90+ free CMOS VLSI design questions on MOSFETs, the inverter, static and dynamic logic, sizing and logical effort, SRAM, power, wires and latch-up. Includes 18 worked answers, 8 solved numericals, a gate sized transistor by transistor and an inverter delay and power calculator.
580 questions with interview-ready answers and tips, 20 SPICE and switch-level exercises, 25 solved numericals and 30 silicon debugging scenarios.
CMOS questions check whether you understand what's underneath the gates you design, place or time. They're the core of circuit, memory and standard-cell interviews, and they show up in physical design and STA rounds whenever delay, leakage, wires or corners come up.
A good CMOS answer always connects a device fact to a circuit consequence: PMOS mobility to gate sizing, subthreshold slope to leakage, wire resistance to repeaters. Start with the formula sheet and the calculator, skim the questions by topic, then attempt each worked answer and numerical before opening it.
MOSFET regions and equations, threshold voltage, body effect, the inverter transfer curve, noise margins and why the PMOS is wider. Expect to draw curves on the whiteboard.
NAND, NOR and complex gates at transistor level, sizing for equal drive, logical effort, pass gates, domino logic, flip-flops and the 6T SRAM cell.
Leakage, power gating, wire delay and repeaters, crosstalk, IR drop, latch-up, ESD, ageing, corners and "the silicon doesn't match simulation".
Before any equation, say which region each transistor is in. Most wrong answers apply the saturation formula to a device that's in the linear region.
Series in the NMOS network means parallel in the PMOS network, and the other way round. Saying this while drawing a gate shows method, not memory.
Treat an on transistor as a resistor and everything it drives as a capacitor. t ≈ 0.69RC answers most "which is faster" questions in seconds.
Wider devices add input capacitance, lower VT adds leakage, dynamic logic adds noise risk and clock power. Name the cost before you're asked.
Slow-slow for setup, fast-fast for hold and leakage, and watch for temperature inversion at low voltage. A number without a corner is half an answer.
Well taps for latch-up, wide wires for electromigration, matched devices for analog, diffusion sharing for speed. Silicon is drawn, not just simulated.
Long-channel equations for intuition, plus the corrections that matter in modern nodes.
Treat each transistor as a resistor driving the load. Change the numbers and the results update.
"Why not just use domino everywhere?" Knowing what each style costs answers that question and most of its follow-ups.
| STYLE | STATIC POWER | OUTPUT SWING | SPEED | WATCH OUT FOR | WHERE IT'S USED |
|---|---|---|---|---|---|
| Static CMOS | Leakage only | Full rail | Baseline | Tall PMOS stacks make wide NORs slow | Almost all standard-cell logic |
| Pseudo-NMOS | Yes, whenever output is low | VOL > 0 (ratioed) | Fast for wide NOR | Static current, VOL sizing | Wide decoders, PLAs |
| NMOS pass transistor | Leakage only | High level drops to VDD − VT | Good, compact | Weak 1, needs a level restorer | Muxes in older designs |
| Transmission gate | Leakage only | Full rail | Good for short chains | No drive: buffer every 3–4 gates | Muxes, XORs, latches, flip-flops |
| Domino | Leakage only | Full rail | Fastest | Charge sharing, leakage, noise, clock power; non-inverting only | High-speed datapaths, wide ORs |
| DCVSL | Leakage only | Full rail, dual rail | Medium | Ratioed while switching; twice the wiring | Dual-rail logic, XOR-heavy blocks |
A cross-section of what interviewers ask most. The 580-question ebook covers every topic in far more depth, with each question answered.
How the transistor works.
What changes when channels get short.
How it's built and drawn.
The circuit every other one is built on.
From one gate to a whole path.
Styles beyond static CMOS.
Storage at transistor level.
The densest circuits on the chip.
Energy and wires.
Keeping silicon working for years.
580 questions in 27 chapters, all answered: 20 SPICE and switch-level exercises, 25 solved numericals, 30 debugging scenarios, plus DRAM and flash, ESD, ageing, analog blocks and clock circuits.
