Opamp works with DC won't amplify AC

Discussion in 'General Electronics Chat' started by kochevnik, Jun 25, 2013.

  1. kochevnik

    Thread Starter New Member

    Jan 7, 2013
    14
    0
    Hi all -

    I'm a beginner so be kind :)

    I have an inverting opamp circuit (similar to attached) that is working just fine with DC on the I- pin but when I set up my function generator to output an AC signal of roughly the same voltage P-P the opamp is outputting a similar sinewave, but it's not amplified very much.

    The opamp is a MCP6241-E/P with 550khz gain bandwidth product (datasheet attached). I'm powering it with a voltage 5.35 V on Vdd and Vdd/ 2 = 2.66v on I+ and this opamp is supposed to go rail to rail (which it does with DC just fine).

    The basic circuit I have is with a 1 M ohm resistor from Vout to I- and then I have a bunch of different resistors I try on Vin 800k 467k 325k 200k etc this lets me easily watch the change on Vout as I change the gain. This all works great with DC - I get almost the exact right voltages on Vout that would be predicted from the opamp formulas.

    So now I hook up a 150khz - not quite a clean sine wave - from my signal generator but fairly close, then I use the EXACT same setup, only adding a 4u7 cap on Vin (I was under the impression this is a DC blocking cap) and then check Vin and Vout on my scope while I change the resistor values on Vin (same values I have listed above).

    The output is a nice clean sinewave, but the peak to peak voltage does not vary much (5% maybe) as I try the different resistor combinations.

    For example - my not so clean sine wave going into Vin is 0.285v P2P and on Vout I get a nice clean sinewave of 0.530 volts P2P - for ALL of the different resistor combos above. I switch them and I get some small changes in Vout, but nothing like the changes I got with the DC testing.

    And a few odd things - If I run Vin directly into pin2 without a resistor or cap, I still get a nice sinewave on Vout, only the voltage is now 0.23v P2P not the 0.53 volts I get when I try all the resistors.

    So what gives ? What am I doing wrong ?
     
  2. Gibson486

    Member

    Jul 20, 2012
    199
    12
    your gain bandwidth is 550Khz.....

    Your freq is 150Khz

    your closed loop gain is around 100....

    I'd give you the answer, but you will learn more if you see where I am getting at...

    the one advice i will give you is that when you have gains as high as yours, you should really consider using multiple gain stages.
     
  3. LvW

    Active Member

    Jun 13, 2013
    674
    100
    kochevnik,
    in principle, you can use amplifiers with gain values of 100 or even larger.
    That is NOT the problem
    However, in your case it is the limited (and for your opamp rather low) slew rate that limits the frequency range of your amplifier. Try some lower frequencies (1kHz) and you will be successful. The large signal bandwidth is approximately 10 kHz (maximum). Above this frequency the output will start to be distorted (triangle) and the amplitude will decrease.
     
  4. kochevnik

    Thread Starter New Member

    Jan 7, 2013
    14
    0
    Thanks - appreciate the replies - but the circuit is just an example - remember that the test resistors I used were 1M ohm for the feedback resistor and 810k ohms for the resistor on Vin - this gives a gain of -0.810 for an inverting amp.

    Or for one of the other resistors I tried (325k ohms) this gives a gain of
    -0.325

    according to this calculator :

    http://www.daycounter.com/Calculators/Op-Amp/Op-Amp-Voltage-Calculator.phtml

    Example :

    r1 = 1000
    r2 = 810
    r3 = 0
    r4 = 1
    v1 = 0.1425 (1/2 of .285v ?)
    v2 = 2.66
    Vp = 5.35
    Vn = 0


    gives a Vout = 4.70v

    changing v to 0.285v (?) still gives 4.58v

    Neither of these is close to what I actually saw on the scope.
     
  5. Gibson486

    Member

    Jul 20, 2012
    199
    12
    See figure 2-15 and 2-5 in the spec sheet.
     
  6. kochevnik

    Thread Starter New Member

    Jan 7, 2013
    14
    0
    OK - I get this, but the output is a very nice clean sine wave.

    Question - if the Gain bandwith is 550khz and I have a 150khz input signal and I'm using a 1M ohm resistor on the feedback and also a 325k ohm resistor on Vin - this should work, right ? gain is -.325 and frequency is 150khz so this is within the gain bandwidth product for this opamp (550khz) ? or not ?
     
  7. kochevnik

    Thread Starter New Member

    Jan 7, 2013
    14
    0
    Ah - thanks Gibson - that makes perfect sense - I missed that.

    And Lvw thanks also - that's what you were trying to get thru to me.

    Thank you very much guys.

    Solved.
     
  8. nigelwright7557

    Senior Member

    May 10, 2008
    487
    71
    If you are amplifying you have to divide the bandwidth by the gain to get the real bandwidth.
     
  9. Gibson486

    Member

    Jul 20, 2012
    199
    12
    Just doing frequency times gain gives you the GBWP of OPEN loop gain. You have a closed loop circuit, so your GBWP goes down.

    Also, you have the gain inversed.
     
  10. crutschow

    Expert

    Mar 14, 2008
    13,014
    3,234
    Note that the closed loop gain you use for an inverting amp to calculate the frequency response is the equivalent non-inverting gain (the gain from the plus input with the inverting signal input at AC ground). Thus for an inverting gain of -1, for example, the equivalent non-inverting gain is 2, so you would divide the GBW by 2 to get the closed loop gain.
     
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