{
 "schema": "learn-deck/1",
 "module": "asic-analog",
 "deck": "asic-analog",
 "title": "Analog CMOS design",
 "lang": "en",
 "cards": [
  {
   "id": "ic:vth-def",
   "type": "flash",
   "q": "What sets the MOSFET threshold voltage VTH?",
   "a": "VTH = ΦMS + 2ΦF + Qdep/Cox: work-function difference, surface potential at inversion, and depletion charge over oxide capacitance.",
   "ex": "Razavi ch. 2. Adjusted in fabrication by channel implant. Cox ≈ 17.25 fF/µm² for tox = 20 Å.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:triode-eq",
   "type": "flash",
   "q": "Drain current in the triode region?",
   "a": "ID = µnCox (W/L)[(VGS − VTH)VDS − VDS²/2], valid for VDS ≤ VGS − VTH.",
   "ex": "Peak of each parabola is at VDS = VGS − VTH, the overdrive.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:ron",
   "type": "flash",
   "q": "On-resistance of a MOSFET in deep triode?",
   "a": "Ron = 1/[µnCox (W/L)(VGS − VTH)], for VDS much smaller than 2(VGS − VTH).",
   "ex": "A voltage-controlled resistor; also equals 1/gm of the same device in saturation.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:sat-eq",
   "type": "flash",
   "q": "Square-law drain current in saturation?",
   "a": "ID = ½ µnCox (W/L)(VGS − VTH)², for VDS ≥ VGS − VTH (pinch-off).",
   "ex": "VD,sat = VGS − VTH: the larger the overdrive, the less headroom for swing.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:region-rule",
   "type": "mcq",
   "q": "An NFET is saturated when...",
   "choices": [
    "VG − VD < VTH",
    "VDS < VGS − VTH",
    "VGS < VTH",
    "VD < VS"
   ],
   "a": 0,
   "ex": "Pinch-off view: the gate–drain difference cannot sustain inversion. Works without knowing VS; PFET: VD − VG < |VTHP|.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:gm-forms",
   "type": "flash",
   "q": "Three forms of gm in saturation?",
   "a": "gm = µnCox (W/L)(VGS − VTH) = √(2µnCox (W/L) ID) = 2ID/(VGS − VTH).",
   "ex": "Rises with overdrive at fixed W/L; falls with overdrive at fixed ID.",
   "tags": [
    "mos",
    "smallsignal"
   ]
  },
  {
   "id": "ic:pmos-mobility",
   "type": "mcq",
   "q": "Relative to NMOS, PMOS µpCox is about...",
   "choices": [
    "Equal",
    "Double",
    "One tenth",
    "Half"
   ],
   "a": 3,
   "ex": "Lower current drive and gm; also lower rO. Use NFETs where possible except when flicker noise matters.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:body-effect",
   "type": "flash",
   "q": "Body effect formula and meaning?",
   "a": "VTH = VTH0 + γ(√(2ΦF + VSB) − √(2ΦF)), γ ≈ 0.3–0.4 V^0.5: raising VSB raises VTH.",
   "ex": "Forward-biasing the bulk lowers VTH: easy for PFETs in their own n-well.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:clm",
   "type": "flash",
   "q": "Channel-length modulation?",
   "a": "ID ≈ ½ µnCox (W/L)(VGS − VTH)²(1 + λVDS); λ ∝ 1/L, so long devices make better current sources.",
   "ex": "Doubling L at fixed W and overdrive cuts the ID–VDS slope by 4; at fixed ID by 2.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:ro",
   "type": "flash",
   "q": "Small-signal output resistance rO?",
   "a": "rO = 1/(λ ID) approximately; the slope of ID vs VDS in saturation.",
   "ex": "rO limits the maximum gain of most amplifiers.",
   "tags": [
    "mos",
    "smallsignal"
   ]
  },
  {
   "id": "ic:subthreshold-slope",
   "type": "mcq",
   "q": "Subthreshold: how much must VGS drop for ID to fall one decade?",
   "choices": [
    "About 8 mV",
    "About 80 mV",
    "About 300 mV",
    "About 1 V"
   ],
   "a": 1,
   "ex": "ID = I0 exp(VGS/(ξVT)), ξ ≈ 1.5. A 0.3 V VTH leaves ~5600× off-current ratio: leakage in big memories.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:subthreshold-gm",
   "type": "flash",
   "q": "gm in weak inversion?",
   "a": "gm = ID/(ξVT): maximum gm per current, still below a bipolar's IC/VT. Strong-to-weak transition near overdrive 2ξVT ≈ 80 mV.",
