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Power for an Op amp from single source

  1. Jun 12, 2012 #1
    Is it possible to power the op amp in this diagram in this manner if it requires +18V and -18V?
    This doesn't seem valid to me, but I'm not sure why it's not exactly.

    Attached Files:

  2. jcsd
  3. Jun 12, 2012 #2


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    sure. voltage is relative anyway. you could choose to ground the negative end of the 36 V supply instead of what is halfway between the positive and negative ends. but then the output of the op-amp would not be bipolar.
  4. Jun 12, 2012 #3
    I'm not seeing what the purpose of the resistors are, whether they are there or not the supply terminals on the opamp will see the positive and negative leads of the battery.
  5. Jun 12, 2012 #4
    Well these resistor create a circuit that is called "virtual ground".
    Every real life "active device" to work properly as an amplifier supplied from a single source need proper bias circuit.
    When you use BJT as a CE amplifier, you use a voltage divider to bias the active device somewhere in the "linear region".
    In case of single supply op amp you have to do the same think. You need to bias the op amp somewhere in the middle of his "linear region". And this resistor provide proper DC bias for Op amp.

  6. Jun 12, 2012 #5


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    The schematic in the first post does set a bias point so to speak but it is not useable if the load is between the output (naturally) and ground. The resistors are not able to carry any significant load current. The schematic in post #4 is definately workable. I have used it or something similar many times.
  7. Jun 12, 2012 #6


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    Yep. Here a Texas Instruments Application Report with a collection of Single-Supply Op-Amp Circuits.

    "www.eng.yale.edu/ee-labs/morse/compo/sloa058.pdf" [Broken]
    Last edited by a moderator: May 6, 2017
  8. Jun 12, 2012 #7
    That's what I have been doing in the last few months!!! I am designing electronics inside guitar and guitar effect pedals. They all use one 9V battery only. I just put the power across the op-amp like your picture and use a voltage divider to get half voltage for biasing the input of the op-amp like what Jony130's post.
  9. Jun 12, 2012 #8


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    sounds like a good gig.

    do you or have you worked on any digital effects devices?

    can you tell us what some of these stomp boxes are?


    r b-j
  10. Jun 12, 2012 #9
    I am familiar with the need for biasing networks when working with active amplifiers, I'm just not seeing how the resistors in the original post provide any type of biasing at all.

    Regardless of the value of R1, the terminals labeled Vcc and Vee are going to sit at the voltages on positive and negative terminals of the battery. How is that providing any form of biasing at all?

    The only thing I can see is that the resistors provide one with a point to ground the circuit.
  11. Jun 12, 2012 #10
    One is a Distortion pedal that I put my own twist on the clipping circuit. The major one is inside the guitar that I cannot review at this point. In fact I am very close of completing the patent application to submit to the USPTO this week or latest next. I am too cheap to pay thousands of dollars to have a patent attorney to do it. I spent the last half a year not only working on the circuits, also learning how to write the patent. It'll still cost me like $1800 just for the whole process if I am lucky enough that they grant me the patent.

    I am not comfortable enough in DSP to do anything in digital. That's where you're the man.:smile:
  12. Jun 12, 2012 #11


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    The original circuit was not meant to be complete.

    The one quoted in your post shows the original two resistors as R1 and R2.

    Because the previously negative line is now grounded then the non-inverting "+" input to the opamp now needs to be held at half the supply voltage.

    Another consequence of this is that there has to be a capacitor in series with the input because the linear region of the opamp is now near the half voltage point of the opamp supply (half of 36 volts) and the input may not be at +18 volts or so.
  13. Jun 13, 2012 #12


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    it defines that ground is midway between the positive and negative terminals of the battery. that's all it does.
  14. Jun 13, 2012 #13
    That's all I needed to know.
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