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These figures are from book by Sedra/smith.

I have a doubt regarding input impedence or resistance (Rin) shown in figure.

In book it is written that "The input resistance (Rin) represents the loading effect of the amplifier input on the signal source. It is found from Rin=Vi/Ii"(in figure).

it started with common emitter configuration (small signal π model shown in 1st figure ) and then author introduces another model( CE amplifier with an emitter resistance Re ) with slight modification shown in figure2 and one of the main motive to add emitter resistance is increase input resistance(Re)

I am not sure ,how increasing input resistance lead to a better amplifier (specially bjt circuits shown in figures )?

Specially after adding emitter resistance , voltage drop across Rπ (π model of CE with emitter resistance) decreases which causes lower amplification that's why I asked this question

There are lot of questions of input impedence on web but i hope someone answer it for that particular amplifier circuit rather than general one.

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    \$\begingroup\$ Ask yourself what would happen if the signal source's output impedance \$R_{sig}\$ was very high, what would the input signal at the base of the transistor be (don't calculate it, use reason, will the signal be small or large?) if the amplifier has a low input impedance and when it would have a high input impedance. \$\endgroup\$
    – Bimpelrekkie
    Oct 29, 2020 at 22:02
  • \$\begingroup\$ This question has very much to do with \$ I_b\$ in opamps \$\endgroup\$
    – Voltage Spike ♦
    Oct 29, 2020 at 22:13
  • \$\begingroup\$ @Bimpelrekkie , yes basic answer would be that we wanted most of the voltage drop across Rπ which makes sense but when we calculate voltage across​ Rπ in second figure it actually comes out to be less than previous one so our motive of better amplification doesn't achieve? And that's where I'm not able to figure it out \$\endgroup\$
    – user215805
    Oct 29, 2020 at 22:15
  • \$\begingroup\$ Not sure what you mean with your comment about Rπ in the second figure, there is no Rπ shown. In the second diagram the small-signal input impedance will be (Beta+1)*(re+Re) which should be much higher than the source impedance. \$\endgroup\$
    – John D
    Oct 30, 2020 at 0:02
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    \$\begingroup\$ Ah, I see the confusion. The emitter resistor greatly increases the input impedance, but it provides negative feedback and reduces the gain of the stage. So if you're talking about the overall voltage gain including loading effects, you have to do the calculations to see if you're better off (and maybe optimize the gain). Obviously if the emitter resistor is too high the voltage gain can drop below unity. \$\endgroup\$
    – John D
    Oct 30, 2020 at 16:21

1 Answer 1

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Constant input impedance implies a FLAT FREQUENCY RESPONSE, very useful in amplifiers with feedback where high_frequency poles causes poor phase_margin.

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