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Op amp Circuit w/current source

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate Vout if
    a) I=1mA, R1=2.2kΩ, and R2=1kΩ;
    b) I=2A, R1=1.1Ω, and R2=8.5Ω.
    c) For each case, state whether the circuit is wired as a non-inverting or an inverting amplifier.

    NOTE: I is set to 1mA in the circuit below.

    oacircuit-2.jpg

    2. Relevant equations
    Req=(R1*R2)/(R1+R2)
    V=IR

    3. The attempt at a solution
    First, I calculated Req for R1 and R3:

    Req=(2.2kΩ*1kΩ)/(2.2kΩ+1kΩ)=687.5Ω

    I then converted the current source to a voltage source:

    V=(1mA)(687.5Ω)=687.5mV

    Am I going in the right direction? This seemed like an easy problem, but I don't know if I'm doing it right.
     
  2. jcsd
  3. Sep 1, 2012 #2
    For opamps in this configuration the difference in voltage is so small between pins 2 & 3 it is usually ignored. Given that assumption, what is the voltage across R1? What is the current through R1?

    Once you've found the voltage through R1, you can calculate the current through R2 and the voltage across R2. What output voltage do you need at the opamp output to develop the calculated voltage across R2?
     
  4. Sep 2, 2012 #3
    Current through R1 (I1): I1=V/R1=687.5mV/2.2kΩ=0.3125mA
    Voltage across R1 (V1): V1=(I1)(R1)=(0.3125mA)(2.2kΩ)=687.5V

    Current through R2 (I2): I2=V/R2=687.5mV/1kΩ=0.6875mA
    Voltage across R2 (V2): V2=(I2)(R2)=(0.6875mA)(1kΩ)=687.5V

    Output voltage (Vout): KCL=0=(I-I1)-(Vout/R3)→Vout=R3(I-I1)=1kΩ(1mA-0.3125mA)=999.6875V

    Are the steps above correct?
     
  5. Sep 2, 2012 #4

    rude man

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    Homework Helper
    Gold Member

    Afraid not.

    Look first at R1. What is the voltage at the minus pin on the op amp if the + pin is at zero V? What is the universal rule for any ideal op amp circuit for the input pin voltage difference, assuming the amplifier is not saturated?

    So then you see you had no cause to parallel R1 and R2, and there is no 687.5 mV.

    Next: armed with the above info, write the KVL at the - pin.

    A final note: no way can an LM324 handle the level of currents suggested by part (b). What does that suggest abot the behavior of this circuit with that input current? (Hint: junk that op amp!).
     
  6. Sep 2, 2012 #5
    There is no voltage difference between the + and - pins, so the voltage at the minus pin should be zero.

    I still don't get what to do about the current source. Do I convert it to a voltage source using a source transformation (V=IR1)? To me, R1 and R2 look parallel, so that's where I'm getting lost.

    If I were to use KCL at the - pin, would it be like this?:

    (I-I1)-(Vout/R3)=0, where I1 is the current through R1


    Thanks
     
  7. Sep 2, 2012 #6

    rude man

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    Homework Helper
    Gold Member

    Well, R1 and R2 are NOT in parallel. They go to different places.
    There is no R3, right?
    Next: if the current thru R1 is ALWAYS ZERO, how about just ditching it?

    So how does that impact your last equation above? BTW there is a sign error in it also.
     
  8. Sep 2, 2012 #7
    By R3, I meant R2 (oops).

    So, the current through R1 is zero? Would that mean I and R2 are parallel? I think the new equation would be:

    I+(Vout/R2)=0

    I feel like I'm going in circles :(. This just isn't making sense to me lol.
     
  9. Sep 2, 2012 #8
    If the current through R1 is zero, which it is, and if the input impedance to the opamp is extremely high, which it is, then then current has no place to go except through R2. Knowing that, you should be able to calculate the voltage across R2. Which side of R2 is positive and which is negative? If the left side of R2 is at ground, what voltage must the right side be?
     
  10. Sep 2, 2012 #9
    V2=(I)(R2)=(1mA)(1kΩ)=1V

    Positive is on left side, negative is on right. So, the voltage on the right would be -1V?

    Also, is the current through R1 zero because it's connected to ground?
     
  11. Sep 2, 2012 #10
    Correct.

    The current through R1 is zero because both sides are at the same voltage. The voltage at pin 2 is called a virtual ground. When an opamp has very high gain and has negative feedback, both inputs are at nearly the same voltage.

    It works like this, if the opamp has a gain of 1 million which is typical, then the difference between the two inputs must be 1 millionth the difference between the output voltage and the non-inverting input, pin 3. That is why there is a virtual ground at pin 2 whenever negative feedback is used.
     
  12. Sep 3, 2012 #11
    So, Vout is -1V then?

    Also, part c of the question asks whether the op-amp is wired as inverting or non-inverting. If Vout is negative, does that simply mean it would be wired as inverting?
     
  13. Sep 3, 2012 #12
    Yes, Vout is -1 V and the op-amp is inverting.
     
  14. Sep 3, 2012 #13
    Thank you so much for your help! I really do appreciate it. This stuff is all new to me, so it's definitely a learning process :).
     
  15. Sep 3, 2012 #14
    I enjoyed it. Just out of curiosity, if you don't mind, what year are you in and where are you studying? And by the way, the profession needs more good analog engineers.
     
  16. Sep 3, 2012 #15
    I'm in my third year studying at St. Cloud State in MN. I was going to just get a degree in computer science, but I decided to go for a degree in computer engineering instead. Software/programming is quite easy for me, so I love the fact that the CE major includes hardware and some stuff from electrical engineering (even if it is a long learning process, since it is all brand-new to me haha).
     
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