0
\$\begingroup\$

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

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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
0
\$\begingroup\$

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

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.