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Need some insight into an inverting op amp example

  1. Feb 13, 2016 #1
    Currently working through a chapter on op amp circuits, fundamentals.

    Came upon this practice problem in the section on inverting op amps

    2DrQLXk.png

    imgur link: http://i.imgur.com/2DrQLXk.png

    7KRKzKC.png

    imgur link: http://i.imgur.com/7KRKzKC.png

    A working of this problem is not given. I got the correct answer the second time, by working through some KCL equations on a more fundamental level, but failed the first attempt based on what is evidently an incorrect assumption. But I don't fully understand why that assumption is wrong.

    Before both solutions, I converted the current source to a voltage source in series with a resistor [itex]R_s[/itex] with the equation [itex]V_s = i_s R_s[/itex].

    I've labeled the node between [itex]R_1[/itex], [itex]R_2[/itex], and [itex]R_3[/itex] as [itex]v_2[/itex]. [itex]v_1[/itex] is between [itex]R_s[/itex] and the inverting input to the op amp.

    My second attempt combined

    [tex]\frac{-v_s}{R_1} = \frac{v_2-v_0}{R_3} + \frac{v_2}{R_2} \\ \text{with} \ \ v_2 = -i_1 R_1 \\ \text{and} \ \ i_1 = i_s[/tex]

    to obtain the correct proof.

    My first wrong attempt made an assumption about calculating the equivalent resistance of [itex]R_1[/itex], [itex]R_2[/itex], and [itex]R_3[/itex], that goes like this:

    -- you can think of [itex]R_1[/itex], [itex]R_2[/itex], and [itex]R_3[/itex] as being a "single resistance" as their combined configuration ("lump") takes [itex]v_0[/itex] to Ground (a zero Volt node). Because the node [itex]v_1[/itex] is (theoretically/ideally) at zero volts because it is short circuited to the common analog ground because we are considering an ideal op amp, and for an ideal op amp we can think of the op amps input terminals as being a short circuit when doing voltage calculations (and as an open circuit when doing current calculations). The presence of the [itex]R_s[/itex] resistor connected to [itex]v_1[/itex] can be ignored, because it will not (ideally) change the fact that [itex]v_1[/itex] is (ideally) always 0V.

    So, using this assumption, [itex]R_3[/itex] goes from [itex]v_0[/itex] into a two parallel resistors [itex]R_1[/itex], and [itex]R_2[/itex] which both (ideally) connect to the common analog ground node...soo

    [tex]R_{eq} = R_3 + \frac{R_1 R_2}{R_1 + R_2}[/tex]

    and

    [tex]v_0 = \frac{-R_{eq}}{R_s} v_s = -R_{eq} i_s[/tex]

    This of course does not work. In that it does not agree with the text or the also much more sound reasoning from just doing the KCL equations outlined initially.

    What I don't understand really, is what's wrong with my reasoning in the assumption I constructed above, keeping in mind that this assumption would only work for ideal op amps (ie, [itex]v_1[/itex] is assumed as always being at 0V).
     
  2. jcsd
  3. Feb 13, 2016 #2

    berkeman

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    Staff: Mentor

    Why would you convert the current source to something else? The basic assumptions of an ideal opamp make this a straightforward problem with the input current source flowing through the resistors....
     
  4. Feb 13, 2016 #3

    jim hardy

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    It incorrectly assumes that current through R3 is equal to Is ?
     
  5. Feb 14, 2016 #4
    Could you help me out there then? Because I can't see how to do it without converting to a voltage source with a series resistor.

    I can see that the current divides to flow into R2 and R3 so

    [tex]i_1 = i_s = i_2 + i_3[/tex]

    and I can see that the current would divide like

    [tex]i_2 = \frac{R_3}{R_2+R_3}(v_1-v_0) \\ \text{and} \ \ i_3 = \frac{R_2}{R_2+R_3}v_1[/tex]

    and

    [tex]v_1 = -i_s R_1[/tex]

    because the node just before R_1 is at zero and the node v_1 is just after R1.

    which can bring me to

    [tex]i_s = \frac{(R_2+R_3)v_1 - R_3 v_0}{R_2 + R_3} = v_1 - \frac{R_3 v_0}{R_2 + R_3}[/tex]

    giving

    [tex]i_s = -i_s R_1 - \frac{R_3 v_0}{R_2 + R_3}[/tex]

    then

    [tex](1+ R_1)i_s = -\frac{R_3 v_0}{R_2 + R_3}[/tex]

    giving

    [tex]\frac{v_0}{i_s} = \frac{-(R_2+R_3)(1+R_1)}{R_3}[/tex]

    I can't see how to do this without changing to a voltage source with a series resistor...I'd love to know though, an easier way would be much better.
     
  6. Feb 14, 2016 #5
    Are you sure that the result is in amps ??
     
  7. Feb 14, 2016 #6

    jim hardy

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    EDIT

    oops Jony got there before me, nicely done Sir !


    sanity check on assumptions and method

    units okay ?

    upload_2016-2-14_6-49-46.png
     
  8. Feb 14, 2016 #7
    Well, that was a dumb mistake :P Thanks :)
     
  9. Feb 14, 2016 #8

    Well, that was a dumb mistake :P Thanks :)
     
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