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Circuits with Feedback - Assignment Example

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The paper "Circuits with Feedback " is purposed to solve the problems when the equation for the gain of a feedback amplifier is obtained, each of the amplifiers incorporate series feedback, the amplifier has a gain of 150 or to show how to produce a voltage series feedback network…
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Circuits with Feedback
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"Circuits with Feedback"

Download file to see previous pages Derivation of Feedback Equation for Inverting Amplifier
Let we consider an inverting amplifier as shown in fig. 2.

Then if ‘A, is the gain then, as vi+ = 0, there fore

vo = A(vin+ – vin¯)

vo = - A vin¯
Also from Ohm’s Law the current is simply the difference in the voltage across R1 divided R1,

i1 = (vin+ – vin¯)/R1 ……(5)

Similarly, if = ( vi¯ – vo)/Rf ……(6)

By Kerchief’s current Law at the inverting input,

i1 = if + i¯ where i¯ is the current entering the amplifier at

it’s inverting point, yet for an idea amplifier it reaches to zero hence,

i1 = if ……… (7)

Substituting the values from equations (5) and (6) we have,

(vin+ – vin¯)/R1= (vi¯ – vo)/Rf ……… (8)

Or vin+/R1 – vin¯/R1= vi¯/Rf – vo/Rf

Solving for ,
= ……… (9)

This is the required voltage gain expression for inverting amplifier with feedback in

terms of open loop gain.

Problem 2: Each of the amplifiers shown below incorporate series feedback. The

gains, input resistances

and output resistances are quoted without feedback. For each amplifier determine: -

a) The feedback fraction.

b) The gain with feedback.

c) The input impedance with feedback.

d) The output impedance with feedback.

Solution: (i) a) If we consider the given cct. then for feedback fraction ‘β, for

given cct. is defined as,




...
terms of open-loop gain A.

Derivation of Feedback Equation for Inverting Amplifier

Let we consider an inverting amplifier as shown in fig. 2 (Bogart, 1997, p. 670).

Then if 'A, is the gain then, as vi+ = 0, there fore

vo = A(vin+ - vin)

vo = - A vin


Fig. 2(a)

Also from Ohm's Law the current is simply the difference in the voltage

across R1 divided R1,

i1 = (vin+ - vin)/R1 (5)

Similarly, if = ( vi - vo)/Rf (6)

By Kerchief's current Law at the inverting input,

i1 = if + i where i is the current entering the amplifier at

it's inverting point, yet for an idea amplifier it reaches to zero hence,

i1 = if (7)

Substituting the values from equations (5) and (6) we have,

(vin+ - vin)/R1= (vi - vo)/Rf (8)

Or vin+/R1 - vin/R1= vi/Rf - vo/Rf

Solving for ,
= (9)

This is the required voltage gain expression for inverting amplifier with feedback in

terms of open loop gain.

Problem 2: Each of the amplifiers shown below incorporate series feedback. The

gains, input resistances

and output resistances are quoted without feedback. For each amplifier determine: -

a) The feedback fraction.

b) The gain with feedback.

c) The input impedance with feedback.

d) The output impedance with feedback.

Solution: (i) a) If we consider the given cct. then for feedback fraction ', for

given cct. is defined as,

= (2ia)

Substituting the values in equation (2ia) from given cct. we have,

=


= 0.175 2ia


Fig. (2i)

(2ib) If 'Avf, be the gain ...Download file to see next pagesRead More
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