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Op-Amp basic problem with a LED diode

  1. Nov 22, 2011 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 22, 2011 #2
    Ah! no:cry:
    V at the -input = +2V (that battery)
    So current through R1 = 2V/2kΩ =1mA
    So 1mA flows from output of amp towards the -input
    making the voltage across R2 = 1mA x 10kΩ =10V
    Voltage at output = 10V + 2V (that battery) = 12V (this is only just OK...it is less than +13V supply)
    So voltage across RL = 12 -1.4 = 10.6V
    So output current through LED = 10.6/2kΩ = 5.3mA which is about right for an LED
    Was this part of a bet?
     
  4. Nov 22, 2011 #3

    Femme_physics

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    Hmm, I think I get it, because NO voltage is listed at the +end, Vout = 2V (based on the minus leg).

    How does it get to the -input? I don't understand the current pathway.

    Like this ...

    http://img64.imageshack.us/img64/12/pathway1.jpg [Broken]

    Or that?...

    http://img195.imageshack.us/img195/7198/pathway2.jpg [Broken]


    Not this time, but as a part of the bet I get to do more physics :wink:
     
    Last edited by a moderator: May 5, 2017
  5. Nov 22, 2011 #4
    Here is another way without talking about the current.
    This is an inverting amplifier and the Voltage gain is given by -R2/R1.
    (you made a mistake with Vout = R2/(R1+R2) xVin.... that is for a non-inverting amplifier:cry:)

    So the voltage gain = 10kΩ/2kΩ =x5
    So Vout = 5xVin. But Vin = +2V (that battery) Vout = 10V BUT this is actually the voltage across R2 So the voltage out must be 10 + 2 = +12V

    Remember NO current can flow into or out of the - or + inputs
    Your last diagram would be perfect if you removed the blue line coming from the battery and going to the right. The blue lines with arrows are perfect.

    Keep winning bets:wink:
     
  6. Nov 22, 2011 #5

    Femme_physics

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    Well, that's just voltage divider, regardless of what type of op-amp, no?

    I'm not sure why I should remove it. It is what happening, no?


    Wait, but you earlier said...

    So shouldn't it be -10kΩ/2kΩ?
     
  7. Nov 22, 2011 #6
    You are asking very very good questions !!
    I will try to deal with your middle question first
    Remember NO current can flow into or out of the - or + inputs
    Your last diagram would be perfect if you removed the blue line coming from the battery and going to the right. The blue lines with arrows are perfect.

    Your blue line from the + on that battery shows a current flowing into the +input.... against the rules !!!

    The voltage gain has a VALUE of 10/2 =5 and I was a bit sloppy not putting the -sign.
    This -sign can be very confusing!! It means that the OUTPUT changes in the OPPOSITE direction to the input voltage. If you look at the voltage on the -input (with that battery connected) it is +2V. The other end of the resistor R1 is connected to 0V (the earth or ground symbol).
    This means that the left hand end of R1 is DOWN compared to the right hand end of R1 (by2V)
    This means that the output must be UP (this is what the -sign means). It is up by a factor of 5 so it is 5x2 greater than2 (I hope that makes sense) = 12V

    The answer to your first question is voltage at - input MUST = voltage at +input (the +2V battery)

    Can't wait for your next question
     
  8. Nov 22, 2011 #7

    berkeman

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    Please do not do students' homework for them here at the PF. You should have pointed out the error and given a hint, and that should have been enough for Femme to continue doing the work.
     
  9. Nov 22, 2011 #8

    I like Serena

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    Hi Fp! :smile:

    I'm trying to see if I can help you, but I don't understand yet what you wrote down.

    You're using the symbols [itex]V_{in}, V_{out}, A_f[/itex], but what are they?
    Can you perhaps mark them in your drawing?
    And where did the [itex]1.4[/itex] come from?
     
  10. Nov 22, 2011 #9

    AlephZero

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    No. The question says use the IDEAL op amp model. That says NO current flows into either of the inputs.

    Of course in real life op amps are not "ideal" and there is a tiny amount of input current, but it is so small you can usually ignore compared with everything else. The resistance between the input pins of a real op amp could be anything between say 5 MOhms for a cheap low spec part like a 741, up to more than 100 MOhms. Even with a 5 MOhm resistance, the input current would be several orders of magnitude less than the milliamps flowing in the rest of the circuit.

    That's better, but still not quite right. You don't have any current flowing into the - input, but you still showed a current flowing into the + input and through the op amp somehow.

    With the ideal op amp assumptions, the 2V battery doesn't have any current flowing through it. What it does is make the + input have a potential of 2V, and the op amp then makes the - input also have a potential of 2V. The output voltage and current from the op amp is "whatever it takes" to keep the - input pin at 2V.
     
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  11. Nov 22, 2011 #10

    Femme_physics

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    Then what's the point of the 2 voltage battery? It has NO WHERE TO GO. It's stuck. It's useless.

    Bit lost you, tbh! Do you mean to tell me that currents flow to the all the grounds in the circuit?

