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Old 3rd November 2010, 05:44 AM   #1
RobertE is offline RobertE  United States
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Default question about this schematic ?

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.

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Old 3rd November 2010, 06:19 AM   #2
RobertE is offline RobertE  United States
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Ok, the RC network is a high pass filter, BUT why is it needed in only some cases, and what are these cases? Also, what would be the difference if it was done with a 22k and 10u ?

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Old 3rd November 2010, 01:44 PM   #3
Leon08 is offline Leon08  Germany
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Quote:
Also, what would be the difference if it was done with a 22k and 10u ?
This would be basicly the same in regard of frequency, but it would be a lighter load.
This filter is needed for blocking some unwanted frequencies.
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Old 3rd November 2010, 01:44 PM   #4
DF96 is offline DF96  England
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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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Old 3rd November 2010, 01:54 PM   #5
macboy is offline macboy  Canada
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The IC has a single supply. My guess is that the output and input pins are biased to about 1/2 of V+. As already stated, the capacitors block this DC bias to allow ground-referenced signals to be input and output.
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Old 3rd November 2010, 02:27 PM   #6
RobertE is offline RobertE  United States
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Thanks everyone !! I think I am getting this...

On the "lighter load using the 10u" aspect, what difference does this make? When driving the amp by the words it would seem to be a better situation???
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Old 3rd November 2010, 02:58 PM   #7
Leon08 is offline Leon08  Germany
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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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Old 3rd November 2010, 03:00 PM   #8
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That schematic is labelled "TEST CIRCUIT" ie how the chip may be tested. It does not necessarily have any value as an APPLICATION circuit!
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Old 3rd November 2010, 03:18 PM   #9
RobertE is offline RobertE  United States
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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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Old 3rd November 2010, 03:33 PM   #10
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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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