general characteristics of negative feedback amplifiers

General Characteristics of Negative Feedback

Amplifiers

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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.

AdAfAfdA

S/

/

Mathematically, the sensitivity of the transfer gain can be written as follows:

We know that, A

AfA

1

2)1(

12)1(

12)1(

)1()1(

So,AA

AA

A

dAdA

AdA

AdA

dAfdA

)1()1(1

12)1(

1AA

fA

AAAA

AdAfdA


AAdA

AfdAS

11

/

/Hence, the sensitivity is

AS

11

For example, if S=0.1 then A

dAA

fdA1.0

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 )713(1 AD

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 )813(/ DAfA

)913(1

1Then,1If

AA

AA

fAA

and the gain may be made to depend entirely on the feedback network.www.assignmentpoint.com

Frequency Distortion:It is seen from Eq. (13-9) [Af=1/] 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.


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

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 =0.01 then

9.9001.010001

10001

fA

5001.01001

1002

fA

That means the closed-loop gain drops from 90.9 to 50.

This means it gets linearized.www.assignmentpoint.com

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

nVsV

NS

We may precede the original amplifier A1 by the clean amplifier A2 and apply negative feedbackwww.assignmentpoint.com

111

21121

AAnV

AAAAsV

oV

So the signal-to-noise ratio in the output is obtained as follows:

22111/

21121 A

nVsV

AAAnV

AAAAsV

NS

It is seen from above equation that the signal-to-noise ration is increased by the use of negative feedback.www.assignmentpoint.com

Increase Bandwidth: By employing the negative feedback the bandwidth can be increased.

Let an amplifier have the upper and lower cutoff frequency H and L without feedback. The gain of high frequency and low frequency gain Ao (=Amid) as

HjAo

HA /1

/1

andLj

AoLA

For high frequency, the overall gain with negative feedback can be obtained as follows:

HjAoAo

HjAo

HjAo

HAHA

HfA

/1

/11

/11

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

AoH

Hf

1www.assignmentpoint.com

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:

AoL

Lf

1

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.

AoH

Hf

1 www.assignmentpoint.com

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+A)=RiD.

If the feedback is returned to the input in shunt with the applied current, it decreases the input resistance.


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+A)=Ri/D.


Output ResistanceNegative 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.


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.

)1113( oViRiIfViRiIsV

)1213(and

iRiIVALRoRLRiRiIvA

LRoRLRiVvA

oV

)1313(where,

LRoR

LRvA

iVoV

VA

From Fig. 13-10 the input impedance with feedback is Rif=Vs/Ii.

Also,


From Eqs. (13-11) and (13-12)

)1( VAiRiIiRiIVAiRiIsV

)1413()1( DiRVAiRiIsV

ifR

VAD 1Here,

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)1513(limlim

LRoRLRvA

LRVA

LRvA


Output Resistance of Voltage-Series Feedback Replacing the output voltage Vo by V of Fig. 13-10 and setting Vs=0, we obtained

VfViVoR

iVvAVI

and

From this equations, we obtained as follows: )2813(

)1(

oRvAV

oR

VvAVI

)2913(1

Hence,

vA

oR

IV

ofR

Note that Ro is divided by the desensitivity factor 1+Av, which contains the open-circuit voltage gain Av (not AV).www.assignmentpoint.com

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

)/(1

)/(

1

1'

LRoRLRvALRoRLRoR

LRvALRoRLRoR

LRvA

oR

LRvA

oR

LRofRLRofR

LRofRofR

)3013(1

'

)/(1

)/('

VAoR

LRoRLRvALRoRLRoR

ofR

where, Ro’=Ro||RL is the output resistance without feedback but

with RL considered as part of the amplifier.www.assignmentpoint.com

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 )1.1113( oIiRiIfViRiIsV

)1.1213(and,

iRiIMGLRoR

oRiRiImG

LRoRoRiVmG

oI

)1813(where,

LRoR

oRmG

iVoI

MG


From Eqs. (13-11.1) and (13-12.1)

