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 enter image description here

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

  • \$\begingroup\$ 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. \$\endgroup\$ – Andy aka Dec 22 '15 at 9:41
  • \$\begingroup\$ Thank you for making a point. I just add the content here. \$\endgroup\$ – binu Dec 23 '15 at 11:23
  • \$\begingroup\$ @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 \$\endgroup\$ – Alex2134 Dec 24 '15 at 15:29
  • \$\begingroup\$ @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. \$\endgroup\$ – Alex2134 Dec 24 '15 at 15:56
  • \$\begingroup\$ @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 \$\endgroup\$ – binu Dec 25 '15 at 8:24

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


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