Ebers-moll model of transistor

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Introduction to Ebers moll model of transistor

Ebers Moll model is a simple and elegant way of representing the transistor  as a circuit model. The Ebers Moll model of transistor holds for all regions of operation of transistor. This model is based on assumption that base spreading resistance can be neglected. It will be obvious that why two diodes connected back to back will not function as a transistor from the following discussion, as dependent current source term will be missing which is responsible for all the interesting properties of transistor.


Transistor in inverted mode of operation

The normal mode of operation corresponds to the use of emitter as source of collector current and inverted mode of operation corresponds to the use of collector as source of emitter current(which is the case when BJT is operated in inverse active region). The BJT when operated in normal mode and inverse mode is shown in the figure below

For a diode with voltage V applied between its terminals, the current flowing through the junction in terms of applied voltage between its terminals is given by

                                                  I=IO *(eV/Vt-1)

Where IO is reverse saturation current of the transistor (Current which flows across junction when it is reverse biased will be –IO as eV/Vt tends to zero).

The collector current in a BJT when operated in normal mode is given as 

                Ic= -αN*IE + Ico *(1-eVCB/Vt)

In inverse mode of operation the collector current can be found by replacing αN by αI,  ICO by IEO, VBC by VBE

IE= -αI*IC+IEO(1-eVBE/Vt)

Description of Ebers-moll model

Ebers-moll model of transistor

The current equations derived above is interpreted in terms of a model shown in the figure. This model of transistor is known as Ebers Moll model of transistor.From the diagram applying Kirchhoff’s current law at the collector node, we get

 IC= -αN*IE + ICO *(1 – eVCB/Vt)

Where αN is the current gain of common base transistor mentioned above in normal mode of operation, VBC is the base to collector voltage, Ico is the reverse saturation current of base collector junction.Similarly at emitter and base node by applying Kirchhoff’s current law

 IE= -αI*IC+IEO(1 – eVBE/Vt), IE+IB+IC = 0

Where αI is the inverted current gain of common base transistor with roles of collector and emitter interchanged, VBE is the base to Emitter voltage, Ico is the reverse saturation current of base Emitter junction. αI  and αI are related through the reverse saturation currents of the diode as

αI *ICO = αN *IEO

The above equations are derived based on the assumption of low level minority carrier injection (the hole concentration injected into the base is very much less compared to the intrinsic electron concentration in base), in such a case emitter or collector current is mainly dominated by diffusion currents, drift current is negligible compared to drift currents. 

The Base to emitter voltage and base to collector voltage in terms of currents can be derived as follows

IE= -αI*IC+IEO(1 – eVBE/Vt) ,  IC= -αN*IE + ICO *(1 – eVCB/Vt)

IEI*IC=IEO(1 – eVBE/Vt) ,  ICN*IE = ICO *(1 – eVCB/Vt)

(IEI*IC)/IEO = (1 – eVBE/Vt) ,  (ICN*IE)/ICO = (1 – eVCB/Vt)

eVBE/Vt = 1 – ((IEI*IC)/IEO) ,  eVCB/V= 1 -((ICN*IE)/ICO)

Applying anti log on both sides we get

VBE= Vt *ln( 1 – ((IEI*IC)/IEO)) ,  VCB = Vt *ln( 1 -((ICN*IE)/ICO) )

For example in cutoff region IE=0 amps and IC = ICO then the base to emitter voltage is

VBE = Vt*ln(1-(αI*ICO)/IEO))

VBE,cutoff = Vt*ln(1-αN)

Two back to back diodes in series, will it function as BJT ?

Now coming to important question of Why two back to back diodes cannot function as a transistor?

Consider two diodes connected back to back in the configuration shown below

back to back diodes

back to back diodes in series

It is obvious that if one junction is forward biased then other junction will be reverse biased consider for example diode D1 is forward biased and diode D2 is reverse biased much like a NPN transistor in active region according to the junction voltages only current order of reverse saturation current flows through the series junctions. This can be explained as follows: the reverse biased diode D2 at most will allow only currents order of reverse saturation currents. Since D1 and D2 are in series same current should flow through both of them then only currents order of reverse saturation currents flow through their junctions. It is obvious that this is not the case with the transistor in active region (because of the internal design of transistor). The forward current entering the base is sweeped across into collector by the electric filed generated by the reverse bias voltage applied across the base collector junction.

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