PRACTICAL DESIGN OF VOLTAGE-SHUNT FEEDBACK AMPLIFIER USING BJT

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

Figure 1 shows a voltage-shunt feedback amplifier. This amplifier is employed when precise gain low values of input and output impedances are required. As in the case of current-shunt feedback here also we require only one stage of amplification for our design.

2. SPECIFICATIONS

·         Voltage gain             :           5 (Exactly)

·         Input impedance      :           Low

·         Output impedance   :           Low

·         Voltage swing           :           4.5 V

·         Current swing           :           1 mA

3. DESIGN STEPS

Steps 1 to 6: Design of the Standard Amplifier without Feedback

Following the Standard-Amplifier design steps 1 to 6, given in an earlier blog, first we complete the design of the amplifier without feedback.

Step 7: Design of Feedback Resistor R_F and Input Resistor R_I

In the voltage-shunt feedback amplifier, the resistor R_F acts as the feedback connection from the output to the input, as shown in Fig. 1. The gain of the voltage-shunt feedback amplifier is

                                           A_f = ‒R_F/R_I    (1)                                                    

The calculation of the input impedance Z_if requires complex expressions. However, since voltage-shunt feedback gives a low Z_if, we just specify only that idea in our design.

Step 8: Design of R_F and R_I

Let R_F = 100 kΩ (arbitrary value to reduce DC base-current drain through R_F). Substituting values in Eq. (1), we get

                        A_f = 5 = 100 kΩ/R_I    (2)         

From which we find

                                                                      R_I = 20 kΩ

Assuming the source resistance to be 2 kΩ, R_I will reduce to 18 kΩ. This is because the actual input resistance is equal to R_s+R_I. However, for fixing the gain accurately, we may use a potentiometer in series with R_I. The value of this pot is

R_s + R_I + R_pot = 22 kΩ

Therefore, we choose

R_I = 15 kΩ

and

R_pot = 10 kΩ

Step 9: Design of Coupling Capacitors

As the input impedance is low, coupling capacitors to be used must have high values. Since the input impedance with feedback is given by

                            R_if = R_i/(1+AB) = h_ie/AB    (3)                                            

where A = gain of the amplifier without feedback, B = 1/A_f = 0.2, and h_ie is input impedance in the hybrid-parameter equivalent circuit. Assuming h_ie = 1.5 kΩ, and A = 125, we get from Eq. (3) by substitution

             R_if  = 1.5 k/(125x0.2) = 60 Ω

We can now calculate the value of C_C using the equation

                                      C_C = 1/2πf_LR_i = 1/2πx30x60 = 88.4 μF

Choose 100 μF as coupling capacitors.

                   

The completely designed amplifier is shown in Fig. 2.