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Ideal op amp problem

  1. Oct 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I have used words here I'm not sure will explain what I mean accurately. If you're confused, ask, and I'll provide as much context as I can.

    Given the below circuit

    ?temp_hash=d7990c5e84d984c9d71d6119b899d0ad.png

    a) What are the three properties of an ideal op amp? These properties provides background for two equations, which equations are these?

    b) Derive the expression for the DC amplifying factor AV = VOUT / VIN. Prove answers in decibels as well.

    c) Let VIN be 100mV DC. A load RL = 1k ohm is attached to VOUT. What's the power amplifying? What's causing this power amplifying?

    d) Derive the expression for the transfer function AV = VOUT / VIN. Use impedance. Write on standard form.

    e) Find the 3dB frequency. Sketch the bode-gain for the frequency response to the circuit. Which type of circuit is this?
    2. Relevant equations


    3. The attempt at a solution

    a) Properties: RIN = infinity. ROUT = 0, AV0 = infinity.

    Equations: V+ = V- and iIN = i- = i+ = 0

    b) Look at my response for d).

    c) I'm not sure how I'm supposed to compute this. I (think I) have lecture notes and examples in my book to help me out with the circuit - but the attached resistor to VOUT is throwing me off here.

    d) We've worked out this equation for another assignment: ?temp_hash=d7990c5e84d984c9d71d6119b899d0ad.png

    For question b), what's the difference? They're still asking for AV = VOUT / VIN. The difference is answers in dB, standard form and use of impedance...

    e) I know what a bode gain is, but where do I get the numbers from to calculate the frequency and the gain?

    I also have formulas to calculate dB from current, voltage or power.
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2014 #2

    mfb

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    Staff: Mentor

    For (b), you can plug in numbers and simplify Av a lot. You should also do this in (d), but the result will stay more complex (pun not intended). This will also help you to answer (e).
    For (c), if you use the assumption of zero output impedance, what is the power you get out if you attach the 1kOhm resistor (with ground at the other side) compared to the power the input side draws?
     
  4. Oct 12, 2014 #3
    I just remembered, the expression in (d) is from a slightly different circuit (I'll see if I can get a picture for that up here if that's needed). Does it still make sense? I mean, Av = VOUT/VIN is sound, but will it still continue to be ZC / (ZC + ZR)? It's getting late...

    As for (c), I'm sorry, but I'm not sure. As I said, without the attached resistor, I have a clue as far as calculations go. Previously, I derived another expression: ?temp_hash=c077bd975dad58a4916cedf7953d5efe.png

    I'm sorry for this incredible stupid question, but believe me when I say it's not because I'm lazy: Will this equation help me here?

    Thanks so much for showing interest in my post and helping me out, I will do whatever I can to make use of your guidance.

    Edit: In that picture, Rf is R3, and Ri would be R4.
     

    Attached Files:

  5. Oct 12, 2014 #4

    mfb

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    I guessed that the equation for (d) needs some modification (it does not include a second resistor!) but I don't know how.
    You are not sure about what? Also, find the short answer for (b) first, it will make (c) easier. What is VOUT? Based on that, what are current in and voltage at the 1kOhm resistor? What is the power it gets? Find the corresponding values for the input resistor as well and compare.
    Which setup did you have where you calculated this? Those equations are always for some specific setup, it does not help to take a random one.
    It looks reasonable, however.
     
  6. Oct 12, 2014 #5
    I will start working with your info above, I just wanted to reply to this first.

    An identical circuit, just used different identifications for the two resistors. Ri is the series one, and Rf is the resistor in parallel with the capacitor.

    So basically, I've been looking at that equation to modify for (d), but I can simplify it for (b) and use that info to solve (c).

    I will give it a go tonight. If I don't understand it, I'll get back to this first thing tomorrow.

    Thanks.
     
  7. Oct 12, 2014 #6
    I probably should've done this tomorrow, but here's what my mind is telling me right now.

    I simplify the expression for exercise (b), and I end up with -100(1/(1+j1)) which really just is -50(1-j).

    I have no clue as to how I can use that. My bet is that I messed up when calculating the omega (not omega c, looking at my picture in post #3). I did 2*pi*f, using this f: 1/(2*pi*Cf*Rf). When omega c is 1/(Cf*Rf) it's pretty obvious that omega/omega c is the same.

    I will try it tomorrow.
     
  8. Oct 13, 2014 #7

    NascentOxygen

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    That is the precise equation applicable here, BUT you definitely must be able to derive from scratch the expression for gain for any OP-AMP circuit. In an exam you won't be able to dredge up a formula from your notes and use that! It is a good opportunity here for you to determine the gain equation from basic principles, without cribbing.
     
  9. Oct 14, 2014 #8

    mfb

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    What is the "frequency" of a DC current?
     
  10. Oct 14, 2014 #9

    psparky

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    Nodal analysis at node 6 will get you your transfer function.

    Your break frequency happens at 1/(2*pi*R*C) as you already know in your bode plot

    Bode plots simply show the GAIN at any given frequency from 0 to infinity.

    The DC thing is simple. You can just look all the way to the left of your bode plot....what it says at zero frequency. Or you can just picture the capacitor as an open circuit in DC.....in other words just remove it and you obviously know the gain is 100.

    Now picture what happens when the frequency is huge....that capacitor approaches a short circuit becuase of its 1/(jwc) effect. Therefore gain will be approaching zero as frequency approaches infinity.

    So now you know the gain at zero frequency (DC), you know the gain at infinity....and you know the break frequency.

    A single pole filter like this drops at a rate of -20db/deca.

    Also keep in mind that you can pick any random frequency (or omega) and plug it into your transfer function, then do the math.
    Your transfer function will give you the exact gain and phase angle for any given frequency.

    Also note that the dB drops -20 per deca as well as you go thru your transfer function. In other words, if you use values for omega that increase like this:
    1, 10, 100 ,1000, etc.....you will see the db dropping as prescribed. Or you will see the gain drop accordingly as well......same, same. Keep in mind this last paragraph is related to your type of filter. A different sort of filter and the dB will climb as you increase frequency!~
     
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