Operational Amplifiers

for

Electronic Circuits

http://cktse.eie.polyu.edu.hk/eie304

by

Prof. Michael Tse

C.K. Tse: Operational Amplifiers

2

Where do we begin?

In Year 1, you have studied the op-amps and simple op-amp circuits, assuming
that the op-amp is an ideal element satisfying the following conditions:

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

Since the gain A ≈ ∞, v

i

≈ 0 if v

o

is infinite,

the two input terminals have same potential if v

o

is infinite

a “virtual” short-circuit exists between the two input terminals

+

–

+
v

i

–

v

o

+
v

i

–

Av

i

±

+
v

o

–

C.K. Tse: Operational Amplifiers

3

Practical op-amp

Input stage:

Differential amplifier of high input resistance
e.g., JFET pair

Amplification:

Common-emitter stage with active load

Output stage:

Emitter follower with low output resistance

DA

CE

EF

input stage

gain stage

output stage

C.K. Tse: Operational Amplifiers

4

The 347 IC op-amp

output stage

+
–

single-ended output

The 347 is a Quad
JFET input op-amp
using biFET
technology.

1
2
3
4
5
6
7

Manufacturer listed spec:
R

in

= 10

12

Ω; A

VOL

=100dB = 10

5

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

14
13
12
11
10
9
8

+

–

+

–

+

–

+

–

V+

V–

C.K. Tse: Operational Amplifiers

5

DC parameters

1

Differential voltage gain A

VOL

For example, A

VOL

= 106 dB = 200,000

For limited output voltage swing = ±13V, the input voltage for open
circuit operation must be less than 13/200000 = ±65 µV.
Usually, op-amp with such a large gain is operated closed loop.

2

Single-ended input resistance
= input resistance measured at either terminal
BJT opamps typically have Rin ≈ 10

6

Ω

biFET opamps typically have Rin ≈ 10

12

Ω

biMOS opamps typically have Rin ≈ 10

15

Ω

3

Output resistance
Typically around 100Ω

4

CMRR usually drops at increasing frequency

120

100

CMRR

10 100 1k 10k 100k 1M

C.K. Tse: Operational Amplifiers

6

Review of the basics

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

TWO GOLDEN RULES:

RULE 1:
The output

attempts

to do whatever is necessary to make the voltage

difference between the two inputs zero.

RULE 2:
The inputs draw no current.

C.K. Tse: Operational Amplifiers

7

Example

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

R

1

R

2

–

+

v

i

v

o

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

i

x

i

x

Apply the Golden Rules:

It first says that the output will

try

to set

itself in order to make the difference
between the inputs zero. That means,

it

will try to make the –ve input 0 V
because the +ve input is 0 V.

0V

Therefore,

This is the inverting amplifier.

C.K. Tse: Operational Amplifiers

8

Warnings

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

+

–

Do Golden rules apply in the following circuits?

+

–

–

+

–

+

–

+

+

–

x

2

x

sq.

–1V

C.K. Tse: Operational Amplifiers

9

Other examples (where Golden rules work)

In Year 1, you have seen a number of useful op-amp circuits.

R

1

R

2

–

+

v

i

v

o

Applying the Golden rules, we get

This is the non-inverting amplifier.

–

+

v

i

Here, simply

This is the voltage follower.

C.K. Tse: Operational Amplifiers

10

Other examples (where Golden rules work)

More examples

R

2

R

f

–

+

v

2

v

o

This is the summing amplifier.

–

+

v

2

This is the difference amplifier.

R

1

R

3

v

1

v

3

R

1

R

1

v

1

R

2

R

2

C.K. Tse: Operational Amplifiers

11

Other examples (where Golden rules work)

More examples

R

–

+

v

i

v

o

This is the integrator.

