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I was going through the AC equivalent circuit of Common Collector Amplifier from ELECTRONIC PRINCIPLES BY MALVINO 3RD EDITION and I came across the analysis for the output impedance of the circuit. Now if we look intuitively in the diagram from FIG 8-4(a) we can easily reach figure 8-4 (b) no problem at that...

Then from Figure 8-4(b) we reach 8-4(c) by Thevenization at the base...

But we cant get the diagram of (d) from (c) by Thevenization at the point Vout ....

where am I wrong? please correct me or show me the logic. They have applied KVL in the base emitter loop and then from the expression of the current they have constructed the circuit of figure (d) which has the beta-ac.

But Thevenization gives (Rs||R1||R2 +re')||RE ....

Please help me... I am new to electronics. Do i need to apply Zout=Vout/Iout... What is the problem in applying Thevenin's theorem?

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Update: I felt that due to the approximations of ic nearly equals ie and hence ib nearly equals ie/ß, we need to remodel the circuit of figure 8-4 (c) a bit as shown

schematic

simulate this circuit – Schematic created using CircuitLab

The modified version is:

schematic

simulate this circuit

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  • \$\begingroup\$ The current source just as infinite impedance (in theory, not in practice here) and so they just removed it and then re-arranged the remaining resistances. I'm assuming here that they are using \$r_e^{'}\$ for the dynamic emitter impedance and not the Ohmic one. \$\endgroup\$ – jonk Dec 20 '19 at 23:22
  • \$\begingroup\$ Yes they are using re' for the emitter diode ac resistance... Now we see that Ic is a current source which is ideal. Which is ok as you pointed out... Now for thevenins theorem, we find equivalent resistance by switching off current source i.e. removing the current source and switching off voltage source by shorting it... We get (re'+Rs||R1||R2)||RE but I am not getting the beta ac in the simple Thevenins approach... \$\endgroup\$ – Abhishek Ghosh Dec 21 '19 at 4:57
  • \$\begingroup\$ Before I consider writing anything, have you read my reply here? Could you go through it for a moment and make sure that it doesn't help at all? If not, I'll see about saying something here. But perhaps you can tell me what's wrong with what I wrote there. Keep in mind that I use \$r_e\$ for the dynamic AC emitter resistance and use \$r_e^{'}\$ for the Ohmic resistance. Just FYI. \$\endgroup\$ – jonk Dec 21 '19 at 7:03
  • \$\begingroup\$ @jonk i went through your well written answer, but the thing is that many of the mathematics and terms used there are unknown to me, as I am a beginner and that's why I sort of picked up an easy book for beginning. Now the thing is that in the book they have found out Zout here in the way as they have shown but in the proof of CE Zout they did using Thevenins. The thing is that I am having problem with the Thevenins proof. I don't feel that the current source being dependent upon the base current we can't simply just remove it for Rthevenin. I hope I could make you understand my problem. \$\endgroup\$ – Abhishek Ghosh Dec 21 '19 at 20:00
  • \$\begingroup\$ Call a small change of \$V_\text{IN}\$ the variable \$\text{d}x\$. In step (b) this becomes a new variable \$\text{d}y=\frac{R_1\mid\mid R_2}{R_\text{S}+\left(R_1\mid\mid R_2\right)}\text{d}x\$ and new Thevenin source resistance \$R_\text{TH}=R_\text{S}\mid\mid R_1\mid\mid R_2\$. This is then in series with \$r_e\$ and \$R_\text{E}\$, with \$V_\text{OUT}\$ taken at the top of \$R_\text{E}\$. So at \$V_\text{OUT}\$, it's \$\text{d}z=\frac{R_\text{E}}{r_e+R_\text{E}+\frac{R_\text{TH}}{\beta}}\text{d}y\$. Now plug in \$\text{d}y\$ there and simplify. The collector current doesn't even show up. \$\endgroup\$ – jonk Dec 22 '19 at 6:02

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