   "ex": "Great for low-power biosignal front ends, poor for speed.",
   "tags": [
    "mos",
    "design"
   ]
  },
  {
   "id": "ic:gmb",
   "type": "flash",
   "q": "Bulk transconductance gmb?",
   "a": "gmb = gm γ/(2√(2ΦF + VSB)) = η gm, η ≈ 0.25. Bulk acts as a second gate with the same polarity.",
   "ex": "Falls as VSB grows.",
   "tags": [
    "smallsignal"
   ]
  },
  {
   "id": "ic:cgs-sat",
   "type": "flash",
   "q": "Gate capacitances in saturation?",
   "a": "CGS = (2/3) W L Cox + W Cov; CGD ≈ W Cov (overlap only). CGB negligible: the channel shields the bulk.",
   "ex": "Triode: CGS = CGD = ½WLCox + WCov. Off: only overlaps, plus CGB through depletion.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:junction-cap",
   "type": "flash",
   "q": "Source/drain junction capacitance model?",
   "a": "Cj (per area) × area + Cjsw (per perimeter) × perimeter, each ∝ 1/(1 + VR/ΦB)^m, m ≈ 0.3–0.4.",
   "ex": "Folding a wide device halves drain junction area and cuts gate resistance by 4.",
   "tags": [
    "mos",
    "layout"
   ]
  },
  {
   "id": "ic:ft",
   "type": "flash",
   "q": "Transit frequency fT of a MOSFET?",
   "a": "fT = gm/[2π(CGS + CGD)] ≈ µn(VGS − VTH)/(2πL²) by square law.",
   "ex": "Lower supplies force lower overdrive, hence lower speed unless L shrinks.",
   "tags": [
    "mos",
    "design"
   ]
  },
  {
   "id": "ic:pmos-model-same",
   "type": "tf",
   "q": "The small-signal model of a PMOS is identical to that of an NMOS.",
   "a": true,
   "ex": "Razavi: drawing PMOS source-up makes the dependent source look inverted, but it is not.",
   "tags": [
    "smallsignal"
   ]
  },
  {
   "id": "ic:intrinsic-gain",
   "type": "flash",
   "q": "Intrinsic gain of a transistor and its typical nanometer value?",
   "a": "gm rO, the most a single device can give; roughly 5 to 10 for short-channel devices.",
   "ex": "gm rO ∝ √(W L/ID): longer, wider, less current.",
   "tags": [
    "smallsignal",
    "design"
   ]
  },
  {
   "id": "ic:finfet-width",
   "type": "mcq",
   "q": "Effective channel width of a FinFET?",
   "choices": [
    "WF only",
    "WF + 2HF, in fixed increments",
    "HF only",
    "Any continuous value"
   ],
   "a": 1,
   "ex": "WF ≈ 6 nm, HF ≈ 50 nm: widths come in ~100 nm steps by adding fins.",
   "tags": [
    "mos"
   ]
  },
  {
   "id": "ic:cs-gain",
   "type": "flash",
   "q": "Gain of a common-source stage with resistive load?",
   "a": "Av = −gm RD (λ = 0); with rO: Av = −gm (rO || RD). Input at gate, output at drain.",
   "ex": "Av = −√(2µnCox (W/L)/ID)·VRD: trades gain against swing, bandwidth and capacitance.",
   "tags": [
    "cs"
   ]
  },
  {
   "id": "ic:cs-diode",
   "type": "flash",
   "q": "CS stage with a diode-connected NMOS load: gain?",
   "a": "Av ≈ −√((W/L)1/(W/L)2)·1/(1 + η): set by ratios, so nearly linear and bias-independent.",
   "ex": "PMOS diode load removes η: Av = −√(µn(W/L)1/(µp(W/L)2)). Gain 5 needs ~12.5× ratio and costs swing.",
   "tags": [
    "cs"
   ]
  },
  {
   "id": "ic:diode-z",
   "type": "flash",
   "q": "Impedance of a diode-connected MOSFET?",
   "a": "1/gm || 1/gmb || rO ≈ 1/(gm + gmb).",
   "ex": "Looking into a source with λ = 0 also gives 1/(gm + gmb).",
   "tags": [
    "smallsignal"
   ]
  },
  {
   "id": "ic:cs-cursrc",
   "type": "flash",
   "q": "CS stage with current-source load: gain and swing?",
   "a": "Av = −gm1 (rO1 || rO2); output high limit VDD − |VOD2|. Bias point needs feedback to be defined.",
   "ex": "Longer L raises rO and gain; W must scale with L to keep the overdrive.",
   "tags": [
    "cs"
   ]
  },
  {
   "id": "ic:cs-active",
   "type": "flash",
   "q": "Complementary (active-load) CS stage, i.e. inverter as amplifier?",
   "a": "Av = −(gm1 + gm2)(rO1 || rO2). Drawback: bias current is a strong function of VDD and VTH, and it amplifies supply noise.",
