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The following CE Amplifier Small-Signal Equivalent Circuit and nodal equations are from this site.

enter image description here

I'm completely lost how they form the above KCL Equations for the nodes B, C and D.

For example, for B node the currents \$\frac{v_i}{R_s}\$ and \$\frac{v_e}{r_{\pi}}\$ does not make sense to me for a KCL equation at node B. \$\frac{v_i}{R_s}\$ is a current when \$V_i\$ and \$R_s\$ is shorted. What is that to do with the KCL?

Similarly for node C what is the current \$\frac{v_e}{r_o}\$? Again this equation doesn't make sense to form a KCL for node C.

What are the current directions in these cases? And in this model \$R_C\$ is shorted but why \$V_i\$ is not shorted to ground?(as done in a any small signal analysis) Can someone explain how these equations form a bit in more detail?

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  • \$\begingroup\$ I think Tim did a fine job, but perhaps he didn't elaborate enough for you. I can walk you through the process for node B:$$\begin{align*} \frac{v_\text{b}}{R_\text{S}}+\frac{v_\text{b}}{r_\pi}&=\frac{v_i}{R_\text{S}}+\frac{v_\text{e}}{r_\pi}\\\\ v_\text{b}\left(\frac{1}{R_\text{S}}+\frac{1}{r_\pi}\right)&=\frac{v_i}{R_\text{S}}+\frac{v_\text{e}}{r_\pi}\\\\v_\text{b}\frac{1}{\frac{1}{\frac{1}{R_\text{S}}+\frac{1}{r_\pi}}}&=\frac{v_i}{R_\text{S}}+\frac{v_\text{e}}{r_\pi}\\\\\frac{v_\text{b}}{R_\text{S}\mid\mid r_\pi}&=\frac{v_i}{R_\text{S}}+\frac{v_\text{e}}{r_\pi}\end{align*}$$ \$\endgroup\$
    – jonk
    Commented Feb 12, 2019 at 6:40

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They're skipping steps. The first step in, for example, line B is: $$(v_b - V_i)\frac{1}{R_s} + (v_b - v_e)\frac{1}{r_{\pi}} = 0$$

If you then note that \$\frac{1}{R_s \parallel r_{\pi}} = \frac{1}{R_s} + \frac{1}{r_{\pi}}\$, can you see how this can be rearranged to get what they have? Does it look more like the KCL that you're familiar with?

\$V_i\$ is not "shorted as in small signal analysis" because this is small-signal analysis, and you should assume that \$V_i\$ is just the AC component of the actual voltage on the base.

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  • \$\begingroup\$ How about for C node what is the current ve/ro? Can you also show where does the equation for node C come from and ve/ro? And why vi is not shorted as done in small signal analysis? \$\endgroup\$
    – GNZ
    Commented Feb 12, 2019 at 1:54
  • \$\begingroup\$ Does line B now make sense to you? In light of it making sense, what do you get when you attempt to do the same thing at line C? \$\endgroup\$
    – TimWescott
    Commented Feb 12, 2019 at 2:05
  • \$\begingroup\$ You didn't write a KCL equation. There is no "=" sign. So it is still implicit. I also dont know how they form the KCL equation for node C.Again Im also asking Why Vi is not shorted to ground as in Thevenin and small signal. \$\endgroup\$
    – GNZ
    Commented Feb 12, 2019 at 2:13
  • \$\begingroup\$ "There is no "=" sign." Pardon me. Clear now? \$\endgroup\$
    – TimWescott
    Commented Feb 12, 2019 at 4:09
  • \$\begingroup\$ "as in Thevenin and small signal" What? I'm sorry, no. Just no. I'm not sure who taught you this stuff or how badly, but that is a proper small-signal representation of a common emitter transistor amplifier. It may be different from what you're used to, but if so, that means you just need to stretch your brain to encompass it. \$\endgroup\$
    – TimWescott
    Commented Feb 12, 2019 at 4:10

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