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I just have solved one of the bjt common emitter question below. However, I'm curious that how can I additionally find those things below:

1) Input and Output Impedance

2) AC Gain

Here is the example and my hand draw solution. However, the question only asked to find Vout/Vin...

Also, my 3rd question is, Does Vout/Vin (I calculated as -3) mean voltage gain?

Another my 4th question is Does voltage gain mean mean AC gain? Or, what is AC gain exaclty? I am not sure

I will be enormously happy if you satisfy my curiosity and show me the right way...

Also, if you have suggestions to solve this question in a more usable way, I am open to listen your suggestions.

I'm just secondary class student, that is why I try to figure out with questioning...

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Emilia
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  • For input impedance: https://electronics.stackexchange.com/questions/159140/how-do-i-find-the-input-impedance-of-a-common-emitter-amplifier – ErikR Jun 03 '21 at 16:50
  • Thank you Erik for that. Also do you know how can I calculate output impedance please? – Emilia Jun 03 '21 at 16:53
  • Yes, - 3 is an AC signal voltage gain. \$A_V = - \frac{\beta R_C}{R_B + r_\pi} \approx \frac{\beta R_C}{R_B} \approx -3 V/V\$ And can be calauclate using this : Rin = Vi/Ii Can you do it? As for the Rout see this https://electronics.stackexchange.com/questions/464496/measuring-input-and-output-resistance-of-the-hybrid-model/464516#464516 or this – G36 Jun 03 '21 at 18:46
  • And I highly recommend you to read this pdf http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/section%205_6%20%20Small%20Signal%20Operation%20and%20Models%20lecture.pdf – G36 Jun 03 '21 at 18:50
  • @G36 I am really sorry but still I could not figure out the technique which you used for calculating Rin and Rout. I really want to understand the logic of this calculation. – Emilia Jun 04 '21 at 03:40
  • @G36 also that was not a homework, do not get me wrong please. It was an example which I took from youtube. Here is a link: https://www.youtube.com/watch?v=mJqDKyEiJrI&list=PL12v34PfMhRSzQrn6fv98Fh4mltiAvv4F&index=5 – Emilia Jun 04 '21 at 03:41
  • @G36 wow, the lecture note which you sent me, it looks so sophisticated and well designed. Today, I am starting to study with reading those lecture notes to understand better. May I ask you, do you have more lecture notes for "BJT Common Emitter" case please? Thank you so much again... ^_^ – Emilia Jun 04 '21 at 03:43
  • But you did manage to find the DC base current. Now to find Rin you need to do exactly the same thing as you did before , but now you need to find \$i_b\$ on a small-signal AC equivalent cirtcuit. For your circuit we have: \$ \Large v_i = i_bR_B + i_b r_\pi = i_b(R_B + r_\pi)\$ Thus, \$\Large R_{in} = \frac{v_i}{i_{in}} = R_B\ + r_\pi\$ is that clear to you? – G36 Jun 04 '21 at 18:34
  • Look here https://electronics.stackexchange.com/questions/355899/how-is-possible-that-with-same-ibase-there-is-more-than-one-vce/355955#355955 and here http://www.ittc.ku.edu/~jstiles/412/handouts/5.6%20Small%20Signal%20Operation%20and%20Models/Example%20A%20Small%20Signal%20Analysis%20of%20a%20BJT%20Amp.pdf or this https://electronics.stackexchange.com/questions/476659/kvl-equations-for-this-small-signal-model/476666#476666 and this – G36 Jun 04 '21 at 18:39
  • Ah yeah now that is clear as drinkable water. You saved me... Finally I understood... – Emilia Jun 04 '21 at 19:06
  • And to be able to find Rout you need to find the resistance "seen" from the "load resistance point of view" In your circuit, there is no load (open circuit RL = 00). Now let us look at Rout. If we are looking from the load perspective we can see two paths for an AC current to flow: https://i.stack.imgur.com/37b8O.png The first one ---> into Rc resistor and the second one into the BJT's collector terminal. If we ignore the Early effect. The Collector AC resistance (dynamic) is equal to infinity. Thus Rout must be equal to Rc resistance. – G36 Jun 04 '21 at 19:45
  • Or we can use this small-signal diagram and solve for Rout. W can do this by placing a test voltage source (Vx) at the output (we set Vin = 0V) and solve for Ix. And Rout = Vx/Ix https://i.stack.imgur.com/o8OiD.png And I hope that you see why Rout = Rc, because Vin = 0V there is no base current. Therefore Vbe = 0V and Ic = gm*Vbe = 0. So, Ix = Vx/Rc ---> Rout = Vx/Ix = Rc. Any additional questions about it? – G36 Jun 04 '21 at 19:45
  • @G36 I clearly understood with thanks to you detailed and kind explanation. I don't know how to thank you... – Emilia Jun 05 '21 at 12:57

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