A question about the exact formula for the collector current

Regarding the collector current Ic equation in the below figure:

The above figure tells:

Ic = Is * (exp(Vbe/Ut)-1)

But I thought it should be:

Ic = α * Is * (exp(Vbe/Ut)-1) + Icbo

1-) Is α and Icbo neglected in the figure?

2-) Is the reverse saturation current Is the reverse saturation current between the base-emitter junction? Is Is that the same thing called Ieo in some texts?

• I'm not familiar with $I_{EO}$, itself. Is it just another name for $I_{EBO}$? In any case, I recommend studying Level 1 BJT Model. Pretty much all the details are there for you. (I am assuming you mean 'area' when you write $a$. You will see that term in the above document.) I find plotting these things on a log chart helps a great deal in understanding the factors, though. For example, $I_S$ is just what you get at the y-axis when you apply a ruler to the flat part in region II curve of the BJT operating mode. – jonk Feb 10 '18 at 18:18
• No that is alpha – floppy380 Feb 10 '18 at 18:27
• Okay. Then I recommend reading the document I gave you. All answers are in there and detailed well. Start with the section on DC operation. – jonk Feb 10 '18 at 18:45
• In that document there is not a single term called "reverse saturation current" The terminology is horrible. – floppy380 Feb 10 '18 at 18:47
• Sorry, then. I started to consider what I wanted to write about this, because what you wrote isn't consistent with the way I think of things. This means we'd be at cross-purposes until we'd established mutual foundations we understood and agreed to share. And then we could move forward. Then I just gave up. Too much work establishing terms. I apologize for my lack of energy. – jonk Feb 10 '18 at 19:04

As per Ebers–Moll Model of an NPN transistor:

$$I_E = I_{ES} (e^{V_{BE}/V_T} -1) -\alpha_RI_{CS}(e^{V_{BC}/V_T}-1)$$ $$I_C = \alpha_FI_{ES} (e^{V_{BE}/V_T} -1) -I_{CS}(e^{V_{BC}/V_T}-1)$$ where:

$I_{ES}$ = reverse saturation current of Base-Emitter Junction.

$\alpha_R$ = reverse current gain

$\alpha_F$ = foward current gain

$I_{CS}$ = reverse saturation current of Base-Collector Junction.

Approximated Ebers-Moll Models look like: $$I_E \approx I_{ES} (e^{V_{BE}/V_T} -1)$$ $$I_C \approx \alpha_FI_{ES} (e^{V_{BE}/V_T} -1)$$ which satisfies the normal convention: $I_C = \alpha_F I_E$

Therefore, whatever source you took the figure from, By $I_S$ they mean, $\alpha_F I_{ES}$.

If you are looking for where is $I_{CBO}$, it is there in the original equation of $I_C$ , the term --> $-I_{CS}(e^{V_{BC}/V_T}-1)$ is nothing but equals to $+ I_{CBO}$.

From the original Ebers-Moll Model equations, you can see that even $I_C = \alpha_F I_E + I_{CBO}$ is an approximation, as it neglects the second term in the expression for $I_E$.