3. operational ampliﬁers - arraytool · 3/3/2016 · 3. operational ampliﬁers s. s. dan and s....

Ideal Op-Amp Inverting Configuration Non-inverting Configuration Integrators & Differentiators Beauty of Miller’s Theorem *** Summary

3. Operational Amplifiers

S. S. Dan and S. R. Zinka

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

February 10, 2016

3. Operational Amplifiers ECE/EEE/INSTR F244, Dept. of EEE, BITS Pilani Hyderabad Campus

Outline

1 Ideal Op-Amp

2 Inverting Configuration

3 Non-inverting Configuration

4 Integrators & Differentiators

5 Beauty of Miller’s Theorem ***

6 Summary

Outline

1 Ideal Op-Amp

6 Summary

Introduction

• An operational amplifier, op-amp, is nothing more than a DC-coupled,high-gain differential amplifier

• The name "operational amplifier" stems from the op-amp’s ability toperform mathematical operations.

• One of the reasons for the popularity of the op amp is its versatility ...one can do almost anything with op amps!

• Op amp is a circuit building block of universal importance

• Early op amps were constructed from discrete components (vacuumtubes and then transistors, and resistors), and their cost wasprohibitively high (tens of dollars)

• IC op amp has characteristics that closely approach the assumed ideal

Introduction

Let’s See Internal Circuitry of an Op-Amp

Non-invertinginput

Offsetnull

50 kΩ

Offsetnull

Invertinginput

39 kΩ

50 kΩ 50 Ω

7.5 kΩ

4.5 kΩ

30 pF25 Ω

Output

Q8 Q9Q12 Q13

Current mirror Current mirror

Current mirrorDifferential amplifier Classs A gain stage

Output stage

Voltage level shifter

A Few Types of Op-Amps

The Ideal Op Amp – Terminals

• Require dc power to operate• No terminal of the op-amp package is physically connected to ground• An op amp may have other terminals for specific purposes

• Require dc power to operate

• No terminal of the op-amp package is physically connected to ground• An op amp may have other terminals for specific purposes

• Require dc power to operate• No terminal of the op-amp package is physically connected to ground

• An op amp may have other terminals for specific purposes

The Ideal Op Amp – Characteristics

A (v2 - v1)+

Inverting input

Noninverting input

Output

A (v2 - v1)+

Inverting input

Noninverting input

Output

_+v1i1 = 0

i2 = 0v2

A (v2 - v1)+

Inverting input

Noninverting input

Output

_+v1i1 = 0

i2 = 0v2

A (v2 - v1)+

A tends to ∞

Summary of Ideal Op Amp Characteristics

• The op amp responds only to the difference signal, v2 − v1, and henceignores any signal common to both inputs (i.e., ∞ CMRR)

• The ideal op amp is not supposed to draw any input current, i.e., theinput impedance of an ideal op amp is supposed to be infinite

• The output terminal is supposed to act as the output terminal of an idealvoltage source, i.e., the output impedance of an ideal op amp issupposed to be zero

• An important characteristic of op amps is that they are direct-coupled ordc amplifiers, where dc stands for direct-coupled

• Ideal op amps will amplify signals of any frequency with equal gain,and are thus said to have infinite bandwidth

• In almost all applications the op amp will not be used alone in aso-called open-loop configuration

Outline

1 Ideal Op-Amp

6 Summary

The Inverting Configuration

• As mentioned in the previous section, op amps are not used alone;rather, the op amp is connected to passive components in a feedbackcircuit

• Here we are using negative feedback. Can you guess how?• Why the above configuration is known as the inverting configuration?

• Here we are using negative feedback. Can you guess how?

• Why the above configuration is known as the inverting configuration?

The Closed-Loop Gain

v0 = −R2R1

?? ∞

v0 = −R2R1

0 0 V ∞

v0 = −R2R1

The Closed-Loop Gain – A Few Observations

• The closed-loop gain is independent of the op-amp gain.

• The fact that the closed-loop gain depends entirely on external passivecomponents (resistors R1 and R2) is very significant. It means that wecan make the closed-loop gain as accurate as we want by selectingpassive components of appropriate accuracy.

• Through applying negative feedback we have obtained a closed-loopgain that is much smaller than A but is stable and predictable. That is,we are trading gain for accuracy.

Effect of Finite Open-Loop Gain

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

0 - vO/A A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

?- vO/A A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

?- vO/A A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

?- vO/A A

i1 = i2 =vI + vO/A

=−R2/R1

1 + (1 + R2/R1) /A

Effect of A on Input and Output Resistances

?- vO/A A

Ri =vII1

vI + vO/A=

< R1 (1)

Ro = 0 (2)

To make Ri high we should select a high value for R1. However, if the requiredgain is also high, then R2 could become impractically large. We may concludethat the inverting configuration suffers from a low input resistance.

?- vO/A A

Ri =vII1

vI + vO/A=

< R1 (1)

Ro = 0 (2)

?- vO/A A

Ri =vII1

vI + vO/A=

< R1 (1)

Ro = 0 (2)

?- vO/A A

Ri =vII1

vI + vO/A=

< R1 (1)

Ro = 0 (2)

?- vO/A A

Ri =vII1

vI + vO/A=

< R1 (1)

Ro = 0 (2)

Now, let’s see two applications of op-amp inverting configuration ...

