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In the saturation region for an NPN BJT,

$$I_C = I_S\cdot e^{\frac{V_{BE}}{V_T}} - I_{SC}\cdot e^{\frac{V_{BE}}{V_T}}$$

Since \$\small I_C\$ is approximately equal to \$\small I_E\$, will \$\small I_E\$ be the same exact equation, or is $$I_E = I_{SC}\cdot e^{\frac{V_{BC}}{V_T}} - I_S\cdot e^{\frac{V_{BE}}{V_T}}$$ ?

ocrdu
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HDY
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1 Answers1

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When considering the common-emitter configuration of a BJT, I directly obtained the analytical link between the collector current Ic and the collector-emitter voltage Vce, considering the base current Ib as a parameter.

enter image description here enter image description here enter image description here

Franc
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  • please cite the source of any 3rd party material you paste into your answer. – Neil_UK Jan 27 '24 at 09:01
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    FYI, all-image answers are discouraged because they aren't searchable, they can't be formatted by site and user style sheets, they are often poorly readable (I notice these are line art in JPG format, a poor choice), and they are difficult to edit or otherwise maintain by other users. – Tim Williams Jan 27 '24 at 09:05
  • The photos published are taken from a MATHCAD worksheet I wrote and then transformed into a pdf file. The formulas obtained are the result of my calculations. – Franc Jan 27 '24 at 09:07
  • The photos are in fact files with jpg extension and saved in Windows Paint. – Franc Jan 27 '24 at 09:40
  • Try to use PNG instead of JPEG. – G36 Jan 27 '24 at 09:59