Timeline for Common base bias reduces input impedance
Current License: CC BY-SA 4.0
13 events
when toggle format | what | by | license | comment | |
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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 |