CURRENT-SHUNT FEEDBACK AMPLIFIER – THEORY, ANALYSIS

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

In the current-shunt feedback amplifier (CShFBA), current proportional to the output current is feedback in shunt with the input. A practical implementation of this configuration is shown in Fig. 1. In the circuit shown, we notice that output (collector) current I_o flowing through emitter resistor R_E develops a voltage I_oR_E across it, which is fed back in shunt with the input voltage V_be through the feedback resistor R_F; hence the name current-shunt feedback.

 

2. ANALYSIS 1: GAIN FACTOR OF THE CShFBA

In the current-shunt feedback circuit, as stated earlier, we feed back a voltage V_f (= BI_o = I_oR_E), which is proportional to the fed-back current I_f, in shunt with the input voltage V_i. Since BI_o is a current, we find that feedback factor B is a mere number and has no units attached to it. The equivalent circuit of the current-shunt feedback amplifier is shown in Fig. 2 and this will be used for the analysis.

 

 

  The equivalent circuit, shown in Fig. 2 can be seen to be similar to the hybrid equivalent circuit with the following modifications incorporated in it:

·         Output section is the Norton equivalent circuit of the hybrid model, with the following equivalences:

                                      a_iI_i = h_feI_b

                                       r_o = 1/h_oe

·         Input section is converted into a Norton equivalent circuit, with

                                      BI_o = -h_reV_or_i   

                                        r_i  = h_ie

·         We find that current-shunt feedback amplifier is a current (current-to-current) amplifier and that we have to prove that current gain A_if gets stabilized in this case. We define current gain A_if  as

                                                A_if = I_o/I_s        (1)

Referring to Fig. 2, we have from the input loop

                                       I_s = I_i + I_f = I_i + BI_o  (2)                                                      

where we have used the relation I_f = BI_o. Similarly, from the output loop, we find that output current

                                                     I_o = a_iI_ir_o/(r_o + R_L) = A_i I_i       (3)                    

where we define a modified current gain A_i as

                                         a_ir_o/(r_o + R_L) = A_i                (4)                                                      

Now, using the above equations, we get

                                                         

                                   A_if = A_i/(1+ A_iB)              (5)

From Eq. (5), we find that current gain is stabilized with CShFB.

3. ANALYSIS 2: INPUT IMPEDANCE OF CShFBA

We define input impedance with current-shunt feedback as

                                                Z_if = V_s/I_s    (6)                                             

Substituting for V_s = V_i, and I_s = BI_o + I_i, we get                                                

           

                       Z_if = V_i/(I_i + BI_o) = V_i/(I_i + BA_iI_i) = Z_i/(1 + A_iB)     (7)  

Inspection of Eq. (7) reveals that input impedance decreases with CShFB.

4. ANALYSIS 3:  OUTPUT IMPEDANCE OF CSHFBA

To get the expression for the output impedance, we draw the equivalent circuit, shown in Fig. 3 based on the same principles stated earlier, viz.,

·         Remove the input current source by open circuiting the input terminals, leaving the internal impedance of the source untouched.

·         Leave all dependent sources and circulating currents untouched.

·     Remove the load resistance from the output side, since the output impedance is defined with this condition imposed on it.

·         Apply an external voltage across the output terminals and find the output current due to this. As stated in the previous case, the direction of output current I_o has become reversed (i.e., anticlockwise) in this process. This change in the direction of I_o will produce a change in the direction of any source dependent on it. In this case, such an action occurs. We find that we have to reverse the direction of the dependent source BI_o, when direction of I_o is reversed. This is shown in Fig. 3.

     

As before, we define output impedance with feedback as

                                                               Z_of = V_o/I_o       (8)                                                 

From the output loop (converted into an equivalent Thevenin), using KVL, we find

                                   I_o = (V_o ‒a_iI_ir_o)/r_o     (9)  

From the input loop, as the direction of I_o has been reversed, we get

                      

                                                                        BI_o = I_i          (10)

                                                                                                          

Substituting for I_i from Eq. (10) into Eq. (9), we have

                                                                                             

                                   I_o = (V_o ‒a_i BI_or_o)/r_o     (11)  

Rearranging Eq. (11), we get

                                  

                                      Z_of = r_o (1+a_iB)    

Equation (11) reveals that output impedance increases with CShFB.