Removing DC offset from a signal

Discussion in 'The Projects Forum' started by JingleJoe, Jan 27, 2012.

  1. JingleJoe

    Thread Starter Member

    Jul 23, 2011
    185
    10
    I have been working on some simple sound making devices recently (4000 series cmos, lunetta synths) Now I know what you're going to say "please provide full circuit schematics etc etc" Unfortunately for you I don't have any, this is just for my devices in general, their output is about 0 to 10 volts, thats rail to rail.


    Short version: How should one combine a voltage divider (for attenuation) and capacitor in order to remove any DC from a signal?


    Long version: I intend to connect my sound machines to amplifiers and mixers but to avoid discombobulating them I intend to capacitively couple their outputs.
    However the output voltage of many of my inventions is almost rail to rail, which is just too much for most amplifiers; they prefer around 100mV.
    To remedy this I have used a voltage divider, if input voltage is about 10V, then a 10k resistor and 100 ohm resistor form a good enough divider to get the voltage down to a level my amplifiers can handle.
    However if the output voltage has some DC offset (which it often does) this is also transfered out with the divider.
    In what arrangement would one combine a voltage divider and capacitor to get rid of this DC offset but not filter the sound too much?
    I was considering perhaps as a high pass filter but then you have a resistor effectively in series with the capacitor which will affect it's reactance, but should one merely adjust the capacitor untill suitable reactance at 20Hz is found?
    I have found that a capacitor after the divider sort of works but often there is too much filtering.
    What would you chaps suggest?
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    You should add a series capacitor (in the signal line, NOT in the the common). A resistor from the amplifier end of the capacitor to common will ensure that the capacitor charges, to avoid a "pop" on connecting.

    The resistor can be a few times the input impedance of the intended amplifier, orb maybe 100kΩ. The capacitor needs to be big enough to couple bass frequencies, and again this will depend on the load* impedance. For 1kΩ, about 10μF, for 10kΩ...

    *Actually, the sum of the source and load impedances, but you are proposing a divider with a 100Ω lower resistor: that's a lot lower than most amplifier input impedances.

    The -3dB frequency is given by f = 1/(2*pi*(Rsource + Rload) * C).

    Thus if you know the -3dB frequency you can find C: C = 1/(2*pi*f*(Rsource + Rload)).
     
  3. JingleJoe

    Thread Starter Member

    Jul 23, 2011
    185
    10
    Fantastic thankyou that was very helpful :)

    P.S. Please clarify if the attached circuit is what you suggested.
     
    Last edited: Jan 27, 2012
  4. PaulEE

    Member

    Dec 23, 2011
    423
    32
    P.S. - That resistor from amplifier side to ground is to ensure that any input bias currents from the amplifier's input terminals (if it's an op-amp) will not charge the capacitor with DC. When this happens, the output disappears because the input is pegged to a constant DC value. Usually, something like 100 K for a typical op-amp will due just fine. If you wanna get technical, you can look up the worst-case bias current in the datasheet and do the maths necessary to derive the "nominal" resistor value...or just stick 100 K in there :)
     
  5. JingleJoe

    Thread Starter Member

    Jul 23, 2011
    185
    10
    Thanks! I added a diagram above which from what has been said, I think may work. I just stuck a 100k resistor on the output :p
     
  6. PaulEE

    Member

    Dec 23, 2011
    423
    32
    Lookin' good! Let us know how it works out.
     
  7. JingleJoe

    Thread Starter Member

    Jul 23, 2011
    185
    10
    The circuit I posted above works great! Even with caps as low as 1uF there was very very little distortion below 100Hz, with 470nF there was significant distortion, despite this the DC was removed and the signal was attenuated perfectly every time :)
    I tried a capacitor before the voltage divider however this lead to significant distortion with values below 10uF.

    Thanks for the advice chaps!
     
  8. PaulEE

    Member

    Dec 23, 2011
    423
    32
    No problem!
     
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