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Virtual ground or not?

Discussion in 'Electronic Basics' started by [email protected], Feb 24, 2005.

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  1. Guest

    Using a classic inverting op amp circuit with the "+" input tied to
    ground, I lift the pin and connected it 1 volt. Will the "+" input
    behave like a virtual ground when a signal is applied to it's "-"
    input? What should I expect?

    boat_ranger
     
  2. Which pin did you lift? And what do you mean by 'lift'?

    If the V+ input is tied to ground, and there is feedback of some kind,
    the opamp will try to make the V- input equal ground. This makes it an
    inverting amplifier (the gain is negative,) so unless you have a
    positive and negative power supply for the opamp, it won't be able to do
    that. The output will just sit at zero.

    This makes sense, right? An opamp's output is A * (V+ - V-), where A is
    some large number. If V+ = 0, then the output is -A*V-, which means it
    can only go below 0 for inputs higher than ground.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  3. Guest

    Hi, the "+" pin was removed from gnd and tied to 1volt, provided
    by a voltage divider. The output then shifted to 1 volt, without a
    signal on the "-" input.

    The op amp has a + and - 12v supply.

    In my case the input signal will swing negative so the output swings
    positive.

    My thinking is that the 1volt provides a offset for the output, I'm
    unsure what to expect if I place a scope on the "+" input while there
    is a signal on the minus.

    What should I expect and why?

    boats_ranger

    ..
     
  4. Well, it depends on how you hook it up. If you leave the negative input
    unconnected, there is no telling what will happen. If you put a signal
    on the V- input without some kind of negative feedback, then the output
    will be A*(1-V-), where A is the open loop gain of the opamp. A is
    typically quite large for low frequencies, more than 100,000. However,
    your opamp clearly can't put the output outside the +-12V supply. Thus,
    almost any signal will cause the output to be at the positive or
    negative extreme of what the opamp can output (which won't be -+12V
    unless you have a 'rail to rail' opamp).

    If you use feedback from the output to the inverting input, then the
    opamp will try to modify it's output so the inverting input equals the
    noninverting input.

    If you have a reference voltage of Vref, an input resistor of Ri, an
    input voltage of Vi, and a feedback resistor of Rf, then it's 'almost
    true' that V- == Vref. Thus,

    (Vref - Vi)/Ri + (Vref - Vo)/Rf = 0

    This just states that the current going into the V- node is equal to the
    current coming out.

    You can solve that equation for what you don't know. In your case, Vref
    = 1, so

    (1 - Vi)/Ri + (1 - Vo)/Rf = 0

    If we assume that Ri = Rf and aren't 0, then they divide out, so

    (1 - Vi) + (1 - Vo) = 0

    so

    Vo = 2 - Vi

    This is another way of saying

    (Vo + Vi)/2 = 1

    which says the average of the output and input voltage is 1V, your
    reference.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  5. Ingvar Esk

    Ingvar Esk Guest

    If you mean that V- is not connected to anything, I think the output is
    unpredictable and very sensitive to noise on the V- input.
    Without feedback to the inputs you will basically see +12V or -12V depending
    on the input level on V- compared to V+ (1V).
    You will see 1V, as that is what you provided. The V+ is not an output so it
    can't be affected by the opamp.
    Ingvar
     
  6. Guest

    The way it works is a little more complicated than you imagine, but
    not too much.

    If the -ve input goes slightly +ve (w.r.t. the +ve input) then, as you
    know, the output goes -ve. And, because the gain of the op-amp is
    huge, the output will slam down -ve as hard as it can...

    ....but as it moves down, the voltage on it will be fed back to the -ve
    input via the feedback resistor. The size of the voltage appearing on
    the -ve input depends on the ratio of the input resistor to the
    feedback resistor (you will have to draw this out and think about it
    maybe: if you get to understand it, then you will also see why how the
    gain depends on this ratio too).

    You also know that the "input impedance" of the op-amps inputs is
    "very high" i.e. it is easy to "move them up or down" but you also
    know that some books refer to "virtual earth" ?!

