common-emitterimpedance
[![CE]](https://i.stack.imgur.com/ovkiZ.jpg)
I don't understand why \$g_{m}\$ survives. Since \$v_{\pi}\$ is equivalent to \$\Delta V\$ , which is perturbation of signal, which is then set as AC ground. Doesn't that mean \$v_{\pi}= \Delta V = 0\$? Thus make \$g_{m}\$ open for calculation? How can I calculate the output impedance like so? With inspection method assuming \$g_{m}\$ is open, I'm getting \$R_{out} = ( r_{\pi} || R_{E} ) + r_{0} \$ also could you show how to calculate the input impedance as well?
The output impedance isn't the series 400 ohm feeding the device, it is the device in parallel with the series resistor. The device is likely to have an output impedance below 1 ohm. Having said that, at high frequencies it'll get worse so I would recommend placing a 100nF cap in parallel to ensure at HF the net output impedance seen by your IA remains low across all frequencies of interest.
The data sheet tells you that the dynamic impedance is typically 0.15 ohms but you should still put a capacitor in parallel because the impedance will rise for significantly higher frequencies.
EDIT
Here's a screen shot of what I mean: -
Above 1MHz the output impedance is about 10 ohm so it's pretty good but what happens at 10MHz+ is just guesswork. 100nF has an impedance of 1.6 ohms at 1MHz and obviously gets lower as frequency rises. A slight change of mind would be to use 1uF because then the composite impedance of device and capacitor should never really rise above 1 ohm for any frequency up to well over 10MHz.
Remember, that it's not just "your" signal that you might be fighting common mode problems with but, broadband noise at your input.
You need to specify what happens in place of Vmeas when you remove it. Do you substitute an open circuit, a short circuit, or something else?
Best Answer
No. What is happening here is the test voltage Vx will create a current Ix. These are used to calculate the output impedance. What you are not taking into consideration here is that Vx will cause a voltage change at node P. Since the other end of r_pi is connected to ground, this cause a change in delta V. Delta V is not zero.