14699775563. feedback amplifiers

Feedback Amplifiers

Dr. Monir HossenECE, KUET

Department of Electronics and Communication Engineering, KUET

Department of Electronics and Communication Engineering, KUET 2

Feedback Concepts

The input signal Vs and a feedback signal Vf is applied to the mixer.

For negative feedback the difference between Vs and Vf equal to Vi is applied to the amplifier.

Vo is the output and a portion of output β connected to the feedback network.

The process of injecting a fraction of output energy of some device back to the input is known as feedback.


Positive Feedback

When the feedback signal is in phase with the input signal and thus aids it, it is called positive feedback.

Both amplifier and feedback network introduce a phase shift of 180°.

The result is a 360° phase shift around the loop, causing the feedback voltage Vf to be in phase with the input signal Vin.

The positive feedback increases the gain of the amplifier. However, it increased distortion and instability.


Negative Feedback

o When the feedback signal is out of phase with the input signal and thus opposes it, it is called negative feedback.

o The amplifier introduces a phase shift of 180° into the circuit while the feedback network is so designed that it introduces 0° phase shift.

o The result is that the feedback voltage Vf is 180° out of phase with the input signal Vin.

o Negative feedback reduces the gain of the amplifier. o However, it provides low distortion, stability in gain, increased

bandwidth and improved input and output impedances.


Advantages of Negative Feedback Higher input impedance. Better stabilized voltage gain. Improved frequency response. Lower output impedance. Reduced noise. More linear operation. Reduce non-linear distortion.


Feedback Connection Types Feedback can be connected in four basic ways. Both voltage and current can be feed back to the

input either in series or parallel.1) Voltage – series feedback.2) Voltage – shunt feedback.3) Current – series feedback.4) Current – shunt feedback.

oSeries feedback connections increase the input resistance.oShunt feedback connections decrease the input resistance.oVoltage feedback decrease the output impedance.oCurrent feedback increase the output impedance.


Voltage Series Feedback Here, a portion of output voltage

feedback in series to the input. Negative feedback results overall

gain reduction. If there is no feedback (vf = 0).

The voltage gain of the amplifier is:

i

o

s

o

vv

vvA

If vf is connected in series with the input, then => vi = vs – vf

Since vo = Avi = A(vs – vf) = Avs – Avf

= Avs – A(βvo) => vo(1+ βA) = Avs

AA

vv

s

o

1 AA

vvA

s

of

1

So overall voltage gain with feedback is


Example Ex. 01) When negative voltage feedback is applied to an amplifier of gain 100, the overall gain falls to 50. (i) Calculate the fraction of the output voltage fedback. (ii) If this fraction is maintained, calculate the value of the amplifier gain required if the overall stage gain is to be 75.


Example


Ex. 2) The gain of an amplifier without feedback is 50 whereas with negative voltage feedback, it falls to 25. If due to ageing, the amplifier gain falls to 40, find the percentage reduction in stage gain (i) without feedback and (ii) with negative feedback.

Example


Ex. 03) An amplifier is required with a voltage gain of 100 which does not vary by more than 1%. If it is to use negative feedback with a basic amplifier the voltage gain of which can vary by 20%, determine the minimum voltage gain required and the feedback factor.

Example


The gain with feedback

Voltage Shunt Feedback

fi

i

s

of II

AIIvA

Here,Vo = AIi Is = Ii+ If

oi

if vI

AIA

ii

if AII

AIA

AAAf

1

Ii

Isi

o

IvA

o

f

vI

of vI

vo RL


Input Impedance with Feedback (1/3) Voltage Series Feedback:

The input impedance of the network can be determined as:

i

is

i

os

i

fs

i

ii z

Avvz

vvz

vvzvI


Input Impedance with Feedback (2/3) isii AvvZI

iiis AvZIv

iiiis ZAIZIv

i

iiii

i

sif I

ZAIZIIvZ

We know,

)1(So, AZZ iif


Input Impedance with Feedback (3/3) Voltage Shunt Feedback:

Ii

Is

If = βvo

Ri

Ro

AIivo RL

o

f

vI

oi

i

fi

i

s

iif vI

vII

vIvZ

Here,Zi = vi /Ii A=vo / Ii

ioii

iiif IvII

IvZ//

/

AZZ i

if

1


Voltage Series Feedback:

o The output impedance is determined by applying a voltage v resulting in a current I with vs shorted out ( vs = 0)

o The voltage v is then – v = IZ0 + Avi

Output Impedance with Feedback (1/3)

for vs = 0, vi = - vf


Output Impedance with Feedback (2/3) fo AvIZv , thatSo

)( vAIZv o

Rewriting the equation as:

oIZAvv

AZ

IvZ o

of

1


Current Series Feedback:

IAZvAv

ZvAv

ZvI

of

oi

o

Output Impedance with Feedback (3/3) By applying a signal voltage v to the output with vs shorted out resulting a current I, the ratio of v to I being the output impedance.

