operational ampliﬁers - city ucktse.eie.polyu.edu.hk/eie209/6.op-amp.pdf · c.k. tse: operational...

Operational Amplifiers

forBasic Electronics

http://cktse.eie.polyu.edu.hk/eie209

byProf. Michael Tse

January 2005

C.K. Tse: Operational Amplifiers 2

Where do we begin?

We begin with assuming that the op-amp is an ideal element satisfying thefollowing conditions:

Output resistance = 0 (perfect output stage)Input resistance = ∞ (perfect input stage)Differential voltage gain = ∞

Since the gain A ≈ ∞, vi ≈ 0 if vo is infinite,the two input terminals have same potential if vo is infinitea “virtual” short-circuit exists between the two input terminals

+

–

+vi–

vo

+vi–

Avi±+vo–


The 347 IC op-amp

output stage

+–

single-ended output

The 347 is a QuadJFET input op-ampusing biFETtechnology.

1234567

Manufacturer listed spec:Rin = 1012Ω; AVOL=100dB = 105

CMRR = 100dBGBW = 4MHz (gain-bandwidth)SR = 13V/µs

141312111098

+–

+–

+–

+–

V+ V–


The basicsAn op-amp is a very high gain differential amplifier. In almost all applications(except in comparator and Schmitt trigger), feedback is used to stabilize thegain.

TWO GOLDEN RULES:

RULE 1:The output attempts to do whatever is necessary to make the voltagedifference between the two inputs zero.

RULE 2:The inputs draw no current.


ExampleConsider the following op-amp circuit. What is the voltage gain?

R1

R2

–

+

vivo

Then, it says that the current flowing intothe inputs are zero.

ix

ix

Apply the Golden Rules:

It first says that the output will try to setitself in order to make the differencebetween the inputs zero. That means, itwill try to make the –ve input 0 Vbecause the +ve input is 0 V.

0V

Therefore,

This is the inverting amplifier.


WarningsThe Golden rules sometimes do not apply. NOTE CAREFULLY that GoldenRule 1 says that “the output attempts to…”. The output attempts, but it mayfail to do what it wants to do!

+

–

Do Golden rules apply in the following circuits?

+

–

–

+–

+

–

+

+

–

x2 xsq.

–1V


Other examples (where Golden rules work)

R1

R2

–

+vi

vo

Applying the Golden rules, we get

This is the non-inverting amplifier.

–

+vi

Here, simply

This is the voltage follower.


Other examples (where Golden rules work)More examples

R2

Rf

–

+

v2

vo

This is the summing amplifier.

–

+v2

This is the difference amplifier.

R1

R3

v1

v3

R1

R1

v1

R2

R2


Other examples (where Golden rules work)More examples

R–

+

vi

vo

This is the integrator.

–

+

This is the differentiator (theoretically).In practice, this circuit won’t work!!!

vi

R

C

C


Examples (where Golden rules do not work)

+

–outv

1v

2v

Comparator

Since the voltage gain typically exceeds 100,000, the inputs must be withina fraction of a millivolt in order to prevent the output from swinging all theway to extreme positive or negative. It is assumed that the supply voltagesare +10 V and –10 V and that the gain is 100,000.

1. If v1 is larger than v2 by more than 0.0001 V, the output willswing to +10 V.2. If v2 is larger than v1 by more than 0.0001 V, the output willswing to –10 V.

The output cannot makethe two inputs equal!!!Golden Rule 1 fails!!!



+

–outv

1v

2v

Comparator

But this simple comparator suffers from a problem if the input signals havenoise! The output may switch (jump up and down) when the signals areclose to each other.

The output cannot makethe two inputs equal!!!Golden Rule 1 fails!!!



+

–outv

1v

2v

Comparator

Suppose v2 = 5V (constant) and v1 is an input.This circuit is supposed to compare v1 with 5V.

But if v2 has noise, the output may jump when v2 is near 5V.