–

+

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

v

i

R

C

C

C.K. Tse: Operational Amplifiers

12

Practical considerations

Finite input currents

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

None of these works!

v

o

–

+

v

i

C

v

o

–

+

v

i

C

x

x

C.K. Tse: Operational Amplifiers

13

Practical considerations

Offset in integrator

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

–

+

v

i

C

C

–

+

In practice we need a discharge path to
prevent saturation. Usually R has to be
big enough, so that the discharge rate
becomes insignificantly slow compared
to the signal frequency.

R

C.K. Tse: Operational Amplifiers

14

Practical considerations

Finite loop gain and bandwidth

The openloop gain of practical op-amps is not infinite and drops as frequency increases.

The

unity gain bandwidth

or

gain-bandwidth product (GBW)

is the upper frequency

at which the gain drops to unity (0 dB). Typical values of GBW are greater than 1 MHz.

A

v

100Hz

10k

10M

100dB

60dB

0dB

f

b

f

t

f (Hz)

The gain is given by

where

ω

b

= 2π f

b

is the

break frequency

.

The

gain bandwidth product

is

ω

t

= 2π f

t

Note:

ω is in rad/s and f is in Hz.

slope = –20 dB/dec

C.K. Tse: Operational Amplifiers

15

Practical considerations

Gain bandwidth product

The gain at

ω is

A

v

20log|A

o

|

0dB

ω

b

f

(Hz)

For example, if the GBW is

f

t

= 10 MHz

then, the gain at 1 MHz is

A

v

(1M) = 10 or 20dB

and the gain at 100 kHz is

A

v

(100k) = 10/0.1 = 100

or 40 dB

We can find the gain anywhere
along the slope region using this
formula: A

v

(f) = f

t

/f .

Obviously, A

o

f

b

= f

t

or A

o

ω

b

=

ω

t

10M

1M

20dB

100k

40dB

slope = –20 dB/dec

C.K. Tse: Operational Amplifiers

16

Effects of first-order A

v

(

ω

) on the closed-loop gain

Example: Inverting amplifier

Suppose the op-amp gain is A

v

(

ω), not infinity! The inverting amplifier gain at ω becomes

R

1

R

2

–

+

v

i

v

o

A

v

(

ω)

Since

and

A

o

ω

b

=

ω

t

we have

C.K. Tse: Operational Amplifiers

17

Effects of first-order A

v

(

ω

) on the closed-loop gain

Example: Inverting amplifier

Clearly the transfer function has a typical first-order response.

where

A

v

20log|R

2

/R

1

|

ω

p

ω

(rad/s)

ω

m

ω

m

=

ω

p

(R

2

/ R

1

)

slope = –20 dB/dec

C.K. Tse: Operational Amplifiers

18

Applications

Current source

+
–

R

I

o

LOAD

v

R

We see that v

R

is fixed by the voltage divider.

The op-amp will make sure that the voltage across
R is also equal to v

R

, which is fixed!

Therefore the current flowing down R must be

which is the load current.

Thus, this circuit provide a constant current
source for the load.

Note: the load is floating for this case!

C.K. Tse: Operational Amplifiers

19

Applications

Current source for grounded load

–
+

R

I

o

LOAD

v

R

Again v

R

is fixed by the voltage divider.

The op-amp will make sure that the voltage at the
lower end of R is also equal to v

R

, which is fixed!

Therefore the current flowing down R must be

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

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

V

cc

C.K. Tse: Operational Amplifiers

20

Applications

Current source for grounded load

(voltage controllable)

–
+

R

I

o

LOAD

v

R

Here, v

R

is controllable/adjustable by v

IN

.

