question about this schematic ?

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RobertE

Member
2010-09-09 3:11 am
  • #1
    In the attached schematic, why are there two Vo (voltage out)?

    I am assuming the top one (without CR) is used if directly coupling to another circuit, e.g. an op amp, and the latter is used if "going out", say to output jacks that then would go into an amp. If this is true, what is the suggested CR doing in this case ?

    Btw, the schematic is for a line level audio switch.

    Thanks! This is one of the best sites on the Internet !!!!
     

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    DF96

    R.I.P.
    2007-05-04 11:17 pm
    myweb.tiscali.co.uk
  • #4
    The capacitor blocks DC and passes AC. The resistor grounds the output and provides bias for the capacitor, as electrolytic capacitors need a suitable voltage across them to maintain the dielectric. Whether you need this network depends on what comes next. Generally, external signals should be referenced to ground but internal signals do not need to be.

    This combination of R and C forms a high-pass filter and so sets a low frequency limit. The corner frequency is given by f = 1/(2 pi R C)
     
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    Leon08

    Banned
    2010-09-06 7:22 pm
    at Home
  • #7
    O.K imagine that this 1k resistor is a short to ground.A short has no resistance at all, so it will be a very "heavy" load for amplifier.Of course it's better for him if there is a big resistance to ground.Then dependent on what circuit you will connect it will have also some impact on the load.
     
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    RobertE

    Member
    2010-09-09 3:11 am
  • #9
    cliff, it;s the only circuit provided in the datasheet - it works - just trying to understand it (obviously the switches, etc. are for testing and not applicable to my app).

    Leon, then why not use a huge resistor and have a very light load? What are the ramifications of this ? Do you want a heavy load to the amp? By wording I would think not, but I don't understand it :)
     
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    cliffforrest

    Member
    2004-09-05 4:39 pm
    South Coast, UK
  • #10
    Source and load resistance levels are compromise - like most of engineering!

    A low > zero output impedance is "good" but very difficult for a tube amplifier and has to be current limit protected otherwise. In practice, up to 1K to 5K is reasonable and easily done.

    But now there is a finite Zout, the roll off of turn-over frequency with a series C has to be considered!

    High Zin seems good but renders the amp much more likely to hum and noise pickup, requiring screening etc. In practice, 10K to 100K is livable with.

    So a 1K out, 50K in would be pretty ideal, other things being equal.
     
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    Leon08

    Banned
    2010-09-06 7:22 pm
    at Home
  • #12
    By wording I would think not, but I don't understand it :)
    Click to expand...
    What do you not understand?
    That's very easy.The current always goes the way of the smallest resistence.So if put a 1k resistor to ground then some current, not all, will go through him to ground, but if you put a 1R resistor to ground then almost all current will go to ground and it will be like a short.
     
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    RobertE

    Member
    2010-09-09 3:11 am
  • #14
    Leon08 said:
    What do you not understand?
    That's very easy.The current always goes the way of the smallest resistence.So if put a 1k resistor to ground then some current, not all, will go through him to ground, but if you put a 1R resistor to ground then almost all current will go to ground and it will be like a short.
    Click to expand...

    I meant why not put a 1M ohm, so all the current goes to the amp? Is it just because the capacitor would be too small ?
     
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    macboy

    Member
    2003-10-22 6:57 pm
    Ottawa, Canada
  • #16
    RobertE said:
    I meant why not put a 1M ohm, so all the current goes to the amp? Is it just because the capacitor would be too small ?
    Click to expand...

    Then only a very small current will flow through the resistor. This could be less than the leakage current through the capacitor, or less than the leakage current of the input pin of the amplifier (or opamp) following this circuit. That can and will cause a DC offset which, if you have no other DC blocking downstream, can blow your speakers, overheat your amp, etc.

    Besides the possiblility of a steady state DC offset, you would likely get "thumps" when switching inputs, etc. This due to the extremely low cutoff frequency of the high-pass filter formed by the R and C. Fc=1/(2*pi*R*C). Even with the values suggested (220uF and 1kOhm) you get a transient lasting about 1/2 second before falling to 10% of the initial value. With a 1 Meg resistor (and 220 uF), it would last 1000 times as long... about 8 minutes. I'm not kidding.

    Any value is a compromise as already pointed out.
     
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    RobertE

    Member
    2010-09-09 3:11 am
  • #18
    Leon08 said:
    But why he needs this filter at all? An amplifier has already an iput filter.
    Click to expand...

    That's kind of my question... Wouldn't an amp always have dc blocking caps??? Why wouldn't it?

    I've tried the circuit with and without it, and I don't see a huge difference. I think without it some dc is getting processed by the amp because at high volumes the bass handling seems a lot worse.
     
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