    The reason for this is: while the actual inputs are "high z" (on their
    own) they will "obviously" have an actual "impedance" equal to
    whatever is connected to them i.e. the resistors (in parallel (to
    "earth")).

    But this "impedance" is not a static thing i.e. "just" the resistors,
    it is dynamic i.e. the "effect" of the "virtual earth" depends on how
    the -ve node is affected when you try to move it up or down from the
    *outside* i.e. before the input resistor - the effect is that it
    "cannot be moved" (because the feedback resistor moves it back again
    (via the output)... So the effect is that it appears to be fixed or
    "virtual earth" (but only if the +ve input is at earth).

    If you move the +ve input to 1V then the -ve input will now become
    fixed at 1V too so it will now become a "virtual 1V".

    The phase "virtual earth" is sloppy and as you can see, misleading.

    Cheers
    Robin
     
  7. John Fields

    John Fields Guest

    I think you're confused as to what a 'virtual ground' is. If an opamp
    is hooked up in the inverting configuration with the non-inverting (+)
    input grounded, like this"


    +--[R2]---+
    | |
    | +12 |
    | | |
    VIN>--[R1]--+--|-\ |
    / | >---+-->VOUT
    / +--|+/
    / | |
    VV | -12
    |
    GND

    The output of the opamp will do whatever it has to to drive the -
    input to whatever voltage GND happens to be, regardless of what Vin
    happens to be. Therefore, the - input, not the +, is the virtual
    ground since it always tries to be at GND potential.

    If R1 = R2 and we make a table with a few values of Vin:

    VIN VV VOUT
    -----|---|------
    -2 0 +2
    -1 0 +1
    0 0 0
    +1 0 -1
    +2 0 -2


    In your example,:


    +--[R2]---+
    | |
    | +12 |
    | | |
    VIN>--[R1]--+--|-\ |
    | >---+-->VOUT
    VREF>-[R3]--+--|+/
    | |
    [R4] -12
    |
    GND

    If the + input is at +1V, then the output will swing to whatever
    voltage it has to in order to drive the - input to +1V, so the - input
    is a "virtual +1V".

    Making a table similar to the first, where VV is the voltage on the
    opamp's - input, we'll have:

    VIN VV VOUT
    -----|---|------
    -2 +1 +4
    -1 +1 +3
    0 +1 +2
    +1 +1 +1
    +2 +1 0V

    So in both cases the voltage on VV stays constant and follows the
    voltage on the + input, so it's "virtually" the voltage on the +
    input.
     
  8. Guest

    In essence whatever the voltage on the "+" input will cause the output
    to drive the
    "-" input to be equal through the feedback resistor.
    It's not clear to me, how the above table was derived, given r1= r2
    and Vref = 1v.

    boats_ranger
     
  9. John Fields

    John Fields Guest

    ---
    It wasn't derived with Vref = 1V, it was derived with the voltage on
    the non-inverting input of the opamp equal to 1 volt.

    Looking at:

    +--[R2]---+
    | |
    | +12 |
    | | |
    VIN>--[R1]--+--|-\U1 |
    | >---+-->VOUT
    VREF>-[R3]--+--|+/
    | |
    [R4] -12
    |
    GND

    It should be clear that if Vref = 1V, then it would be impossible for
    the voltage on U1+ to also be equal to 1V, and I don't believe I
    referred to the voltage on U1+ as "Vref" anywhere in my earlier post.

