Based on theory when you want to find the output resistance of a circuit with dependent sources you can't turn them off so you have to add another source of known value in the points of interest and just compute the resistance. So far so good.

My issue is with the computation of the input resistance of a BJT transistor, whose equivalent circuit includes a dependent current source as seen below.

enter image description here

Why whenever I see a textbook computing the input resistance of a BJT it always just sums the resistances connected between BE? This is a procedure for when there are only independent sources.

For example in the case of the single transistor illustrated above the input resistance is \$R_{in}=r_{\pi}\$.

In all the other circuits without transistors we have to add a new source as I have mentioned and compute the resistance but in here we just work without caring about the dependent source. Why is that?

Basically if we consider a voltage source of \$1V\$ as an input then the current through BE would be \$i_{be}\$ so it is easily realized, as E is usually grounded, that the input resistance is

$$ R_{in} = \frac{u_{be}}{i_{be}}=r_{\pi} $$

So is this procedure with the addition of a new source necessary but omitted from the textbooks because it is obvious (as the voltage in the input would be seen in the BE terminals making the resistances connected there easily computable)? Or they actually just add the resistances without caring about the dependent source for some other reason?

  • \$\begingroup\$ Can you give an example of a circuit where you have to "add a new source"? \$\endgroup\$ – The Photon Jun 8 '19 at 0:03
  • \$\begingroup\$ Also, your first sentence talks about finding the output resistance of a circuit, but the rest of the question seems to be asking about finding the input resistance of a common-emitter transistor circuit. Which one do you actually want an answer to address? \$\endgroup\$ – The Photon Jun 8 '19 at 0:06
  • \$\begingroup\$ Sorry, resistance. Edited. \$\endgroup\$ – The Photon Jun 8 '19 at 0:10
  • \$\begingroup\$ @ThePhoton Oh. I just mentioned the easiest example of an input resistance of a circuit with a transistor. I actually know all the input resistance formulas so it hasn't got to do with the actual formulas but the way they are computed. This question is about circuit theory mostly not about specific examples and actual transistor circuits. \$\endgroup\$ – Adam Jun 8 '19 at 0:13

Or they actually just add the resistances without caring about the dependent source for some other reason?

The \$g_m\$ source produces current in the C-E branch, not in the B-E branch. So it doesn't contribute to determining the B-E equivalent resistance.

If you wanted the equivalent input resistance of a common-emitter amplifier with emitter degeneration (rather than just the transistor itself) ,

enter image description here

then you would certainly need to consider the effect of the \$g_m\$ source.

In comments you mentioned you are interested in a circuit theory answer. One thing to notice is that the transistor model you presented is the DC form of the hybrid pi model. It is designed so that you can write

$$\begin{bmatrix}v_b \\ i_c\end{bmatrix} = \begin{bmatrix}r_\pi & 0 \\ g_m & 1/r_o\end{bmatrix} \begin{bmatrix}i_b \\ v_c\end{bmatrix}$$

(using lower case variables because these are small signal variations around the bias point)

That is, the model is designed so that \$V_b\$ only depends on \$I_b\$, while \$I_c\$ depends only on \$I_b\$ and \$V_c\$. More to the point, the \$r_\pi\$ element only affects the B-E port, and \$g_m\$ and \$r_o\$ only affect the C-E port.

  • \$\begingroup\$ The matrix is a great addition and explains everything. So basically what is actually happening is that they do follow the normal procedure of placing a source and determining the input resistance but they just omit it due to the nature of the transistor and the independence of the current of the BE branch. Thanks that makes sense! \$\endgroup\$ – Adam Jun 8 '19 at 1:15

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