    And how come you determined it's a factor of 5? By doing Af = 1 - 10000/2 = 5?





    http://img72.imageshack.us/img72/1812/voutin.jpg [Broken]

    But i know now it can't be true because you said no current goes into the Op-Amp.

    I can't mark Af because it's the amplification factor of the Op-Amp.

    1.4 is the voltage across the LED.
     
    Last edited by a moderator: May 5, 2017
  12. Nov 22, 2011 #11
    I'll add that because you are assuming that the op amp is ideal, then
    [tex]V_+=V_-[/tex]
    and the 2V appears at both of the inputs.
     
  13. Nov 22, 2011 #12

    gneill

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    The 2V battery has the job of setting the potential with respect to ground at the + input of the opamp. No current need flow to have a particular potential. This potential is used as a reference level by the opamp. (Imagine a little guy inside the opamp looking out a closed porthole at a voltmeter outside connected to the + lead. He takes the reading and passes it on to the rest of the gang inside the opamp. No current comes in through the window!)

    You'll notice that the circuit has a path from the opamp output back to its - input via resistor R2. This is known as a feedback loop. The opamp will adjust its output voltage until the voltage on the - input matches what it sees on its + input. When this is accomplished (and very quickly indeed with an infinite gain ideal opamp!) both inputs will be at +2V. This is a typical mode of operation for an opamp.
     
  14. Nov 23, 2011 #13

    Femme_physics

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    Right! I thought it's because a comparator op-amp....in all op-amps it's like that?

    What? How? Are we living in a sci-fi universe where electrical components can pass messages without passing current? telekinesis?

    OK, let me see the math for it again. I'll start with the TYPE of amplifier because I got the feeling this is a NON-INVERTING amplifier. Look,

    http://img36.imageshack.us/img36/585/typeof.jpg [Broken]

    In both of them voltage goes through the plus
     
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  15. Nov 23, 2011 #14

    I like Serena

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    Voltage doesn't "go". It "is".
    Current "goes".

    Voltage is like "height" on a mountain.
    Current is like water that "flows" from the mountain.


    By telekinesis the little electrical men in the op-amp can see how "high" things look outside of the op-amp.
    Didn't you get the memo?

    When the little men see that there is a difference in height on the inputs, they start creating current that they push out of the output, hoping that will reduce the difference in height.



    In this particular case you can choose.
    You can either choose the INVERTING amplifier or the NON-INVERTING amplifier.
    This is because you don't have an actual signal, just a constant voltage.

    It does help if you write which one you're using and write the corresponding formula down. :wink:
     
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  16. Nov 23, 2011 #15

    Femme_physics

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    But I feel like I'm getting the little kids explanation instead of what's really going on explanation. I'm still not sure how the Op-Amp knows, but I do know there are no little men in it. Should I just accept it as is and go with the little men explanation?

    I'm afraid I don't know what is the difference between "actual signal" and "constant voltage"..?
     
  17. Nov 23, 2011 #16

    I like Serena

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    I'd go with the little men explanation.

    But I'll try to explain anyway.

    An op-amp is like a transistor.
    A small base current on a transistor opens the gate for a much larger current to flow from collector to emitter.
    An op-amp is built from a lot of transistors in such a way that this effect becomes much stronger.

    The inputs do draw current in practice, just like the base current of a transistor.
    But it is negligibly small compared to the current that starts flowing out of the op-amp, which is drawn from the independent power supply of the op-amp.



    With an actual signal I mean a voltage that varies in time, giving rise to different (negative or positive) amplifications.
    A constant voltage is equivalent to the ground, and indeed can be used instead of a ground.
     
  18. Nov 23, 2011 #17

    Femme_physics

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    Fair enough. So, if an ideal op-amp was real no current would've flown throughout the circuit at all, right? Because that tiny bit of current that escapes really makes the entire process run.

    That has to do with an AC source, no?

    hmm...well, it's not a question that's modeled on time or AC current, so that's the part that makes it obvious that it's a constant current?
     
  19. Nov 23, 2011 #18

    I like Serena

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    Yep!


    For instance.

    But I'll give a better example.
    In your other thread with the block diagram, you have a temperature sensor.
    It measures the temperature and transmits it on an electrical wire.
    High temperature is high voltage, low temperature is low voltage.

    This really is a signal.


    Yes.
     
  20. Nov 23, 2011 #19

    I like Serena

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    Now that I know what your symbols are, I can tell you that your original work was fine, except for the first formula.
    Your first formula does not relate Vin properly to Vout.
     
  21. Nov 24, 2011 #20

    Femme_physics

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    Wait, I'm still not sure how in this case I can choose between inverting and non-inverting, and in other cases I can't choose.

    Are you saying that in those two I have a signal?

    http://imageshack.us/photo/my-images/36/typeof.jpg/

    I don't see an AC current source, and I don't see some temperature sensor in those.

    How can I tell?

    So according to this:

    http://img805.imageshack.us/img805/2741/shouldberight.jpg [Broken]


    I just do this:

    http://img521.imageshack.us/img521/1165/solsolsolsol.jpg [Broken]

    Yes?
     
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