(1/31/99) (1/31/99)

Physics 623
The Operational Amplifier
Jan 31, 1999

1  Purpose

2  Discussion

Modern integrated circuit operational amplifiers (op amps) have characteristics so closely approaching the ideal op amp that they are components of almost all signal processing and instrumentation equipment. The properties of a commonly used 741 op amp are compared to the ideal values listed in the table below:

Parameter Ideal Value mA741 Value 1990 best avail. value
(not all in same device)
Input current Iin = 0 ~ 80 nA 0.1 pA
Voltage gain Av = ¥ 200,000 2×106
Bandwidth BW = ¥ 1 MHz at Av = 1 500 MHz
Balance Vo = 0 for Vi = 0 Vo = 0 at Vi = 1-2 mV mV
Temperature none (dV/dT)offset typically (dV/dT)offset 0.1 mV/C
dependence 15 mV/° C

Op amps can be obtained commercially for $0.50 to $3.00 each. This and their versatility are the reasons for their great popularity. As you perform the suggested investigations, be alert to observe and note those cases in which the real op amp fails to match the ideal properties. Try to find the relevant properties in the ``spec sheets'', and take note of the ``typical applications'' section - this is a prime spot for getting circuit ideas or finding out how new devices might be used.

Almost all applications of op amps involve negative feedback as indicated in the diagram below:

opampfig1.gif
Figure 1:

Assume that point A is a virtual ground (i.e. V- = V+), and that Ii = 0, and prove the general result that

Avf = Vo
Vi
= - Z¢
Z
.

3  Procedure

  1. Connect the op amp as an ordinary amplifier as shown below.

    opampfig2.gif
    Figure 2:

  2. Connect the op amp as a non-inverting amplifier.

    opampfig3.gif
    Figure 3:

  3. Construct an adder. Verify, using three or four pairs of input values, that:
    Vo = - ( v1 + v2 )
    It is most convenient to use the square blue 10k potentiometers with pins on the back side as adjustable voltage sources for v1 and v2.

    opampfig4.gif
    Figure 4:

  4. As shown below, connect the op amp as an integrator. Use positive pulses 1 msec wide at a frequency of 40 Hz as the input.

    opampfig5.gif
    Figure 5:

  5. Now modify the integrator to be a charge sensitive amplifier. This is the electronic analog of a ballistic galvanometer.

    opampfig6.gif
    Figure 6:

  6. Construct a logarithmic amplifier as shown below.

    opampfig7.gif
    Figure 7:

  7. Note that, while an Op-Amp has differential inputs, the gain stabilized amplifier circuits in parts 1 and 2 above do not.

    opampfig8.gif
    Figure 8:

    (This particular circuit has several drawbacks. Check Horowitz & Hill for a nifty ``3 Op-Amp Instrumentation Amplifier'' that has:

  8. The final circuit will be used in the next lab, so build it on the circuit board that is marked with your name and save it. This is a high gain circuit and some care in the layout is needed; the instructor should have an example to show you before you start.

    The circuit consists of three identical stages, which should be arranged from left to right across the upper breadboard block, leaving about 1/3 of the block free on the left. Clip the component leads short and bend them with your pliers so that the component bodies lie flat against the breadboard - this minimizes stray capacitances. Use a single bus on the socket for all the grounds - for example, the inner strip just below the op amps. It is a good idea to put 0.1 mF ceramic capacitors from the ± 15 V supply pins to the ground bus as close as possible to each op-amp. The instructor will show you a sample layout before you start.

    The circuit for a single tuned amplifier stage is:

    opampfig9.gif
    Figure 9:

    Ignoring the capacitors for a moment, this is just an inverting amplifier with a gain of RF/R1 = 30. The addition of C1 introduces a high-pass corner at f1 = 1/2pR1C1 that reduces the gain at low frequencies, ie f << f1:

    opampfig10.gif
    Figure 10:

    The addition of Cf reduces the gain at high frequencies, i.e. f2 = 1/2pRfCf

    opampfig11.gif
    Figure 11:

    With both capacitors, the gain is just the product of these two curves. Since f1 = f2 º f°, the result is :

    opampfig12.gif
    Figure 12:

    Therefore, the maximum gain of the stage is 1/2 (R1/RF) = 15 at f°, and drops off µ f for f << f° and as 1/f for f >> f° - both 6 db per octave. What is f° for the given component values?

    To make a high gain amplifier with a fairly narrow bandwidth, we cascade three (3) identical stages. The gain is then:

    Av = ( 30 )3 æ
    ç
    è
    f/f°
    1+(f/f°)2
    ö
    ÷
    ø
    3

     
    The peak gain is now Av = 303/8 = 3375 at f = f° and drops off as 1/f3 for f >> f° , and as f3 for f << f°.

    opampfig13.gif
    Figure 13:


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On 31 Jan 1999, 17:06.