1st Application – The Weighted Summer

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

itot =v1R1

+ · · ·+ vn

vO = −(

+ · · ·+ vn

)Rf = −

R1v1 +

R2v2 + · · ·+

2nd Application – Generalized Weighted Summer

vO =Ra

Rbv1 + · · ·+

Rbvn−

Rn+1vn+1 −

Rn+2vn+2 − · · · −

vO =Ra

Rbv1 + · · ·+

Rbvn−

Rn+1vn+1 −

Rn+2vn+2 − · · · −

vO =Ra

Rbv1 + · · ·+

Rbvn−

Rn+1vn+1 −

Rn+2vn+2 − · · · −

Outline

1 Ideal Op-Amp

6 Summary

The Non-inverting Configuration

• As mentioned in the previous sections, op amps are not used alone;rather, the op amp is connected to passive components in a feedbackcircuit

• Here also we are using negative feedback. Can you guess how?• Why the above configuration is known as the non-inverting

configuration?

• Here also we are using negative feedback. Can you guess how?

• Why the above configuration is known as the non-invertingconfiguration?

configuration?

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vIR1

vO = vI +vIR1

R2 = vI

i1 = i2 =vI − vO/A

=1 + R2/R1

1 + (1 + R2/R1) /A

i1 = i2 =vI − vO/A

=1 + R2/R1

1 + (1 + R2/R1) /A

0 AvI - vO/A

i1 = i2 =vI − vO/A

=1 + R2/R1

1 + (1 + R2/R1) /A

0 AvI - vO/A

i1 = i2 =vI − vO/A

=1 + R2/R1

1 + (1 + R2/R1) /A

0 AvI - vO/A

i1 = i2 =vI − vO/A

=1 + R2/R1

1 + (1 + R2/R1) /A

0 AvI - vO/A

Ri → ∞

Ro = 0

0 AvI - vO/A

Ri → ∞

Ro = 0

0 AvI - vO/A

Ri → ∞

Ro = 0

0 AvI - vO/A

Ri → ∞

Ro = 0

Now, let’s see an application of op-amp non-inverting configuration ...

An Application—The Voltage Follower

• The property of high input impedance is a very desirable feature of thenon-inverting configuration

• Also known as unity-gain amplifier

• The circuit is said to have 100% negative feedback

• Since the non-inverting configuration has a gain greater than or equal tounity, depending on the choice of gain, some prefer to call it “a followerwith gain”

Outline

1 Ideal Op-Amp

6 Summary

The Inverting Configuration with General Impedances

= −Z2Z1

The Op-Amp Integrator

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

− 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

vO = − 1C

−∞

−∞vIdt− 1

0vIdt =

−VC −1

vO = − 1C

−∞

−∞vIdt− 1

0vIdt = −VC −

The Op-Amp Integrator in Frequency Domain

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

We can observe that

at ω = 0, the above circuit is operating with an open

So, it is clear that the integrator circuit will suffer deleterious effects from thepresence of the op-amp input dc offset voltage.

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

We can observe that

at ω = 0, the above circuit is operating with an open

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

We can observe that at ω = 0, the above circuit is operating with an open

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

The Miller Integrator

vO (t)vI (t) +

= −Z2Z1

1+jωRFC

= − RF/R1 + jωRFC

So, dc problem of the integrator circuit can be alleviated by connecting aresistor RF across the integrator capacitor C.

vO (t)vI (t) +

= −Z2Z1

1+jωRFC

vO (t)vI (t) +

= −Z2Z1

1+jωRFC

vO (t)vI (t) +

= −Z2Z1

1+jωRFC

The Op-Amp Differentiator

vI (t) vO(t)

vO = −iR = −CdvIdt× R = −RC

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

Differentiator circuits behaves like noise magnifiers and also sufferer fromstability problems. So, they are generally avoided in practice.

vI (t) vO(t)

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

vI (t) vO(t)

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

vI (t) vO(t)

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

vI (t) vO(t)

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

Outline

1 Ideal Op-Amp

6 Summary

Inverting Configuration

-AviR2/(1+A)

_+v1+v2

R2/(1+A-1)

_+v1+v2

= − AR2R1 (1 + A) + R2

Non-inverting Configuration (Wrong!)

-AviR2/(1+A)

R2/(1+A-1)

R1 (1 + A) + R2

Miller’s Theorem

V2= KV1

1 2Z II

V2= KV1Z1 Z2

1 2I2= II1= I

Z1 = Z/ (1− K)

Z2 = Z/(1− 1/K)

Miller’s theorem needs a common ground (or reference voltage) on bothsides ...

Miller’s Theorem

V2= KV1

1 2Z II

V2= KV1Z1 Z2

1 2I2= II1= I

Z1 = Z/ (1− K)

Z2 = Z/(1− 1/K)

Miller’s Theorem

V2= KV1

1 2Z II

V2= KV1Z1 Z2

1 2I2= II1= I

Z1 = Z/ (1− K)

Z2 = Z/(1− 1/K)

Outline

1 Ideal Op-Amp

6 Summary

=−R2/R1

1 + (1 + R2/R1) /A

Ri =R1

Ro = 0

=1 + R2/R1

1 + (1 + R2/R1) /A

Ri → ∞

Ro = 0

(log scale)

6 dB/octave-

= −Z2Z1

= −1/jωCR

= − 1jωRC

ωRC∠90

vO (t)vI (t) +

= −Z2Z1

1+jωRFC

So,the dc problem of the integrator circuit can be alleviated by connecting aresistor RF across the integrator capacitor C.

vI (t) vO(t)

= −Z2Z1

= − R1/jωRC

= −jωRC = ωRC∠− 90

3. operational ampliﬁers - arraytool · 3/3/2016 · 3. operational ampliﬁers s. s. dan and s....

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