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Feb 9, 2021 at 17:01 vote accept electronicah
Feb 9, 2021 at 16:20 history edited Andy aka CC BY-SA 4.0
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Feb 9, 2021 at 16:18 comment added Andy aka @electronicah not quite so fast, this Q and A should be closed down so, you need to consider accepting my answer. The link shows you how to do it but, if you want the simple version try this
Feb 9, 2021 at 15:31 comment added electronicah I think I got what I need for now, I will need to go back and check math model to derive different loop gains.
Feb 9, 2021 at 14:44 comment added Andy aka @electronicah what to do next for completing this question and answer session
Feb 9, 2021 at 14:39 comment added electronicah Thanks for feedback.
Feb 9, 2021 at 14:31 comment added Andy aka Q1 is an emitter follower hence any voltage on its base will be 99% transferred to its emitter hence, virtually 100% negative feedback.
Feb 9, 2021 at 14:13 comment added electronicah I do understand that it will be virtual ground, but I am trying to understand the reason why? actually simulation supports your opinion. I just did not see the feedback in my analysis and trying to understand how can I correlate small signal analysis "KCL" to the feedback loop.
Feb 9, 2021 at 13:58 comment added Andy aka @electronicah did you forget something in your comment? Anyway, you can do KCL to your hearts content but it still won't alter the fact the the input is connected to a virtual earth and that has theoretically zero ohms input impedance. The emitter follower formed by Q1 will have an output impedance (at the emitter) of only a few tens of ohms so, I'm not sure what else you might expect.
Feb 9, 2021 at 13:56 comment added electronicah I am talking about the small signal model of both transistors, Doing KCL on Q1 emitter, shows impedance as follows 1) R1 2) Rpi of Q2 3) re of Q1+ reflected R3 into the emitter. all should be in parallel. Is my understanding is correct or not.
Feb 9, 2021 at 12:27 comment added Andy aka Link to virtual ground. Regards this: I am assuming that I am connecting to ground through Rpi2 - I have no idea what this means @electronicah
Feb 9, 2021 at 12:07 comment added electronicah Initially I was tying to create a common base topology with Q1, and installed Q2 for biasing purposes to generate a constant current across R1 with diode drop. It seems the circuit behavior different now. Could you please further explain the virtual ground, I am assuming that I am connecting to ground through Rpi2. Further explanation will be greatly appreciated.
Feb 9, 2021 at 10:42 history answered Andy aka CC BY-SA 4.0