Issues relates capacitance in a high pass filter/ac coupling

Discussion in 'General Electronics Chat' started by trytolearn, Apr 10, 2010.

  1. trytolearn

    Thread Starter Active Member

    Mar 1, 2008
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    Hi

    I am building up an amplifier with an active high pass filter at the input of the op-amp,

    [​IMG]

    The amplifier will run at low frequency, approximately 1khz to 2khz
    the mean purpose of this high pass filter is to remove the DC-offset in the signal.

    I have been advised to use a large capacitor, and I have realised that the with small capacitors, the output square wave is more likely to be distorted. something like this

    [​IMG]

    Could anyone please explain me why this is happening? is that because the capacitor is too small, so that the capacitor is charges up too quickly? or because it is too small, it is discharged too quickly?

    I have also been advised that the value of the capacitor is related to the impedance, for my case the lower the input impedance, the better it is. (hence, large capacitor )

    This confuses me a lot, because low input impedance means short circuit, then what's the difference between having a wire directly connected to the input or have this high pass filter circuit?

    Many thanks for your kind help !
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    It's because the capacitor is too small.

    If the capacitor is made even smaller and the input square wave has very low rise/fall times, you'll notice that eventually the output will be just positive and negative "spikes", the peak-to-peak values of which will be approximately twice the input peak-to-peak voltage. This type of circuit is sometimes called a differentiator.

    If you increase the size of the capacitor, the output will better approximate the input, but will be independent of the DC level of the input.
     
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  3. trytolearn

    Thread Starter Active Member

    Mar 1, 2008
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    so is it true to say that the charging time and discharging time is too small?

    Thanks
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, the RC time constant is too low.

    You're somewhat limited as to the range of resistances that you can use for R1. You really want R1 to be around 10k Ohms to prevent loading of the input signal.

    I suggest that you use metal film resistors, as they do not generate as much noise as carbon or carbon film resistors.
     
  5. trytolearn

    Thread Starter Active Member

    Mar 1, 2008
    33
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    Thanks for your reply

    this application is for a photo-diode sensor and i am using 100 SMD ohms for R1, because the signal is very weak, and the R2 is 2.2M ohms very large gain :D

    Thank you again.
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You say you want to amplify 1 - 2 kHz.
    You say R2 = 2M2 and R1 = 100.
    The gain will be -R2 / R1 = 22000.
    To have this gain upto the 2 kHz the opamp must have a gainproduct of 22000 * 2000 = 44.000.000 = 44 MHz.
    Better place two amplifiers with a gain of 150 afther eachother.
    This will give you a gain of 22500 total.

    Bertus
     
  7. trytolearn

    Thread Starter Active Member

    Mar 1, 2008
    33
    0
    Hi Bertus

    Thank you for the advice.

    I will give it a go
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Along with what Bertus is saying, if the signal is really quite weak as you say it is, then you should have an initial "voltage follower" or "unity-gain buffer" stage so as to present a high-impedance input and a low-impedance output; so as to not load the input signal.

    [eta]
    Here's what I'm talking about:

    [​IMG]

    C1 isolates the DC level of the input, but passes the effects of the AC signal.
    R1/R2 are a voltage divider that establishes C2 at a voltage level of Vcc/2.
    R3 presents a high impedance, but maintains the noninverting input to the opamp at approximately Vcc/2.
    U1 is an "ideal" opamp.
    The output of the opamp is connected to the noninverting input. This causes the opamp to closely follow the voltage level that is on the noninverting input. Since gain is unity (=1), the opamp will be operating with it's rated bandwidth available.
     
    Last edited: Apr 10, 2010
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