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A couple inverting op-amp questions

  1. Feb 28, 2015 #1
    For the simplest case you have a feedback resistor and an input resistor. The non-inverting input is connected to ground.

    1. If the gain is only dependent on the ratio of the resistors, is there some sort of advantage in picking higher valued resistors (in the k range) rather than 10 or 20 ohms? The current drawn from the output will be completely dependent on the load resistance anyways.
    300px-Op-Amp_Inverting_Amplifier.svg.png

    2. What is the point of connecting a resistor in between the non-inverting input and ground?
    basic_opamp.gif

    Thanks!
     
  2. jcsd
  3. Feb 28, 2015 #2

    LvW

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    @1) The resistors should be (a) much larger than the finite (small) output impedance of the opamp and (b) much smaller than the (large) input resustance of the opamp. This is desired because otherwise the opamp cannot be trated anymore as a ideal device during all calculations. As another aspect: The input resistance of the invering circuit should be not too small (otherwise the signal source is loaded too much).

    @2) Although neglected during calculations - there is a small dc input bias current into the opamp input terminmals. This input current causes a small dc offset voltage at the inverting terminal across the resistors Rin and Rf. However, if app. the same dc voltage exists at the non-inv. terminal, the offset at the inv. terminal is compensated. For this purpose, the resistor between non-inv. terminal and ground should have a value of app. Rin||Rf.
     
  4. Mar 1, 2015 #3

    Baluncore

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    As LvW wrote, but also ...

    The op-amp output must supply the current to Rf and to the load of the next stage. Because the inverting input is a virtual earth, If = Vout / Rf. So use higher value resistors to reduce current consumption.
    Try to avoid using resistors over 100k, that way the surface leakage on a damp or contaminated PCB should not be a problem. High value resistors may also result in bandwidth problems where high track or terminal capacitance is involved. The bandwidth problem is reduced when a virtual earth is employed because there is very little change in terminal voltage.

    There was a time when op-amp input bias currents were quite high. If the resistance seen by the op-amp inputs were not similar, there would be an input offset voltage resulting from the bias currents. Input bias currents are now so close to zero that there is no advantage to be gained by using the bias current compensation resistor.
     
  5. Mar 1, 2015 #4
    Alright, makes sense. Also, is there a simple way to power an op amp using only one power source? Everything I found online was really complicated.

    Thanks
     
  6. Mar 1, 2015 #5

    berkeman

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    Staff: Mentor

    If you take into account that there are parasitic capacitances associated with a real circuit implementation of that schematic, what would some of the bandwidth considerations be in choosing the range of those resistors...? :smile:
     
  7. Mar 1, 2015 #6

    davenn

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    some are designed to work only from split rail PSU's +V 0V -V, some of those will work on single rail supplies
    Others are designed to work off single rail supplies

    The datasheet for any given Op-Amp will state its operating conditions

    cheers
    Dave
     
  8. Mar 2, 2015 #7

    LvW

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    In general, ALL opamps are able to work with single or double supply. Therefore, the question arises why some devices are described as "especially suited" for single supply?
    The answer is connected with the ability of the device to work "rail-to-rail". If the particular application requires only positive output signals, in many cases one is interested to allow outputs down to (nearly) zero. In this case, it is possible (but not necessary!) to use devices especially designed for single supply operation. I think, that is the only point which matters. In all other cases, you can (can !) use a single supply voltage only - however, you must be aware of the consequences:
    (1) A special input biasing is necessary, and
    (2) An output coupling capacitor is necessary (because of a dc output voltage at app. half of the supply voltage).
     
  9. Mar 2, 2015 #8

    jim hardy

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    To state it more simply,

    Look at the humble 741:
    250px-LM741_Pinout_Round.svg.png

    there's no pin for "Circuit Common"
    so walk a mile in Mr opamp's shoes......
    supply voltage is the difference between Vs+ and Vs-.
    You have no idea whether there's a third 0V terminal on your supply that's connected to Circuit Common...
    So whether the supply is single or dual is not known to the opamp. Only the guy who built the circuit knows for sure..


    As LvW said, you might think of it in terms of how much of the range between Vs+ and Vs- is useable?
    Remember voltage is potential difference
    Mr 741 wont tolerate his input pins closer than a couple volts to either supply, and can't drive his output closer than that either.
    For that reason we establish a "Circuit Common for signal" in the vicinity of halfway between Vs+ and Vs- . Usually we do that with a split supply so that the circuit can handle signals of either polarity. Junction of Vs+ and Vs- becomes signal common. But observe it doesn't connect to the opamp, there's no pin for it.

    "Single Supply" opamps like LM324 will tolerate inputs within millivolts of Vs- and can drive their output pin there so long as the current demanded is small. BUt even the LM324 needs about 1.5 volts of "headroom" from Vs+ on both inputs and output.
    "Rail to Rail" opamps can work all the way to both supplies which is really handy in today's battery powered world.

    Took me a long time to realize that circuit common is just a handy place to hook your voltmeter's black wire.


    Hope that helped.


    old jim
     
    Last edited: Mar 2, 2015
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