Connect with us

Calculate Op Amp Gain

Discussion in 'General Electronics Discussion' started by steve harris, Jun 21, 2017.

Scroll to continue with content
  1. steve harris

    steve harris

    2
    0
    Jun 21, 2017
    Hi,

    How do I calculate the gain of this attached circuit?

    Basically do i have to take the R1 into account?

    I am measuring the output to the right hand side of R1.

    is it

    A) A = 1 + (R2 / R3)

    or

    B) A = 1 + ( (R1 + R2) / R3)

    Thanks

    Steve
     

    Attached Files:

  2. davenn

    davenn Moderator

    13,657
    1,888
    Sep 5, 2009
    B would be the logical choice
    but it really shouldn't be done that way. The feedback should be taken directly from the output pin
     
  3. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    I would go for A since the feedback comes from the output.

    R1 throttles the output of the op-amp so the amp has to work harder. The frequency response and the output swing will be restricted.
    An ideal op-amp has a zero output impedance.
     
  4. steve harris

    steve harris

    2
    0
    Jun 21, 2017
    thanks all,

    i would/should have mentioned.

    this is a dc circuit thing.

    this is from a board i am repairing and came across this part of the circuit which is not something i have seen before.

    they are feeding a 5v reference voltage (low current) into the +ve input of the op-amp and using the output as a 15v output.

    thanks
     
  5. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    R1 limits the output current so may be to protect the amp from a short.
     
  6. davenn

    davenn Moderator

    13,657
    1,888
    Sep 5, 2009
    NO, the feedback comes after the resistor on the output
    this will significantly change the gain value
     
  7. ramussons

    ramussons

    365
    70
    Jun 10, 2014
    A = [1 + ( (R1 + R2) / R3)] * (R2+R3)/(R1+R2+R3)
     
  8. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    The gain is determined by the feedback which comes from the circuit output, not from the op-amp output. The calculations you have shown deals with the voltage at the op-amp output which is of no concern, it will be greater than the circuit output voltage and will depend on the load.
     
  9. LvW

    LvW

    604
    146
    Apr 12, 2014
    The voltage at the opamp output is of "no concern"??
    I rather think, we have an active device (called opamp) which has a low-resistive voltage output and it is THIS output only which provides voltage feedback!
    The only difference to the classical non-inverting circuit is that we do not further process the ouput voltage at the opamp output (for which reason ever) but the voltage between R1 and R2 (simple voltage division).
    Referenced to this node (between R1 and R2) the gain is as given by ramussons (post#7).

    And - yes, this output voltage (and the gain) will depend on a possible load resistor; however, thius was not part of the question.
     
    Last edited: Jun 22, 2017
  10. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    No, no The feedback comes from the output so the output voltage will be controlled by R2 and R3 so long as the op-amp can drive it through R1.

    Perhaps someone can spice it. I cannot do it at the moment due to a computer failure.
     
    Harald Kapp likes this.
  11. Ratch

    Ratch

    1,081
    331
    Mar 10, 2013
    Why not just calculate it?

    Harris.JPG
    The gain is calculated from the loop equation. If you want the op-amp out, then two equations are needed. The op-amp will have to work harder to supply vout because of r1.

    Ratch
     
    duke37, Harald Kapp and davenn like this.
  12. AnalogKid

    AnalogKid

    2,393
    665
    Jun 10, 2015
    Referring to the original schematic, the opamp voltage gain from pin 3 to pin 6 is as in equation A, with a series feedback resistance of (R1+R2). Pin 6 is a zero-ohm point at voltage (Vin times the gain). Pin 3 is a zero-ohm point at voltage Vin, driven there by the opamp's closed loop. So R1 and R2 form a voltage divider between two zero-ohm points of two signals with identical frequency and phase, but different amplitudes. Enter Ohm's Law.

    All of that assumes a perfect opamp and an infinite impedance load at Vout.

    ak
     
  13. Ratch

    Ratch

    1,081
    331
    Mar 10, 2013
    I have a hard time understanding your explanation. Do you agree that my calculation are correct? If not, what are your calculations?

    Ratch
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

    10,391
    2,271
    Nov 17, 2011
    R1 has no influence on Vout as long as the opamp can drive enough current, see post #10.
    This kind of circuit is sometimes used with capacitive loads. Most opamps are limited in their capability to drive capacitive loads. The resistor R1 helps to stabilize the amplifier in that case while the feedback via R2/R3 from Vout is required to set the gain correctly.

    As much as I deSPICE the use of a simulator when a thoretical calculation can give the result, I follow @duke37 's request:
    upload_2017-6-23_8-1-3.png

    As expected gain is 2 (1+R2/R3).
    Changing R1 has a rather small influence on gain (in the 1 % range, depending on how much R1 is changed). This is a frequency dependend effect which at e.g. 10 Hz signal frequency is almost not visible. Possibly the opamp model and some implicit parasitic capacitances in the simulation cause this effect.
    In real life the real parasitics and the real frequency response of the opamp will have even more impact.
     
    duke37 likes this.
  15. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    Thank you Harald.
    With 1V in, the voltage at the junction of R2,R3 will be 1V.
    The output will be 2V.
    U1 out will be 3V wih no load.
     
  16. LvW

    LvW

    604
    146
    Apr 12, 2014
    Yes - of course, this is correct.
    But this result was given in general form in post#7 already:

    A = [1 + ( (R1 + R2) / R3)] * (R2+R3)/(R1+R2+R3)

    Harald: How can you say that R1 has "no influence"?
     
  17. Ratch

    Ratch

    1,081
    331
    Mar 10, 2013
    It must be stipulated that the values of the output and U1 will be 2 v and 3 v respectively only if Vin = 1 v and R1=R2=R3.

    Ratch
     
  18. Ratch

    Ratch

    1,081
    331
    Mar 10, 2013
    The above equation simplifies to A = (R2+R3)/R3 as I calculated in post #11. R1 cancels out.

    I believe he meant no influence on the gain at vout.

    Ratch
     
  19. Ratch

    Ratch

    1,081
    331
    Mar 10, 2013
    No, the gain is 1+R2/R3 . See post #11

    Ratch
     
  20. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    Take R1 = R2 = R3
    The simple equation A = (R2+R3)/R3 with 1kΩ, get 2
    Complex equation A = [1+((R1+R2/R3]*(R2+R3(R2+R3)/(R1+R2+R3) = 2
    Take R1 = 0.1k , R2,R3 = 1k
    A= 2.1k*2k/2.1k = 4.2k/2.1k = 2

    so it appears that R1 is cancelled and does not affect the answer.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-