Operational Amplifiers

Operational Amplifiers for Basic Electronics http://cktse.eie.polyu.edu.hk/eie209

by Prof. Michael Tse

January 2005

Where do we begin? We begin with 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 = ∞

+ vi –

+ –

vo

+ vi –

± Avi

+ vo –

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

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The 347 IC op-amp

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

+ –

output stage single-ended output

C.K. Tse: Operational Amplifiers

1 2 3 4 5 6 7

–

–

+

+

V+

14 13 12 11 10 9 8

V–

+

+

–

–

Manufacturer listed spec: Rin = 1012Ω; AVOL=100dB = 105 CMRR = 100dB GBW = 4MHz (gain-bandwidth) SR = 13V/µs 3

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.

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Example Consider the following op-amp circuit. What is the voltage gain? R2

vi

R1

0V –

Apply the Golden Rules: ix

ix

vo

+

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

Therefore,

This is the inverting amplifier. C.K. Tse: Operational Amplifiers

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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? + –

+ – – +

– +

–

–1V

+ –

+ x2 C.K. Tse: Operational Amplifiers

sq.

x 6

Other examples (where Golden rules work) R2 R1

Applying the Golden rules, we get

– vi

+

This is the non-inverting amplifier.

Here, simply – vi

+

vo

This is the voltage follower.

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Other examples (where Golden rules work) More examples Rf v1 v2 v3

R1 R2

–

R3

+

R2

R1

v1 v2

This is the summing amplifier.

–

R1

+ R2

vo This is the difference amplifier.

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Other examples (where Golden rules work) More examples C

vi

R

– + This is the integrator.

R

C

vi

– +

vo

This is the differentiator (theoretically). In practice, this circuit won’t work!!! C.K. Tse: Operational Amplifiers

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Examples (where Golden rules do not work) Comparator v1

+

v2

–

vout

The output cannot make the two inputs equal!!! Golden Rule 1 fails!!!

Since the voltage gain typically exceeds 100,000, the inputs must be within a fraction of a millivolt in order to prevent the output from swinging all the way to extreme positive or negative. It is assumed that the supply voltages are +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 will swing to +10 V. 2. If v2 is larger than v1 by more than 0.0001 V, the output will swing to –10 V. C.K. Tse: Operational Amplifiers

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Examples (where Golden rules do not work) Comparator v1

+

v2

–

vout

The output cannot make the two inputs equal!!! Golden Rule 1 fails!!!

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

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Examples (where Golden rules do not work) v1

Comparator 5V v1

+

v2

–

v2 t

vout

vout

5V ±

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.

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Examples (where Golden rules do not work) Schmitt Trigger — a better comparator How does it work?

–

vin

vout

+

A

R2

R1

Assume the op-amp is powered by ±10V, and now vout = +10V. Obviously, vin must be less than vA: v in <

10R1 = vA R1 + R2

What happens if vin moves just above 10R1/(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: v in > C.K. Tse: Operational Amplifiers

†

–10R1 = vA R1 + R2 13

Examples (where Golden rules do not work) Schmitt Trigger –

vin

vout

+

A

We have a situation similar to hysteresis.

R2

R1

upper trip point =

10R1 R1 + R2

lower trip point =

–10R1 R1 + R2

† vin

†

10R 1 R1 + R 2 – 10R 1 R1 + R 2

t

vout +10 t –10

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Examples (where Golden rules do not work) Schmitt Trigger 10V –

vin

vout

+ –10V

What are the upper and lower trip points?

90kW 10kW + –

8V

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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! C

vi

– +

x

vo

vi

– +

x

vo

C

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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. C

vi

– +

R 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.

C

– + C.K. Tse: Operational Amplifiers

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Applications Current source

We see that vR is fixed by the voltage divider. The op-amp will make sure that the voltage across R is also equal to vR, which is fixed!

vR + –

Therefore the current flowing down R must be LOAD

R

Io

which is the load current. Thus, this circuit provide a constant current source for the load. Note: the load is floating for this case!

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Applications Current source for grounded load Vcc R

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

– +

vR

Again vR is fixed by the voltage divider.

Therefore the current flowing down R must be Io LOAD

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.

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Applications Current source for grounded load (voltage controllable) Vcc R R2

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

– +

Ix

vIN

Here, vR is controllable/adjustable by vIN.

vR + –

Vcc - (Vcc - R2 Ix ) R R v = 2 IN R1 R

Io = Io LOAD

R1

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

†

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Other non-ideal behaviour Example of input bias current R2

vi

R1

–

ib

vo

+

ib

Problem: Since ib flows into both inputs, the negative input side will have a slightly negative dc voltage even when vi = 0, whereas the positive input side is still 0V because there is no resistor there! Therefore, vi ≠ 0, i.e., some unwanted offset! R2

Practical solution:

vi

R1

–

ib R1||R2

C.K. Tse: Operational Amplifiers

+

vo

ib

21

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

–

R1 = 2kΩ C1 = 4.7µF the cutoff frequency is approx 17 Hz. C.K. Tse: Operational Amplifiers

R1

R2

C1 22

Summary We have studied the basics of op-amps, and some applications. Basic rules of op-amp circuit analysis Some practical considerations Some applications

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