The small-signal impedance looking into the emitter when base voltage is not fixed

Question

In section 2.3.2 of the 3rd edition of The Art of Electronics, the small-signal impedance looking into the emitter, \$r_\text{e}\$, for the base held at a fixed voltage is defined as "the derivative of \$V_\text{BE}\$ with respect to \$I_\text{C}\$," which approximately equals \$25/I_\text{C}\text{(mA)}\$ using Ebers-Moll equation. But it seems to me that the authors use the same relation for \$r_\text{e}\$ even if the base voltage is not fixed, for example in section 2.3.3, where they calculate the output impedance of the circuit in figure 2.43B.

So my question is: Is \$r_\text{e}\$ defined as \$\frac{\text{d}V_\text{BE}}{\text{d}I_\text{C}}\$, regardless of whether the base voltage is fixed or not?

Answer 1

Well to be more precise \$r_e\$ is defined as \$r_e = \frac{dV_{BE}}{dI_E} = \frac{V_T}{I_E}\$

If for example, you take and plot the \$I_E = f(V_{BE})\$ or \$V_{BE} = f(I_E)\$

You will get this

As you can see \$r_e\$ is the slope of \$V_{BE} = f(I_e)\$.

This means that the \$r_e\$ is a "property" of a transistor alone.
And we can "replace" when doing small-signal analysis the nonlinear base-emitter diode with an ordinary \$r_e\$ resistor. And analyze the circuit using the linear network analysis technique we well know and understand.

Here is an example: pi model of common collector

Additional because \$\frac{I_C}{I_E} = \frac{\beta*I_B}{I_B + \beta * I_B}=\frac{\beta}{\beta +1}\$

And that the typical \$\beta\$ value is much large then \$50\$ we can say that \$I_E \approx I_C\$ and use the simplfid equation \$r_e \approx \frac{25mV}{I_C} = \frac{1}{g_m}\$

Answer 2

Short answer:

The output impedance is always defined under the condition that there is no input signal. The output impedance can be measured/simulated by applying a test signal (voltage or current) into the output node.

For this measurement, the output must not contain any signal portion derived from the (amplified) input. Therefore, the input signal source must be shorted.

Answer 3

I'm looking at 2.43B of the 3rd edition and I think the authors are clearer in the text than in the diagram.

Note that the text says \$\frac{R_S}{\beta+1}+r_e^{\;'}\$. Note that the diagram says \$Z_{_\text{OUT}}=r_e^{\;'}+\frac{Z_{_\text{IN}}}{\beta}\$. Not quite the same thing, but almost. Perhaps the authors might have better written in the diagram that \$Z_{_\text{OUT}}=r_e^{\;'}+\frac{Z_{_\text{IN}}}{\beta+1}\$ and that \$G=\frac{R_{_\text{LOAD}}}{R_{_\text{LOAD}}+Z_{\text{OUT}}}\$. In the diagram, I think they were neglecting the term \$\frac{Z_{_\text{IN}}}{\beta+1}\$, as being negligible, when writing out the equation for \$G\$. That isn't always the case. In the text, they got the details in place but then didn't follow up to compute \$G\$.

(Here, I'm using \$r_e^{\;'}\$ instead of \$r_e\$, as \$r_e\$ is often the model parameter indicating Ohmic resistance at the emitter pin for a BJT and this is not the same thing as the dynamic resistance, \$r_e^{\;'}\$.)

The upshot is this: The source impedance can be reflected to the emitter by dividing by \$\beta+1\$, just as the emitter impedance can be reflected back to the base by multiplying by \$\beta+1\$. It's just two different perspectives of the same thing, depending on what you are looking to figure out. \$r_e^{\;'}\$ resides on the emitter side of this reflection and in series with it. So if you are reflecting backwards from the emitter back to the base, then you take \$r_e^{\;'}\$ in series with whatever is appropriately in series with it at the emitter before reflecting backwards. (Once reflected back, the term \$r_e^{\;'}\$ is often then called \$r_\pi\$, rather than \$\frac{r_e^{\;'}}{\beta+1}\$ which looks more complicated.) And if you are reflecting forwards from the base towards the emitter then you first reflect the base source impedance by dividing by \$\beta+1\$ and then add \$r_e^{\;'}\$.

\$r_e^{\;'}\$ itself is always computed using \$r_e^{\;'}=\frac{\text{d}\,V_{_\text{BE}}}{\text{d}\,I_{_\text{E}}}\$ from the active mode BJT equivalent of the Shockley diode equation (or better, as found in one of the three non-linear Ebers-Moll models: transport, injection, or hybrid-\$\pi\$.)