Operational Amplifier-Basic Electronics-Lecture 06-Electronic and Information Engineering, Lecture notes of Basic Electronics

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Operational Amplifiers

for Basic Electronics http://cktse.eie.polyu.edu.hk/eie 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 = ∞ 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

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.

Example

Consider the following op-amp circuit. What is the voltage gain? R 1 R 2

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

Other examples (where Golden rules work)

R 1 R 2

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 R 2 R f

v 2 vo 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

Examples (where Golden rules do not work)

v out 1 v 2 v Comparator 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 v 1 is larger than v 2 by more than 0.0001 V, the output will swing to +10 V.
2. If v 2 is larger than v 1 by more than 0.0001 V, the output will swing to – 10 V. The output cannot make the two inputs equal!!! Golden Rule 1 fails!!!

Examples (where Golden rules do not work)

v out 1 v 2 v Comparator 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. The output cannot make the two inputs equal!!! Golden Rule 1 fails!!!

C.K. Tse: Operational Amplifiers 13

Examples (where Golden rules do not work)

Schmitt Trigger — a better comparator

vout R 1 R 2 vin A How does it work? Assume the op-amp is powered by ±10V, and now v out = +10V. Obviously, v in must be less than vA : What happens if v in moves just above 10 R 1 /( R 1 + R 2 )? Clearly, v out falls to

• 10V because of comparator action. Therefore, vA drops to – 10 R 1 /( R 1 + R 2 ), and v in must be greater than vA : † vin < 10 R 1 R 1 + R 2 = vA † vin > –10 R 1 R 1 + R 2 = vA

Examples (where Golden rules do not work)

Schmitt Trigger

vout R 1 R 2 vin A We have a situation similar to hysteresis. † upper trip point = 10 R 1 R 1 + R 2 † lower trip point = –10 R 1 R 1 + R 2 t t vin vout

10 R 1 R 1 (^) + R 2 10 R 1 R 1 (^) + R 2

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! vo

vi C vo

vi C

x

x

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.

vi 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

Applications

Current 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 the lower 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 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

Applications

Current 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 the load current Io , must be Thus, this circuit provide a controllable constant current source for the grounded load. V cc

vIN R 2 † Io = Vcc - ( Vcc - R 2 Ix ) R =

R 2

R 1

vIN R R 1 Ix