Answer out loud first, or sketch the circuit, then open it. Each answer shows what a strong reply covers, what the interviewer is listening for and the slip that costs candidates most often.
linear : I_D = µC_ox(W/L)·[(V_GS − V_T)·V_DS − V_DS²/2] saturation : I_D = ½·µC_ox(W/L)·(V_GS − V_T)²·(1 + λ·V_DS)
In short-channel devices, velocity saturation makes the current closer to linear in (VGS − VT), which is why the alpha-power model with α between 1 and 1.5 is used for estimates.
As Vin rises from 0 to VDD, the output stays high (NMOS off), falls steeply around the switching threshold VM where both devices are saturated and the gain is highest, then settles at 0 (PMOS off).
| Input | NMOS | PMOS | Output |
|---|---|---|---|
| Below VTn | Off | Linear | VDD |
| Just above VTn | Saturated | Linear | Starts to fall |
| Around VM | Saturated | Saturated | Steep drop |
| Above VM | Linear | Saturated | Near 0 |
| Above VDD − |VTp| | Linear | Off | 0 |
At either stable input, one transistor is off, so there's no DC path from VDD to ground. The only static current is leakage. Current flows from supply to ground only briefly while the input passes through the middle region.
Holes have roughly half the mobility of electrons in silicon, so a PMOS of the same size conducts about half the current and has about twice the on-resistance. Making it about twice as wide gives the pull-up and pull-down equal strength, so:
It isn't a fixed rule. The exact ratio depends on the process (strain engineering and FinFETs change hole mobility a lot), and for minimum average delay the best ratio is smaller, about the square root of the mobility ratio, because a wider PMOS also loads the previous stage.
A NAND puts its NMOS in series and its PMOS in parallel. A NOR does the opposite: the slow PMOS go in series. For equal drive to a unit inverter (NMOS 1, PMOS 2):
| Gate | NMOS width | PMOS width | Input capacitance | Logical effort |
|---|---|---|---|---|
| NAND2 | 2 (series) | 2 (parallel) | 4 | 4/3 |
| NOR2 | 1 (parallel) | 4 (series) | 5 | 5/3 |
The NOR is larger, has more input capacitance and is slower for the same drive, and the gap grows quickly with more inputs (NOR3 needs PMOS of width 6). That's why libraries and synthesis favour NAND-based logic.
Leakage rises steeply with temperature and with lower VT, so in advanced nodes it can be a third or more of total power, and it's the dominant term in standby.
In bulk CMOS, the PMOS source, n-well, p-substrate and NMOS source form a parasitic PNPN structure: a vertical PNP and a lateral NPN connected as a thyristor. If a current injection (an I/O overshoot, an ESD event, a supply glitch) turns one transistor on, its current flows through the well or substrate resistance and turns the other on. If the loop gain exceeds one, the structure latches into a low-resistance path from VDD to ground, which can only be cleared by removing power and may destroy the chip.
Take the unit inverter as NMOS width 1 and PMOS width 2. The worst-case pull-down and pull-up paths of each gate must match those resistances.
The general results are (n + 2)/3 for an n-input NAND and (2n + 1)/3 for an n-input NOR. The NOR's logical effort grows twice as fast with fan-in, which is the quantitative version of "avoid wide NORs".
For a chain of inverters, path effort F = H = 64. The delay is lowest when every stage has the same effort, and the best effort per stage is about 4.
N = log₄(64) = 3 stages f = 64^(1/3) = 4 per stage Sizes: 1 → 4 → 16 (each stage drives 4× its own input) Delay D = N·f + N·p_inv = 3 × 4 + 3 × 1 = 15 τ
A single inverter driving the load directly would take about 64 + 1 = 65 τ. If the output must not be inverted, compare 2 and 4 stages: 2 stages (f = 8) give 18 τ, 4 stages (f ≈ 2.83) give about 15.3 τ, so 4 is the better even-stage choice.
A dynamic gate has a clocked PMOS that precharges the dynamic node high while the clock is low, and an NMOS network plus a clocked foot transistor that conditionally discharges it while the clock is high. Domino adds a static inverter after the dynamic node, so outputs start low and can only rise during evaluation. That monotonic behaviour lets domino gates drive each other in a chain within one clock phase.
The cell is two cross-coupled inverters (pull-ups PU, pull-downs PD) and two NMOS access transistors (PG) to the bit lines.