)1( MGiRiIiRiIMGiRiIsV

)1613()1( DiRMGiRiIsV

ifR

MGD 1Here,

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

)1713(0

lim0

lim

LRoR

oRmG

LRMG

LRmG


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

IfViVIiVmGoR

V and

From this equations, we obtained as follows: IImGoR

V

)1.38.13()1( mGoRIV

ofR

Note that Ro is multiplied by the desensitivity factor 1+Gm, which contains the transconductance Gm (not GM).www.assignmentpoint.com


LRofRLRofR

LRofRofR

'

LRmGoRLRmGoR

ofR

)1(

)1('

oRmGLRoRmGLRoR

ofR

)1('

)/(1

)/()1('

LRoRoRmGLRoRmGLRoR

ofR

)2.38.13(1

1''

MGmG

oRofR


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 )1913( oIiIfIiIsI

)2013(and

iIIALRoRoRiIiA

oI

)2113(where,

LRoR

oRiA

iIoI

IA


From Eqs. (13-19) and (13-20)

)2213()1( IAiIsI

From Fig 13-11, Rif=Vi/Is and Ri=Vi/Ii. Using (13-22), we obtain

IAiIiV

IAiIiV

sIiV

ifR

1

1

)1(

)2313(1

DiR

IAiR

ifR

IAD 1Here,

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 )2413(0

lim0

lim

LRoR

oRiA

LRIA

LRiA


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

IoIfIiIIiIiAoR

V and

From this equations, we obtained as follows: IIiAoR

V

)3513()1( iAoRIV

ofR

Note that Ro is multiplied by the desensitivity factor 1+Ai, which contains the short-circuit current gain Ai (not AI).



LRofRLRofR

LRofRofR

'LRiAoR

LRiAoRofR

)1(

)1('

oRiALRoRiALRoR

ofR

)1('

)/(1

)/()1('

LRoRoRiALRoRiALRoR

ofR

)3713(1

1''

IAiA

oRofR

For RL=, AI=0 and , so that Eq. (13-37) reduces to

ofRofRiAoRofR 'meansthat)1('


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 )1.19.13(oViIfIiIsI

)1.20.13(and iIMRLRoRLRiImR

oV

)2613(where,

LRoR

LRmRMR


From (13-19.1) and (13-20.1)

)1( MRiIiIMRiIfIiIsI

From Fig 13-11, Rif=Vi/Is and Ri=Vi/Ii. Using (13-22), we obtain

MRiIiV

MRiIiV

sIiV

ifR

1

1

)1(

)2513(1

DiR

MRiR

ifR

MRD 1Here,

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 )2713(limlim

LRoR

LRmR

LRMR

LRmR


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

VoVfIiIoR

iImRVI

and

From this equations, we obtained as follows:oR

mVRVI

)1.3213(1

mR

oR

IV

ofR

Note that Ro is divided by the desensitivity factor 1+Rm, which contains the transresistance Rm (not RM).



LRofR

LRofRLRofRofR

'

LRmR

oR

LRmR

oR

ofR

1

1'

LRmRLRoRLRoR

ofR

'

)/(1

)/('

LRoRLRmRLRoRLRoR

ofR

)2.3213(1

''

MRoR

ofR

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+RM which contains the voltage gain RM that takes RL into account.


Table 13-4 summarizes the different components for different topologies.


Method of Analysis of a Feedback AmplifierIt is desirable to separate the feedback amplifier into two blocks, the basic amplifier A and the feedback network , because with a knowledge of A and , 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 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

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

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 = Xf / Xo.

6. Evaluate A by applying KVL and KCL to the equivalent circuit obtained after step 4.

7. From A and , find D, Af, Rif, Rof, and Rof ’. www.assignmentpoint.com

general characteristics of negative feedback amplifiers

Documents

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transfer gain

feedback network

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gain of high frequency

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