Op Amp Calculations

Discussion in 'Homework Help' started by pianolife, May 21, 2013.

  1. pianolife

    Thread Starter New Member

    Nov 15, 2012
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    Dear members of All About Circuits,

    I am still stuck on a (I guess) quite easy question about Op Amps.

    I must find Vout on the circuit in the picture.
    I've found that
    - On the red and the blue branches the current is 20microA.
    - On the positive terminal I have 2 V, and, from the principles of the Op Amp, I must have 2 V on the negative terminal as well.
    - On the feedback loop my current is 40microA (which multiplied by the 100k resistor gives 4V, which apparently is not the result).

    Using a circuit simulator, I've found that Vout is 6V.
    I can't demonstrate that, though.
    I think I need to use KCL for the red and the blue nodes, but I don't know how to show that.

    Can anyone please help me?

    Thank you so much.

    All the Best.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
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    Whenever you specify a current, you need to also specify what direction that current is flowing. So add current arrows onto your circuit diagram with a Ib being the current in the blue brance, Ir being the current in the red branch, and Ip being the current in the purple branch. You pick the direction for each. Then determine the current in each branch with the result being positive if it is flowing in the direction of that branch's current arrow and negative otherwise.

    Applying KVL at the junction then becomes trivial since KCL simply says that the sum of the currents entering a node must equal the sum of the currents leaving the node.
     
  3. rahdirs

    New Member

    May 22, 2013
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    By concept of virtual ground,the voltage at inverting input is approximately 2V,
    then by KCL,
    Vout-2V/100K = (2V-1V/50K) + (2V-1V/50K).
    this gives,
    Vout = 6V
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Please don't just work people's homework for them. Consider that, in most cases, the person has seen examples in class and in the text book that have showed them this stuff. Obviously that wasn't enough; it is unlikely that having someone show it to them one more time is unlikely to have some magical effect that the others didn't (not impossible and it does happen, but not likely).

    In many things and for most of us, we need to struggle and fight with some concepts in order to get past some misconception and that misconception only gets exposed and confronted during said struggle. Our goal is to help the person work their own way through the problem step-by-step in order to discover that stumbling block and have their "Ahah!" moment.

    Aside from that, the following is wrong:

    Vout-2V/100K = (2V-1V/50K) + (2V-1V/50K).

    because this is the same as

    Vout - 2V/100K = (2V - 1V/50K) + (2V - 1V/50K)

    Do you see the difference?

    All I did was add some spaces. How can spaces change the equation? They can't. Don't get sloppy with order of operations. It's easy to make mistakes like this when writing an expression in text. But since we are always having to write expressions in text when programming or entering parameters into a simulator or similar things, we need to get in the happen of paying particular attention to these things when we write expressions in text.
     
  5. pianolife

    Thread Starter New Member

    Nov 15, 2012
    28
    0
    Dear WBahn, I am sorry if I gave you the impression that I "wanted someone to do my homework". It's not like this, please believe me, I am working as hard as I can to understand this stuff. I am struggling, indeed, but I never asked someone else to do my stuff. And I am sorry if my post was sounding like that.

    Sorry again

    Best
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    Though many people do come here asking, sometimes even offering to pay, for someone to do their work for them, I was one implying that this was the case here at all. Nor was I implying that rahdirs was giving you the answer with anything other than the best of intentions. But neither of those change my point that the best way to learn something is to struggle through it yourself and we, here, need to balance the desire to give you enough to keep your struggle progressing to a successful conclusion and giving you too much so that, while you end up with an answer to the problem, you don't gain any of the comprehension that was the point of working the problem in the first place.
     
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