In this case, I wanted to show how the AC analysis looks like with the complication of an R_E. The analysis would've been much easier without R_E, by having a bypass capacitor across it.

Thanks David. 😊 I'm going to assume that "internal resistance of the AC generator" is what most textbooks refer as r_o (internal resistance between the Collector and Emitter), and the "source resistance" is a series resistance to V_i, mostly indicated as R_i. For R_i, it's presence would create a voltage divider between itself and the parallel resistance R_B and Z_b, where V_i becomes the main source and V_(R_B||Z_b) (i.e. voltage across the parallel of R_B and Z_b), becomes the smaller voltage. If R_i is large compared to R_B||Z_b, then most of V_i will not appear across the input to the amplifier and you'll have a very poor amplifier. This is part of what we call impedance mismatch. The presence of R_i will affect the A_v and Z_i. For r_o, it's going to end up being parallel to R_E and therefore affect all 4 parameters. I don't include r_o to simplify the analysis since in most cases r_o tends to be more than 10 times larger than R_E. Hope that helps!

THANK YOU!

Assuming that your load resistor is R_L and it's connected between the V_o and gnd, then I_o would have to be redefined as the current going through this new resistor. The definition of I_o will change because it is now only a part of beta x I_b, as the rest of beta x I_b flows through R_C. You'll have to apply the Current Divider rule to get a new expression for I_o, which will now contain R_L. BTW, a change in I_o will also affect V_o and therefore A_v. 🙄

THANK YOU PROFESSOR!

Thanks David! Looking what R_E1 affects, you can see that it's in Z_b (which in turn affects Z_i), A_v, and A_i. So depending on your overall design parameters (e.g. required Z_i, A_v, and A_i), then you'll have to set R_E1 to achieve your desired goals. A_i has an additional problem though as it's dependent on beta (which varies with devices of the same type), and r_e (which depends on your biasing currents). But since we don't generally use CE amplifiers as current amplifiers, that's probably a moot point.

THANK YOU!

Hi David. Glad that you liked it! 😊 Do help to fill the survey so that I know what I should cover. Someone had already suggested microcontroller programming, which I'd love to do!

Hi there. Do you mean this: codevillagesolutions.gumlet.io/CC-Amplifier-with-Emitter-Resistor-andf-bootstrapped.png In the meantime, can you fill the survey so that I get a much better idea of what I should focus on: kzbin.info?event=comments&redir_token=QUFFLUhqa2JTaTVNbVJ4TDFDWjJSLUotLWZZSVZqSmlpQXxBQ3Jtc0tsS3lYTXpHeENGcHZvSTJrc1RYbUVuZHVwZGxiZ3BySG5VNEtzNHQ2QXhzcldpdk1rek5fOEI5N1hrbmVsUVFpR2I4YmVnMU5TZ1NZdkF6UEM0U2dpWUVqUUtCd21oNnVUV0dLLVVWRk5OT1BaV0xVTQ&q=https%3A%2F%2Fsurvey.qwary.com%2Ff%2FAABlew%3Fidentifier%3DS_wSzSPnasH9Wc_FT15X0J1BuEcPl5gILyVTYFwsvWc%3D

Hi! In a theoretical sense, they can be whatever you want. So if V1=10V and V2=20V, then, as an unscaled summer, V0=-30V. And as an average circuit, V0=-15V. However, in a physical circuit, the limits of the inputs and the resultant output would depend on what physical OpAmp you end up using. For example, a typical 741 OpAmp can only output up to around +/-20V (see LM741A datasheets). Hope that helps! 😁

Hi Salman. Apologies for getting back so late. If you use Miller's Theorem, you have to understand that the assumption is that no current flows into the input of the amplifier. Obviously that's not the case here as I_B (for biasing) and i_B (from the signal), are not zero amps. So the derived expression via Miller's Theorem will not be the same as per my derivation, as I do not assume zero amp input current. Hope that helps. 😊

Hi David. I've just uploaded one version of it here: kzbin.info/www/bejne/baOmZHqNqqyonbMsi=jTGA-JlxrPck_PNr