   "ex": "Nanometer example: gain ≈ 2.5 with 1 V supply.",
   "tags": [
    "cs"
   ]
  },
  {
   "id": "ic:cs-degen",
   "type": "flash",
   "q": "Source degeneration RS: effect on Gm and gain?",
   "a": "Gm = gm/(1 + gm RS); Av = −RD/(1/gm + RS). Linearises ID vs Vin at the cost of gain and noise.",
   "ex": "Gain by inspection: resistance at the drain over total resistance in the source path.",
   "tags": [
    "cs"
   ]
  },
  {
   "id": "ic:degen-rout",
   "type": "flash",
   "q": "Output resistance of a degenerated transistor?",
   "a": "Rout = [1 + (gm + gmb)RS] rO + RS: rO boosted by the degeneration factor.",
   "ex": "Equivalently RS boosted by the intrinsic gain plus rO. Basis of the cascode.",
   "tags": [
    "cs",
    "cascode"
   ]
  },
  {
   "id": "ic:gm-rout-lemma",
   "type": "flash",
   "q": "Razavi's gain lemma?",
   "a": "Av = −Gm Rout: Gm with the output shorted, Rout with the input zeroed. Usually both are found by inspection.",
   "ex": "Degenerated CS with ideal current-source load: gain returns to −gm rO, independent of RS.",
   "tags": [
    "smallsignal"
   ]
  },
  {
   "id": "ic:sf-gain",
   "type": "flash",
   "q": "Source-follower gain with ideal current source?",
   "a": "Av = gm/(gm + gmb) = 1/(1 + η) < 1 because of body effect; with RL: Av = Req/(Req + 1/gm), Req = 1/gmb || rO || RL.",
   "ex": "PMOS follower with bulk tied to source removes η but has lower gm.",
   "tags": [
    "follower"
   ]
  },
  {
   "id": "ic:sf-rout",
   "type": "flash",
   "q": "Output resistance of a source follower?",
   "a": "1/(gm + gmb): body effect lowers it, since a falling source also lowers VTH.",
   "ex": "Followers cost a VGS of headroom and add nonlinearity from VTH(VSB) and rO(VDS).",
   "tags": [
    "follower"
   ]
  },
  {
   "id": "ic:sf-vs-cs-50ohm",
   "type": "mcq",
   "q": "Driving an external 50 Ω load: which gives more gain for the same gm?",
   "choices": [
    "Source follower",
    "Common source",
    "Neither: both ≈ gm RL if gm RL << 1",
    "Depends on VTH"
   ],
   "a": 2,
   "ex": "SF: gm RL/(1 + gm RL); CS: −gm RL. With gm RL << 1 both are ≈ gm RL.",
   "tags": [
    "follower",
    "cs"
   ]
  },
  {
   "id": "ic:diff-swing",
   "type": "flash",
   "q": "Differential vs single-ended swing?",
   "a": "Two outputs each ±V0 give 4V0 peak-to-peak differentially; a 1 V supply can deliver 1.6 V pp.",
   "ex": "Also: supply-noise rejection, simpler biasing, cancelled even-order distortion.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:tail-role",
   "type": "flash",
   "q": "Why the tail current source in a differential pair?",
   "a": "It makes ID1 + ID2 independent of the input common-mode level, so gm, gain and output CM stay fixed.",
   "ex": "Without it, a low input CM turns the pair off and clips.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:cm-range",
   "type": "flash",
   "q": "Allowed input common-mode range of a resistively loaded NMOS pair?",
   "a": "VGS1 + (VGS3 − VTH3) ≤ Vin,CM ≤ min(VDD − RD ISS/2 + VTH, VDD).",
   "ex": "Low limit keeps the tail saturated; high limit keeps the pair out of triode. Choose CM low for swing.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:diff-largesignal",
   "type": "flash",
   "q": "Differential pair large-signal current?",
   "a": "ID1 − ID2 = ½µnCox(W/L)ΔVin √(4ISS/(µnCox W/L) − ΔVin²), until one side turns off at ΔVin1 = √(2ISS/(µnCox W/L)).",
   "ex": "ΔVin1 = √2 × the equilibrium overdrive. Linearity vs gm: raise ISS or shrink W/L.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:diff-gain",
   "type": "mcq",
   "q": "Differential gain of a symmetric pair with resistive loads?",
   "choices": [
    "gm RD/2",
    "gm RD",
    "2 gm RD",
    "gm RD/(1+η)"
   ],
   "a": 1,
   "ex": "Equilibrium Gm = √(µnCox(W/L)ISS) = gm of each device at ISS/2. Single-ended output halves it.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:half-circuit",