At, vs = 0, vi = - vf

IAZvIZ oo vIAZo )1( )1( AZ

IvZ oof


Effect of Negative Feedback on Gain and Bandwidth (1/2)

We know overall gain with negative feedback is:

For a practical amplifier the open-loop gain drops off at high frequencies due to the active device and circuit capacitance.

Gain also drop off at low frequencies for capacitively coupled amplifier stages.

When open-loop gain A drops enough low then the βA is no longer larger than 1, so Af = 1/ β is not true.

AA

AAAf

1

1So, fA

1for A


The figure shows that the amplifier with negative feedback has more bandwidth Bf than the amplifier without feedback B.

The feedback amplifier also has a higher upper 3-dB frequency and smaller lower 3-dB frequency.

However, the product of gain and frequency for both cases are same value.

Ao*f1 = Afo*f1f and Ao*f2 = Afo*f2f

Effect of Negative Feedback on Gain and Bandwidth (2/2)


Gain Stability with Feedback In addition to the β factor setting a precise gain value, we are

also interested in how stable the feedback amplifier is compared to an amplifier without feedback.

This shows that the magnitude of the relative change in gain

is reduced by the factor compared to that without

feedback f

f

AdA

A

AdA


Example

The improvement is 100 times. Thus, while the amplifier gain changes by 20%, the gain with feedback changes by only 0.2%.

Ex. 04) If an amplifier with gain of 1000 and feedback of 0.1 has a gain change of 20% due to temperature, calculate the change in gain of the feedback amplifier.

Soln: We know


Practical Feedback Circuits (1/2) Here, R1 and R2

resistors are used as a feedback network.

A part of output signal is obtained from R2 to ground.

vf is connected in series with the source signal vs.

Without feedback the amplifier gain is:

A = vo/vi = -gmRL

where RL is the parallel combination of RD, Ro and (R1+R2).

Voltage series feedback:


The feedback factor:

21

2

RRR

vv

o

f

Practical Feedback Circuits (2/2)

We know the gain with negative feedback is:

mL

Lmf

gRRR

RRg

AAA

21

211

If βA>>1 we have:

2

211R

RRAf


Example Exam: 05) Calculate the gain without and with feedback for the FET amplifier circuit of figure bellow and the following values: R1 = 80 kΩ, R2 = 20 kΩ, Ro = 10 kΩ, RD = 10 kΩ, and gm = 4000 µS,


Example Soln:


Exam. 06) Calculate the amplifier gain of the circuit of Fig. bellow for op-amp gain A 100,000 and resistances R1 1.8 k and R2 200 .

Example


Barkhausen Criterion (1/3) Typically, the feedback network is composed of

passive lumped components that determine the frequency of oscillation.

So, the feedback is complex-transfer function, hence denoted as β.

We can derive the requirements for oscillation as follows:

o Initially, sinusoidal source is xin

o The condition of oscillation is even xin is zero the output xout can be non-zero.


β

Axin +

-

xin + βxout

βxout

xout

Barkhausen Criterion (2/3)

The output of the amplifier block can be written as ][ outinfout xxAx

infoutfout xAxAx

inffout xAAx )1(

f

infout A

xAx

1


Barkhausen Criterion (3/3) If xin is zero, the only way the output can be

nonzero is to have βAf = 1. This condition is known as Barkhausen

criterion.


Example/ Problem Ex/Pb 07): An amplifier has an open-loop gain A=100,000. A

negative feedback of 10 dB is applied. Find (i) voltage gain with feedback, (ii) value of feedback fraction β.

Soln: (i) open-loop gain without feedback in dB = 20log10 100,000 = 20log10105

= 100 dB voltage gain with feedback = 100-10 = 90 dB Now, 20log10(Af) = 90 => log10 (Af) = 90/20 =4.5 so, Af = antilog 4.5 = 31622 (ii) A

AAf

1 *000,1001000,10031622

510*17.2


Nonlinear Distortion in Feedback Amplifier (1/3)

Nonlinear

distortion

Fundamental &

Harmonics

In nonlinear distortion, elongate +ve half cycle and compress –ve half cycle.


Nonlinear Distortion in Feedback Amplifier(2/3)

Nonlinear distortion produces harmonics of the input signal.

If a input signal has a 1KHz frequency then the output signals frequencies are 1, 2, 3 … KHz.

Here, 1KHz is fundamental frequency, and all others are harmonics.

Sometime nonlinear distortion is called harmonic distortion.

Total Harmonic Distortion (THD) without feedback is:

%100VoltagelFundamentaVoltageHarmonicTotal

THD

Total Harmonic Distortion (THD) with feedback is:

So THDf <THD


Nonlinear Distortion in Feedback Amplifier (3/3)

vf A

THDTHD

1

Here, Av is gain

without feedback. β is the feedback

fraction

Thanks for Your Kind Attention

Department of Electronics and Communication Engineering, KUET

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14699775563. feedback amplifiers