Small Signal - Low frequency - Analysis of common collector amplifier-continue

Analysis of amplifier circuit includes derivation of equation for 1)input resistance 2)output resistance 3)voltage gain and 4)current gain. Derivation and analysis of input resistance is given in - 'Analysis of Common Collector amplifier (Emitter Follower) -begins here' . Remaining parameters will be discussed here.

Before beginning, recollect the common collector amplifier circuit and it's small signal model. It was discussed in detail in the -'begins here' post. For your convenience once again giving the circuits.
Here is the amplifier circuit:

fig 1

This is the small signal equivalent of amplifier circuit:

fig 2

Next step is to find out second parameter, i.e output resistance.

b)Output resistance (R_o ) 

Output resistance of the circuit can be calculated by 'looking into the output side of the circuit' from the mark as given in fig 3. We can see that R_B, R_S , r_e, and r_o  resistances are contributed for R_o. Among these R_B and R_S  are in input side and their impact on output resistance is marked as R_ob, in fig 3. 

fig 3

Looking towards R_B and R_S from arrow mark shows that they are parallel to each other( R_B ‖ R_S ). It's presence in base terminal affects the emitter side resistance and is given as R_ob = ( R_B ‖ R_S ) / (1+ β).

(To know how to write this equation, Please see 'Analysis of Common Collector amplifier (Emitter Follower) -begins here'.

Now we can complete the derivation of  R_o. Look into the circuit through the arrow marked along with R_o. We can see that r_o is parallel with (r_e + R_ob ). ##   r_e and R_ob are in series.  

Then equation for  R_o =  r_o ll (r_e + R_ob ) ------(1)    substitute for R_ob.

                               R_o =  r_o ll { r_e + [( R_B ll R_S ) / (1+ β)] } -------(2) Equation for output resistance.

Note: Output resistance of Common collector amplifier(Emitter follower) is low.

Reason behind it: r_o is much larger than{r_e + [( R_B ll R_S ) / (1+ β)]}. So result of parallel combination of these two terms is less than the smaller quantity. So this result is nearly equal to the smaller quantity i.e, { r_e + [( R_B ll R_S ) / (1+ β)] }.

                               R_o =̃   r_e+ [( R_B ll R_S ) / (1+ β)]  -------(3)
Next parameter we have to find is voltage gain.

c): Voltage gain (Av)

Voltage gain for an amplifier is output voltage/input voltage. For the circuit in fig 2,
    Voltage gain, Av = Vo/Vs
We cannot directly write an equation for voltage gain using fig2. Therefore consider it's Thevenin's equivalent and thus make it simpler. Two steps are there in this simplification:
First, consider the input side only which includes Vs, Rs and R_B (fig 4a) and find it's Thevenin's equivalent. Then the circuit in fig 4a can be replaced with it's Thevenin's equivalent.

fig 4

 In order to calculate R_TH, voltage source Vs is short circuited(fig 4b). R_TH = Rs ll R_B. As in fig4c, V_TH is the voltage across R_B. V_TH  = R_BVs / (Rs + R_B).

The second step is to take all resistances in fig 2 to one side- either to input side or to output side. Here resistances in the output(emitter) side are take to base side, using the shortcut method described in 'Analysis of Common Collector amplifier (Emitter Follower) -begins here'. The effect of Emitter side resistances on base side is obtained by multiplying emitter side resistances with (1+ β).

Now apply the result of above two steps  in fig 2.

fig 5

Here in fig 5, input side(as in fig 4a) is replaced with V_TH and R_TH. Also output resistances are transformed in a way so that now we can consider the circuit as a resistance network(no more worry about amplification factor between base and emitter terminals). Rearranging circuit in fig 5 give fig6.

fig 6

  
Now write equation for voltage gain.  VoVTH=(1+β)(rollRL)(RsllRB)+(1+β)re+(1+β)(rollRL)--

                                                                                                              -------(4)

Substitute for V_TH. V_TH  = R_BVs / (Rs + R_B).
VoVs=RB(Rs+RB)(1+β)(rollRL)(RsllRB)+(1+β)re+(1+β)(rollRL)                                                                                                                                      ------(5)

Observing equation(5) for voltage gain, we can see that the denominator has a greater value than denominator. Therefore the voltage gain is less than unity. But under certain conditions it becomes very close to unity. The conditions are 1)Rs << R_B and 2) (R s ll RB) << (1+ β) re + (1+ β) (ro ll R L ). Voltage gain is nearly equal to unity means - voltage at the output side(Emitter) follows very closely the voltage at the input side. Thus the name Emitter follower.

d): Current gain (A _is)

Short circuit current gain = ie/ib =  (1+ β) . Current gain is high.

 Voltage buffer

Because of high input resistance and low output resistance, Emitter follower act as a voltage buffer. It is used for connecting a high resistance source to a low resistance load.
 

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