The current flowing down R, which is close to the
load current I

o

, must be

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

V

cc

+
–

v

IN

R

2

I

o

=

V

cc

− (V

cc

− R

2

I

x

)

R

=

R

2

R

1

v

IN

R

R

1

I

x

C.K. Tse: Operational Amplifiers

21

Applications

Active filters (example: low-pass)

v

o

–

+

v

i

R

2

R

1

R

f

This is a low-pass filter (allowing signals of
low frequency to pass through). The nodal
equations are:

i

i´

(1)

(2)

(3)

From (2) and (3), we get

Hence, we have

C.K. Tse: Operational Amplifiers

22

Applications

Active filters (example: low-pass)

v

o

–

+

v

i

R

2

R

1

R

f

From (1), the transfer function can be
written as

i

i´

gain

3dB

cutoff frequency or
3dB corner frequency

ω

C.K. Tse: Operational Amplifiers

23

Applications

Active filters (example:high-pass)

v

o

–

+

v

i

R

2

R

1

R

f

We can easily derive the transfer function as

i

i´

gain

3dB

cutoff frequency or
3dB corner frequency

ω

C.K. Tse: Operational Amplifiers

24

Interim summary

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

Basic rules of op-amp circuit analysis
Some practical considerations
Some applications

Our next task is to look into the internal details of an op-amp and some practical
problems arising from the non-ideal properties of op-amps.

C.K. Tse: Operational Amplifiers

25

The 741 operational amplifier

C.K. Tse: Operational Amplifiers

26

The 741 operational amplifier

What is inside the op-amp?

Basically we may divide an op-amp into three main stages:

Main functional parts
1.

Input stage

2.

Second stage (gain stage)

3.

Output stage

Other parts
•

Bias circuit

•

Short-circuit protection circuit

C.K. Tse: Operational Amplifiers

27

The bias circuit in 741

Widlar mirror

current sources

C.K. Tse: Operational Amplifiers

28

The bias circuit in 741

Widlar mirror

current sources

current copiers

proportional to I

REF

I

REF

bias for Q17

bias for Q18,Q19

C.K. Tse: Operational Amplifiers

29

Short-circuit protection circuit in 741

These transistors are
usually off, but will be
on when the output
draws a large current.

C.K. Tse: Operational Amplifiers

30

Input stage

differential

amp

active load

C.K. Tse: Operational Amplifiers

31

Input stage

Q1 and Q2 are emitter followers, providing high input resistance.

Q3 and Q4 serve as differential amplifier, providing high CMRR.

Q5, Q6, Q7 and R1, R2, R3 provide the load (active load) for the differential
amplifier.
This loading circuit also provides a single-ended output for the next stage.

Q3 and Q4 also serve as DC level shifter.

Q3 and Q4 are pnp transistors, hence protect the input stage Q1 and Q2 against
emitter-base junction breakdown, since the emitter base junction of npn may
break down at about 7V of reverse bias. Usually, pnp has emitter-base
breakdown around 50 V!!

C.K. Tse: Operational Amplifiers

32

Second stage (gain stage)

high gain
common-
emitter

EF

CE

active load

C.K. Tse: Operational Amplifiers

33

Second stage (gain stage)

Q16 is an emitter follower, providing high input resistance for the stage so as
to minimize its loading effect.

Q17 is a common-emitter amplifier, loaded by Q13B.

Q13B is a current mirror and serves as active load.

The gain is

GAIN

second stage

≈ g

m,17

r

o,13B

Note the capacitor Cc. This is called internal compensation cap and is used for
maintaining stability in when the op-amp is used in a feedback configuration.
We’ll revisit this issue later when we study feedback and stability.

C.K. Tse: Operational Amplifiers

34

Output stage

Class AB output stage

diodes

complementary
push-pull

C.K. Tse: Operational Amplifiers

35

Output stage

Provides low output resistance.

Class AB output stage provided by Q14 and Q20 as complementary push-
pull.

Q19 and Q18 maintain a V

BE

drop to smooth out the crossover distortion,

as in the class AB amplifier.

C.K. Tse: Operational Amplifiers

36

Non-ideal performance

Slew rate limitation

One of the very significant limitations of op-amps is the limited rate at
which the output can follow the input, when the input has a relatively
large swing. But op-amps are often used for swings as wide as ±10V, for
example, with 741 having ±15V as supply voltages.

This causes distortion in the output even when the signal frequency is
well below the 3dB corner frequency!

Op-amp circuit, e.g.,
inverting amp, voltage
follower, low-pass filter,
etc.

distortion!

