# 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. – Andy aka Dec 22 '15 at 9:41
• Thank you for making a point. I just add the content here. – binu Dec 23 '15 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 – Alex2134 Dec 24 '15 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. – Alex2134 Dec 24 '15 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 Dec 25 '15 at 8:24

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