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Operational amplifiers question...

Discussion in 'Electronic Basics' started by [email protected], Oct 8, 2006.

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  1. Guest

    My knowledge of circuits ends at Ohm's Law.

    This is a question from my instrumental analytical chemistry text,
    which oviously hasn't helped me. I looked through my physics text but
    it doesn't cover operational amplifiers.

    Design a circuit having and output given by -v_o = 3v_1 + 5v_2 - 6v_3.
    I'm assuming there are many solutions to this problem and I'd like to
    come up with my own.

    Here is one solution:
    http://static.flickr.com/103/264187072_a04bbc6001.jpg?v=0

    Could someone walk me throught the steps that result in the above
    equation so I can make up my own circuit?

    Thanks,
    Ashley
     
  2. Well, since your knowledge ends at Ohm's, you don't understand an
    opamp? I think you will need that part, first.

    Ignoring power supply voltages for the moment, the opamp has two
    inputs and one output. The opamp controls the output, but can only
    observe the inputs. It's only desire is to adjust the output in such
    a way that the inputs match each other. It _hopes_ that this is
    possible to do, but it has no idea if that's the case. The only rule
    it applies is that when the + input is more positive than the - input,
    it drives the output to be more positive and if the opposite is true,
    it drives the output to be more negative. If the condition remains,
    the opamp keeps moving the output in the appropriate direction until
    it runs out of voltage compliance.

    The first opamp in the diagram is an inverting buffer. Since the +
    input is grounded, the - input must also be at ground potential if and
    only if the opamp's output is somehow connected to allow that to take
    place. Otherwise, all bets are off. Assume for the moment that
    things are arranged that way (a detail you will see in a minute), so
    that the output can slew the - input around to set it at the same
    potential as ground. This node is then said to be a "virtual ground"
    since it is at ground, but not actually connected to ground. Now
    consider the two 1k ohm resistors. Let's say the input is at -2V.
    Assuming that the virtual ground really is at ground potential, then
    2mA must flow from that virtual ground node towards the -2V source.
    Where does that current come from? Not the opamp's - input. That
    input should just be observing, not supplying current. So the current
    must come through the other 1k ohm resistor. To do that, the opamp's
    output must be at +2V. If so, then 2mA will flow from the output via
    that resistor into that virtual ground node and then to the -2V input
    source. So it's kind of like a see-saw. When the input goes down,
    the output of the first opamp must go up in order to supply the needed
    current so that the - input can actually be at ground potential.

    There's another issue, though. Assume just for a moment that the -2V
    input actually tried to pull down the virtual ground node just a
    little below ground, so that it became slightly negative. The opamp
    would "see" that the + input was then above the - input and would
    attempt to raise the voltage on the output to compensate. It turns
    out that this is exactly what's needed, as when the output goes a bit
    more positive, it will similarly pull that virtual ground node towards
    a more positive value, which is exactly what is needed to get it back
    to ground potential. In other words, a negative-going excursion or
    spike on the input to the first 1k ohm resistor would be immediately
    opposed by a positive-going output, bring the virtual ground node back
    to ground. If the opamp had been wired with the inputs the other way
    around, the output would have gone in the negative direction, making
    the 'virtual ground' node even more negative, making the output go
    even more negative, etc., until the opamp simply railed at the max
    negative voltage it could handle. Which is why it is hooked up the
    way it is.

    Once you understand that, you can see that the first opamp is just a
    way of inverting the voltage. The ratio of the feedback resistor (the
    one going from the output back to the - input) to the input resistor
    sets the "gain." With 1k and 1k, the gain is 1k/1k or just 1. It
    inverts, but doesn't change the magnitude.

    The second opamp has a 3k feedback resistor and there are three
    different input resistors. The gain ratios will be 3k/1k=3, 3k/600=5,
    and 3k/500=6. 3, 5, and 6. If you supply all three inputs at the
    same time with three different voltages, each of these voltages will
    set up a current flow you can easily predict knowing that the virtual
    ground node must be 0V. Each of those currents must be supplied
    through the 3k feedback, though. So the output will go to whatever
    voltage is needed in order to supply all three currents at the same
    time. That will be the sum of those currents and to do it. Also, the
    output will be inverted. With the gains involved, that means
    Vo=-3*V1-5*V2-6*V3'. But since V3 is inverted by the first opamp,
    that means: Vo=-3*V1-5*V2+6*V3. Negating the whole thing gives
    -Vo=3*V1+5*V2-6*V3.

    Jon
     
  3. OK, lets be "idealistic".
    An operational amplifier (ideal) has infinite input impedance
    (current=0), zero output impedance, and infinite amplification.
    So the feedback and input resistors define the equilibrium reached when
    currents at junction of the marked (-) terminal of the amp equal _zero_.
    Sooooo the voltage applied to input resistor divided by Rin is one side
    of the equation, and the resulting output voltage divided by Rfb the
    other side. And they are equal as NO current flows into the third path
    (into amplifier).
    So^2 the overall amplification is given by the ratio Rfb/Rin.
    (Life is not ideal so make allowances).

    In your cirquit you have input amplifier with amplification of one
    connected to switch which transfers the output to selected resistor(s).
    Now do your homework to find what each connection gives as overall
    output. Primary school stuff.

    HTH

    Stanislaw
    Slack user from Ulladulla.
     
  4. The simplifying assumption for these simple opamp circuits
    is that the amplifier has infinite differential (the
    difference between the two inputs) gain, so that, when the
    negative feedback is controlling the output voltage, the two
    inputs of the amplifier must be at the same voltage (zero
    differential input voltage). This means that zero voltage
    difference times infinite gain = finite output. No real
    opamp has infinite gain, but a low frequency gain on the
    order of 1 million is common, so the approximation can be
    quite useful. A second simplifying assumption is that the
    inputs of the opamp have infinite resistance, so do not draw
    any current. Actual input resistances might vary from 100k
    ohms to billions of ohms, depending on the design of the opamp.

    In the circuit you show, the left opamp has equal input and
    feedback resistors connected at the - input. With the above
    first simplifying assumption, since the + input is grounded
    the - input must also balance at ground, in order to hold
    the input difference at zero. The only way this can happen
    is if the output is the negative of the input signal, so
    that equal resistors connecting those two voltages always
    average the two to zero. This is the basic gain -1
    amplifier configuration. If Ri is the input resistor and
    Rfb is the feedback resistor connecting output to - input,
    the formula for gain (using the simplifying assumption) is
    gain = -Rfb/Ri
    The right amplifier is another inverting amplifier, except
    that it has 3 input resistors. Two of these are fed
    directly by input signals, and one is fed with the negative
    of an input signal, produced by the left opamp. Since the -
    input always stays at approximately zero volts, Each of the
    input signals is independent of the others, driving a
    current through its input resistor to zero volts (you can
    use Ohm's law to calculate those 3 currents). The feedback
    resistor has to carry the sum of those 3 currents to the
    output (since the second simplifying assumption makes it
    impossible for that current to go anywhere else).

    So the gain for the top input is -3000/1000=-3,
    the gain for the middle input is -3000/600=-5,
    and the gain for the bottom input is
    (-1000/1000)*(-3000/500)=6

    Lots of other resistor pairs have the same ratios, and
    produce the same gains.

    The whole job can also be performed with a single opamp, but
    then, the signal from one input will change the current
    loading on the other inputs. The interaction also makes the
    math a bit messier.
     
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