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I have a slight problem trying to understand how to determine the output impedance of a emitter-follower configuration using the re model for small signal analysis. I also added the contribution of \$r_o\$.

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From my understanding of circuit theory and analysis theorems, to determine the output impedance, I have to set the input \$V_i = 0\text{ V}\$ (\$R_b\$ has no influence now) and keep \$V_o\$ at a constant value (please correct me if I am wrong, I mostly study on my own without having someone to give me feedback on my understanding). I keep the controlled current source (for there still is a current going though the base node), \$\beta r_e\$, \$r_o\$ and \$R_e\$.

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My first instinct would be to calculate $$Zo = \beta r_e \parallel r_o \parallel R_e$$

But I find in the book that $$Zo = \frac{\beta r_e}{\beta + 1} \parallel r_o \parallel R_e$$

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What exactly is it that I don't understand?

Null
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Stefan Tudora
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  • Little r e (aka \$r_E\$) is the output impedance at the emitter and it is in parallel with \$R_E\$. – Andy aka Feb 17 '21 at 17:25
  • This problem was discussed recently, see here: https://electronics.stackexchange.com/questions/331671/how-to-calculate-the-output-impedance-of-the-bjt-in-a-common-collector-amplifier?rq=1 – LvW Feb 17 '21 at 17:29
  • @LvW, that is a different problem. I am using small signal analysis using re model. The link you gave me is about h-parameters. – Stefan Tudora Feb 17 '21 at 17:34
  • OK - the question contains h-parameters. But my answer - for example - uses the re model. (By the way: You can express h-parameters with re=1/gm and vice versa). – LvW Feb 17 '21 at 17:45
  • @LvW, It's just that the book I am reading right now started modeling BJTs using the re model first, though I understand your point of view for I have seen a lot of books using the hybrid parameters first. Maybe I shall skip right to the part where my book addresses them and return later to this part. I asked the question because I was assured that I can do the analysis using basic circuit theorems (like source transformation). – Stefan Tudora Feb 17 '21 at 17:53
  • look here https://electronics.stackexchange.com/questions/429716/arriving-at-a-wrong-output-impedance-for-a-bjt-emitter-follower-configuration-ci/429726#429726 any additional qestions? – G36 Feb 17 '21 at 21:06
  • And this https://electronics.stackexchange.com/questions/512131/frequency-response-of-bjt-frequencies-with-ce-bypassed#comment1313075_512131 – G36 Feb 17 '21 at 21:16
  • @G36, thank you! The first link you provided did answer my question (now that I think about it, it was just a problem of applying Thevenin's theorem). Should I delete my question considering it's based on a problem already solved? – Stefan Tudora Feb 18 '21 at 09:45
  • @G36, with reference to your explanations on [link](https://electronics.stackexchange.com/questions/429716/arriving-at-a-wrong-output-impedance-for-a-bjt-emitter-follower-configuration-ci/429726#429726), I agree with the **Emitter Follower** calculations, but what I found strange is the calculation about **CE amplifier**...Specifically, my question is how do you suppose that Ib=0 (which should involve that VE=0)? Could you please elaborate this point? – barrow Feb 19 '21 at 10:41
  • @barrow Hi, you need to distinguish between real-world situations from the small-signal approximation. In this case (AC analysis case), I used the simplified BJT's model that does not include the base-width modulation (Early effect) thus, no \$r_o\$ resistance on the schematic was shown. So, any change in \$v_{ce}\$ voltage has no effect on the base current \$i_b\$. Thus, the AC component of a base current is \$0A\$ due to a change in \$v_{ce}\$ voltage for Vin = 0V. Do you see it? – G36 Feb 19 '21 at 19:43
  • But if you want to include the Early effect then please read this https://electronics.stackexchange.com/questions/377737/finding-output-resistance-of-cb-amplifier-with-ro/377769#377769 – G36 Feb 19 '21 at 19:43
  • @G36, from the small-signal model you used in the picture, what I see is that effectively you did not take into account the Early effect, and infact the **ro** is absent, but that does not have any effect on **ib**. Therefore, following a change for **vce**, there will be a change on the test current, which will be splitted on **RC** and on **ic**. Agreed? – barrow Feb 19 '21 at 23:35
  • @G36, On the other hand, during small-signal we have to follow the AC load line, and with a vce change, there will be a change on ic, and therefore a change on ib. – barrow Feb 19 '21 at 23:44
  • @barrow No, I disagree. Can you explain to me how any change in "test voltage" can change the base current? If the BJT without **ro** resistance is a unidirectional device. This means that to change the collector current you need to change the **Vbe** (V_pi) voltage in the first place (BJT is a dependent current source). How can you have a non-zero **Vbe** in this circuit? https://i.stack.imgur.com/KDCuw.png For simplicity assume **Vx = 1V, Rc =1k, Re1 = r_pi = 1k** – G36 Feb 20 '21 at 12:32
  • @@G36, Thanks a lot, you're right! and I was way off base! – barrow Feb 20 '21 at 17:57

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