C.K. Tse: Operational Amplifiers

37

Non-ideal performance

Slew rate limitation

Let’s consider a simple voltage follower.

v

o

–

+

v

i

Suppose we send a step to the input. What is the expected output?

C.K. Tse: Operational Amplifiers

38

Non-ideal performance

Slew rate limitation

To find the output, let’s derive the transfer function, assuming that the op-
amp causes a first-order response as explained before.

We expect the output to be exponentially rising step when a step is applied to the
input. If f

t

is 1.3MHz (for 741), the output should reach 90% of 5V in about 0.3

µs.

v

o

–

+

v

i

5V

0.3µs

C.K. Tse: Operational Amplifiers

39

Non-ideal performance

However, the output is much slower, reaching 5V in 5µs!

v

o

–

+

v

i

5V

5µs

Slew Rate = rate of change of output in the region of constant slope

in this case

5V

C.K. Tse: Operational Amplifiers

40

Origin of slew rate limitation

Large and fast increase in the input drives the input differential stage completely
out of the linear range of operation.

CE

Class
AB

I

x

I

1

I

1

2I

1

Q1

Q2

–

v

i

+

v

o

C

high gain

Q2 takes up all the bias current and Q1 is cutoff when v

i

is increased to 5V very rapidly.

The current in Q2 is 2I

1

, and this current passes on to the

second (CE) stage.

But due to internal compensation,
the second stage has a capacitor
C, forming an integrator. Hence,

I

x

= 2I

1

Thus,

C.K. Tse: Operational Amplifiers

41

Origin of slew rate limitation

That means, the output has a constant rate of change given by

This is, by definition, the

slew rate.

Note:

-The slew rate limitation is due to the limited current available to charge
the capacitor.
-The circuit does not have an overall feedback. In fact, during the slewing
period, the input is being cutoff. Thus, the slew rate is the same whether
or not feedback is applied.

C.K. Tse: Operational Amplifiers

42

Can we alleviate slew rate limitation?

If we can reduce the transconductance (gain) of the DA stage, it is less
likely to be driven out of the linear range.

The transconductance can be reduced by

emitter degeneration

.

Suppose the emitter resistance in the DA is R

E

. Then the

transconductance is reduced to

Hence, slew rate can be improved.

g

mI

=

g

m

2

1

1

+

g

m

R

E

2

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

C.K. Tse: Operational Amplifiers

43

Other non-ideal behaviour

Input bias current

Due to the input stage requiring biasing, a small current is needed in the
input.

I

EE

+
v

i

–

i

b

i

b

i

b

= bias current

This small bias current is dc and
causes voltage drops in the
resistors connecting to the
inputs of the op-amp, e.g., the
feedback resistors.

Hence, we see

small dc

residual voltage at the op-amp
output even if the input is 0.

C.K. Tse: Operational Amplifiers

44

Other non-ideal behaviour

Example of input bias current

R

1

R

2

–

+

v

i

v

o

i

b

i

b

Problem: Since i

b

flows into

both inputs, the negative input
side will have a slightly negative
dc voltage even when v

i

= 0,

whereas the positive input side is
still 0V because there is no
resistor there! Therefore, v

i

≠ 0,

i.e., some

unwanted offset

!

R

1

R

2

–

+

v

i

v

o

i

b

i

b

Practical solution:

R

1

||R

2

C.K. Tse: Operational Amplifiers

45

Other non-ideal behaviour

Input offset voltage

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

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

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

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

+

–

R

1

= 2kΩ

C

1

= 4.7µF

the cutoff frequency is
approx 17 Hz.

R

1

R

2

C

1

C.K. Tse: Operational Amplifiers

46

Summary

Op-amp circuits

Ideal analysis
Practical problems

Frequency response

Finite loop gain and bandwidth

Applications

Current source
Active filters

µA741 circuit design (looking inside)

Non-ideal behaviour

Slew rate
Input bias current
Offset voltage