Not getting how cut-off, Xc=R, and dB relate

Discussion in 'General Electronics Chat' started by dfro, Jan 7, 2011.

  1. dfro

    Thread Starter Active Member

    Feb 6, 2006
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    I am trying to get clear how the cut-off frequency, Xc=R in a filter, and -3 dB gain loss relate.

    This page (http://www.allaboutcircuits.com/vol_2/chpt_8/2.html) says that the -3dB cut-off frequency is where the Vout is .707 the Vin. But, then it also says that it is where Xc=R in the RC network.

    I am not getting this. Doesn't Xc=R make a voltage divider that drops the Vout to .5, not .707?

    I think I need to understand this before I try to understand the -3dB connection.

    Help would be appreciated,
    Dave
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    An RC filter that has the capacitance reactance equal to the resistance drops 0.707 times (-3dB) instead of 0.5 times (-6dB) because it has a phase-shift.
     
  3. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    I never did get all that down pat. Ten log this and 20 log that? I wind up deriving which one applies, every time I get into that kind of circuit. So...don't feel stupid if it's confusing. It bothers me, too.
     
  4. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    ...thats not saying much...

    ;)

    JUST KIDDING

    AWWW BOY.. I dont know why I make myself laugh so hard... ;)

    phew...

    ok.
     
  5. bertus

    Administrator

    Apr 5, 2008
    15,648
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    Hello,

    Take a look at this page of the eBook:
    http://www.allaboutcircuits.com/vol_2/chpt_4/3.html

    When the Xc=R the two vectors have the same amplitude.
    Due to the 90° phase shift between the R and C , the voltage accross the R (Er) or C (Ec) is 1/SQR(2) from the applied voltage (Et).

    Bertus
     
    Last edited: Jan 7, 2011
  6. #12

    Expert

    Nov 30, 2010
    16,284
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    This means war, you know!

    Just kidding. Everybody looks up stuff they have used in the past and everybody reads things they have read before. You can't keep it all in your head.
     
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  7. #12

    Expert

    Nov 30, 2010
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    Hey, Bertus, that page has the RC figures labeled, "capacitor inductor circuit". You might want to change the label.

    I left a note in Feedback and Suggestions.
     
    Last edited: Jan 7, 2011
  8. dfro

    Thread Starter Active Member

    Feb 6, 2006
    37
    0
    Not that phase shift stuff!! I had a hunch it was that.

    I will hit the books (webpages) for a while before asking more questions on this.

    A reference to the 'Series resistor-capacitor circuits' section and some reminders about phase shift would be a good thing to add on the 'Low pass filter' page. Just a suggestion. In fact a section that really fleshes out the Xc=R, Vout=.707*Vin, -3dB, f(-3db) cut-off properties would be a great addition to the ac book.

    I am currently learning how to design a tube power amp, and this is THE language of tube amp design. I am grappling with ac equivalent circuits, thevenin equivalent circuits, etc. so I can make very accurate computations of how one stage of the amp is tied to the next through coupling networks. I actually think this would fit in the book well - how one section of a circuit is coupled to the next through capacitors/resistors, low output impedance, high input impedance, etc.

    Every beginning electronics book I have read shows examples of simple RC or RL filters, and Thevenin equivalent circuits. But then I try to understand tube amp circuits and transistor circuits, and I do not know what to make of all the extra resistors before and after the RC filter - how to pick it apart. I know the surrounding circuit is effecting a simple RC filter, but I cannot find an explanation of how to analyze it.

    I think I am getting close to understanding this with pdf page 144 (scanned book pg 131) of BasicTheoryAndApplicationOfElectronTubes-Army1952.pdf:

    http://www.archive.org/details/BasicTheoryAndApplicationOfElectronTubes

    (Please, disregard these comment if these things are already in the book.)

    Thanks,
    Dave
     
    Last edited: Jan 10, 2011
  9. dfro

    Thread Starter Active Member

    Feb 6, 2006
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    I realize I did not understand the difference between u (mu) and Av.

    Amplification (Av) = change in anode voltage/change in grid voltage along a loadline.
    The value of gain predicted for this stage uses the loadline which is based on a chosen RL value, and already includes the attenuation caused by the potential divider formed by rp and RL.

    This is done by drawing a loadline on an anode characteristics graph and charting the changes along this sloped line.
    Av is predicted along a loadline, WHERE THE ANODE CURRENT CHANGES.

    u (mu), on the other hand, is called the amplification factor.
    u = change in anode voltage/change in grid voltage WITH ANODE CURRENT HELD CONSTANT.

    The formula that relates these two variables is:

    Av = u*(RL/(rp+RL)), where RL/(rp+RL) is the attenuation caused by the voltage divider.

    Therefore,

    If, rp=68k and RL=100k

    Av = 20(100,000/(68,000+100,000)) = 11.9 gain
     
    Last edited: Jan 11, 2011
  10. dfro

    Thread Starter Active Member

    Feb 6, 2006
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