electronics- negative feedback in operational...

Op amp output

ElectronicsNegative Feedback in Operational Amplifiers

Terry Sturtevant

Wilfrid Laurier University

July 29, 2013

Terry Sturtevant Electronics Negative Feedback in Operational Amplifiers

Op amp output Op amp output with negative feedbackExample: non-inverting amplifier

The equation for the output of an op amp is always true as long asthe output is not saturated

Vo = A (V+ − V−)

For negative feedback, make the voltage at V− be someportion of Vo.

This ratio of V− to Vo is called the feedback factor, β.

V−Vo

For negative feedback, make the voltage at V− be someportion of Vo.This ratio of V− to Vo is called the feedback factor, β.

V−Vo

For negative feedback, make the voltage at V− be someportion of Vo.This ratio of V− to Vo is called the feedback factor, β.

V−Vo

For instance, with just a wire from V− to Vo, the feedabckfactor is 1.

This is the case for a voltage follower.

β = 1

V−Vo

� �

For instance, with just a wire from V− to Vo, the feedabckfactor is 1.This is the case for a voltage follower.

β = 1

V−Vo

� �

For instance, with just a wire from V− to Vo, the feedabckfactor is 1.This is the case for a voltage follower.

β = 1

V−Vo

� �

For feedback

V− = βVo

whereβ ∈ [0, 1]

So the op amp equation

Vo = A (V+ − V−)

becomesVo = A (V+ − βVo)

Vo = AV + − βAVo

For feedbackV− = βVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

β ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

becomes

Vo = A (V+ − βVo)

Vo = AV + − βAVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

whereβ ∈ [0, 1]

Vo = A (V+ − V−)

Vo = AV + − βAVo

Vo + βAVo = AV+

Vo (1 + βA) = AV+

Vo = V+

( A1 + βA

Vo = V+

1A + β

Vo + βAVo = AV+

Vo (1 + βA) = AV+

Vo = V+

( A1 + βA

Vo = V+

1A + β

Vo + βAVo = AV+

Vo (1 + βA) = AV+

Vo = V+

( A1 + βA

Vo = V+

1A + β

Vo + βAVo = AV+

Vo (1 + βA) = AV+

Vo = V+

( A1 + βA

Vo = V+

1A + β

Vo + βAVo = AV+

Vo (1 + βA) = AV+

Vo = V+

( A1 + βA

Vo = V+

1A + β

If the gain, A, is much greater than 1, then

A� β

and so1A � β

Vo = V+

��1A + β

Vo ≈ V+

)so Vo only depends on β.

A� β

and so1A � β

Vo = V+

��1A + β

Vo ≈ V+

A� β

and so1A � β

Vo = V+

��1A + β

Vo ≈ V+

A� β

and so1A � β

Vo = V+

��1A + β

thusVo ≈ V+

A� β

and so1A � β

Vo = V+

��1A + β

Vo ≈ V+

so Vo only depends on β.

A� β

and so1A � β

Vo = V+

��1A + β

Vo ≈ V+

So for a voltage follower, where β = 1,

Vo ≈ V+

as expected.

Vo ≈ V+

)So for a voltage follower, where β = 1,

Vo ≈ V+

as expected.

Vo ≈ V+

as expected.

Vo ≈ V+

as expected.

Non-inverting amplifier

This can be redrawn

Rf and Ri form a voltage divider, so the voltage at V− is afraction of Vo

By definition, this is the feedback factor, β.Thusβ = Ri

Rf +Ri

sogain = 1

β = Rf +RiRi

as expected

By definition, this is the feedback factor, β.

Thusβ = Ri

Rf +Ri

sogain = 1

β = Rf +RiRi

as expected

By definition, this is the feedback factor, β.Thus

β = RiRf +Ri

sogain = 1

β = Rf +RiRi

as expected

Rf +Ri

sogain = 1

β = Rf +RiRi

as expected

Rf +Ri

gain = 1β = Rf +Ri

as expected

Rf +Ri

sogain = 1

β = Rf +RiRi

as expected

Rf +Ri

sogain = 1

β = Rf +RiRi

as expected

This can be redrawn

Vi��

VA��

��

= VAVB

= 1 + RfRi

This can be redrawn

Vi��

VA��

��

= VAVB

= 1 + RfRi

This can be redrawn

Vi��

VA��

��

= VAVB

= 1 + RfRi

If Ri is not grounded, then it looks like this.

Vi��

��

��VC

V0 is VA - VC

Vi is VB - VC

so VoVi

= VA−VCVB−VC

= 1 + RfRi

Vi��

VAVo��

��VC

V0 is VA - VC

Vi is VB - VC

so VoVi

= VA−VCVB−VC

= 1 + RfRi

Vi��

VAVo��

��VC

V0 is VA - VC

Vi is VB - VC

so VoVi

= VA−VCVB−VC

= 1 + RfRi

Vi��

VAVo��

��VC

V0 is VA - VC

Vi is VB - VC

so VoVi

= VA−VCVB−VC

= 1 + RfRi

Vi��

VAVo��

��VC

V0 is VA - VC

Vi is VB - VC

so VoVi

= VA−VCVB−VC

= 1 + RfRi

We can redraw the circuit slightly differently.

Since VA−VCVB−VC

= 1 + RfRi

This can be rewritten VA −(

1 + RfRi

)VB = −Rf

= 1 + RfRi

1 + RfRi

)VB = −Rf

= 1 + RfRi

1 + RfRi

)VB = −Rf

= 1 + RfRi

1 + RfRi

)VB = −Rf

The derivation goes like this:

VA − VCVB − VC

= 1 +RfRi

VA − VC =

)(VB − VC )

VA −��VC =

)VB −��VC −

VA −(

1 +RfRi

)VB = −Rf

VA − VCVB − VC

= 1 +RfRi

VA − VC =

)(VB − VC )

VA −��VC =

)VB −��VC −

VA −(

1 +RfRi

)VB = −Rf

VA − VCVB − VC

= 1 +RfRi

VA − VC =

)(VB − VC )

VA −��VC =

)VB −��VC −

VA −(

1 +RfRi

)VB = −Rf

VA − VCVB − VC

= 1 +RfRi

VA − VC =

)(VB − VC )

VA −��VC =

)VB −��VC −

VA −(

1 +RfRi

)VB = −Rf

VA − VCVB − VC

= 1 +RfRi

VA − VC =

)(VB − VC )

VA −��VC =

)VB −��VC −

VA −(

1 +RfRi

)VB = −Rf

Finally, we can set VB to ground.

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

and VB = 0Then VA = −Rf

RiVC which is an inverting amplifier!

So the feedback factor, β, is the same as for a non-invertingamplifier.

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

and VB = 0

Then VA = −RfRi

VC which is an inverting amplifier!So the feedback factor, β, is the same as for a non-invertingamplifier.

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

which is an inverting amplifier!So the feedback factor, β, is the same as for a non-invertingamplifier.

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

VB = 0

So if VA −(

1 + RfRi

)VB = −Rf

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