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General Characteristics of Negative Feedback Amplifiers

General Characteristics of Negative Feedback Amplifiers. www.AssignmentPoint.com. Sensitivity of transfer Amplification : The fractional change in amplification with feedback divided by the fractional change without feedback is called the sensitivity of the transfer gain.

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General Characteristics of Negative Feedback Amplifiers

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  1. General Characteristics of Negative Feedback Amplifiers www.AssignmentPoint.com www.assignmentpoint.com

  2. Sensitivity of transfer Amplification: The fractional change in amplification with feedback divided by the fractional change without feedback is called the sensitivity of the transfer gain. Mathematically, the sensitivity of the transfer gain can be written as follows: We know that, www.assignmentpoint.com

  3. Hence, the sensitivity is For example, if S=0.1 then That means the percentage change in gain with feedback is one-tenth the percentage variation in amplification if no feedback is present. Desensitivity of transfer Amplification: The reciprocal of the sensitivity is called the desensitivity D, or From Eq. (13-4) [Af=A/(1+bA)], it is seen that the transfer gain is divided by the desensitivity after feedback is added. Thus and the gain may be made to depend entirely on the feedback network. www.assignmentpoint.com

  4. Frequency Distortion: It is seen from Eq. (13-9) [Af=1/b] that if the feedback network is purely resistive, the gain with feedback is not depend on frequency even though the basic amplifier gain is frequency dependent. So the frequency distortion arising because of varying gain, A, with frequency is considerably reduced in a negative feedback amplifier. www.assignmentpoint.com

  5. Nonlinear Distortion: Suppose that a large amplitude signal is applied to a stage of an amplifier so that the operation of the device extends slightly beyond its range of linear operation, and as a consequence the output signal is slightly distorted. Negative feedback now introduced, and the input signal is increased by the same amount by which the gain is reduced, so that the output signal amplitude remains the same. It is clear from Eq. (13.9) that the negative feedback reduces the dependence of the overall closed-loop gain (Af) on the open-loop gain (A) of the amplifier. www.assignmentpoint.com

  6. Let’s understand this through this example: Say the open loop gain drops from 1000 (A1) to 100 (A2) had we a negative feedback gain with b=0.01 then That means the closed-loop gain drops from 90.9 to 50. This means it gets linearized. www.assignmentpoint.com

  7. Reduction of Noise: Negative feedback can be employed to reduce the noise in an amplifier (i.e. to increase the signal-to-noise ratio, SNR). The signal-to-noise (SNR) for the amplifier A1 is We may precede the original amplifier A1 by the clean amplifier A2 and apply negative feedback www.assignmentpoint.com

  8. So the signal-to-noise ratio in the output is obtained as follows: It is seen from above equation that the signal-to-noise ration is increased by the use of negative feedback. www.assignmentpoint.com

  9. Increase Bandwidth:By employing the negative feedback the bandwidth can be increased. Let an amplifier have the upper and lower cutoff frequency wH and wL without feedback. The gain of high frequency and low frequency gain Ao (=Amid) as For high frequency, the overall gain with negative feedback can be obtained as follows: The upper cutoff frequency with negative feedback is the frequency at which the real and imaginary parts of the denominator of above equation are equal, thus www.assignmentpoint.com

  10. It is seen from the previous equation that the upper cutoff frequency is increased by applying negative feedback. Similarly, the lower cutoff frequency is obtained as follows: It is revealed in the above equation that the lower cutoff frequency is decreased by applying negative feedback. From the above discussion it can be stated that negative feedback increases the bandwidth of an amplifier. www.assignmentpoint.com

  11. Input Resistance If the feedback signal is returned to the input in series with the applied voltage, it increases the input resistance. Since the feedback voltage Vf oppose Vs, the input current Ii is less than it would be if Vf were absent. Hence the input resistance Rif=Vs/Ii is greater than the input resistance without feedback Ri. For this type of feedback topology Rif=Ri(1+bA)=RiD. If the feedback is returned to the input in shunt with the applied current, it decreases the input resistance. www.assignmentpoint.com

  12. Since Is=Ii+If, then the current Ii (for a fixed value of Is) is decreased from what it would be if there were no feedback current. Hence the input resistance Rif=Vi/Is= RiIi/Is is decreased because of this type of feedback. For this type of feedback topology Rif=Ri/(1+bA)=Ri/D. www.assignmentpoint.com

  13. Output Resistance Negative feedback which samples the output voltage, regardless of how this output signal is returned to the input, tends to decrease the output resistance (Rof<<Ro). Negative feedback which samples the output current, regardless of how this output signal is returned to the input, tends to increase the output resistance (Rof>>Ro). The output resistance for the feedback amplifier can be defined as the resistance with feedback Rof looking into the output terminals with RL disconnected. The output resistance is determined by applying a voltage V, resulting in a current I, with Vs (Vs =0 where input source is voltage) shorted out or Is (Is =0 where input source is current) opened out. So, Rof=V/I. www.assignmentpoint.com

  14. Input Resistance of Voltage-Series Feedback Fig. 13-10 shows the topology of Fig. (a) in which the amplifier is replaced by its Thevenin’s model. From Fig. 13-10 the input impedance with feedback is Rif=Vs/Ii. Also, www.assignmentpoint.com

  15. From Eqs. (13-11) and (13-12) Whereas Av represents the open-circuit voltage gain without feedback, Eq. (13-13) indicates that AV is the voltage gain without feedback taking the load RL into account. Therefore www.assignmentpoint.com

