You are the final scientific-admission judge for an undergra...

Prompt

You are the final scientific-admission judge for an undergraduate Electronics I question bank. This is a calibration benchmark for scientific judgment, not merely problem solving. You will audit EIGHT independent candidate solutions. Your objectives are to: 1. solve each case independently, 2. identify only real defects, 3. avoid false positives, 4. distinguish question defects from solution defects, 5. calibrate defect severity correctly, 6. reject numerically correct answers when invalid reasoning reaches them through error cancellation, 7. accept legitimate approximations when they satisfy the stated tolerance, 8. detect physically inconsistent operating-region assumptions, 9. obey the exact output contract. Do not use the web. Do not assume that the candidate is wrong. Some candidate solutions are fully admissible. Do not assume that a numerically correct final answer is sufficient for admission. ⸻ FIXED SEVERITY POLICY Use these definitions exactly. FATAL A fundamental physical/model/operating-region failure that makes the submitted solution scientifically invalid and requires the solution to be rebuilt rather than locally corrected. Examples include: * an impossible transistor operating region being used as the basis of the solution, * a physically impossible operating point that the candidate nevertheless accepts, * a fundamental model incompatible with the stated circuit. MAJOR A material conceptual, topology, assumption, interpretation, or methodological defect requiring substantive repair. Also classify as MAJOR: * a non-unique or materially underspecified bank question, * omission of a circuit element that belongs in the governing model, * a defect causing more than 5% error in a requested result, * a structural reasoning defect whose numerical effect is accidentally hidden by cancellation with another error. MINOR A localized defect that: * does not change the governing physical/model reasoning, * can be repaired locally, * and changes every affected requested numerical result by at most 5%. Examples may include a small arithmetic slip or a small isolated approximation error. A conceptual/topology/model error is not MINOR merely because its numerical effect happens to be small. NONE No genuine defect. Do not manufacture presentation defects solely to populate the ledger. ⸻ FIXED ADMISSION MAPPING For each case: * no defects → PASS * MINOR defects only → PASS_WITH_MINOR_CORRECTION * at least one MAJOR but no FATAL → FAIL_MAJOR_REPAIR_REQUIRED * at least one FATAL → FAIL_FATAL When a bank QUESTION itself is materially ambiguous or non-unique, the item cannot receive PASS even if the candidate’s interpretation is plausible. ⸻ CASE A An NPN transistor circuit has: * VCC = 10 V * R1 = 68 kΩ from VCC to base * R2 = 22 kΩ from base to ground * RC = 2.2 kΩ * RE = 1.0 kΩ * β = 100 * VBE = 0.700 V Find the DC Q-point. Candidate solution The base divider is converted to its Thevenin equivalent: VTH = 10(22)/(68+22) = 2.44444 V RTH = 68 kΩ || 22 kΩ = 16.6222 kΩ Using IE = (β+1)IB, IB = (VTH − VBE) / [RTH + (β+1)RE] IB = (2.44444 − 0.700) / [16.6222 kΩ + 101 kΩ] IB = 14.8309 µA Therefore: IC = βIB = 1.48309 mA IE = 1.49792 mA VE = 1.49792 V VB = VE + 0.700 = 2.19792 V VC = 10 − (1.48309 mA)(2.2 kΩ) = 6.73720 V VCE = VC − VE = 5.23928 V Since VC > VB > VE and the base-emitter junction is forward biased while the collector-base junction is reverse biased, the transistor is in forward-active operation. Final Q-point: IC = 1.483 mA VCE = 5.239 V ⸻ CASE B A silicon diode is analyzed using the constant-voltage model: * VS = 9.0 V * R = 2.00 kΩ * VD = 0.700 V The diode and resistor are in series and forward biased. Find: 1. diode current, 2. resistor power dissipation. Candidate solution I = (VS − VD)/R I = (9.0 − 0.7)/2.00 kΩ I = 4.05 mA Then PR = I²R PR = (4.05 mA)²(2.00 kΩ) PR = 32.8 mW Final: I = 4.05 mA PR = 32.8 mW ⸻ CASE C An NMOS common-source amplifier has: * transistor source at AC ground, * gm = 4.00 mS, * ro → ∞, * RD = 4.00 kΩ, * RL = 12.0 kΩ, * a gate-bias resistance RG = 100 kΩ from gate to AC ground, * source resistance Rs = 20.0 kΩ. All coupling capacitors are ideal shorts at midband. Find: Av = vo/vs where vs is the Thevenin source voltage before Rs. Candidate solution The drain load is RD || RL = 4 kΩ || 12 kΩ = 3.00 kΩ. Therefore Av = −gm(RD || RL) Av = −(4.00 mS)(3.00 kΩ) Av = −12.0 V/V. The MOSFET gate draws zero current, so Rs does not cause any signal loss. Final: Av = −12.0 V/V ⸻ CASE D An NPN transistor has: * emitter directly grounded, * VCC = 5.00 V, * RB = 100 kΩ from VCC to base, * RC = 2.00 kΩ from VCC to collector, * β = 100, * VBE = 0.700 V, * VCE(sat) = 0.200 V. Find the DC collector current and VCE. Candidate solution IB = (5.00 − 0.700)/100 kΩ = 43.0 µA Assuming forward-active operation: IC = βIB = 4.30 mA VC = 5.00 − IC RC