OrCAD and Active LPF

Discussion in 'Homework Help' started by JasonL, Mar 25, 2014.

  1. JasonL

    Thread Starter Active Member

    Jul 1, 2011
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    I want to use pSpice to simulate the frequency response for the transfer function, but the results from the simulation conflicts with what I calculated. I'm not sure what is wrong.

    The frequency response for the transfer function should be a LPF and the minimum should approach 0, but in pSpice I get strange results. Images are attached.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Those "strange results" are due to the finite bandwidth of the op amp. The 741 (a very old, low performance op amp) has a gain-bandwidth-product of about 1MHz, thus, as the frequency approaches that value the opamp gain approaches 1, and there is little feedback to provide the LPF function. Then the signal feeds more or less directly from the input through the capacitor to the output.

    Also note that, in real life, the 741 cannot drive the very low 16Ω feedback resistor you are using. If you look at the data sheet you will see that it can't drive a load of much less than 2kΩ.
     
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  3. LvW

    Active Member

    Jun 13, 2013
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    Yes - as mentioned by chrutschow, a parallel combination of R=16 ohm and C=1µF is rather "strange". Use larger resistors (kohm range) and smaller capacitors to get the same time constant.
    And, of course, watch the frequency capability of the opamp used.
    For good performance (filtering) the gain-bandwidth product should be larger than the filter pole at least by a factor of 50...100. And there is another opamp parameter you should be aware of for frequencies in the kHz range: Slew rate.
     
    Last edited: Mar 26, 2014
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  4. JasonL

    Thread Starter Active Member

    Jul 1, 2011
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    Thanks for the help! I'm still unsure what parameters to look for when selecting op amps. Are there any sources you would recommend I should start reading to learn? I'm a bit confused on where to look at data sheets to check what would be the minimum load an op amp can drive.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    If you look at the listing for Output Voltage Swing you will see that they use a minimum of 2kΩ for the test. The would be the smallest load (including the feedback resistor) that the op amp should drive.

    You should look at the Bandwidth or Gain-Bandwidth-Product and the slew rate if you are operating at frequencies above a few tens of kilohertz.

    If you want to operate from a single supply then you want a single-supply or rail-rail type opamp.

    If you are concerned about DC signal accuracy then you need to look at the input current bias and input volt offset parameters.

    Here are some references that may help your further.
     
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  6. JasonL

    Thread Starter Active Member

    Jul 1, 2011
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    I am looking at this link http://www.physics.unlv.edu/~bill/PHYS483/op_amp_datasheet.pdf

    Slew Rate: How fast the output can change (measured in V/us). This gives you an idea of the maximum
    frequency and amplitude signal the output can handle without distortion. The LM741A's output typically
    can only slew at 0.7V/us. If you had a 10KHz ,10Vpp sine wave on the output the fastest point at which
    the voltage changes is at the zero crossing. The rate of change is d/dt (10sin(2*pie*10,000t)) = 0.63V/us.
    Since the LM741A only guarantees a slew rate of 0.3V/us there is a chance that the 10Vpp, 10KHz sine
    wave will have distortion at the zero crossings. To operate without distortion you could lower the voltage
    to say 3Vpp or lower the max frequency to say 3Khz. Again, the LM741A is considered a slow op-amp.
    You can get op-amps with slew rates in excess of 1000V/us (1V/ns).

    I don't know how 10sin(2*pie*10,000t)) = 0.63V/us.
     
  7. jjw

    Member

    Dec 24, 2013
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    The slew rate is d/dt ( U*sin(2*pi*f*t))=U*2*pi*f*cos(2*pi*f*t)
    which has maximum U*2*pi*f
    in your example 10V*2*pi*10000 1/s ~ 628000 V/s = 0.628 V/us
     
    Last edited: Mar 29, 2014
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  8. LvW

    Active Member

    Jun 13, 2013
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    It happens rather often that I read such a requirement. But why?
    For my opinion, each opamp can be operated with single supply (and, in fact, I did very often in the past). I see no reason why single supply operation is connected with rail-to-rail properties (in case of linear amplification).

    EDIT: According to my recent experience I like to add that the above is a question only - nothing else. For my opinion, the clarification of this question could be helpful for all beginners.
     
    Last edited: Mar 30, 2014
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