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I'm confused.

In different sources sometimes I find the equation

\$ I_C = I_S\cdot\left(e^{\frac{U_{BE}}{U_T}}-1\right) \$

or like wikipedia

\$ I_C = I_S\cdot e^{\frac{U_{BE}}{U_T}} \$

Can someone explain which equation is used in which case? Or is it just a simplification?

Thank you

Kind regards Thomas

Martin Zabel
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Thomas
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1 Answers1

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Under FORWARD bias case, since the exponential term dominates, you take \$I_C=I_S\cdot e^{\frac{U_{BE}}{U_T}}\$ neglecting reverse current.

However at lower values of UB, its better to take the first expression as both are comparable to some extent.

Under reverse bias you can even approximate the first equation to :

\$I_C = -I_S\$ since the exponential term is negligible(example: \$e^{\frac{-5}{0.026}} \approx 0\$.

In short,

\$I_C=I_S\cdot \left(e^{\frac{U_{BE}}{U_T}}-1\right)\$ (lower positive/negative values of UB) .

\$I_C=I_S\cdot e^{\frac{U_{BE}}{U_T}}\$ (high positive values of UB).

\$I_C=I_S\cdot (−1)\$ (high negative values of UB).

Martin Zabel
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Ashik Anuvar
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    And for the sake of accuracy I would write VBE instead of VB. In the correct expression, the "-1" is the reason that the classical set of ciurves Ic=f(VCE) does NOT cross the origin (although in most of the graphical representations this effect cannot be seen because of the scaling). – LvW Dec 04 '15 at 09:50
  • What is UB stands for? Upper Bias or higher voltage? – KMC Apr 07 '20 at 08:17