    However, I think what you really had in mind was this:

    +--[R2]---+
    | |
    | +12 |
    | | |
    VIN>--[R1]--+--|-\U1 |
    | >---+-->VOUT
    Vref>--+----+--|+/
    | |
    +1V -12

    Such being the case, and remembering that the opamp's output will do
    whatever it has to to make the voltage on U1- be equal to the voltage
    on U1+, then we can model the circuit as a simple voltage divider,
    like this:


    Vout
    |
    [R2]
    |
    Vref---+---1.0V
    |
    [R1]
    |
    Vin

    Now, the game becomes trying to find out what Vout has to be to make
    sure that Vref always stays at 1.0V as Vin varies, if R1 and R2 are
    equal resistances. Since, if they're equal, it won't matter what
    value they are, let's set them at one ohm for convenience and redraw
    the circuit like this with Vin = 0V:


    Vout
    |
    [1R] R2
    |
    +---1.0V Vref
    |
    [1R] R1
    |
    0V Vin


    From Ohm's law,

    E
    I = ---
    R

    and, since there's 1V across R1, we can say:


    Vref - Vin 1V
    I = ------------ = ---- = 1A
    R1 1R


    Since R1 and R2 are in series the same current has to be flowing
    through both of them, so if R2 has a resistance of one ohm there'll be
    one volt dropped across it as well. That one volt will be on top of
    the voltage across R1, so we now have:


    2V Vout
    |
    [1R] R2
    |
    +---1V Vref
    |
    [1R] R1
    |
    0V Vin

    And if we start a new table it'll look like this:


    Vin Vref Vout
    ------|------|------
    -2 +1
    -1 +1
    0 +1 +2
    +1 +1
    +2 +1

    Proceeding in the same vein, if we make Vin equal to 1V and we want
    Vref to also be at 1V, then zero current must flow in R1, and the only
    way to also make zero current flow in R2 will be to make Vout equal to
    Vref, so we now have:


    1V Vout
    |
    [1R] R2
    |
    +---1V Vref
    |
    [1R] R1
    |
    1V Vin

    And our new table will have a new entry:

    Vin Vref Vout
    ------|------|------
    -2 +1
    -1 +1
    0 +1 +2
    +1 +1 +1
    +2 +1


    Now, by setting the voltage at Vin to be whatever you want it to be
    and calculating the current in R1 and R2 (remembering that they have
    to be the same) you ought to be able to figure out what Vout has to be
    to make that happen, and to fill out the rest of the table.
     
  10. Guest

    I must have lost something if, Vin = -2 , Vref = +1, R1 and R2 =
    1ohm
    and "+" is a virtual +1v.

    Given the above the current is I = (1 -(-2))/1 = 3amps
    The IR drop across R1 is 3 volts, If the R2 current = 3 amps. the
    voltage
    drop would be 3volts. Why is the answer zero? I must be missing
    somthing.

    A confused boat_ ranger.
     
  11. Rich Grise

    Rich Grise Guest

    If you had included a little context, it'd be much easier to figure
    out what you're talking about.

    1. If you insist on using the new fucked-up google groups, then at least
    learn how to copy/paste context.

    2. Or, get a real newsreader and a real newsserver, and post like someone
    who knows their elbow from a hole in the ground.

    Good Luck!
    Rich
     
  12. Guest

    If you had included a little context, it'd be much easier to figure
    That's a fair statement. If there a method which make's
    questions/answers easier to interpret then I'm for that. Just ask.
    Cut and paste is not a problem, some of the illustrations are
    difficult to
    decipher.
    I never heard of a newsreader, As far as the post, I appreciate those
    who took the time to answer and I do read each reply carefully.

    "Rich", if what I written upset's you don't read the thread, It's not
    worth the time.

    boat Ranger
     
  13. John Fields

    John Fields Guest

    ---
    I don't understand what you're talking about.

    If this is what you're referring to:


    Vout
    |
    [1R] R2
    |
    +---1V Vref
    |
    [1R] R1
    |
    -2 Vin

    Then there will be a 3V drop across R2 if Vref is to remain at 1V,
    which makes Vout equal to +4V.

    Here's the table I originally posted, where VV is Vref:

    VIN VV VOUT
    -----|---|------
    -2 +1 +4
    -1 +1 +3
    0 +1 +2
    +1 +1 +1
    +2 +1 0V

    Notice that the "answer" is zero when Vin = +2V, not when it's -2V.
     
  14. Rich, you've been in a foul mood lately, not your usual sunny self.

    If we scare off all the beginners, there won't be anything to do except
    whine about politics and religion. Next thing you know, Larkin and
    Thompson will start posting recipies again.