   "type": "flash",
   "q": "Half-circuit concept?",
   "a": "For differential inputs on a symmetric pair the tail node is a virtual ground; analyse one CS half. Gain −gm(RD || rO).",
   "ex": "Any inputs split into CM and DM components; superpose.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:degen-diff",
   "type": "flash",
   "q": "Degenerated differential pair?",
   "a": "Linear range widens by ±RS ISS; gain RD/(1/gm + RS). Split the tail into two sources joined by 2RS to save headroom.",
   "ex": "Cost: differential noise and offset from the two tails.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:cm-gain",
   "type": "flash",
   "q": "Common-mode gain with tail resistance RSS (symmetric pair)?",
   "a": "Av,CM = −(RD/2)/(1/(2gm) + RSS): the two halves merge into one device of 2gm.",
   "ex": "Shifts the output CM and eats swing but does not create a differential error.",
   "tags": [
    "diffpair",
    "cmrr"
   ]
  },
  {
   "id": "ic:cm-dm",
   "type": "flash",
   "q": "Common-mode to differential conversion?",
   "a": "With ΔRD or Δgm mismatch a CM input yields a differential output: ACM−DM ≈ −Δgm RD/[(gm1 + gm2)RSS + 1].",
   "ex": "Gets worse at high frequency: tail capacitance shorts RSS. Transistor mismatch dominates over RD mismatch.",
   "tags": [
    "cmrr"
   ]
  },
  {
   "id": "ic:cmrr-def",
   "type": "flash",
   "q": "CMRR of a differential pair?",
   "a": "ADM/ACM−DM ≈ (1 + 2gm RSS) gm/Δgm ≈ 2gm² RSS/Δgm.",
   "ex": "Even an ideal tail leaves CM-to-DM conversion if gmb1 ≠ gmb2.",
   "tags": [
    "cmrr"
   ]
  },
  {
   "id": "ic:diff-gm-vs-cs",
   "type": "mcq",
   "q": "For the same total current, a differential pair's Gm relative to one CS device at ISS?",
   "choices": [
    "Equal",
    "Double",
    "Half",
    "1/√2 of it"
   ],
   "a": 3,
   "ex": "Each device carries ISS/2 and gm ∝ √ID.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:cm-swing",
   "type": "flash",
   "q": "Max output swing of a pair vs a CS stage?",
   "a": "Pair: VDD minus two overdrives per side, 2VDD − 4VD,sat differentially. CS: VDD minus one overdrive.",
   "ex": "Nanometer pairs rarely exceed gain 5, so the input swing also limits the output.",
   "tags": [
    "diffpair"
   ]
  },
  {
   "id": "ic:opamp-gain-error",
   "type": "flash",
   "q": "Closed-loop gain error of a feedback amplifier with open-loop gain A1?",
   "a": "Relative error ≈ (1 + R1/R2)/A1: a gain-of-10 stage with 1 % error needs A1 > 1000.",
   "ex": "Open-loop CS gain gm RD varies >20 % with process; feedback fixes that.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:settling",
   "type": "flash",
   "q": "Settling time of a one-pole op amp in closed loop?",
   "a": "τ ≈ (1 + R1/R2)/(A0ω0); 1 % settling takes 4.6τ.",
   "ex": "Gain-of-10, 5 ns to 1 %: needs 1.47 GHz unity-gain bandwidth.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:one-stage-gain",
   "type": "flash",
   "q": "Gain of the simplest one-stage op amp (pair with current-source loads)?",
   "a": "gmN (rON || rOP), hardly above 10 in nanometer nodes; bandwidth set by CL.",
   "ex": "At least four devices (two inputs, two loads) always contribute noise.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:closed-loop-zout",
   "type": "flash",
   "q": "Closed-loop output impedance of a unity-gain buffer built from a one-stage op amp?",
   "a": "Open-loop rO divided by 1 + loop gain ≈ 1/gmN, nearly independent of the open-loop Rout.",
   "ex": "So raise Rout for gain without losing drive; output pole ≈ gmN/CL.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:telescopic-gain",
   "type": "flash",
   "q": "Telescopic cascode op amp: gain and swing?",
   "a": "Av ≈ gmN[(gmN rON²) || (gmP rOP²)]; differential swing 2[VDD − (VOD1 + VOD3 + VISS + |VOD5| + |VOD7|)].",
   "ex": "Input CM, Vb1 and Vb2 must be set tightly; shorting input to output leaves only one VTH minus one overdrive of range.",
