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We were asked to get the β’ (the β of the single BJT equivalent), VB’E’on (the voltage from B’ to E’ when conducting),VC’E’sat (the minimum voltage that can exist between C’ and E’), Isat (max possible value of I), and the largest value of resistor R for Isat to flow using this circuit:

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The given was β = 25, VBEon = 0.7 V, and VCEsat = 0.2 V

Somebody said that the β of both transistors are equal so inner β = 25 and outer β = 25, but I don’t know why they are equal.

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  • \$\begingroup\$ mama, In reading what you wrote about what's being asked, it appears to me that you also have already been prepared for the simplified version of the Ebers-Moll large scale model that uses a version of the Shockley diode equation. (You mention Isat.) Are you supposed to apply a model where Vbe varies with collector current? \$\endgroup\$ Apr 20, 2023 at 19:42

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The assumption is that both BJT transistors are matched, hence they have similar characteristics.

If you go from B' to E', you'll find 2 VBE junctions in series. We're told that VBEon is 0.7V, that's the minimum voltage required such that both transistors start conducting. You can think of these junctions as DC voltage sources. Hence, it's as if 2 voltage sources are in series from B's to E', i.e. you have 2VBEon or 1.4V, which is the answer for VB'E'on.

The VC'E'sat of the pair of BJTs is equal to VBEon + VCEsat. Why? Going from E', we need only a VCEsat towards to collector of the outermost transistor. However, This is not enough to turn on the inner most transistor, as its emitter voltage is raised by the VBE of the outermost transistor. Hence, we need to go the longer path, from E' through the innermost transistor emitter towards C': this is effectively VCE+VBE or 0.9V.

This arrangement is commonly known as a Darlington pair, which is used in order to boost current gain from B' to E', which is approximately beta^2 (assuming both betas are equal). In your case, total beta = 25*25 = 625.

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  • \$\begingroup\$ So the β‘ of the circuit is just β^2? So that’s β‘ = 625? \$\endgroup\$
    – mama b
    Apr 20, 2023 at 17:02
  • \$\begingroup\$ Yes - the typical properties of this Darlington combination are (1) beta=beta1*beta2 . Hence, the only advantage: Higher input resistance if compared with a single transistor having the same transconductance (resp. voltage gain). \$\endgroup\$
    – LvW
    Apr 20, 2023 at 17:18
  • \$\begingroup\$ @mamab does my answer satisfy your question? Is there any other doubt? \$\endgroup\$
    – Designalog
    Apr 20, 2023 at 20:37
  • \$\begingroup\$ The maximum current will be approximately (10-0.9)/(1k+100) = 8.27 mA. You can then determine the resistances of the voltage divider that will supply the required voltage at B'. The base current will be practically negligible. \$\endgroup\$
    – PStechPaul
    Apr 20, 2023 at 21:13
  • \$\begingroup\$ I tried researching for darlington pair and the equation for beta is B = B1 + B2 + BIB2. So i got 675 \$\endgroup\$
    – mama b
    Apr 21, 2023 at 4:35
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The question data are poorly written. A better way is

$$\beta_1=\beta_2=25$$

The \$\beta\$s need not be the same but because only one is given they are assumed to be the same.

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