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I was reading different texts about input and output resistance of the above circuit and trying to understand the Rin and Rout without using any transistor models such as pi-model or T-model. I think I understand how can we comprehend what input resistance is. Here is the summary from what I understand for Rin without using any model:

Lets assume the voltage at the point Vb is increased by ΔV.

This means the voltage at the point Ve also will increase by ΔV.

If we want to write an equation for the input resistance Rin(the resistance seen from the point Vb); we can use the β relation between the base current Ib and the emitter current and the fact that the same ΔV will appear both at Vb and Ve. So Rin can be derived by using the following steps:

ΔIb = ΔV / Rin

ΔIe = ΔV / (Re//Rload)

since Ie = (1+β)*Ib

Rin = (1+β)*(Re//Rload)

So Rin input resistance can be written in terms of Re, Rload and β.


I cannot find a similar step by step comprehension derivation of the Rout without using a model. I mean I would like to clearly write down how Rout is obtained as I have written for Rin above.

The texts say that the Rout(looking back at the emitter) is:

Re // (Rsource/β)

But I'm very confused at this point how this is derived How they obtain Rsource/β conceptually.

How can we step by step explain Rout here as in Rin case?


1 Answer 1


How they obtain Rsource/β conceptually.

let Rsource=Rb

  • if Rb is biased from some voltage Vcc
  • such that the value of Rb controls Ie using KVL;

    • Ib=(Vcc-Vbe-Ve)/Rb
    • thus Ie/β =(Vcc-Vbe-Ve)/Rb
    • or (Vcc-Vbe-Ve)/Ie = Rb/β
    • let Vcc and Vbe be constants and Zout=Ve/Ie
  • the change in emitter voltage from an external current can be nulled by an opposing base current amplified by Rb/β thus Zout = Rb/β // Re


in conclusion the Emitter Follower Zin = β Re and ( for any Ze ) Zout = Rb/ β ( or any complex Zb parts)

For future reference

  • a transistor amp with negative feedback using Rf from collector to base with Rseries to a source voltage input,
  • we get voltage gain Av= -Rf/Rb and it turns out that the input and output impedance becomes a function of negative feedback gain such that Zin is reduced by feedback ratio and Zout at collector is reduced by feedback R ratio //Rc

The same is true for Op Amps but the gain is so high we consider the differential input impedance to be zero and likewise for output impedance until current limiting saturates the driver.

  • \$\begingroup\$ I dont get this. 1-) Why Ib = Vcc/Rb ? Isn't Ib = "Vcc/(Rb+Re)" neglecting Vbe? Is that because u re assuming Re much bigger than Rb? 2-) I still dont understand where Rb/β comes from and the meaning of it at all. You wrote 1/Ie=Rb/(β *Vcc) I think theres a very hidden implicit meaning here. Can you explain it for a novice level? thanks \$\endgroup\$
    – user1245
    Dec 30, 2016 at 14:23
  • \$\begingroup\$ let me re-arrange \$\endgroup\$ Dec 30, 2016 at 14:26
  • \$\begingroup\$ When cascading stages it is important to remember source and load impedance effects. This is why if you put a Cap on the base the output lowers the base impedance by β essentially a capacitance booster seen from emitter. \$\endgroup\$ Dec 30, 2016 at 14:49
  • \$\begingroup\$ equations more clear now. but you wrote " Zout=Ve/Ie" "the change in emitter voltage from an external current can be nulled by an opposing base current amplified by Rb/β thus Zout = Rb/β // Re" Nulling vs output impedance here what is that about? I think it will take a year to understand this for me. This Zout is very implicit. Texts dont explain it clearly. \$\endgroup\$
    – user1245
    Dec 30, 2016 at 15:06
  • 1
    \$\begingroup\$ Also understand Emitter Follower as a current amplifier is the basic LDO as a buffered voltage regulator. Never drive external with load above source V (pullup R) because Emitter follower cannot pull down. . Then realize Common Emitter , CE is a current source at collector not a voltage source like CC, so external loads must be same or HIGHER Z than Rc collector to avoid distortion, current "starvation" \$\endgroup\$ Jan 2, 2017 at 16:15

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