   "tags": [
    "opamp",
    "cascode"
   ]
  },
  {
   "id": "ic:design-flow",
   "type": "flash",
   "q": "Razavi's op-amp design procedure?",
   "a": "Power budget first, then allocate overdrives for swing, size W/L from ID and overdrive at min L, then check gain and lengthen devices.",
   "ex": "Five knobs per transistor: ID, VGS − VTH, W/L, gm, rO.",
   "tags": [
    "design",
    "opamp"
   ]
  },
  {
   "id": "ic:linear-scaling",
   "type": "flash",
   "q": "Linear scaling of an op amp?",
   "a": "Double every width (not length): ID and gm double, rO halves, overdrive, swing and gain unchanged; power scales.",
   "ex": "Scaling widths down 10× costs 10× speed into CL and √10 more input noise.",
   "tags": [
    "design"
   ]
  },
  {
   "id": "ic:vb1-tracking",
   "type": "flash",
   "q": "How to make the cascode bias Vb1 track the input CM level?",
   "a": "Generate it with a diode-connected device sitting on the tail node: Vb1 = Vin,CM − VGS1,2 + VGS,b1, with Mb1 narrow and long.",
   "ex": "Otherwise a small CM shift pushes the input pair into triode.",
   "tags": [
    "opamp",
    "bias"
   ]
  },
  {
   "id": "ic:folded-why",
   "type": "flash",
   "q": "Why choose a folded cascode over a telescopic one?",
   "a": "Input and output CM levels can be equal without losing swing, and the input CM range is wider.",
   "ex": "Costs: more power (extra branch), 2–3× lower gain, lower folding-node pole, more noise.",
   "tags": [
    "opamp",
    "cascode"
   ]
  },
  {
   "id": "ic:folded-gain",
   "type": "flash",
   "q": "Folded-cascode op amp gain?",
   "a": "|Av| ≈ gm1 {[(gm3 + gmb3) rO3 (rO1 || rO5)] || [(gm7 + gmb7) rO7 rO9]}: Gm ≈ gm1, Rout the two cascoded branches.",
   "ex": "rO1 || rO5 hurts because M5 carries both input and cascode currents.",
   "tags": [
    "opamp",
    "cascode"
   ]
  },
  {
   "id": "ic:folded-swing",
   "type": "mcq",
   "q": "Peak-to-peak swing per side of a folded cascode with cascoded loads?",
   "choices": [
    "VDD − 4 overdrives",
    "VDD − 5 overdrives",
    "VDD − 2 overdrives",
    "VDD − VTH"
   ],
   "a": 0,
   "ex": "Telescopic loses one more: the tail current source's overdrive.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:two-stage",
   "type": "flash",
   "q": "Why a two-stage op amp?",
   "a": "It separates gain (stage 1, often cascoded) from swing (stage 2, a plain CS): output swing VDD − |VOD5| − VOD7.",
   "ex": "More than two stages is rare: each stage adds a pole and compensation becomes hard.",
   "tags": [
    "twostage"
   ]
  },
  {
   "id": "ic:mirror-pole",
   "type": "flash",
   "q": "What is the mirror pole and why avoid it?",
   "a": "The pole at the mirror gate node of a differential-to-single-ended converter; usually the first nondominant pole, so it limits PM.",
   "ex": "Fully differential topologies avoid it (and get a zero at 2ωp,mirror for free).",
   "tags": [
    "opamp",
    "stability"
   ]
  },
  {
   "id": "ic:gain-boost",
   "type": "flash",
   "q": "Gain boosting (regulated cascode)?",
   "a": "An amplifier A1 drives the cascode gate to pin its source: Rout ≈ (A1 + 1) gm2 rO2 rO1, gain ≈ gm1 gm2 rO1 rO2 (A1 + 1), no headroom lost.",
   "ex": "Two views: gm2 boosted by A1 + 1, or A1 regulates the current by holding the source node.",
   "tags": [
    "gainboost"
   ]
  },
  {
   "id": "ic:gain-boost-aux",
   "type": "flash",
   "q": "Which auxiliary amplifier for a gain-boosted NMOS cascode?",
   "a": "A folded-cascode with PMOS inputs, so the minimum at nodes X/Y stays VOD1 + VISS, not VGS3 + something.",
   "ex": "An NMOS CS as A1 forces Vout > VGS3 + VOD2; a PMOS CS pushes its own device into triode.",
   "tags": [
    "gainboost"
   ]
  },
  {
   "id": "ic:gain-boost-poles",
   "type": "flash",
   "q": "Frequency response of a regulated cascode?",
