CURRENT-SERIES FEEDBACK AMPLIFIER- THEORY, ANALYSIS

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

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

It may be noted here that the emitter-follower circuit is a voltage-series feedback amplifier even though it may look like the current-series feedback amplifier shown in Fig. 1. This is because, in the emitter follower, the output voltage and the feedback voltage are same (=I_oR_E). But, in the case of the current-series feedback amplifier shown above, the fed-back voltage is dependent only on the output current I_o and not on the output voltage V_o.

2. ANANLYSIS 1:GAIN FACTOR OF CSFBA

For the current-series feedback circuit, as stated earlier, we feedback a voltage BI_o, which is proportional to output current I_o and in series with input voltage V_i. Since BI_o is a voltage, we find that feedback factor B has units of resistance (i.e., ohms). From the discussions given in above, we find that BI_o = I_oR_E, from which we get B = R_E. The equivalent circuit of the current-series feedback amplifier is shown in Fig. 2, and this will be used for the analysis. This equivalent circuit can be seen to be similar to the hybrid equivalent circuit, with the following modifications incorporated in it:

·   Output side is the Norton equivalent circuit and the input side is the Thevenin equivalent circuit of the hybrid model, with the following equivalences:

                                   g_mV_i ≡ h_feI_b

                                       r_o  ≡  1/h_oe

                                    BI_o  ≡  h_reV_o

                                        r_i ≡ h_ie

·          We find that the current-series feedback amplifier is a transconductance (voltage-to-current) amplifier and hence its gain factor is transfer conductance (transconductance) g_m. Therefore, we have to prove that with CSFB, transconductance g_m gets stabilized.

We define transconductance with current-series feedback as G_mf as

G_mf = output current/input voltage = I_o/V_s    (1)

                                                             

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

                                              V_s = V_i + BI_o       (2)                                           

Similarly, from the output loop, we find that output current

                                                                                                                                                

                      I_o = g_mV_ir_o/(r_o+R_L) = G_mV_i  (3)

where we define a modified transconductance G_m as

                                             G_m = g_mr_o/(r_o+R_L)  (4)                                              

 Now, using Eqs. (1), (2), and (3), we get

                                                                  G_mf = G_m/(1+ G_mB)    (5)

                                               

From Eq. (5), we find that transconductance gets stabilized with CSFB.

3. ANALYSIS 2: INPUT IMPEDANCE OF CSFBA

We define input impedance with feedback as

Z_if = V_s/I_i                (6)  

where I_s = I_i. Substituting for V_s and I_i, we get

                                

                              Z_if = (V_s+BI_o)/I_i = (V_s+BG_mV_i)/I_i = r_i(1+G_m)         (7)

where we have used the expression for the input impedance of the amplifier without feedback as

r_i = Z_i = V_i/I_i                (8)

                                                                                          

Equation (7), tells us that input impedance gets increased with CSFB.

4. ANALYSIS 3: OUTPUT IMPEDANCE OF CSFBA

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 voltage source by short circuiting the input terminals and leaving the internal impedance of the source untouched.

·       Leave all the 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 measure the output current due to this. As stated in the previous case, direction of the 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 does happen. We find that direction of the dependent source BI_o has to be reversed when the 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  (9)                                                                  

From the output loop, using KVL, we find

                                    I_o = (V_o  ‒ g_mV_i r_o)/r_o   (10)                                                  

From the input loop, we get

  V_i  = BI_o   (11)                                                       

Substituting for V_i from Eq. (11) into Eq. (10), we have

                                                        

                          I_o = (V_o  ‒ g_mBI_o r_o)/r_o   (12)                                    

Rearranging Eq. (12), we get

                                                     I_or_o(1+ g_mB) = V_o   (13)   

                                                                                              

From Eq. (13), we find

Z_of = V_o/I_o   = r_o(1+g_mB)   (14)         

                                 

Equation (14) reveals that output impedance increases with CSFB.