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When shorting the Vin and Cs and get Thevenin Resistance Rth using the test source from Ce's point of view.

$$ Rth = R_E || \frac{r\pi+ R_s||R_B}{\beta_o + 1} $$

I wonder how the test source was used and how these results came out.

  • Take a look here https://electronics.stackexchange.com/questions/429716/arriving-at-a-wrong-output-impedance-for-a-bjt-emitter-follower-configuration-ci/429726#429726 And you should start reading from "But for the emitter follower, we have a different situation". Any additional questions? – G36 Dec 13 '20 at 07:31
  • I gave an answer but I just noticed that your expression contains \$\beta_0\$ while your model uses a transconductance \$g_m\$. Some cleanup seems necessary and the controlled current source needs to become \$\beta_0i_b\$ with \$i_b\$ entering \$r_\pi\$ from the top. – Verbal Kint Dec 13 '20 at 11:22

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In this type of exercise, you are asked to find the resistance \$R\$ driving capacitor \$C_E\$ to determine the pole and, later on, the zero associated with it. Generally speaking, when you are tasked with a resistance or impedance determination, you have several choices: a) can work by inspection meaning you can infer the resistance or impedance without writing a line of algebra just by "reading" the schematic diagram or b) you install a test current source \$I_T\$ connected across the port at which the impedance must be determined. This is called a driving port impedance determination or DPI. This current source stimulates the port and generates a response voltage \$V_T\$ across its terminals. The resistance \$R\$ or the impedance \$Z\$ you want is simply \$Z(s)=\frac{V_T(s)}{I_T(s)}\$.

In your case, you will determine the natural time constant of this circuit by zeroing the excitation \$V_{in}\$ meaning you replace the source by a short circuit. Then, you temporarily remove capacitor \$C_E\$ and install the test current source across its connecting terminals:

enter image description here

The key is really rearranging the circuit in its simplest form by grouping series-parallel resistors and lump them into a single element. Then, the cool thing to spot is the fact the resistance \$R\$ you want will be in the form of \$R=R_{int}||R_E\$ as \$R_E\$ is in parallel with capacitor \$C_E\$. You can therefore temporarily disconnect \$R_E\$ from the circuit, determine \$R_{int}\$ as shown in the picture and bring \$R_E\$ back for the final result. A few lines of algebra should you should lead there but I leave it to you from now.

Verbal Kint
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