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.
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.
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:
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.