# How is the solution for $\alpha$ and $\beta$ current gain of a BJT derived?

I'm just an enthusiast, and not too good at maths; I'm exploring constant current source circuits, in doing so I came across two equations for BJT current gain:

Could someone help me understand how $\alpha$ and $\beta$ are derived to the equations below:

$\alpha=\frac{\beta}{\beta+1}$ and $\beta=\frac{\alpha}{1-\alpha}$

I know that:

$I_c=\alpha I_e$ and $I_c=\beta I_b$

Any help would be much appreciated.

Thanks Alex

Check this out. Found some rich content here electronics-tutorials-transistors

α and β Relationship in a NPN Transistor

The value of Beta for most standard NPN transistors can be found in the manufactures data sheets.

• Why don't you copy and paste the vital parts of the linked page? This answer is correct at the moment but if that website dies this answer becomes useless. Commented Dec 22, 2015 at 9:41
• Thank you for making a point. I just add the content here.
– binu
Commented Dec 23, 2015 at 11:23
• @binu Could you tell me mathematically, I know this is probably basic algebra, but I'm not familiar, how $I_B=I_E - \alpha I_E$ solves to $I_B=I_E(1 - \alpha)$ ? Really helpful insight this for me. Thanks Alex Commented Dec 24, 2015 at 15:29
• @binu I presume, thinking about this if $I_B=I_E - \alpha I_E$ which is effectively $I_B=1I_E - \alpha I_E$ to give $I_B=I_E(1 - \alpha)$ makes sense. Commented Dec 24, 2015 at 15:56
• @Alex2134 Yes it is. After getting out the common factor IE out, IB=IE−αIE can be factorize to IB=IE(1−α). Further : factoring an algebraic expression a(b + c) = ab + ac , ab + ac = a(b + c) , in here a = IE, b = 1, c = α
– binu
Commented Dec 25, 2015 at 8:24

You are missing one more equation, which is that $I_e = I_c + I_b$