VOLTAGE-SHUNT FEEDBACK AMPLIFIER – THEORY, ANALYSIS

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

In the voltage-shunt feedback amplifier (VShFBA) configuration a voltage proportional to the output voltage is fed back in shunt with the input. A practical implementation of this circuit is shown in Fig. 1. In the circuit shown, a portion of the output voltage V_o is fed back in shunt with the input voltage V_be through the feedback resistor R_F; hence the name voltage-shunt feedback.

 

2. ANALYSIS 1: GAIN FACTOR OF VShFBA

For the voltage-shunt feedback circuit, as stated earlier, we feedback a current I_f (= BV_o), which is proportional to the output voltage V_o, in shunt with the input voltage V_i. Since BV_o is a current, we find that feedback factor B has the units of conductance (i.e., siemens) attached to it. The equivalent circuit of the voltage-shunt feedback amplifier is shown in Fig. 2, and this will be used for the analysis.

It can be seen that the voltage-shunt feedback amplifier is a transresistance (current-to-voltage) amplifier, and that we have to stabilize the transresistance (or, transfer resistance) with voltage-shunt feedback.

 We define transresistance as

                                                            R_mf = V_o/I_s       (1)      

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

                                                                                 

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

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

                                   I_o = r_mI_i/(r_o+ R_L)    (3)                                    

Using the above equation, we find the output voltage as

                                                V_o =I_oR_L = r_mI_i R_L/(r_o+ R_L)= R_mI_i   (4)

where R_m = r_m/(r_o+R_L) is the modified transresistance. Now, using the above equations, we get                                    

                                                R_mf = V_o/I_s = V_o/(I_i+BV_o) = R_m/(1+R_mB)   (5)

From Eq. (5), we find that transresistance gets stabilized with VShFB.

3. ANALYSIS 2: Input Impedance of VShFBA.

We define input impedance with voltage-shunt feedback as

                                       Z_if = V_s/I_s   (6)                                                                   

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

                           Z_if = Z_i/(1+R_mB)         (7)          

      

From Eq. (7), we find that input impedance decreases with VShFB.

4. ANALYSIS 3: OUTPUT IMPEDANCE OF VShFBA

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 source 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 does not happen since the dependent source is BV_o.

 

 As before, we define output impedance with feedback as

                                             Z_of =V_o/_o   (8)                                                             

 From the output loop, using KVL, we find

                                                                                             

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

From the input loop, we get

BV_o  = ‒/_I   (10)                                                                                                       

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

                                                                                                                                                                          I_o = (V_o + r_m BV_o)/r_o    (11)

Rearranging Eq. (11), we get

                              Z_of = r_o/(1+r_mB)           (12)                                                                                                                                              

From Eq. (12), we obtain that output impedance decreases with VShFB.

Note: It is interesting to note that the inverting configuration of opamp is a voltage-shunt operation and the non-inverting configuration is a voltage-series operation.