Naive question about sound levels ( technical )

[gilesgoat]

Ok .. "I know almost not much apart this I found" : https://en.wikipedia.org/wiki/Line_level

So "suppose I make something that makes an audio sound" .. WHAT should be the Vpp audio level "on the Line Out" let's say ?

Also "does it make diffence" ( and what kind ) if let's say the audio signal comes out ( let's say for example I pick up the +1.095V thing ) swinging betwen 0V and +1.095V OR between -1.095V and +1.095V ?

I mean yes in the first case you have effectively 1/2 amplitude but "what happens in terms of equipement" if the signal is not swinging between -/+ but only 0/+ ?

Sould not be any input "DC decoupled" ?

Or not ?

I mean a signal swinging -1/+1 V is "equivalent" of a signal swinging 0/+2V once is DC decoupled.

How does it work ?

[Dio]

Lots of answers. I recommend Doug Self's "Small Signal Audio Design" for all of this discussed quite exhaustively. It's a good read too, although $$$. His poweramp book is also excellent, the crossover one is not so fun.

Very roughly: 'consumer' line level is likely 1V peak-to-peak unbalanced. Balanced outs on pro gear are likely to be 2V peak-to-peak since the two outputs are simply driven in antiphase. That however is the 0dB level and most gear will happily drive up to within a couple of volts of the rails if requested ("headroom, dude") - probably +-15V or 18V on pro gear. Both nominally should have no DC component (so -ve-to-+ve) but on unbalanced gear there is no guarantee of a blocking capacitor.

Brief summary: use a nonpolarised DC blocking capacitor on inputs and rebias yourself. 35V if you are really paranoid and/or connecting to pro gear due to the aforementioned potential 18V rails. Clamp zeners if you are further paranoid.

A minor note is also that ground levels are likely to be different on modern gear - sometimes very different (like 100V different) as switchmode gear often results in their 0V actually being half the supply voltage due to leakthrough capacitance in the input transformer. This is not dangerous since it comes from a very high impedance which naturally current limits it to microamps but it can quite noticeable to the touch. Doesn't usually require any design cleverness apart from the usual ground loop concerns though.

[gilesgoat]

I'll go for +1/-1 balanced ( symmetric vs 0 ).

[rowan]

I always thought line level was 0.775V. Anyway I checked my book, what it actually says is

"Line level signals are normally quoted with reference to a level of 0.775V, given the designation dBu [dBv in the US]".
It then goes on to give some example of typical levels found in kit, such as consumer hifi components being -10dBu, professional kit being +4dBu and mics being -60dBu.

What it doesn't specifically say, is if this is peak to peak or RMS but I suspect RMS.

Now if you go balanced.. Is it opamps or transformers?


Also: You should avoid creating a DC offset you nasty pasty

[gilesgoat]

I always thought line level was 0.775V. Anyway I checked my book, what it actually says is

"Line level signals are normally quoted with reference to a level of 0.775V, given the designation dBu [dBv in the US]".
It then goes on to give some example of typical levels found in kit, such as consumer hifi components being -10dBu, professional kit being +4dBu and mics being -60dBu.

What it doesn't specifically say, is if this is peak to peak or RMS but I suspect RMS.

Now if you go balanced.. Is it opamps or transformers?


Also: You should avoid creating a DC offset you nasty pasty

Sure, the only "problem" is that to make it DC free it needs DUAL power supply, otherwise it's impossible.

[Dio]

You don't need dual-rail. It's DC-free if you pass it through a blocking capacitor, as the laws of physics demand (the capacitor cannot pass any fixed voltage, it charges up to the nominal level and then the electrons flow in and out. Yes, this adds a time constant, which is why you make this capacitor a moderate-sized electrolytic). Capacitors and inductors store energy and the voltages at their terminals can move outside the rails.

You know how a charge-pump works, right? Same thing - the capacitor decouples the DC levels on either side, and the movement of electrons in and out of the capacitor allow the voltage to move outside the rail.

0.775V RMS == 0 dBu/dBv.

[gilesgoat]

You don't need dual-rail. It's DC-free if you pass it through a blocking capacitor, as the laws of physics demand (the capacitor cannot pass any fixed voltage, it charges up to the nominal level and then the electrons flow in and out. Yes, this adds a time constant, which is why you make this capacitor a moderate-sized electrolytic). Capacitors and inductors store energy and the voltages at their terminals can move outside the rails.

You know how a charge-pump works, right? Same thing - the capacitor decouples the DC levels on either side, and the movement of electrons in and out of the capacitor allow the voltage to move outside the rail.

0.775V RMS == 0 dBu/dBv.

I tought about it ... I'll ask to "the expert" ( my bro ) ...

Yes in theory "add a 10uF cap" ( possibly unpolarized ) and you should be done .. but are you really ?

[edit] - TESTED / WRONG !

YES one thing is true, you put the cap, "you eliminate the DC component" .. *BUT* ... buy your signal, no matter what (*), is still oscillating between 0 Vdc and +vPeak and NOT -vPeak/2 +Vpeak/2

To do that, the only way you SHOULD have, at the input a summer thing that SUMS that stuff with -Vpeak/2, then it would work.

If you don't believe me, I just simulated the whole thing with LTSpice

"No a simple cap won't do what you think should".

Also you create a "differentiator" so at all effects a HiPass filter in doing so, that may lead to some distortion.

[Dio]

There's no differentiator if the time constant is large enough. 30uF is sufficient in most cases.

As to the DC conditions - well, something has to set them somewhere. Any resistor, even a high value one, will peg the point to whatever the other end of the resistor is at. If you imagine the completely isolated case, where one output with a capacitor feeds another output with a capacitor and there is no DC bias, then you just get current flowing down the cable of opposite polarity, right? Like two capacitors connected back to back.