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Quick question regarding op amps

Discussion in 'General Electronics Discussion' started by CiaranM, Oct 9, 2012.

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  1. CiaranM

    CiaranM

    74
    1
    May 19, 2012
    An op amp with an AC supply uses at at least two capacitors. One from +V to 0V, and one from -V to 0V. [See bottom of this page http://www.birthofasynth.com/Thomas_Henry/pdf/VCO-1/vco1_schem2.pdf ]

    Does an op amp with a DC supply really require a voltage divider to create three rails? Can't I just use two caps like for the AC version?

    If a voltage divider is required, what are good values of resistance to use? Can I get away with using high values to limit the short-circuit current?

    Thanks!
     
  2. Raven Luni

    Raven Luni

    798
    7
    Oct 15, 2011
    I think you have your terms mixed up. Op amps do not use AC. AC stands for alternating current ie. current which changes direction periodically.

    You're referring to a split rail (DC) supply. You can use a standard voltage divider yes. In this case you are not creating 3 rails you are offsetting the input signal. The chosen values should reflect your required input impedance - and thats where my knowledge of the subject ends.

    This is why I tend to use elaborate rail splitters because I cant get my head around that part :p
     
  3. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Using a voltage divider to create a virtual ground does not create a third 'rail'. What it creates is a high-impedance voltage source that is only suitable for driving the reference inputs of op-amps and comparators. If your circuit requires a low-impedance ground, then you need to use two separate power supplies (one for the +V and one for the -V) or use one power supply with active voltage divider circuitry (rail splitter) to provide the low-impedance virtual ground.

    Good values of resistance to use for the voltage divider depend upon the load presented to the virtual ground. Add up the input bias currents of all reference inputs connected to the virtual ground then use that total current with the allowable offset voltage to calculate the largest equivalent resistance for the virtual ground. As a circuit designer, one of your jobs is to determine what the allowable offset voltage is.
     
  4. CiaranM

    CiaranM

    74
    1
    May 19, 2012
    oh my.. thanks for the help.
    I'm still a bit confuddled though. would 100k be appropiate to split the supply for an op amp used to amplify audio?
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,267
    Nov 28, 2011
    If you're generating your derived ground rail using a two-resistor voltage divider (usually with a capacitor from the rail to the negative supply), the values of the resistor are a tradeoff between avoiding wasted supply current (this pushes for higher value resistors, so less current flows through them) and having a "firm" rail, i.e. a rail that stays half way between the positive and negative supply voltages regardless of the state of other signals in the circuit that may be connected to that rail.

    In some cases, your derived ground rail is only used to feed inputs of op-amps (typically, non-inverting inputs of op-amps used in inverting configuration), which themselves draw negligible current. In this case, your rail is not going to get "pulled around" by current feeding into or out of it, and you can use 100k resistors with no problems.

    In other cases, current will be drawn from, or fed into, the rail by current paths in the circuit. Sometimes a resistor is connected from the derived ground rail to a signal point, to which a signal is coupled. The resistor acts to "bias" the circuit point at a DC voltage equal to the derived ground rail.

    But the signal then causes an AC signal to appear across that bias resistor, which causes current to feed through the bias resistor and into the derived ground rail, and this can pull the rail around. Actually, some of the signal is coupled into the rail, and from there, can feed to other parts of the circuit, possibly causing problems.

    You can describe the derived ground rail as being "not stiff enough" to hold its voltage steady in the face of varying currents feeding into/out of it. For this reason in AC applications such as audio, it's common to connect a capacitor (typically something like 10 uF) from the derived ground rail to the negative supply rail, which makes the derived ground rail "stiffer" at AC (signal) voltages, while allowing the two resistors to set the DC voltage.

    If your derived ground rail needs to be able to supply significant DC and/or AC current while remaining steady, there are other solutions. If you have a spare op-amp in a multi-op-amp package, you can use it as a buffer (with its output looped back to inverting input) to buffer a derived ground voltage created with a pair of resistors as before. The op-amp's output becomes the derived ground rail, and the op-amp supplies all the current, steady and alternating, to everything connected to the ground rail.

    The op-amp has a very low output impedance, so the derived ground rail remains very "firm" within a wide range of currents that may be drawn from it at different times. In other words, the derived ground rail remains rock steady at half way between the positive and negative supply voltages regardless of what's connected to it, and what currents those connections may carry, within the limits of the op-amp's output capability.

    In unusual cases, you can add a buffer stage of some kind. There's also a very convenient device called the TLE2426 from Texas Instruments, which is a 3-terminal device in a transistor-style package that connects to the positive and negative supplies and generates a regulated and stable derived half-supply-voltage output that behaves very much like the op-amp example.
     
    Last edited: Oct 13, 2012
  6. CiaranM

    CiaranM

    74
    1
    May 19, 2012
    yet another extremely helpful post from yourself. Thanks Kris!
     
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