How do I find the input impedance of this common emitter amplifier?
\$V_{CC}\$ is supposed to be at the top of \$R_5\$. I forgot how to do the analysis to find the input impedance. The input goes in through the blocking capacitor, \$C_8\$, the output comes out from the collector of the NPN.
Best Answer
\$V_{CC}\$ is small signal ground so \$R_3\$ is connected to small signal ground along with \$R_4\$ (and therefore they are in parallel with one another). This resistance is in parallel with the impedance looking into the base of the transistor, which is \$r_{\pi} + (\beta + 1)Z_E\$, where \$Z_E = R_6 \parallel C_{19}\$. This total impedance is in series with \$C_8\$. Summarizing:
$$Z_{\text{in}} = \frac{1}{sC_8} + R_3 \parallel R_4 \parallel \left(r_{\pi} + (\beta + 1)\frac{R_6}{sR_6C_{19}+1}\right)$$
Generally, at frequencies of interest the capacitors can be regarded as short circuits, in which case \$Z_E = 0\$ and the input impedance simplifies to
$$Z_{\text{in}} = R_3 \parallel R_4 \parallel r_{\pi}$$