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Op amp confusion

Discussion in 'Electronic Basics' started by Chris, Jul 25, 2005.

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  1. Chris

    Chris Guest

    I'm using a lf442cn

    8 to +5v
    4 to gnd
    0.1uf cap between 8 and 4

    2 tied to ground for now

    3 has a 100ohm resistor
    1 has a 100ohm resister

    a 330pf cap between 3 and 1 before resistors

    3 and 1 tied after the resistors

    pin 2 tested against ground = 0v
    pin 1 tested against ground = 3.97v ?? shouldn't this be 0???

    basically this is a buffering circuit

    also get the same results with lf411 single supply amp, but as i can
    only supply a 5v/gnd to the op amp what would be best?

    Thanks
     
  2. JeffM

    JeffM Guest

  3. Is this saying you have a 100 ohm resistor between 3 and 1? If so,
    this is positive feedback.
    What is "before resistors"? Is it in series
    What is "after resistors"?
    A buffer would have the input connected to pin 3 (+ input) and pin 1
    connected to pin 2 (output to - input)

    The LF411 and LF412 have an input voltage range from about the
    positive supply voltage (pin 8) to about 3 volts more positive than
    the negative supply voltage (on pin 4). With a single 5 volt supply
    this leaves you an input range from about 3 to 5.

    They have an output swing range from about 1.5 volts below the
    positive supply voltage to about 1.5 volts above the negative supply
    voltage. With a 5 volt supply, this includes about 1.5 to 3.5 volts.

    but if connected as a buffer, both the input range and output range
    must cover the signal, so this amplifier will only operate for inputs
    between about 3 and 3.5 volts. Blah!

    Get an LM358 (dual) or LM324 (quad). Both input and output ranges
    cover about 1.5 volts below the positive supply to the negative
    supply, or 0 to 3.5 volts.
    http://cache.national.com/ds/LM/LM158.pdf

    If you need more signal range than this, you need what is called a
    rail-to-rail type whose input and output can operate from the negative
    to the positive supply voltage. An example of this would be an
    LMC6482, but there are lots of others.
    http://cache.national.com/ds/LM/LMC6482.pdf
     
  4. Chris

    Chris Guest

    The LF442 is a slightly later, slightly different version of the LF412.
    It has slightly lower bandwidth, and half the power supply current.
    They're part of the same useful family of parts, and the same rules
    apply. None of the LF411 JFET-input family is suitable for single
    supply operation. A hearty seconding on all of Mr. Popelish's
    suggestions.

    By the way, if you want to describe a simple circuit like this (view in
    fixed font or M$ Notepad):

    ___
    .---|___|---.
    | 1K |
    | VCC |
    | 2|\|8 |
    '---|-\ 1 | Vout
    Vin ___ | >----o---o
    o---|___|---|+/
    1K 3|/|4 |
    ===
    GND
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    you might want to try downloading the freeware/beerware (if you ever
    see Andy, you're required to buy him a beer) Andy's ASCII Circuits at
    the above site. It's a lot easier than listing nodes or pins, and
    allows newsgroup readers to easily visualize your problem.

    Once you get your choice of op amp taken care of, you might want to
    look closely at the above op amp voltage follower (buffer) circuit. I
    think this might be a better way to get a voltage follower. The
    diagram uses the standard dual op amp pinouts (like the LF442, LM358,
    &c.) The other end of the feedback resistor from pin 1 (output) should
    be tied to pin 2 (inverting input). Pin 2 should not be grounded.
    That would cause the output to rail positive for any op amp with any
    Vin above GND.

    Also, do yourself a favor, and read the data sheet. It's a free
    education in itself.

    http://cache.national.com/ds/LF/LF442.pdf

    Good luck
    Chris
     
  5. Chris

    Chris Guest

    |\
    -----|+\
    | \-+-/\/\-+--
    | / | |
    +-|-/ === |
    | |/ === |
    | | |
    |______+_/\/\-+

    is basically how it's running, threw in an 7812/7912 all is fine but
    will probably run a 7660 at +5/-5 v i really would of liked not to need
    a -v i did not realise about the limitations on running 5v/gnd but now
    that i think about it i think i get it..

    Thanks guys
     
  6. Kitchen Man

    Kitchen Man Guest

    Someone will correct me if I am wrong, but I see an integrator with no
    input conditioning or noise compensation. You said you wanted a
    buffer? Take a look at this:

    http://www.physics.ucdavis.edu/Classes/Physics116/Lab02_rev.pdf

    ....and rethink your design.
     
  7. Kitchen Man

    Kitchen Man Guest

    I believe that if you want that circuit to be a voltage follower, then
    the feedback path should be a short. The circuit you have printed has
    a gain of 2. With Rf shorted, the gain will be 1.
     
  8. Chris

    Chris Guest

    sorry voltage follower
     
  9. Chris

    Chris Guest

    Hi, Al. The OP said "basically this is a buffering circuit". I was
    going from his description of his circuit to try and determine what he
    had.

    If you'll look carefully at my ASCII diagram above, you'll see it is a
    voltage follower with a gain of 1. Since op amp bias current is
    usually much greater than offset current, many will use identical input
    resistors to cancel out the voltage drops due to the former. I guess I
    kind of assumed that was what he was doing from his description. Of
    course, with a 50pA bias current JFET input, it's kind of a waste of
    resistors.

    But, in the diagram above, since there's no feedback to the input
    terminal, and no gain divider on the negative feedback, the gain is 1.

    Thanks for taking the time to respond. It's always good to know
    someone's checking. If bad information were to ever get out on the
    internets, we'd all be done for. ;-)

    Good luck
    Chris
     
  10. Kitchen Man

    Kitchen Man Guest

    Yes, you're right, the non-inverting feedback default gain of one is
    if the 1K input resistor is tied to return as part of the feedback
    network. Oops!
     
  11. Chris

    Chris Guest

    is one of these methods better than the other, would one produce a
    cleaner signal?

    .-----------.
    | 1K |
    | |
    | |\ |
    '---|-\ | Vout
    Vin | >----o---o
    o------|+/
    |/

    ___
    .---|___|---.
    | 1K |
    | |
    | |\ |
    '---|-\ | Vout
    Vin ___ | >----o---o
    o---|___|---|+/
    1K |/



    Vin |\
    o----|+\ Vout
    | \-o-/\/\---o---o
    | / | 100 |
    o-|-/ === |
    | |/ === 330pf |
    | | |
    |______o_/\/\---o
    100


    supply +5v/-5v
     
  12. Simple and clean, but if there is any input bias current, and the
    input signal comes from a high resistance source, the bias current
    will produce a voltage going through that resistance that will be
    added to the input signal.
    This version attempts to cancel the above mentioned bias current
    voltage drop by causing an equal one across the feedback resistor.
    But you don't actually put another resistor in series with the input.
    The feedback resistor is supposed to be the same as the resistance
    that is unavoidably in the signal source. The only reason to add to
    that would be if you wanted to include a low pass filter into the
    design, or to add over voltage clamping diodes at the + input.
    This variation is used if the output is intended to drive a large
    amount of capacitance, which tends to make the above versions unstable.
     
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