Why does sine wav to cap center the trace?

Discussion in 'General Electronics Chat' started by querylous, Nov 24, 2014.

  1. querylous

    Thread Starter New Member

    Nov 24, 2014
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    Hi there- I'm a general electronics hobbyist and experimenter (not a very advanced one but a very curious one!). Yesterday I noticed that when I feed a 0-2 V sine wave in to a capacitor, and trace it on the other side of the cap, that over time the wave recenters itself over 0, becoming a -1 to +1 (approx) sine wave. Small cap happens instantly, but bigger cap I can see it happening over time... I'll attach a screen shot of the beginning, where it's slowly dropping lower. This is 50v 22uF cap. I've googled this all over and thought about it all day (I'm sure it has to do with the cap charging; but, why does it even charge (since I'm feeding it AC shouldn't it charge/discharge constantly?)). And once the cap is charged, why is it going up and down across 0 instead of just continuing to go 0-2? The sine wave generator and scope are one in the same device connected to my computer via USB. Thanks! Chris
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,140
    1,790
    The series coupling cap eliminates any DC component the signal might have.
     
  3. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Congratulations, you have just found out how the "coupling" capacitor works!

    The coupling capacitor 'couples' an AC signal with a DC offset to another point in the circuit with perhaps another DC offset so that the AC signal can be transferred from one place to the other without upsetting the DC bias in the circuit.

    When you connect the circuit, the cap gets subject to two different excitation signals: one is the AC signal and the other appears as a step change in DC level. The cap has to have time to react to the new DC level, because of the RC time constant of the circuit. The time constant comes from the value of the cap and the impedances of both the driver signal generator and the load impedance. So with a bigger cap it will take longer to charge up to the DC level.

    In a typical audio circuit, one side of the cap might be at 2vdc while the other side is at 4vdc, but the cap is used to pass the AC part of the signal from point A to point B even though the DC levels are not the same at those places in the circuit.

    For a scope that has "AC Coupling", the scope will insert a capacitor in series with the input signal so that you only view the AC part of the signal. That's the same thing happening there but the cap is built right into the scope. It's invaluable for looking at very small AC signals.
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,440
    3,361
    Whatever you connect after the capacitor (such as your scope probe or scope input) has resistance.
    Together, the capacitance C and the resistance R create a high pass filter. It blocks DC and low frequency signals and passes high frequency signals.

    The time constant tau is equal to R x C, the units being seconds. For example 1Ω x 1F = 1s.

    Let us suppose that the capacitance is 1μF and the scope has an input resistance of 1MΩ. Then R x C = 1M x 1μ = 1s.
    i.e. a signal with any DC bias will drift to an average 0VDC after about 1 second.
    (Actually, it gets to within 37% of 0VDC in one time constant. So allow about 3 times the time constant to get to close to 0VDC.)

    The larger the capacitance, the longer the time constant. A 22μF cap will take 22 seconds with the same 1MΩ load,
    while a 1nF cap will take 1ms.
     
    Last edited: Nov 25, 2014
  5. crutschow

    Expert

    Mar 14, 2008
    13,006
    3,232
    The short answer was stated by Papabravo. A capacitor cannot pass DC so any AC waveform that passes through a capacitor to a resistive load to ground has to have an average value of 0V at the output.

    Thus, for example, if you put a pulse train through the capacitor and vary the duty-cycle of the train then you will see the peaks of the pulses drift up or down so that the output average value always settles to 0V.
     
  6. querylous

    Thread Starter New Member

    Nov 24, 2014
    2
    0
    Thank you all for your excellent responses! I'm reading more about this, thinking more about it, and doing more experiments, and then I might get back to you with some further questions!
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    When you come back, it will be helpful to post a schematic (even if just a quick sketch in Paint) of the circuit you are using so that we can provide answers that are more specific to your particular setup.
     
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