Small Signal-Low frequency- Analysis of Common Collector amplifier (Emitter Follower) -begins here

We are very familiar with common collector circuits- otherwise known as Emitter follower. Collector terminal is common to both input side and output side. During analysis, we are finding out the nature(characteristics) of the circuit by observing it's input and output resistances, voltage gain and current gain. Here discussing the analysis of CC amplifier using low frequency - small amplitude signals.

Before beginning, please go through the analysis of CE and CB amplifiers. Better have a look into CE analysis . In all these analysis, we use similar steps -only parameters will vary. Once you understand CE analysis, CB and CC analysis will become easy.

Let's begin with Circuit ....
1. Consider the CC amplifier circuit.

 Here we are starting with a simple - common collector amplifier circuit. Any way there may be some unfamiliarity but will clear as we go on.  Since collector to be 'common', it is tied into a power supply and now it's available to the entire circuit. Base is taken as input terminal - an input source Vs is attached to it through Rs and C1. Output is taken across R_L, connected to the emitter of the transistor. A constant current biasing is provided at the output terminal-it consists of a constant current source(I) and -V_{EE power supply.}

fig 1

 So the first part is over - we got the circuit. Second part is to draw the equivalent circuit. Here it is recommended to use 'T-model'(this will easily lead us to the answer). 


2. Obtain small signal equivalent model (Step by step explanation was given in 'Small signal model of a circuit' ).
Steps to obtain small signal model of the circuit in fig1 are as follows:
     Steps 1: D C voltage sources are replaced by ground
               2: D C current sources are open circuited
               3: Capacitors are shorted

fig 2

             4,5: Rearrange the components  to have a common ground and insert transistor model. Step by step explanation  is  in transistor model .

fig 3

 3. Find out the parameters
   In order to analyze behavior of a transistor we depend mainly on four parameters:  
      a): Input resistance (R_in)

In the input side there is a resistance R_S, but R_S is taken as source resistance and usually not consider for input resistance calculation. Input resistance of the circuit can be calculated by 'looking into the input side of the circuit' from the mark as given in fig 3. We can see that R_B, r_e, R_L and r_o  resistances are contributed for R_in. If rearrange the circuit in fig 3 as in fig 4,  calculation of R_in become easy. 

fig 4

Consider only that part of the circuit which is  relevant in R_in calculation.

fig 5

Equivalent resistance offered by  r_e, R_L and r_o  is marked as R_iB.
In effect,       R_in  = R_B ll R_iB       ---------(1)
 Then we have to find an equation for R_iB and substitute it.  
From figure 5, R_iB = V_B / I_b . In terms of I_e ,
                        R_iB = V_B(1+ β) / I_e .---------(2)     #recollect equation connecting I_b and I_e.
 An equation can be used to substitute for V_B / I_e using resistances as,
                            I_e= V_B / [(r_o ll R_L) + r_e] .-----------(3)
Substitute for V_B / I_e in equation (2).
                         R_iB = (1+ β) [(r_{o ll} R_L) + r_e] .-----------(4)

Note:{{ Here output side (Emitter side) resistances r_e, R_L and r_o have impact on input(base) side resistance R_in . 
There is a Shortcut for finding this:
Impact of output side resistances on input resistance= (1+ β) * equivalent resistance in output side. It's clear from fig4 /fig5 that, equivalent resistance in output side = [(r_o ll R_L) + r_e]. Now have a look on equ(4). Using short cut method, after equ(1), we can directly find equation for R_iB and thus will easily reach at equ(5).
    The term (1+ β) arises due to the relation, I_b = I_e /(1+ β) . Conclusion :

"Reflection of emitter side resistance into base side =  emitter side resistance * (1+ β) ".    Similarly, 

"Reflection of base side resistance into emitter side = base side resistance / (1+ β) ". }}

As per equ (1), input resistance  R_in  = R_B ll { (1+ β) [(r_o ll R_L) + r_e] }     ---------(5)

CC amplifier has a high input resistance. It happened because of the presence of (1+ β) in the term     { (1+ β) [(r_o ll R_L) + r_e] }. 'When two or more resistances are connected in parallel, the resultant resistance will always be less than the lowest among them' - as per this, R_in will be high only if R_B is also high.

This analysis is in the text book: Micro Electronics by Sedra & Smith. I am garnishing it so that you can Easily flip into. 

  Remaining parameters will be in the next post as this become too lengthy.

b): Output resistance (R_o )
c): Voltage gain (Av)
d): Current gain (A _is)