   "a": "Dominant pole ≈ 1/(A0 gm2 rO2 rO1 CL); a second pole and a zero near (A0 + 1)ω0, the aux amp's unity-gain bandwidth.",
   "ex": "Most signal bypasses A1 through the cascode; only the error is slowed down.",
   "tags": [
    "gainboost",
    "stability"
   ]
  },
  {
   "id": "ic:cmfb-why",
   "type": "flash",
   "q": "Why do high-gain fully differential op amps need common-mode feedback?",
   "a": "Current-source loads and tail mismatch leave the output CM undefined: a small current error drives loads or tail into triode.",
   "ex": "CMFB senses the output CM and adjusts a bias current to hold it.",
   "tags": [
    "opamp"
   ]
  },
  {
   "id": "ic:comparison-table",
   "type": "flash",
   "q": "Razavi's topology comparison: gain / swing / speed / power / noise?",
   "a": "Telescopic: med/med/highest/low/low. Folded: med/med/high/med/med. Two-stage: high/highest/low/med/low. Gain-boosted: high/med/med/high/med.",
   "ex": "Pick by the spec that binds first: swing pushes to two-stage, speed to telescopic.",
   "tags": [
    "opamp",
    "design"
   ]
  },
  {
   "id": "ic:swing-check",
   "type": "flash",
   "q": "How to verify an op amp's usable output swing?",
   "a": "Sweep input amplitude and plot gain vs output amplitude; swing is where gain drops ~10 % (1 dB). Or measure closed-loop distortion.",
   "ex": "Triode/saturation borders blur in nanometer devices, so simulate.",
   "tags": [
    "design"
   ]
  },
  {
   "id": "ic:barkhausen",
   "type": "flash",
   "q": "Barkhausen's oscillation criteria?",
   "a": "|βH(jω1)| = 1 and ∠βH(jω1) = −180° (total loop phase 360° with the inversion). Only the loop transmission matters.",
   "ex": "Weaker feedback (smaller β) is more stable: GX moves down, PX stays.",
   "tags": [
    "stability"
   ]
  },
  {
   "id": "ic:bode-rules",
   "type": "flash",
   "q": "Bode plot rules for poles?",
   "a": "Magnitude: −20 dB/dec per pole. Phase: starts at 0.1ωp, −45° at ωp, −90° by 10ωp. Phase is hit far earlier than magnitude.",
   "ex": "So extra poles hurt phase much more than gain.",
   "tags": [
    "stability"
   ]
  },
  {
   "id": "ic:pm-def",
   "type": "flash",
   "q": "Phase margin definition and target?",
   "a": "PM = 180° + ∠βH at the gain crossover. 45° gives a 30 % peak; 60° is the usual optimum (fast, little ringing); 90° is slow.",
   "ex": "PM > 45° requires the unity-gain bandwidth to sit below the second pole.",
   "tags": [
    "stability"
   ]
  },
  {
   "id": "ic:pm-second-pole",
   "type": "mcq",
   "q": "If the loop gain hits unity exactly at the second pole (two-pole system), PM is...",
   "choices": [
    "45°",
    "60°",
    "90°",
    "0°"
   ],
   "a": 0,
   "ex": "Phase there is −135°. So the compensated bandwidth cannot exceed the first nondominant pole.",
   "tags": [
    "stability"
   ]
  },
  {
   "id": "ic:compensation-idea",
   "type": "flash",
   "q": "What does frequency compensation do?",
   "a": "Moves the dominant pole toward the origin so the gain crossover falls well below the phase crossover, trading bandwidth for margin.",
   "ex": "Raising Rout does not help (only low-frequency gain changes). First minimise the number of poles.",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:beta-relax",
   "type": "flash",
   "q": "Compensated for 60° at β = 1; how much can CC shrink for β < 1?",
   "a": "By about 1/β: the magnitude curve drops by −20 log β, so the dominant pole may rise by 1/β. Closed-loop speed stays about the same.",
   "ex": "Do not compensate for unity gain if the closed-loop gain is always higher.",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:cascode-node-pole",
   "type": "flash",
   "q": "Does the internal node of a PMOS cascode load add a pole?",
   "a": "No: its time constant rO7 CN merges into the output pole (1 + gm5 rO5) rO7 CL. The signal does not see it.",
   "ex": "Fully differential telescopic cascodes have essentially one nondominant pole (the NMOS cascode source).",