  16. Output Resistance of Voltage-Series Feedback Replacing the output voltage Vo by V of Fig. 13-10 and setting Vs=0, we obtained From this equations, we obtained as follows: Note that Ro is divided by the desensitivity factor 1+bAv, which contains the open-circuit voltage gain Av (not AV). www.assignmentpoint.com

  17. The output resistance with feedback which includes RL as part of the amplifier is given by Rof in parallel with RL, or where, Ro’=Ro||RL is the output resistance without feedback but with RL considered as part of the amplifier. www.assignmentpoint.com

  18. Input Resistance of Current-Series Feedback Fig. 13-10.1 shows the topology of Fig. (b) in which the amplifier is replaced by its Thevenin’s input model and Norton’s output model. From the above figure the input impedance with feedback is Rif=Vs/Ii. Also www.assignmentpoint.com

  19. From Eqs. (13-11.1) and (13-12.1) Whereas Gm represents the short-circuit transconductance without feedback, Eq. (13-18) indicates that GM is the transconductance without feedback taking the load RL into account. Therefore www.assignmentpoint.com

  20. Output Resistance of Current-Series Feedback Replacing the output voltage Vo and Io by V and-I, respectively, of Fig. 13-10.1 and setting Vs=0, we obtained From this equations, we obtained as follows: Note that Ro is multiplied by the desensitivity factor 1+bGm, which contains the transconductance Gm (not GM). www.assignmentpoint.com

  21. The output resistance with feedback which includes RL as part of the amplifier is given by Rof in parallel with RL, or www.assignmentpoint.com

  22. Input Resistance of Current-Shunt Feedback The topology of Fig. (c) is indicated in Fig. 13-11, with the amplifier replaced by its Norton’s model. From the Fig. 13-11 the input impedance with feedback is Rif=Vi/Is. Also www.assignmentpoint.com

  23. From Eqs. (13-19) and (13-20) From Fig 13-11, Rif=Vi/Is and Ri=Vi/Ii. Using (13-22), we obtain Whereas Ai represents the short-circuit current gain without feedback, Eq. (13-23) indicates that AI is the current gain without feedback taking the load RL into account. Therefore www.assignmentpoint.com

  24. Output Resistance of Current-Shunt Feedback Replacing the output voltage Vo and Io by V and-I, respectively, of Fig. 13-11 and setting Is=0, we obtained From this equations, we obtained as follows: Note that Ro is multiplied by the desensitivity factor 1+bAi, which contains the short-circuit current gain Ai (not AI). www.assignmentpoint.com

  25. The output resistance with feedback which includes RL as part of the amplifier is given by Rof in parallel with RL, or For RL=, AI=0 and , so that Eq. (13-37) reduces to www.assignmentpoint.com

  26. Input Resistance of Voltage-Shunt Feedback Fig. 13-11.1 shows the topology of Fig. (d) in which the amplifier is replaced by its Norton’s input model and Thevenin’s output model. From the Fig. 13-11.1 the input impedance with feedback is Rif=Vi/Is. Also www.assignmentpoint.com

  27. From (13-19.1) and (13-20.1) From Fig 13-11, Rif=Vi/Is and Ri=Vi/Ii. Using (13-22), we obtain Whereas Rm represents the open-circuit transresistance without feedback, Eq. (13-26) indicates that RM is the transresistance without feedback taking the load RL into account. Therefore www.assignmentpoint.com

  28. Output Resistance of Voltage-Shunt Feedback Replacing the output voltage Vo and Io by V and-I, respectively, of Fig. 13-11.1 and setting Is=0, we obtained From this equations, we obtained as follows: Note that Ro is divided by the desensitivity factor 1+bRm, which contains the transresistance Rm (not RM). www.assignmentpoint.com

  29. The output resistance with feedback which includes RL as part of the amplifier is given by Rof in parallel with RL, or where, is the output resistance without feedback but with RL considered as part of the amplifier. Note that is now divided by the desensitivity factor D=1+bRM which contains the voltage gain RM that takes RL into account. www.assignmentpoint.com

  30. Table 13-4 summarizes the different components for different topologies. www.assignmentpoint.com

  31. Method of Analysis of a Feedback Amplifier It is desirable to separate the feedback amplifier into two blocks, the basic amplifier A and the feedback network b, because with a knowledge of A and b, we can calculate the important characteristics of the feedback system, namely, Af, Rif, and Rof. The basic amplifier configuration without feedback but taking the loading of the b network into account is obtained by applying the following rules: To find the input circuit: 1. Set Vo=0 for voltage sampling. In other words, short the output node. 2. Set Io=0 for current sampling. In other words, open the output loop. To find output circuit: 1. Set Vi=0 for shunt comparison. In other words, short the input node. 2. Set Ii=0 for series comparison. In other words, open the input loop. www.assignmentpoint.com

  32. The complete analysis of a feedback amplifier is obtained by carrying out the following steps: 1. Identify the topology. (a) Is the feedback signal Xf a voltage or a current? In other words, is Xf applied in series or in shunt with the external excitation? (b) Is the sampled signal Xo a voltage or current? In other wards, is the sampled signal taken at the output node or from the output loop? 2. Draw the basic amplifier circuit without feedback, following the rules listed above. www.assignmentpoint.com

  33. 3. Use the Thevenin’s source if Xf is a voltage and a Norton’s source if Xf is a current. 4. Replace each active device by the proper model (for example, hybrid- model for a transistor at high frequency or the h-parameter model at low frequency). 5. Indicate Xf and Xo on the circuit obtained by carrying out steps 2, 3, and 4. Evaluate b= Xf / Xo. 6. Evaluate A by applying KVL and KCL to the equivalent circuit obtained after step 4. 7. From A and b, find D, Af, Rif, Rof, and Rof’. www.assignmentpoint.com

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