impedance question

Discussion in 'General Electronics Chat' started by tpny, Jun 18, 2012.

  1. tpny

    Thread Starter Member

    May 6, 2012
    216
    0
    I have a noninverting opamp config with 10x gain. When input is 0V, output is > 0V, ie, there is voltage offset. I want to eliminate this offset by injecting some millivolts to the (-)input using a pot. As the picture shows, I'm connecting the swipe lead of the pot to the (-)input of opamp. This works as I can adjust the pot so that 0V appears at the output when (-)input is tied to ground. My question is: Should I put a series resistor between swipe lead of pot and (-)input of opamp? And what size resistor, thanks!

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    2. [FONT=Arial]                            10k[/FONT]
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    6. [FONT=Arial]Vin ---/\/\/\------------------|-\   |[/FONT]
    7. [FONT=Arial]         1k        |          |   \_|__Vout[/FONT]
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    9. [FONT=Arial]                     | GND--|+/[/FONT]
    10. [FONT=Arial]            resistor?       |/[/FONT]
    11. [FONT=Arial]                here?[/FONT]
    12. [FONT=Arial]                     |[/FONT]
    13. [FONT=Arial]                     |[/FONT]
    14. [FONT=Arial]           5V---/\/\/\---------(-12V)[/FONT]
    15. [FONT=Arial]                  20k pot[/FONT]
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  2. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    One thing you might do is to place a resistor from the +ive input to ground that is equal to the parallel combination of the input and feedback resistors. In general, each input needs a small input bias current. Assuming everything is otherwise balanced, when there is zero input voltage, the input bias current on the -ive input is going through these resistors to something that is ideally 0V on the other end of both and which then results in a voltage at the -ive input. You want to develop the same voltage on the +ive input, which can be done by duplicating the load on the -ive side but just hardtying the other ends of the two resistors to ground. But since this places both of these resistors in parallel, you can just use a resistor that is the parallel equivalent of the two.

    This may well reduce the remaining offset voltage so that it is tolerable or make it easier to null it out with a pot.

    Putting a resistor in series with the pot wiper isn't a bad idea because it will prevent you from applying either supply voltage directly to the -ive input. You can probably use a pretty good size resistor. I would recommend looking up the input bias current for the amp and sizing the resistor so that you can push or pull a couple times that amount with your bias compensation circuit.
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,395
    1,607
    1. Yes.

    2. 909 ohms - the source resistance of your divider.

    909? Yep, you want to have the same resistance in each input leg so any input bias current gives the same offset voltage to both inputs. The (-) sees the 1K and the 10K as parallel resistors, and 1K || 10K = 909 ohms.

    BTW, you only need to match the "order of magnitude" here, getting a .001% match doesn't give you .001% offset

    Now what's the souce resistance? It would be lots easier to do if you had it to equal voltages, it would just be half the pot value. But as you have it you first have to figure out the resistance in each leg.

    The wiper is about zero, so the 5V side is set to about 5/(5+12)*R, the other has 12/(5+12)*R. Bang thru this gives 4.15K.

    So your POT value is too large for this application. A 2K pot should work, Req = .415 ohms, so the series R would be 909-415 = 494.

    493 and 495 are both standard values, so flip a coin and pick one.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    But notice that his offset network is going into the inverting input, not the none inverting input, which is still hard tied to ground.
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,395
    1,607
    It's hard to know exactly where the resistor is going in an ASCII schematic.

    I would place an additional resistor between the pot wiper and the (+) input and use a smaller pot value.
     
  6. tpny

    Thread Starter Member

    May 6, 2012
    216
    0
    why half? is it because we turn the wiper somewhere in between?
     
    Last edited: Jun 19, 2012
  7. tpny

    Thread Starter Member

    May 6, 2012
    216
    0
    Something like this?
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    6. [FONT=Arial]Vin ---/\/\/\-----------------|-\   |[/FONT]
    7. [FONT=Arial]      1k     |                |   \_|__Vout[/FONT]
    8. [FONT=Arial]              |                 |   /[/FONT]
    9. [FONT=Arial]              |     GND---- |+/[/FONT]
    10. [FONT=Arial]              |             |   |/[/FONT]
    11. [FONT=Arial]           495        495   [/FONT]
    12. [FONT=Arial]              |-------------|         [/FONT]
    13. [FONT=Arial]              |[/FONT]
    14. [FONT=Arial]     5V---/\/\/\---------(-12V)[/FONT]
    15. [FONT=Arial]              2k pot[/FONT]
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    by the way, should i use 10k||100k instead of 1k||10k? What's the rationale?
     
    Last edited: Jun 20, 2012
  8. absf

    Senior Member

    Dec 29, 2010
    1,493
    372
    If you use the correct program and correct font, the schematics would be easier to understand. Try this program...

    http://www.tech-chat.de/download.html

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    7.               |        +Vcc       |
    8.   Vin  ___    |       |\|         |
    9.     o-|___|---+-------|-\         |
    10.        1K     |       |  >--------+------o
    11.               |    +--|+/              Vout
    12.               |    |  |/|
    13.              .-.  .-.  -Vcc
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    15.          495 | |  | |495
    16.              '-'  '-'
    17.               |    |
    18.               |   GND
    19.               |
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    21.              _V_
    22.        +5V -|___|- -12V
    23.               2K
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    27. (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
    28. [/FONT]
    29.  
    Allen
     
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