a common-mode step, a loop-gain simulation of the CM loop and large differential swings, where some sensing schemes stop being linear.
Pass device: size the PMOS so that at maximum load and the slowest corner it still conducts the current with only the dropout voltage across it. That's a very wide device with a large gate capacitance, which the error amplifier must drive.
Compensation depends on where the dominant pole sits:
Then check stability over the full load range and corners, the load-step response, PSRR across frequency and the start-up behaviour.
Loop bandwidth decides which noise source wins:
So the best bandwidth is roughly where the two contributions cross: wider if the VCO is noisy, as ring oscillators are; narrower if the reference is noisy or reference spurs must be filtered. A practical ceiling is about one-tenth of the reference frequency, because the loop is sampled once per reference cycle, and continuous-time analysis stops being valid near that point.
Then set the loop-filter zero and poles for about 60° of phase margin and check lock time, which shortens with wider bandwidth.
Then confirm with post-layout extraction and Monte Carlo on the extracted netlist.
Loop gain is T = Aβ. In a gain-of-10 configuration β = 0.1, so the loop crosses unity at about GBW/10, well below the op-amp's second pole. As a buffer, β = 1: the crossover moves up to the full GBW, closer to the non-dominant pole, and phase margin falls.
The output device's VDS follows the supply while the reference device's stays fixed, so channel-length modulation changes the current. A 10% change for a 0.2 V step means ro = ΔV/ΔI = 0.2 V/(0.1·I) = 2 V/I, an Early voltage of only about 2 V, which points to a very short channel.
Also check that the reference current itself doesn't come from a resistor to the supply, which would move it directly.
Unchanged DC gain means the operating point is fine; the loss is in capacitance or resistance on high-impedance nodes.
Compare node capacitances between schematic and extracted netlists, back-annotate the worst ones into the schematic and rerun the AC analysis. Fix it by shortening and thinning critical routes, adding fingers, or moving a sensitive wire away from large nets.
A frequency-independent error points to a static cause, usually the op-amp's finite DC gain. With feedback factor β = Cf/(Cs + Cf + Cp), the closed-loop gain error is about 1/(Aβ).
Cs = 4 pF, Cf = 1 pF, Cp = 0.5 pF -> β = 1/5.5 ≈ 0.18 error 0.5% -> Aβ ≈ 200 -> A ≈ 1100 (61 dB)
A settling error, by contrast, gets worse as the clock speeds up.
The first-order design cancels the linear part of VBE's temperature dependence with a PTAT term. But VBE also has a curvature term (roughly T·ln T), which a linear PTAT term can't cancel. The output therefore has a parabolic shape, highest near the temperature where the slopes cancel, falling at both ends. A few millivolts over the industrial range is typical for an uncorrected bandgap.
Errors that grow with input frequency come from the sampling instant or the sampling switch:
Separate them by looking at the FFT: a raised noise floor points to jitter, harmonics point to the switch or front-end. Then measure with a cleaner clock source.
Solve each on paper first. Write the formula before the numbers, keep the units, and say what the answer means for the design.
Rin = (ro + RD) / (1 + gm·ro)
= (50k + 20k) / (1 + 2m × 50k)
= 70k / 101 ≈ 693 Ω
With RD → ∞ (ideal current-source load):
Rin → very largeNote: the textbook “1/gm” (500 Ω here) only holds when RD is small compared with ro. With a high-impedance load, a common-gate stage stops looking like a low-impedance input.
ID = 100 µA per side gm = 2·ID / Vov = 2 × 100 µA / 0.2 V = 1 mS Adm = gm·RD = 1 mS × 10 kΩ = 10 (20 dB) Full steering at |Vid| = √2 · Vov ≈ 0.28 V
Note: the pair is only reasonably linear well inside that limit. Raising Vov widens the linear range but lowers gm for the same current.
ΔRD = 1% × 10 kΩ = 100 Ω
Acm→dm ≈ ΔRD / (1/gm + 2·RSS)
= 100 / (1k + 200k) ≈ 5.0 × 10^-4
Adm = 10
CMRR = 10 / 5.0e-4 ≈ 20,000 ≈ 86 dBNote: CMRR is set by mismatch and tail impedance together. A cascoded tail raises RSS; at high frequency the tail's capacitance lowers it, so CMRR falls with frequency.
Max slope of Vp·sin(2πft) = 2πf·Vp Set 2πf·Vp = SR f = SR / (2π·Vp) = 10 V/µs / (2π × 1 V) ≈ 1.6 MHz
Note: above this the output turns into a triangle wave no matter what the small-signal bandwidth says. Large-signal and small-signal limits are separate specifications.
Noise density = √(4kTR) = √(4 × 1.38e-23 × 300 × 1000)
≈ 4.07 nV/√Hz
ENBW (single pole) = (π/2)·f3dB ≈ 1.57 MHz
Vrms = 4.07 nV × √(1.57e6) ≈ 5.1 µVNote: integrating to f3dB alone underestimates the noise by about 20%. A single-pole roll-off lets noise through well past the corner.