   "tags": [
    "stability",
    "cascode"
   ]
  },
  {
   "id": "ic:pole-zero-doublet",
   "type": "flash",
   "q": "Cancel a nondominant pole with a zero?",
   "a": "Possible (parallel fast path), but mismatch leaves a pole–zero doublet that causes slow settling tails.",
   "ex": "Also true for the Rz trick in Miller compensation when CL varies.",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:miller-comp",
   "type": "flash",
   "q": "Miller compensation of a two-stage op amp?",
   "a": "CC across the second stage looks like (1 + Av2)CC at the first-stage output: dominant pole ≈ 1/(gm9 RL CC RS) with a small capacitor.",
   "ex": "Pole splitting: the output pole rises to ≈ gm9/(CE + CL), roughly gm9 RL times higher.",
   "tags": [
    "compensation",
    "twostage"
   ]
  },
  {
   "id": "ic:cc-estimate",
   "type": "flash",
   "q": "First estimate of CC for a two-stage op amp with 45° PM?",
   "a": "CC ≈ (gm1/gm9) CL (β = 1); including the second pole's magnitude, CC ≈ gm1 CL/(√2 gm9). Use more for 60°.",
   "ex": "Second pole ≈ gm9/CL, so a big output gm helps.",
   "tags": [
    "compensation",
    "twostage"
   ]
  },
  {
   "id": "ic:rhp-zero",
   "type": "flash",
   "q": "Right-half-plane zero in Miller-compensated op amps?",
   "a": "ωz = gm9/(CC + CGD9) from the feedforward path through CC. It adds negative phase and slows the magnitude roll-off: both hurt PM.",
   "ex": "Typically ωp1 < ωz < ωp2. Setting CC = CL does not cancel the pole: the zero is RHP.",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:rz-trick",
   "type": "flash",
   "q": "Series resistor Rz with CC?",
   "a": "ωz ≈ 1/[CC(1/gm9 − Rz)]. Rz = 1/gm9 sends the zero to infinity; Rz = (CL + CC)/(gm9 CC) puts it on the first nondominant pole.",
   "ex": "Rz is a triode MOSFET: it varies with swing; bias it to track gm9 (ratioed diodes or a resistor-defined gm).",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:two-stage-cl",
   "type": "mcq",
   "q": "Adding load capacitance to a Miller-compensated two-stage op amp...",
   "choices": [
    "Improves PM",
    "Changes nothing",
    "Raises the RHP zero",
    "Lowers the second pole, hurting PM"
   ],
   "a": 3,
   "ex": "Opposite of a one-stage op amp, where a bigger CL lowers the dominant pole and improves PM.",
   "tags": [
    "twostage",
    "stability"
   ]
  },
  {
   "id": "ic:slew-rate",
   "type": "flash",
   "q": "Slew rate of a two-stage Miller-compensated op amp?",
   "a": "SR ≈ ISS/CC (positive); negative side limited by the second-stage current source: (I1 − ISS)/CL if the output device turns off.",
   "ex": "During slew the first stage is fully unbalanced: small-signal analysis does not apply.",
   "tags": [
    "slew"
   ]
  },
  {
   "id": "ic:slew-cl",
   "type": "flash",
   "q": "Slewing into a load CL through CF (Miller): what current must the output device supply?",
   "a": "I1 + ISS + (CL/CF) ISS; if I1 < (1 + CL/CF) ISS the device turns off and SR drops to (I1 − ISS)/CL.",
   "ex": "Class-AB second stages can source large I1 on demand.",
   "tags": [
    "slew"
   ]
  },
  {
   "id": "ic:sf-comp",
   "type": "flash",
   "q": "Source follower in series with CC?",
   "a": "Blocks the feedforward path, so the RHP zero moves to a left-half-plane zero (≈ gm2/CC); poles stay ≈ Miller values.",
   "ex": "Costs headroom: output floor VGS2 + VI2. The common-gate variant avoids that and boosts ωp2 by gm2 RS.",
   "tags": [
    "compensation"
   ]
  },
  {
   "id": "ic:nyquist-why",
   "type": "flash",
   "q": "Why Nyquist when Bode exists?",
   "a": "Bode only tests s = jω. A loop can have no unstable sinusoid yet still grow a σ > 0 waveform; Nyquist counts RHP zeros of 1 + βH.",
   "ex": "Phase from a pole at s1: minus the angle of the vector from the pole to s1.",
   "tags": [
    "stability"
   ]
  },
  {
   "id": "ic:bootstrap-bias",
