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Ebers-Moll model

Short answer

The Ebers-Moll model represents the BJT as two back-to-back diodes with coupled current sources. Terminal currents: IE = −IF + αR IR, IC = αF IF − IR, where IF = IES(e^VBE/VT − 1), IR = ICS(e^VBC/VT − 1), with αF IES = αR ICS = IS.

Explain Ebers-Moll model with necessary figure and mathematical expressions.

The Ebers-Moll (EM) model is a large-signal model valid in all regions of operation (active, saturation, cut-off). It treats the BJT as two back-to-back diodes coupled by current sources, because each junction injects carriers that are collected by the other.

Equivalent circuit (injection version)

  C o---+-----------+
        |           |
       ( ) αF·IF   DC | IR
        |           |
  B o---+-----------+
        |           |
       ( ) αR·IR   DE | IF
        |           |
  E o---+-----------+
  • DE: base-emitter diode carrying IF; DC: base-collector diode carrying IR.
  • A fraction αF of the emitter diode current IF reaches the collector (forward operation), and a fraction αR of IR reaches the emitter (reverse operation).

Equations

Diode currents:

IF = IES [exp(VBE/VT) - 1]
IR = ICS [exp(VBC/VT) - 1]

Terminal currents:

IE = -IF + αR IR
IC =  αF IF - IR
IB = -(IE + IC) = (1 - αF) IF + (1 - αR) IR

(with IE and IC taken as flowing into the transistor)

Reciprocity relation: αF IES = αR ICS = IS.

Written in terms of IS, the commonly used form is:

IC = IS [exp(VBE/VT) - exp(VBC/VT)] - (IS/βR)[exp(VBC/VT) - 1]

Operating regions

  • Active: BE forward, BC reverse: IC ≈ αF IF ≈ IS exp(VBE/VT).
  • Cut-off: both junctions reverse biased: only small leakage currents flow.
  • Saturation: both junctions forward biased: IC is reduced and VCE(sat) is about 0.2 V.
  • Reverse active: BE reverse, BC forward, with βR much smaller than βF.

Typical values: αF = 0.98 to 0.998, αR = 0.02 to 0.5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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