PRACTICAL DESIGN OF CURRENT-SHUNT FEEDBACK AMPLIFIER USING BJT

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

When low input impedance and high output impedance with fixed gain are required, we employ current-shunt feedback amplifier (CShFBA). Figure 1 shows the circuit of a current-shunt feedback amplifier constructed using BJTs. This is a very rarely used feedback connection. One application would be in connecting a low-impedance device such as a dynamic microphone to a high-input impedance circuit. Usually, the lab circuit consists of two stages of amplification before the application of feedback. It can be seen that the feedback connection is made by using a feedback resistor R_F from the emitter of the second stage to the base of the first stage.

 

1. SPECIFICATIONS

·         Current gain                         :           100 (sharp)

·         Voltage gain                         :           50

·         Voltage swing                       :           4.5 V

·         Current swing                       :          1 mA

2. DESIGN PROCEDURE

Steps 1 to 6: Design of the Standard Amplifier

Complete the design of the Standard Amplifier. In this case, we use two identical stages of CE amplifiers in cascade. The overall open-loop gain will then be the square of the gain of individual stages.

Step 7: Design of Feedback Resistance R_F

After applying the feedback, as in the previous cases, the gain will be reduced to the required value. The equations used in this case are:

                                      B = R_E2/(R_E2+ R_F)   (1)

where B is the feedback factor, R_E2 is the emitter resistor of Stage 2, and R_F is the feedback resistor. The current gain of the amplifier with feedback is given by

                                                                                                                                             A_if  = 1/B = 1+(R_F/R_E2)   (2)

The voltage gain is now given by

    

                                                                A_vf  = A_ifR_C2/R_i1            (3)

                                                                                                          

where R_C2 is the collector resistance of Stage 2, and R_i1 is the input resistor of Stage 1. Usually, R_i1 is chosen to be large compared with the source resistance so that the latter can be neglected from calculations. Using Eq. (2) and substituting the values

       A_if  = 100 = 1+(R_F/R_E2)   (2)

from which, we obtain

R_F = 99 R_E = 99 kΩ, Choose the nearest value 100 kΩ.

Step 8: Design of Input Resistance

From Eq. (3) by substituting values, we get

                                                      A_vf  = 50 = 100X4.7 k/R_i1          

Solving yields R_i1 = 9.4 kΩ; hence use a 4.7-kΩ fixed resistor in series with an 8.2-kΩ potentiometer.

The completely designed current-shunt feedback amplifier is shown in Fig. 2.