So the strength order is PD > PG > PU. At low voltage and with random variation these margins shrink, which is why assists (word-line underdrive, negative bit line) and 8T cells with a separate read port exist.
A wire's resistance and capacitance both grow linearly with length L. Its delay is roughly proportional to R×C, so it grows with L2. For a distributed RC line, the Elmore delay is about 0.38·r·c·L2, where r and c are per unit length.
Splitting the wire into k segments with a repeater after each one makes each segment's delay proportional to (L/k)2, and the total is k times that plus k repeater delays. The wire part falls as L2/k, so for long wires the total grows only linearly with L once the repeater count is optimised.
With the gate at VDD, an NMOS passing a 1 charges its output until VGS falls to VT, so the output stops at about VDD − VT, made worse by the body effect. Passing a 0, VGS stays at VDD and the output reaches 0. A PMOS is the opposite: a strong 1, but a 0 that stops at |VTp|.
A transmission gate puts an NMOS and a PMOS in parallel with complementary gate signals, so the pair passes both levels at full swing. It's used in multiplexers, XOR gates, latches and flip-flops.
This is a read-stability failure. During a read, the 0 node rises against the pull-down transistor. At low VDD the read static noise margin shrinks, and random VT mismatch pushes the weakest cells, far out in the tail of millions, past the point where the bump flips them.
Suspect charge sharing. Suppose the top transistor of the pull-down stack turns on while the lower ones stay off. The dynamic node, precharged high, now shares its charge with the internal node below, which may have been left at 0 from an earlier evaluation. The node voltage drops to VDD·Cdyn/(Cdyn + Cint). If that falls below the inverter's switching threshold, the output switches falsely even though the logic says it shouldn't.
It depends on the history of the internal nodes, which is why it appears only for certain sequences.
Measure leakage against temperature and supply voltage. Each mechanism has a different signature, which separates them.
Check the extracted netlist node by node against the schematic to find where the capacitance was added, then fix the layout: share diffusion, shorten critical nets, widen or add parallel vias, and move away from well edges where possible.
This is simultaneous switching noise, or ground bounce. When many drivers switch together, the total current changes very quickly through the package and bond-wire inductance, creating V = L·di/dt on the internal supply and ground. Quiet outputs ride on the bounced ground and look like glitches to the receiver, and internal logic sees a temporarily reduced supply.
When the header switches turn on, the block's discharged internal capacitance charges all at once. The resulting rush current pulls the shared supply down, and the always-on neighbour sees a droop deep enough to fail timing or corrupt state.
Quick estimates come up in every circuit interview. Work each on paper, then tap to check.
V_ov = 1.0 − 0.4 = 0.6 V
I_D = ½ × 250 µA/V² × 6 × 0.6²
= 750 µA/V² × 0.36 = 270 µANote: state the region first. This equation only applies if VDS ≥ 0.6 V.
V_ov = 0.5 V > V_DS = 0.2 V → linear region
I_D = 500 µA/V² × [0.5 × 0.2 − 0.2²/2]
= 500 µA/V² × (0.10 − 0.02) = 40 µANote: the saturation formula would give 62.5 µA, an overestimate of more than 50%.
√(0.7 + 0.6) = √1.3 = 1.140 √0.7 = 0.837 ΔV_T = 0.5 × (1.140 − 0.837) = 0.152 V V_T = 0.45 + 0.152 ≈ 0.60 V
Note: this is why the upper transistors in a series NMOS stack are slower: their sources sit above ground.
Decades = 150 mV / 75 mV = 2 Increase = 10² = 100×
Note: this is why low-VT cells are used only on critical paths, and why VT can't simply keep scaling down with VDD.
r = √0.25 = 0.5
V_M = (0.3 + 0.5 × (1.0 − 0.3)) / (1 + 0.5)
= (0.3 + 0.35) / 1.5 ≈ 0.43 VNote: a PMOS four times weaker moves VM only about 70 mV below VDD/2. The VTC is robust to sizing.
G = (4/3) × (5/3) × 1 = 20/9 ≈ 2.22 F = G × H = 2.22 × 27 = 60 f = 60^(1/3) ≈ 3.91 per stage P = 2 + 2 + 1 = 5 D = 3 × 3.91 + 5 ≈ 16.7 τ
Note: stage effort near 4 confirms that three stages is about right for this path.