±5V

v1

5V v2

t

vout



Schmitt Trigger — a better comparator

–

+outv

R1

R2

vin

A

How does it work?

Assume the op-amp is powered by±10V, and now vout = +10V.

Obviously, vin must be less than vA:

What happens if vin moves just above10R1/(R1+R2)? Clearly, vout falls to–10V because of comparator action.Therefore, vA drops to –10R1/(R1+R2),and vin must be greater than vA:

†

vin <10R1

R1 + R2= vA

†

vin >–10R1

R1 + R2= vA



Schmitt Trigger

–

+outv

R1

R2

vin

A

We have a situation similar to hysteresis.

†

upper trip point = 10R1R1 + R2

†

lower trip point = –10R1R1 + R2

t

t

vin

vout

+10

–10

10R1R + R1 210R1

R + R1 2

–



Schmitt Trigger

What are the upper and lower trip points?

–

+outv

vin

10V

–10V

10kW

90kW

8V+–


Practical considerations

Finite input currents

Very small currents are in fact needed to bias the op-amp input stage. Circuitsthat have no DC path to inputs won’t work!

None of these works!

vo

–

+

vi

C

vo

–

+

vi

C

x x


Practical considerationsOffset in integrator

The op-amp integrator is very easily saturated if there is a small lack ofsymmetry in the input signals. This is because the error gets integrated quicklyand the output will soon move towards the maximum voltage.

–

+

vi

C

C

–

+

In practice we need a discharge path toprevent saturation. Usually R has to bebig enough, so that the discharge ratebecomes insignificantly slow comparedto the signal frequency.

R


ApplicationsCurrent source

+–

RIo

LOAD

vR

We see that vR is fixed by the voltage divider.

The op-amp will make sure that the voltage acrossR is also equal to vR, which is fixed!

Therefore the current flowing down R must be

which is the load current.

Thus, this circuit provide a constant currentsource for the load.

Note: the load is floating for this case!


ApplicationsCurrent source for grounded load

–+

R

Io

LOAD

vR

Again vR is fixed by the voltage divider.

The op-amp will make sure that the voltage at thelower end of R is also equal to vR, which is fixed!

Therefore the current flowing down R must be

which is very close to the load current (if basecurrent is small and op-amp draws very smallcurrent).

Thus, this circuit provide a constant currentsource for the grounded load.

Vcc


ApplicationsCurrent source for grounded load(voltage controllable)

–+

R

Io

LOAD

vR

Here, vR is controllable/adjustable by vIN.

The current flowing down R, which is close to theload current Io, must be

Thus, this circuit provide a controllable constantcurrent source for the grounded load.

Vcc

+–

vIN

R2

†

Io =Vcc - (Vcc - R2Ix )

R

=R2

R1

vIN

R

R1

Ix


Other non-ideal behaviourExample of input bias current

R1

R2

–

+

vivo

ib

ib

Problem: Since ib flows intoboth inputs, the negative inputside will have a slightly negativedc voltage even when vi = 0,whereas the positive input side isstill 0V because there is noresistor there! Therefore, vi ≠ 0,i.e., some unwanted offset!

R1

R2

–

+

vivo

ib

ibPractical solution:

R1||R2


Other non-ideal behaviourInput offset voltage

Due to imperfect symmetry, some voltage has to be applied to the input to get theoutput to zero. Typical value ≈ 5 mV.

The input offset voltage is a function of temperature (due to temperature drift ofdevice parameters).

Practical solution: In many applications, the dc gain is not needed. We cansimply drop the dc gain to 1.

For example, for the non-inverting amplifier, if we set+

–R1 = 2kΩC1 = 4.7µF

the cutoff frequency isapprox 17 Hz.

R1R2

C1


Summary

We have studied the basics of op-amps, and some applications.

Basic rules of op-amp circuit analysisSome practical considerationsSome applications

operational ampliﬁers - city ucktse.eie.polyu.edu.hk/eie209/6.op-amp.pdf · c.k. tse: operational...

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