06.Op_Amp.pdf

8/9/2019 06.Op_Amp.pdf

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Operational Amplifier

(Op-Amp)

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Circuit Diagram of an Op-Amp (IC 741)Current Mirror

D i f f e r e n

t i a l A m p

l i f i e r

Gain Stage

Level Shifter

O u t p u t S t a g e

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Operational Amplifier (Op-Amp)

• An Op-Amp is a very high gain amplifier having anumber of differential amplifier stages

• It has high input impedance (typically a few Megaohm)

• It has a low output impedance (less than 100 Ω)

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Symbol

VO

V+

V-

Op-Amp Model

vo= Av(v+-v-)

In a good Op-Amp

Av→∞ R in →∞ R o →0

+

-

R in

V +

V -

R O

V o

+V cc

-V cc

If v+ is even slightly higher than v -, v o=+V ccIf v+ is even slightly lower than v -, v o=-Vcc

Cannot be usedas an amplifierby itself!

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+

-

R in

V +

V -

R O

V o

+V cc

-V cc

V in

V out

R 1

R 2

Consider the circuit shown below

1 2

2 1 2 1

2 1 2 1

0

1 1 1 1

1 1 1 1

o v

in o

O in

oin

O in o

o oin

v O in o

v v A v

v v v v v R R R R

vv v

R R R R R R R

v v v A R R R R R R R

For A v→∞, we get2 1

1 ino

O

vv

R R R

and 0o

v

vv

A

02

2

out

O

v R v

R R

2

1

out

in

v Rv R

Gain

The (-) terminal is effectivelyat ground. This is referred toas “Virtual Ground”

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+

-

R in

V +

V -

R O

V o

-V cc

V in

V out

R 1

R 2

Interesting Points With A v→∞,

Gain = -R 2/R 1(does not depend on A v)

(+) is at Ground(-) is at Virtual Ground

Zin Zo

Zin =R 1Zout =R O ||R 2≈R O

Concept of Virtual Ground1. V - ≈ 0 V ( -) terminal is at ground potential2. No Current Enters the (-) terminal of the Op-Amp

Valid only when there is feedback connectionbetween the output and the (-) terminal.

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• Input Resistance R i = ∞• Output Resistance R o= 0• Voltage Gain A v = ∞• Bandwidth = ∞ (i.e. can work over a wide range

of frequencies)

• Perfect balance i.e v o = 0 when v 1 = v 2

• Characteristics do not drift with temperature

Characteristics of an ideal OP-Amp

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i) R i = ∞,No current enters into op-amp

Voltage Gain Av

= ∞ or, vo/ v

d= ∞

or, v d = v o/∞ = 0 [since v o is finite]Therefore, v 1 - v2 = 0or,

ii) v1 = v 2

Ideal Op-Amp analysis

+

-

V d = V 1 -V 2

V 1

V 2

V o

i1 =0

i2 =0

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Simple OP-AMP Circuits

Using KVL,v1 – i1R1 = 0

i1 = v 1/R 1

&0 – i1R f – vo = 0or, v o = -i1R f = -v 1R f /R 1

v0/v1 = -R f /R 1

1. Inverting Amplifier

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2. Non Inverting Amplifier

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i

i

11

Input Impedance = ∞

Output Impedance = R o

1

11

1

11

1

1 1

1

o

f

o f f

f o

o

v v v

v vvi R R

v R v R R

Rvv R

Gain

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3. Voltage Follower

vo = v 1

Ideal circuit to test Op-Amp

High Input Impedance and LowOutput Impedance also makes ituseful for interfacing transducersand sensors to other circuits

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V 1

V o+

-

V 2

R 1

R 1

R 2

R 2

Difference Amplifier (or Voltage Subtractor)

Use Superposition(can also be done directly)

If v2=0, 21 1

1o

Rv v

R

If v1=0,2 2

2 21 2 1

22

1

1o R Rv v

R R R

Rv

R

Therefore,2

1 2 2 11

( )o o o

Rv v v v v

R

Difference Gain = R 2/R 1

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Integrator

1

0

1 with (0) 0

t

o ov

i v idt v R C V 1

V o+

-R

C

i

i

Therefore,

ov

10

1 with (0) 0

t

o ov v dt v RC

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V 1

V o+

-

R

C

Differentiator

i

i

1 10

1 with (0) 0t

o

v idt vC

v iR

10

1

1 t

o

o

v v dt RC

dvv RC

dt

Therefore,

or

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ExampleFind v o in the given

circuit.+

-v o+

-

20K Ω

10K Ω

2V

10K Ω

20K Ω 3V

v+

v-

0

20 42

30 3( 3)20 10

1 1 320 10 20 10

3 6 10 V

o

o

v v

v v v

vv

v v

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ExampleFind the gain v o/vi in the

given circuit.

v

1 1

2

2

2

1

2

1

0 Virtual Ground at (-)

2

21

21

2

ii

o

o

i

i

v v

v v Rv v

R R R

v v v v v R R R R

Rv v

R

R Rv

R R

R Rv

R

Gain 2

1

2oV

i

v R R A

v R

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Voltage Controlled Current Source(with grounded load)

1

2

22 2

32 2

3 2

IN OO IN

L L

O L O O L L

v v v vi v v v

R Rv vv

i v v R R

v v v v v v vv vi R R R R

-

+

v IN

v

v

vO

vL

R L

iL

R 1

2R

R3

2R

R 2

2R

R 4

R

R 5 R

i1i1

i2

i2

Therefore, IN L

vi

R

Note that i L does not dependon R L, implying that we havegot a current source!

General Condition:

R 1=R 2 & R3=R 4+R 5

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Creating an effectively “Negative Resistance”

What is “Negative Resistance”?

NormalResistance

ii

i

v R

iNegative

Resistancei

ii

v R

i

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Creating an effectively “Negative Resistance”

NegativeResistance

-

+

R

R

R i

v i

+

-

ii

Let v O = opamp output voltageand v +=v-=v i

1i O iO i

i i

i O ii

i

v v v Rv v

R R R

v v vi R R

Therefore, ii

i

v R

i

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Using Capacitors to make Inductors(practical to do only small values of inductances)

-

+

R

ii

C

R L

v i

+

-

for 2 π fCR L<<1

Using phasors, show that

(1 )i L L

L Li L

v R j RCR R j RCR

i j CR

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Precision Rectifier

The simple half-wave and full-wave rectifiers wesaw earlier have one big drawback – They do notwork for small voltages (say a few millivolts). Theinput voltage must cross the threshold which

forward biases the diode for rectification to occur.

Using Op-Amps, we can design rectifiers which donot have this disadvantage.

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Half-Wave Precision Rectifier

-

+vi

vO

R 1

R 2

D1

D2

v i

v O

You will be making this circuit in EC102 Lab next semester

Slope = - R 2/R 1

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What happens when you reverse the diodes?

v i

v O

Build a full wave rectifier using these two half-waverectifiers and a difference amplifier – needs three opamps!

Slope = - R 2/R 1

-

+ v O

R 1

R 2

D 1

D 2

v i

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Full-Wave Precision Rectifier

-

+v i

v O

R

R

D 1

D 2

-

+

R

2R

2R

Use good resistors with tighttolerances for this circuit to work,i.e. 1% resistors,

v i

vO

slope = -1

06.Op_Amp.pdf

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