VOLTAGE-SERIES FEEDBACK AMPLIFIER - THEORY, ANALYSIS, AND PRACTICAL DESIGN

EDITOR: B. SOMANATHAN NAIR

1. THEORY AND PRACTICAL EXAMPLES OF VSFBA

We had introduced the basic format of the voltage-series feedback amplifier (VSFBA) in a previous blog. As stated in therein, in a voltage-series feedback amplifier, a portion of the output voltage (BV_o) is fedback in series with the input voltage V_i of the amplifier. The block diagram shown in that blog may be explained further with the help of Fig. 1, which shows a voltage-series feedback amplifier constructed using an opamp and Fig. 2, which shows a voltage-series feedback amplifier constructed using a BJT.

            In Fig. 1, we find that the output voltage V_o is developed across resistors R₂ and R₁. A portion of this voltage V_f (=BV_o), developed cross R_1, is fed back in series with V_i, the voltage applied across the input terminals of the amplifier. This action is denoted by the blue-colored arrow in Fig. 1. It may be noted that V_s is the externally applied input signal.

From Fig. 1, we find that

                                        V_i = V_s – V_f                    (1)                                                                             

where the fed-back voltage 

                                                V_f = V_o R₁/(R₁+R₂) = BV_o                           (2)

                                                        

and the feedback factor

                                                                    B = R₁/(R₁+R₂)                   (3)

The output voltage is now given by

                                                       V_o = AV_i = A(V_s – BV_o)                  (4)

Rearanging Eq. (4), we get the gain with voltage-series feedback

                                               A_vf = A/(1+AB)                 (5)                              

            As stated earlier, Fig. 2, shows a voltage-series feedback circuit using BJT. In this case, the output voltage V_o is developed across resistors R₄ and R_E (= R₃) and a portion of this voltage V_f is developed across R₃. This is indicated by the blue-colored arrow in Fig. 2.

We now find from Fig. 2

                                                   V_f = V_o R₃/(R₃+R₄) = BV_o                       (6)

      

            It may be noted that input voltage V_i in this case is the base-emitter voltage V_be, and this comes in series with the fed-back voltage V_f, as shown by the blue-colored arrow in Fig. 2.

            We also find that emitter follower (common-collector amplifier) is a voltage-series feedback amplifier. In Fig. 3, which shows the circuit of an emitter follower, we find that entire output voltage V_o (and not a small portion of it) is fed back in series with the input voltage V_be (= V_i). Since the entire output voltage is fed back, the gain of the amplifier is unity.

2. ANALYSIS 1: VOLTAGE GAIN OF VSFBA

In the voltage-series feedback amplifier circuit, as stated earlier, a portion of the output voltage is fedback in series with the input. This is illustrated in the equivalent circuit, shown in Fig. 4, which can be seen to be the hybrid-parameter equivalent circuit, with the following modifications incorporated in it:

·         Output section is converted into a Thevenin equivalent circuit.

·         Input resistance is designated as r_i instead of h_i.

·         Output resistance is designated as r_o instead of 1/h_o.

·         Transfer gain is represented by a voltage source a_v instead of the current source h_f. In fact, a_v may be taken as the Thevenin equivalent voltage obtained as the product of current source h_f and output impedance 1/h_o [i.e., h_f ×(1/h_o)].

·     Reverse transfer-gain source h_rV_o is replaced with a feedback voltage source BV_o.

·         Input current is designated as I_i and its direction is clockwise in the input loop.

·         Output current is designated as I_o, and its direction is also clockwise in the output loop. Please note that the direction of voltage source a_v is such as to make this current flow in the clockwise direction. However, in the actual hybrid equivalent circuit, the output current flows in the counterclockwise direction, with h_f shown downwards.

            A voltage-series feedback amplifier is a voltage-to-voltage amplifier and we have to prove that the voltage gain gets stabilized when negative feedback is employed in it. We define voltage gain with feedback

                                                         

                                    A_vf =V_o/V_s                (7)

To find this gain, we have from the input loop, using KVL

                                                                                            

                                    V_o =V_i + BV_o              (8)

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

                             I_o = a_vV_i/(r_o+R_L)                    (9)                                                

Using Eq. (9), we find output voltage

                                       V_o  = I_oR_L = a_vV_i R_L/(r_o+R_L) = A_vV_i      (10)

                                                                                  

where we define a modified voltage gain A_v as

                                    A_v = a_vR_L/(r_o+R_L)             (11)

Now, from Eqs. (8), (10) and (11), we get

                                         

                        A_vf = V_o/V_s = A_v/(1+A_vB)            (12)

            From Eq. (12), when A_vB >>1, we find that A_vf = 1/B, and hence conclude that the voltage gain gets stabilized in voltage-series feedback amplifier.

3. ANALYSIS 2: INPUT IMPEDANCE OF VSFBA

We define input impedance of the amplifier with feedback as

                                    Z_if = V_s/I_s                     (13)                                                 

From Fig. 4, we find I_s = I_i. Substituting for V_s and I_i, we get                        

               

         Z_if = (V_i + BV_o)/I_i = (V_i + BA_vV_i)/I_i

              =V_i (1+ B A_v)/I_i= Z_i (1 + BA_v)            (14)

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

                                                     Z_i = V_i/I_i                                 (15)                                                                                  

Inspection of Eq. (14) reveals that input impedance increases with VSFB.

4. ANALYSIS 3: OUTPUT IMPEDANCE OF VSFBA

To get an expression of the output impedance, we use the equivalent circuit shown in Fig. 5, which is drawn based on the following steps:

·        Remove the input voltage source by short circuiting the input terminals (it may be noted that if the input is a current source, to make it zero, we open circuit it). In either case, the internal impedance of the source must be left 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 resulting output current. Notice that 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(voltage or current) source dependent on it. Fortunately, in this case, no sources are dependent on I_o, and hence the change in its direction will not affect our analysis.

We define output impedance of the amplifier with feedback as

                                  Z_of = V_o/I_o                         (16)

From the output loop, using KVL, we find

                                     I_o = (V_o ‒ a_vV_i)/r_o               (17)

From the input loop, we get

                                    BV_o   = ‒V_i                        (18)  

Substituting for V_i from Eq. (18) into Eq. (17), we have

                                         I_o = (V_o + a_vBV_o)/r_o  = (V_o (1+ a_vB)/r_o (19)

From Eq. (19), we get                                          

                                  Z_of = V_o/I_o = r_o/(1+ a_vB)   (20)

Equation (20) reveals that output impedance decreases with VSFB. Now, we state the following general rules related to negative feedback:

·        Impedance will increase with series (or current) connection.

·        Impedance will decrease with shunt (or voltage) connection