MSBs: 2^4 − 1 = 15 unit elements, each worth 64 LSB LSBs: 6 binary-weighted elements (1, 2, 4 ... 32) Total: 21 elements (versus 1023 fully thermometer) 511 = 7×64 + 63 -> 512 = 8×64 + 0 1 MSB element turns on, 6 LSB elements turn off: 7 switch (A fully binary DAC switches all 10 bits here)
Note: fewer elements switching at major transitions means smaller glitches and better DNL, which is why the MSBs are thermometer-coded.
Per cycle: Vout[n] = Vout[n−1] + (Cs/Cf)·Vin
Cs/Cf = 0.25
For f << fs: ωu ≈ (Cs/Cf)·fs
fu = 0.25 × 1 MHz / 2π ≈ 40 kHzNote: the integrator's time constant is set by a capacitor ratio and the clock, both accurate on chip, rather than by an absolute RC product that can vary by ±20%.
TC = (Vmax − Vmin) / (Vnom × ΔT) = 4 mV / (1.2 V × 165 °C) ≈ 20 ppm/°C
Note: the box method describes the worst case across the range, not the slope at room temperature. A curved bandgap can have zero slope at 27 °C and still score 20 ppm/°C.
“How do you compensate a two-stage op-amp?” is among the most common analog questions. Here is the reasoning interviewers want to hear, in order.
For a two-stage op-amp, a larger CL lowers the second pole (gm2/CL), which reduces phase margin. That's the opposite of a single-stage OTA, where the output node is dominant and a larger CL makes it more stable.
During a large step the first stage delivers its full tail current into CC, so the internal slew rate is about Itail/CC. A larger CC buys phase margin and costs both GBW and slew rate unless the current goes up too. The output stage must also be able to charge CL that fast.
It raises the gain like a normal zero but adds phase lag like a pole, the worst of both. It comes from CC feeding the signal forward around the second stage. A series resistor near 1/gm2, or a buffer in the compensation path, removes it.
Break the loop without disturbing the DC operating point, for example with a stability analysis that injects a test signal (Middlebrook or Tian's method), and read the phase of the loop gain where its magnitude crosses 0 dB. The Analog VLSI ebook covers compensation, verification and the sign-off report in depth.
Phone screens and written tests love these. Answer each in a few seconds, then tap to check.
About 1 MHz.
60 dB.
About −6 dB.
−45°.
About −84°.
It adds phase lead.
The source follower, about 1/gm.
The common-gate stage.
−40 dB per decade.
How much supply ripple is rejected relative to the signal gain, in dB.
Two non-overlapping phases.
A missing code.
Deviation of the transfer curve from a straight line.
Offset and 1/f noise, by modulating them out of the signal band.
So every active finger sees the same surroundings.
Global shifts all devices together (corners); local makes neighbours differ (mismatch).
V/√Hz.
The voltage left across a device to keep it in saturation.
Slewing first, then linear (small-signal) settling.
Collects stray substrate carriers: isolation and latch-up protection.
A design with perfect gain on paper fails if the stacked devices don't all stay in saturation at the lowest supply and the extreme input common-mode. Count Vovs before quoting gain.
Interviewers expect you to say what happens at slow/hot and fast/cold, and with Monte Carlo mismatch. Margins exist for the corners, not for typical.
GBW, phase margin and slew rate all depend on the load capacitance. An answer without CL is incomplete.
Mixing hertz and radians per second, or millisiemens and siemens, is the most common way to get a numerical wrong by a factor of 6 or 1000.
Jumping to formulas without naming which device sees which impedance leads to wrong answers on unfamiliar circuits. Draw it first, then calculate.
Matching, parasitics and coupling decide whether an analog block works. Designers are expected to specify and review layout, not just hand over a schematic.
The MOSFET in saturation and its small-signal model, the common-source stage with different loads, current mirrors, the differential pair, frequency response and the Miller effect, feedback and phase margin, and basic op-amp structures. Then noise, references and data converters.
Many analog teams prefer an M.Tech or MS because the work builds on deep circuit coursework and a thesis project, but strong B.Tech candidates with a solid project, simulation experience and good hand analysis do get hired, especially into layout and characterisation roles first.
Any SPICE you can run regularly. LTspice and ngspice are free, and Xschem with ngspice works with open PDKs such as SKY130 if you want real transistor models. What matters is checking your hand calculations against simulation.
Mostly algebra and simple complex-frequency reasoning: small-signal models, poles and zeros, decibels and logarithms, and noise integration. Interviewers want quick estimates with stated approximations more than long derivations.
Parts of it are being automated: sizing optimisers, layout generators and AI-assisted tools are improving quickly. But specifications, architecture choices, trade-offs and silicon debug still depend on engineers who understand circuits, so the skills interviewers test remain in demand.
The CMOS round covers devices, digital gates, power, SRAM and reliability. The analog round assumes that background (see the CMOS questions) and asks you to analyse and design circuits that process continuous signals: gain, bandwidth, stability, noise and matching.
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