Since the gain A ≈ ∞, v

  

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 = ∞

  v v v v Av i i ± o o i

  • – – –
  • – Since the gain A ≈ ∞, v ≈ 0 if v is infinite,

  i o

  the two input terminals have same potential if v is infinite

  o

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

  • + + + +

  Ω; A

  8

  9

  10

  11

  12

  13

  14

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

  5 CMRR = 100dB

  =100dB = 10

  VOL

  12

  The 347 IC op-amp

  = 10

  in

  R

  7 Manufacturer listed spec:

  6

  5

  4

  3

  2

  1

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

  output stage

  V+ V– 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:

  attempts

  The output 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

2 Apply the Golden Rules:

  i x R

  1

  0V try

  It first says that the output will to set

  • v i i
  • x

      itself in order to make the difference

      

    v

    o it

      between the inputs zero. That means,

    • 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. 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
      • – +
      • – + sq.

      2 x x Other examples (where Golden rules work) R

    2 Applying the Golden rules, we get

      R

      1

    • – +

      v i This is the non-inverting amplifier.

      Here, simply

    • v
    • o This is the voltage follower

      • v i
      Other examples (where Golden rules work)

      More examples

      R f

      R 1 v

    1 R

      2 v

    • – 2

      R 3 v

      3

    • This is the summing amplifier.

      R

    2 R

      1

    • v

      1 R 1 o v v

    • 2

      R 2 This is the difference amplifier. Other examples (where Golden rules work)

      More examples

      C R v

    • i
      • This is the integrator.

      R C

    • v i

      v o

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

      Examples (where Golden rules do not work)

      2

      by more than 0.0001 V, the output will swing to –10 V.

      1

      is larger than v

      2

      2. If v

      by more than 0.0001 V, the output will swing to +10 V.

      is larger than v

      v

      1

      1. If v

      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.

      Comparator

      2 v

      1 v

      The output cannot make the two inputs equal!!! Golden Rule 1 fails!!! Examples (where Golden rules do not work) Comparator

      The output cannot make

      v

    • 1

      the two inputs equal!!!

      v out

      Golden Rule 1 fails!!!

      v

    • – 2

      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.

      Examples (where Golden rules do not work) v

      1 Comparator v

      5V

      2 v

    • t

      1 v out v

    • – 2

      v out

      ±

      5V Suppose v = 5V (constant) and v is an input.

      2

      1 This circuit is supposed to compare v with 5V.

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

      2

      2 Examples (where Golden rules do not work) Schmitt Trigger — a better comparator

      How does it work? v in v out

    • – + A
    • R Assume the op-amp is powered by 2 ±10V, and now v = +10V. R out

        Obviously, v must be less than v : A 1

        in

        10R 1

        v in A < = v R 1 + R 2 What happens if v moves just above in

        10R /(R +R )? Clearly, v falls to

        1

        1 2 out –10V because of comparator action.

        Therefore, v drops to –10R /(R +R ), A

        1

        1

        2

        and v must be greater than v : A

        in

      • –10R
      • 1

          v in A > = v R 1 + R 2 Examples (where Golden rules do not work) Schmitt Trigger

        • – + out v
          • R

        • –10R
          • R

          R 1 R 2 v in A We have a situation similar to hysteresis.

          upper trip point =

          10R

          1 R

          1

          2

          lower trip point =

          1 R

          1

          2 t t

        • +10 v in v out –10
        • 10R

          1 R + R 1 2

          10R

          1 R + R 1 2

            Examples (where Golden rules do not work) Schmitt Trigger

            10V v in v out

          • – + –10V
          • 90k What are the upper and lower trip points? 10k W 8V + W 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

            • – –

              v i v i v o v o x x

              C 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

            • – +
            Applications Current source We see that v is fixed by the voltage divider. R

              The op-amp will make sure that the voltage across

              R is also equal to v , which is fixed! + v R R

              Therefore the current flowing down R must be

              LOAD

              I o R

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

              Note: the load is floating for this case! Applications Current source for grounded load Again v is fixed by the voltage divider. R

              V cc R

              The op-amp will make sure that the voltage at the lower end of R is also equal to v , which is fixed! R + Therefore the current flowing down R must be

              v R

              I o 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. Applications Current source for grounded load

              (voltage controllable) Here, v is controllable/adjustable by v . R IN

              V cc R

              The current flowing down R, which is close to the

              R 2

            • load current I , must be
            • o
              • + I x

                V I cc cc x - (V - R ) 2 I o = v R

                R v IN +

                I o R v 2 IN

                = LOAD R R 1 R 1 Thus, this circuit provide a controllable constant current source for the grounded load.

                Other non-ideal behaviour Example of input bias current

                Problem: Since i flows into b R

                2

                both inputs, the negative input side will have a slightly negative

                R

                1

                dc voltage even when v = 0, i

              • v i

                i b v whereas the positive input side is o

              • still 0V because there is no

                i b

                resistor there! Therefore, v ≠ 0, i i.e., some unwanted offset !

                R

                2 R

                1

              • v

                i

                i Practical solution: b v o

              • i b

                R

                ||R

                1

                2 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

                = 2kΩ

              1 C

                = 4.7µF

                1 R R

                2

                1 the cutoff frequency is approx 17 Hz. C

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

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