Small-signal emitter diode resistance (re) approximation At room temperature, which formula gives the approximate AC resistance of the base–emitter junction (often denoted re) in a BJT small-signal model?

Introduction / Context:In small-signal BJT analysis, the dynamic resistance of the base–emitter junction, re, is a key parameter. It captures how the junction voltage changes with current for small AC perturbations. A quick mental formula at room temperature simplifies hand calculations for gain and input resistance in common-emitter and emitter-degeneration stages.

Given Data / Assumptions:

• Room temperature so VT ≈ 25 mV (thermal voltage).
• Emitter current IE is the DC operating current through the junction (IC ≈ IE for large beta).
• Small-signal linearization around the Q-point.

Concept / Approach:The diode small-signal resistance is r = dV/dI. For a PN junction, r ≈ VT / I at the operating point. In BJTs, re is taken as VT/IE (or approximately VT/IC for large beta). Thus, at 25 °C, re ≈ 25 mV / IE, yielding ohms when IE is in amperes.

Step-by-Step Solution:Use VT ≈ 25 mV at room temperature.Write re ≈ VT / IE.If IE = 1 mA, re ≈ 25 mV / 0.001 A = 25 Ω (illustrative).Therefore, the correct proportionality is division by current, not multiplication.

Verification / Alternative check:Compare to transconductance gm ≈ IC/VT. Note that re ≈ 1/gm (using IC ≈ IE), confirming re decreases as current increases.

Why Other Options Are Wrong:25 mV × IE or × IC: implies re increases with current, which is opposite to physics.25 mV / IC: close numerically when beta is large, but the canonical expression uses IE.

Common Pitfalls:Forgetting that currents must be in amperes when using 25 mV to get re in ohms; mixing milliamps directly yields results off by 1000.

Final Answer:re ≈ 25 mV / IE