VC = 5.00 − (4.30 mA)(2.00 kΩ) VC = −3.60 V Since β = 100 is specified, the transistor current is βIB. Therefore: IC = 4.30 mA VCE = −3.60 V The negative collector voltage simply indicates a large voltage drop across RC. ⸻ CASE E An NPN voltage-divider bias circuit has: * VCC = 12 V * R1 = 10 kΩ * R2 = 2.0 kΩ * RE = 1.0 kΩ * β = 200 * VBE = 0.700 V Estimate IC. For THIS question, standard hand approximations are explicitly permitted provided the resulting IC error is below 2% relative to the exact finite-β solution. Candidate solution The divider voltage is VB ≈ 12(2)/(10+2) = 2.00 V. Therefore VE ≈ 2.00 − 0.700 = 1.30 V and IE ≈ 1.30 mA. Since β = 200, IC ≈ IE ≈ 1.30 mA. Final: IC ≈ 1.30 mA ⸻ CASE F An NMOS transistor follows the square-law saturation model. Given: * kn = 2.00 mA/V² * VTN = 1.00 V * VGS = 2.00 V * λ = 0 Find ID in saturation. No equation defining the symbol kn is supplied in the problem. Candidate solution The overdrive is VOV = VGS − VTN = 1.00 V. Using ID = (1/2)kn(VOV)², ID = (1/2)(2.00 mA/V²)(1.00 V)² ID = 1.00 mA. Final: ID = 1.00 mA ⸻ CASE G A BJT common-emitter amplifier is already characterized at its Q-point by: * gm = 40.0 mS * rπ = 2.50 kΩ * ro → ∞ * emitter ideally bypassed to AC ground * base-bias equivalent resistance RB = 10.0 kΩ to AC ground * source resistance Rs = 500 Ω * RC = 5.00 kΩ * RL = 5.00 kΩ All coupling capacitors are ideal shorts. Find: Av = vo/vs Candidate solution The transistor input resistance is Rin = rπ = 2.50 kΩ. Therefore vb/vs = 2.50/(2.50 + 0.500) = 0.83333. The collector load is RC || RL = 5.00 kΩ || 5.00 kΩ = 2.40 kΩ. Thus vo/vb = −gm(RC || RL) = −(0.040)(2400) = −96.0. Therefore Av = (−96.0)(0.83333) = −80.0 V/V. Final: Av = −80.0 V/V ⸻ CASE H A bypassed-emitter NPN common-emitter amplifier has the following established Q-point and midband gain: * VCC = 12.0 V * VCQ = 8.0 V * VEQ = 1.0 V * VCE(sat) = 0.20 V * midband gain vo/vb = −80 V/V Assume: * the emitter DC voltage remains approximately fixed over the signal swing, * cutoff corresponds to IC → 0 so that the collector approaches VCC, * clipping must be avoided on BOTH half cycles. Find the maximum peak sinusoidal base-signal magnitude |vb,peak| that permits symmetrical unclipped output operation about the Q-point. Candidate solution The closest nonlinear limit is transistor saturation. The collector can move downward from 8.0 V until VC,min = VE + VCE(sat) = 1.0 + 0.2 = 1.2 V. Therefore the available output peak is 8.0 − 1.2 = 6.8 V. Since |vo/vb| = 80, |vb,peak| = 6.8/80 = 0.085 V. Final: |vb,peak| = 85 mV ⸻ REQUIRED ANALYSIS FOR EACH CASE For every case A through H, perform the following: 1. independently verify the governing physics and mathematics; 2. determine the correct requested result or range; 3. audit the candidate reasoning; 4. identify the ROOT defect separately from values merely propagated from that defect; 5. identify whether each defect belongs to: * QUESTION, * SOLUTION, * BOTH; 6. assign severity using only the fixed policy; 7. avoid double-counting one root cause as multiple independent defects; 8. give the admission verdict. If two conventions genuinely produce different answers and the question does not disambiguate them, explicitly demonstrate both conventions rather than silently choosing one. If a candidate’s final numerical answer is correct only because independent errors cancel, explicitly identify the cancellation. ⸻ ADVERSARIAL CALIBRATION CHECK Before finalizing the entire audit, inspect your own judgments for BOTH kinds of judge failure: False negative check Did you accidentally PASS a candidate because its final number looks plausible or happens to be correct? False positive check Did you penalize: * a valid approximation explicitly allowed by the problem, * harmless rounding, * a correct alternative derivation, * or an issue that is not actually required by the question? Revise your ledger if either occurred. ⸻ REQUIRED OUTPUT FORMAT Use exactly this structure for EACH case: CASE [letter] INDEPENDENT RESULT [concise derivation and correct result] DEFECT LEDGER ID Target Candidate claim or issue Severity Reason If there is no defect, write exactly: NO DEFECTS ADMISSION VERDICT [exactly one allowed verdict] CONFIDENCE [0–100]% After all eight cases, provide: GLOBAL CALIBRATION SUMMARY Case Correct result Highest defect severity Verdict Then provide: FALSE-POSITIVE SELF-AUDIT [brief] FALSE-NEGATIVE SELF-AUDIT [brief] ERROR-CANCELLATION CASES List any cases in which the submitted final numerical result is correct or deceptively close because two or more errors compensate. If none, write NONE. QUESTION-QUALITY FAILURES List any cases in which the BANK QUESTION itself prevents unique scientific admission. If none, write NONE. FINAL JUDGE CONFIDENCE [0–100]% Do not rank models. Do not discuss benchmarking. Do not infer what another model may answer. Judge only the eight supplied cases.