    Boat_ranger was the guy who wanted to build a 500V 200mA SMPS a few
    weeks ago. Lighten up, he is obviously learning.

    From "The Life of Brian". As you recall, this was sung by the guys
    hanging on the crosses at the end:

    Always Look on the Bright Side of Life

    words and music by Eric Idle

    Some things in life are bad
    They can really make you mad
    Other things just make you swear and curse.
    When you're chewing on life's gristle
    Don't grumble, give a whistle
    And this'll help things turn out for the best...

    And...always look on the bright side of life...
    Always look on the light side of life...

    If life seems jolly rotten
    There's something you've forgotten
    And that's to laugh and smile and dance and sing.
    When you're feeling in the dumps
    Don't be silly chumps
    Just purse your lips and whistle - that's the thing.

    And...always look on the bright side of life...
    Always look on the light side of life...

    For life is quite absurd
    And death's the final word
    You must always face the curtain with a bow.
    Forget about your sin - give the audience a grin
    Enjoy it - it's your last chance anyhow.

    So always look on the bright side of death
    Just before you draw your terminal breath

    Life's a piece of shit
    When you look at it
    Life's a laugh and death's a joke, it's true.
    You'll see it's all a show
    Keep 'em laughing as you go
    Just remember that the last laugh is on you.

    And always look on the bright side of life...
    Always look on the right side of life...
    (Come on guys, cheer up!)
    Always look on the bright side of life...
    Always look on the bright side of life...
    (Worse things happen at sea, you know.)
    Always look on the bright side of life...
    (I mean - what have you got to lose?)
    (You know, you come from nothing - you're going back to nothing.
    What have you lost? Nothing!)
    Always look on the right side of life...

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  15. JeffM

    JeffM Guest

    If you had included a little context, it'd be much easier to figure
    He shouldn't have to ask. Here's the rule book for Usenet.
    http://66.102.7.104/search?q=cache:...gh+at-*-top-*-*-message+do-not-*-*-*-original

    boats_ranger,
    Easy way: Don't click the Reply link that is in plain sight.
    Click the **show options** link then click THAT Reply link.

    To view ASCII art, click the **Fixed font** link at the top of the
    page.

    Forgive Rich. This is his week to be incensed
    by those who post without having lurked long enough
    to glean a clue about how posting is done properly.

    Like that's not obvious.
    http://www.google.com/search?&q=gravity+forte+xnews+40tude+mozilla

    If your ISP doesn't have a good news server:
    http://66.102.7.104/search?q=cache:...notated/newsservers.php+news-servers+giganews

    ....or YOU could take the time to learn how to post properly.
     
  16. Guest

    Yes, I'm a beginner, Again I'm a beginner. But I'm willing to
    learn and
    in the future will be able to pass that knowledge to someone else :)

    John, I build up a circuit this morning, Allow me to double my efforts,
    I will repost
    with the proper etiquette (hopefully).

    To all other, I appreciate your patience in this matter.

    Beginner boat ranger
     
  17. John Fields

    John Fields Guest

    ---
    Your etiquette's OK, but it _would_ be nice if you left a snippet from
    the post you're responding to in your response to it. That way,
    everybody'll know what you're talking about without having to
    backtrack the thread.

    One other thing, just in case you don't know, is that it isn't
    necessary for R1 and R2 to be one ohm each, they just both have to be
    the same resistance. That way you won't have to use such a large
    power supply to try it out.
     
  18. JeffM

    JeffM Guest

  19. Rich Grise

    Rich Grise Guest

    Yeah, sorry, I came down on you a little hard, as has been pointed out
    by others.

    I guess I need to learn some patience.

    Thanks,
    Rich
     
  20. Fred Abse

    Fred Abse Guest

    And from the same team:

    "Pray that there's intelligent life, somewhere out in space
    'Cos there's bugger-all here on earth."

    (From Monty Python's "The Meaning of Life")
     
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