   "type": "flash",
   "q": "Supply-independent bias: idea and flaw?",
   "a": "Mirror Iout back to make IREF (bootstrapping). With λ = 0 the loop supports any current: add RS in one source to pin it.",
   "ex": "Iout = 2/(µnCox(W/L)N RS²)·(1 − 1/√K)². Use long channels to suppress λ and flicker noise.",
   "tags": [
    "bias"
   ]
  },
  {
   "id": "ic:startup",
   "type": "flash",
   "q": "Start-up problem?",
   "a": "The bootstrapped loop also supports zero current. Add a diode device that injects current at power-up and turns off afterwards.",
   "ex": "Check with a DC sweep and a transient supply ramp; complex loops can have several degenerate points.",
   "tags": [
    "bias"
   ]
  },
  {
   "id": "ic:constant-gm",
   "type": "flash",
   "q": "Constant-Gm biasing?",
   "a": "With the RS-defined loop, gm1 = (2/RS)(1 − 1/√K): set by a resistor, independent of µCox, VTH and VDD.",
   "ex": "Temperature then follows the resistor's TC; also used to make Rz track gm9 in compensation.",
   "tags": [
    "bias"
   ]
  },
  {
   "id": "ic:vbe-tc",
   "type": "mcq",
   "q": "Temperature coefficient of a base–emitter voltage at 300 K?",
   "choices": [
    "About +2 mV/K",
    "About −1.5 mV/K",
    "About −0.087 mV/K",
    "Zero"
   ],
   "a": 1,
   "ex": "∂VBE/∂T = [VBE − (4 + m)VT − Eg/q]/T, m ≈ −1.5. Depends on VBE itself, hence curvature.",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:ptat",
   "type": "flash",
   "q": "PTAT voltage source?",
   "a": "ΔVBE = VT ln n between two bipolars at current densities differing by n: TC = (k/q) ln n = 0.087 mV/K × ln n.",
   "ex": "Ratio both currents (n) and emitter areas (m): ΔVBE = VT ln(nm).",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:bandgap-value",
   "type": "flash",
   "q": "Zero-TC combination and its value?",
   "a": "VREF = VBE + 17.2 VT ≈ 1.25 V ≈ Eg/q + (4 + m)VT: the silicon bandgap, hence the name.",
   "ex": "ln n = 17.2 is impossible directly, so amplify the PTAT drop: (1 + R2/R3) ln n = 17.2, e.g. n = 31, R2/R3 = 4.",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:bandgap-topology",
   "type": "flash",
   "q": "Classic CMOS bandgap (Fig. 12.9) operation?",
   "a": "Op amp forces X = Y; R3 carries VT ln n/R3; Vout = VBE2 + (1 + R2/R3) VT ln n. Resistor TC cancels in the ratio.",
   "ex": "Uses substrate pnp: p+ emitter in n-well, collector = substrate (grounded).",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:bandgap-offset",
   "type": "flash",
   "q": "Effect of op-amp offset on a bandgap?",
   "a": "Vout error ≈ −(1 + R2/R3)VOS, and VOS drifts with temperature. Mitigate: big matched devices, ratio currents by m, stack two VBE.",
   "ex": "Stacking gives 2.5 V, hard at low supply.",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:bandgap-feedback",
   "type": "flash",
   "q": "Feedback polarity check in the bandgap loop?",
   "a": "Negative path βN = (1/gm2 + R3)/(1/gm2 + R3 + R2) must exceed positive βP = (1/gm1)/(1/gm1 + R1), ideally by ~2×.",
   "ex": "Otherwise transient response with capacitive load misbehaves.",
   "tags": [
    "bandgap",
    "stability"
   ]
  },
  {
   "id": "ic:curvature",
   "type": "flash",
   "q": "Bandgap curvature?",
   "a": "Zero TC at one temperature only: VBE nonlinearity, PTAT currents and offset drift bend it. Rarely corrected in CMOS.",
   "ex": "PTAT collector currents make the VBE TC slightly less negative: (3 + m) instead of (4 + m).",
   "tags": [
    "bandgap"
   ]
  },
  {
   "id": "ic:ptat-current",
   "type": "flash",
   "q": "PTAT current generator?",
   "a": "Two bipolars at 1 : n areas with matched PMOS mirrors and R1 between emitters: I = VT ln n/R1.",
   "ex": "Add R2 in a third branch with a VBE to get a low-voltage bandgap: VREF = |VBE3| + (R2/R1) VT ln n.",
   "tags": [
    "bandgap",
    "bias"
   ]
  }
 ],
 "version": "1.0.0",
 "updated": "2026-09-14T22:11:35Z"
}