Operational Amplifiers

Transcription

1 Operational Amplifiers for Basic Electronics by Prof. Michael Tse January 2005

2 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 = v o ± A v o Since the gain A, 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 C.K. Tse: Operational Amplifiers 2

3 The 347 IC op-amp The 347 is a Quad JFET input op-amp using bifet technology. output stage V V single-ended output Manufacturer listed spec: R in = Ω; A VOL =100dB = 10 5 CMRR = 100dB GBW = 4MHz (gain-bandwidth) SR = 13V/µs C.K. Tse: Operational Amplifiers 3

4 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 4

5 Example Consider the following op-amp circuit. What is the voltage gain? R 2 Apply the Golden Rules: R 1 i x 0V i x v o 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 5

6 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 x 2 sq. x C.K. Tse: Operational Amplifiers 6

7 Other examples (where Golden rules work) R 2 Applying the Golden rules, we get R 1 This is the non-inverting amplifier. Here, simply v o This is the voltage follower. C.K. Tse: Operational Amplifiers 7

8 Other examples (where Golden rules work) More examples R f v 1 R 1 v 2 R 2 v 3 R 3 This is the summing amplifier. R R 2 1 v 1 v 2 R 1 R 2 v o This is the difference amplifier. C.K. Tse: Operational Amplifiers 8

9 Other examples (where Golden rules work) More examples C R This is the integrator. C R v o This is the differentiator (theoretically). In practice, this circuit won t work!!! C.K. Tse: Operational Amplifiers 9

10 Examples (where Golden rules do not work) Comparator v 1 v 2 v out 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, If v 1 is larger than v 2 by more than V, the output will swing to 10 V. 2. If v 2 is larger than v 1 by more than V, the output will swing to 10 V. C.K. Tse: Operational Amplifiers 10

11 Examples (where Golden rules do not work) Comparator v 1 v 2 v out 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. C.K. Tse: Operational Amplifiers 11

12 Examples (where Golden rules do not work) v 1 Comparator 5V v 2 5V v 1 v 2 ± v out v out t Suppose v 2 = 5V (constant) and v 1 is an input. This circuit is supposed to compare v 1 with 5V. But if v 2 has noise, the output may jump when v 2 is near 5V. C.K. Tse: Operational Amplifiers 12

13 Examples (where Golden rules do not work) Schmitt Trigger a better comparator n A R 2 v out How does it work? Assume the op-amp is powered by ±10V, and now v out = 10V. R 1 Obviously, n must be less than v A : n < 10R 1 R 1 R 2 = v A What happens if n moves just above 10R 1 /(R 1 R 2 )? Clearly, v out falls to 10V because of comparator action. Therefore, v A drops to 10R 1 /(R 1 R 2 ), and n must be greater than v A : n > 10R 1 R 1 R 2 = v A C.K. Tse: Operational Amplifiers 13

14 Examples (where Golden rules do not work) Schmitt Trigger We have a situation similar to hysteresis. n A v out upper trip point = 10R 1 R 1 R 2 R 2 lower trip point = 10R 1 R 1 R 2 R 1 n 10R 1 R 1 R2 10R 1 R 1 R2 t v out 10 t 10 C.K. Tse: Operational Amplifiers 14

15 Examples (where Golden rules do not work) Schmitt Trigger 10V n 10V 90kW 10kW v out What are the upper and lower trip points? 8V C.K. Tse: Operational Amplifiers 15

16 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 x v o x v o C C.K. Tse: Operational Amplifiers 16

17 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 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 17

18 Applications Current source We see that v R is fixed by the voltage divider. v R LOAD 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 R I o 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 18

19 Applications Current source for grounded load v R I o R V cc LOAD 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. C.K. Tse: Operational Amplifiers 19

20 v IN Applications Current source for grounded load (voltage controllable) R 2 I x v R I o R V cc LOAD 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 I o = V cc - (V cc - R 2 I x ) R = R 2 v IN R 1 R R 1 Thus, this circuit provide a controllable constant current source for the grounded load. C.K. Tse: Operational Amplifiers 20

21 Other non-ideal behaviour Example of input bias current R 1 i b i b R 2 v o Problem: Since i b flows into both inputs, the negative input side will have a slightly negative dc voltage even when = 0, whereas the positive input side is still 0V because there is no resistor there! Therefore, 0, i.e., some unwanted offset! R 2 Practical solution: R 1 i b v o R 1 R 2 i b C.K. Tse: Operational Amplifiers 21

22 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. C 1 R 1 R 2 C.K. Tse: Operational Amplifiers 22

23 Summary We have studied the basics of op-amps, and some applications. Basic rules of op-amp circuit analysis Some practical considerations Some applications C.K. Tse: Operational Amplifiers 23