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schematic

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In a BJT common base amplifier configuration the input impedance is ~\$r'\$ (emitter resistance), right?

If I understand correctly, there is no current flowing from emitter to base (as might be assumed by transistor models like hybrid \$\pi\$). This input impedance is a measure of how big of a current change corresponds to a change in input voltage (\$\frac{\Delta V_{\text{in}}}{\Delta I_{\text{out}}}\$). That means that if \$\Delta V_{\text{in}}<0\$, then \$V_{\text{BE}}\$ increases. That in turn corresponds to an increase in \$I_C\$ and also \$I_E\$, meaning that for \$\Delta V_{\text{in}}>0\$ also \$\Delta I_{\text{out}}>0\$ and other way around when \$\Delta V_{\text{in}}<0\$. Is this relation \$\frac{\Delta V_{\text{in}}}{\Delta I_{\text{out}}}\$ the one that sets input impedance \$r'\$ because it's the only one corresponding to this? Am I missing something?

I don't really get the concept, because current and also input voltage shouldn't go from emitter to base? Do I not understand something?

Null
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TheUnknown
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  • Try to analyze this answer https://electronics.stackexchange.com/questions/410972/why-is-there-no-phase-shift-in-cc-and-cb-configuration/411171#411171 Any additional questions? – G36 Jul 01 '22 at 20:14
  • https://electronics.stackexchange.com/questions/573328/what-is-the-small-signal-impedance-looking-into-the-emitter/573393#573393 and small-signal analysis https://electronics.stackexchange.com/questions/429716/arriving-at-a-wrong-output-impedance-for-a-bjt-emitter-follower-configuration-ci/429726#429726 – G36 Jul 01 '22 at 20:21
  • @G36 Yes I understood. So I am correct, right? The only other thing is - why is current reversed in the small signal analysis example? – TheUnknown Jul 01 '22 at 20:52
  • a) input z is delta VIN / delta I_IN (you have delta I_OUT which is not correct). b) "if delta V_IN < 0 then Ic and Ie increase" - well Ic increases *but* bear in mind that although I_E also increases, the current thru R2 actually *decreases*. Input Z in a common base is "low" because if V_In changes, I_in changes very rapidly as the input structure is essentially a clamped forward-biased "diode". – Atomique Jul 03 '22 at 14:52
  • Yes, Zin is re||RE ≈ re. As for the question about "why is current reversed in the small-signal analysis example". The answer is that the Test voltage Vx forces a negative base current thus, β x Ib will flow in the opposite direction. And the small-signal analysis doesn't care that is not possible in real transistors. Because we are dealing here with a small AC signal. So, if the quiescent base current is 10µA. So the AC signal will cause the base current to increase and decrease in the rhythm of an input signal. So we can have a negative and positive base current in the AC domine. – G36 Jul 03 '22 at 18:10

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