Small Signal - Low frequency - Analysis of CE amplifier using hybrid π model

 Here we are stepping to a zone mostly liked by questioners for essay - Low frequency analysis of CE amplifier . Small signal analysis of common emitter amplifier can be done using hybrid π model. Here low frequency analysis is considered.

In order to do the analysis of certain circuit, first step is to draw the model of that circuit. This model will help us to easily find equations for certain parameters of the circuit. Usually, input and output resistances, voltage gain and current gain are these four parameters. The equations will give an idea about the approximate value of these parameters and this in turn will help us to comment about characteristics of the circuit.  

Let's start...

1. Consider the CE amplifier circuit.

Circuit of a common emitter amplifier is given below. It consists of two bias dc voltages - Vcc and

V_{E E}  . An input voltage V_S is applied to the transistor through a resistor R_S. C₁,C₂  are coupling capacitors and C_E is bypass capacitor.

fig 1

 2. Obtain small signal equivalent model (Please see 'Small signal model of a circuit' )
     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: Rearrange the components such as to have a common ground 

fig 3

               5: Simplify the circuit and insert transistor model

fig 4

3. Find out the parameters
   For analyzing behavior of a transistor we depends mainly on four parameters:  
      a): Input resistance (R_in)

Input resistance of the circuit can be calculated by 'looking into the input side of the circuit' from the mark as given in fig 5. This is because R_S is taken as a source resistance and usually not consider for input resistance calculation.

fig 5

Part of the circuit considering for R_in calculation is

fig 6

From the above circuit,    R_in  =    R_b  ‖ r_π        ---------(1)  
Normally, we select R_b such that R_b >>r_π . Then R_in  becomes

                                           R_in ≈ r_π     ----------- (2)
We can conclude that input resistance of CE amplifier is nearly equal to r_π and its value is typically few kilo ohms. So CE amplifier has a moderate value of input resistance.
Note: In fig 6, it is clear that    Vi = V_π . (Voltage across two parallel paths)
       b): Output resistance (R_o )
Again to get the output resistance,  look into the circuit from right side of arrow.

fig 7

Then from the circuit, R_o = Rc ‖ r_O    ----------(3)
 Typically, Rc << r_O .  Therefore, R_o ≈ Rc   ----------(4)
       c): Voltage gain (Av)
Voltage gain (Av) is defined as

                                Av = Vo / Vi               where,
                                Vo = output voltage    and
                                 Vi = input voltage appear across the input terminal(here Base) of  the amplifier.

Let's find out expression for Vi :
In fig 6, we get    R_in  =   R_b ‖ r_π . Using this the input side of fig 5 can be redrawn as

fig 8

As in fig 8, Vi is the voltage across Rin. So,

                  Rin  
Vi =  Vs  -----------            --------(5)
                Rin + Rs                                           
Substitute for Rin using equ (2) and take Vi = V_π   then

        
   --------(6)


Keep this equ in our account and try to get an equ for Vo. Consider the output side of the circuit given in fig 4 or fig5. It can be redrawn as below.

fig 9

In fig 9,resistances  r_o ,Rc and R_L are parallel and -g_mV_π is the current flowing through this parallel combination. Using these facts we can write the equ for output voltage as

                          Vo  =  - g_mV_π ( r_o ‖ Rc ‖ R_L )    ----------(7)   but we have Vi = V_π

  Therefore, Vo / Vi  =  - g_m ( r_o ‖ Rc ‖ R_L )           which is the voltage gain Av = Vo / Vi . 
   So,  Av = Vo / Vi   =  - g_m ( r_o ‖ Rc ‖ R_L )     -------(8)

Another parameter related to this is overall voltage gain. The overall voltage gain for the circuit in fig 4 is given as  Gv = Vo / Vs . We can easily get it as

Gv = Vo / Vs = (Vo / Vi ) x (Vi / Vs) . Here Vo/Vi is in our hand. We already have an equ for Vi/Vs (refer equ 5 & 6).Then 

                                                                        

    Gv =   - g_m ( r_o ‖ Rc ‖ R_L )      r_π

                                                   -----------            --------(9)
                                                    r_π + Rs                                                                                                                                                                                                                                                  
Small signal parameter can be written as  r_π  =  β / g_m   where  β is the current gain.

 Above equation becomes 

             -  β ( r_o ‖ Rc ‖ R_L )     
  Gv =     ---------------------                 --------(10)
                       r_π + Rs                                                                                                                                                                                                                                      
As per equation, voltage gain for CE amplifier is high.
       d): Current gain (A _is)
 For the CE amplifier circuit, short circuit current gain is considered. That is the current gain when is R_L shorted. From fig 9, when R_L is shorted, output current is

                        i_os =  - g_mV_π .

We can write an equation for input current as i_i = Vi / R_in = V_π  / R_in  . 
Short circuit current gain   (A _is) =  i_os  / i_i  .
                                         
                                             A _is =   - g_m R_in    ------(11)

From the above equ, we can realize that CE amplifier has a large current gain.

This analysis is given in the text book: Micro Electronics by Sedra & Smith. It is presented here with fine descriptions of each step.