V = 1.0 × 20 / (20 + 5) = 0.80 V Droop = 200 mV
Note: if the domino inverter switches near 0.5 V this is safe, but noise or leakage on top of it may not be. Precharging the internal node removes the problem.
R = 0.15 × 1500 = 225 Ω C = 0.2 fF × 1500 = 300 fF t ≈ 0.38 × 225 × 300 fF ≈ 25.7 ps At 3 mm: R and C both double → t ≈ 103 ps (4×)
Note: the quadratic growth is the reason long routes get repeaters.
Y = (A·B + C)'. A favourite whiteboard question, because it tests duality, sizing and logical effort in one go. Target: the same drive as a unit inverter with NMOS width 1 and PMOS width 2.
Because the worst case is a single one of them conducting, in series with C. Each must be strong enough on its own to give, with C, the resistance of the inverter's PMOS. Two devices in series each need twice the width: 2 × 2 = 4.
C, for two reasons: it has the lowest logical effort (5/3), and in the pull-down it acts alone instead of through a two-transistor stack. If the late signal must go to A or B, put it on the transistor nearest the output so the lower node is already discharged when it arrives.
AOI21 uses 6 transistors in one stage. An AND2 (NAND2 plus inverter) feeding a NOR2 needs 4 + 2 + 4 = 10 transistors in three stages, with more delay and more capacitance. Complex gates like AOI and OAI are why standard-cell libraries are so effective.
Yes: widen the NMOS and keep the PMOS small. That speeds up the falling edge and lowers input capacitance at the cost of a slower rising edge. Skewed gates are common in domino logic, where only one edge is critical. The ebook has full chapters on gate sizing and logical effort.
Screening calls and written tests often start like this. Answer each in a few seconds, then tap to check.
No. It stops at about VDD − VT.
Six.
4/3.
5/3.
About 4.
About 60 mV/decade.
Leakage only.
α·C·V2·f.
One quarter.
Electrons, so NMOS is stronger.
Quadratic.
Droop on a dynamic node from leakage and charge sharing.
To make outputs monotonically rising, so gates can cascade.
Read stability.
Fast process, high voltage.
A parasitic PNPN thyristor turning on.
NBTI.
1/√(W·L).
1 / (2·N·tpd).
Leakage.
It has no DC path in a stable state, but leakage is a large share of power in modern nodes and dominates in standby.
Check VDS against VGS − VT before choosing a formula. The wrong region can be off by 50% or more.
Only series transistors scale with stack height. Parallel transistors each need only the unit width.
Diffusion and wiring capacitance, well taps, matching and electromigration all change the answer. Mention them.
Setup, hold, leakage and power each have different worst corners, and low-voltage designs can be slowest when cold.
Wider devices, lower VT and dynamic logic all buy speed with power, area or noise margin. Say what you're giving up.
Circuit design, memory, standard-cell and analog-layout roles ask the most, but physical design, STA and DFT rounds also use CMOS to check that you understand delay, leakage, wires, corners and reliability. Most core-company written tests include CMOS basics.
Enough to explain the three operating regions with their equations, threshold voltage and body effect, velocity saturation, subthreshold conduction and the main short-channel effects. Interviews rarely go into band diagrams unless the role is device or process engineering; for that, see our semiconductor questions.
For circuit and memory roles, being able to describe how you would simulate an inverter's transfer curve, delay or a ring oscillator helps a lot. For PD and STA roles it is a bonus, not a requirement.
The MOSFET regions, the CMOS inverter transfer curve and switching threshold, NAND and NOR at transistor level, sizing for equal drive, RC delay, the three kinds of power, and latch-up. Then logical effort, dynamic logic and the 6T SRAM cell.
Yes, in circuit and high-speed design rounds. Expect to compute the logical effort of NAND and NOR gates, size a buffer chain for a large load, or estimate a path delay. The rule of about four per stage is the key result.
Digital electronics stays at the gate and block level: Boolean logic, flip-flops, counters, FSMs and timing. CMOS goes one level down, into transistors, sizing, delay, power, memory cells, wires and reliability. Our digital electronics questions cover the level above.
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