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This is about the 'TEE' Model in BJT small signal analysis on CB configuration. During the lectures, the one I learn is this model;

enter image description here

But in some books I found a different model with a base resistance. enter image description here

If I do the calculations for common base configuration without the base resistance, I end up having a short circuit path from Base to ground. Even in examples this base resistance is not small so that we can neglect (Where the model uses a base resistance). Can anyone just say what is the importance in having a base resistance in this Tee model? :)

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  • \$\begingroup\$ Ok, the first picture came from Sedra Smith. The latter? (Note that the first is a npn and the second is a pnp) \$\endgroup\$
    – Antonio
    Jun 9, 2016 at 12:13
  • \$\begingroup\$ Next is from schaum's outline series! \$\endgroup\$
    – Padmal
    Jun 10, 2016 at 14:38

1 Answer 1

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Blogger - I am not surprised that you feel confused. I only can recommend NOT to use the T-model because it is not a good basis for analyzing transistor circuits (for my opinion). Better use the hybrid- or y-parameter small-signal model.

Disadvantage of the T-model (first picture): (1) In common-emitter configuration it seems that the input resistance would be re (which is identical to 1/gm). However, that is not the case. Hence, there is a contradiction between formulas and equivalent signal diagram. (2) A similar problem does exist in common-base configuration. The base node must be grounded (output node of the current source) - nevertheless, only a part of the current (base current) goes into the grounded node. This is NOT in accordance with Kirchhoffs current law.

As far as the base resistance (second diagram) is concerned, I cannot answer your question without knowing the value (expression) of this part. The input resistance in common base configuration is 1/gm=re. Hence, rb=0. Of course, there is a small (ohmic) path resiostance between the base node and an inner point of the transistor body. But this value is always neglected.

Finally, I do not know any advantage of this (bloody) T-model if compared with all other equivalent circuit diagrams. The basic problem (and the cause of the inconsistencies) of the T-model is the fact that the transconductance gm is modeled as an ohmic (dynamic) resistance re=1/gm. But that`s not correct: It is not a resistive element. The transconductance connects the input voltage with the output current, and the ratio is NOT identical to a resistance with two nodes.

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  • \$\begingroup\$ Thank you for this perfect explanation @LvW! One more thing to get clarified. In T model we're using r_e for the input resistance while in hybrid-pi model we're using r_pi. Other than in the T model, we wouldn't get to use r_e would we? \$\endgroup\$
    – Padmal
    Jun 10, 2016 at 14:43
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    \$\begingroup\$ In the pi-model the role of r_e is taken over by the transconductance gm which is the controlling factor for the controlled current source at the output (remember: r_e=1/gm). That is the reason, I consider the pi-model much more correct and realistic than the T-model which has the mentioned deficiencies. \$\endgroup\$
    – LvW
    Jun 10, 2016 at 15:35
  • \$\begingroup\$ So we're setting r_pi as (1+b)r_e \$\endgroup\$
    – Padmal
    Jun 10, 2016 at 15:36
  • \$\begingroup\$ Yes - that`s correct. \$\endgroup\$
    – LvW
    Jun 10, 2016 at 16:52
  • \$\begingroup\$ It was confusing to me! Now it's all clear! :) You're better than my lecturer! \$\endgroup\$
    – Padmal
    Jun 11, 2016 at 1:06

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