Output voltage of media

[[email protected]]

Some pieces of audio gear have specified ranges of output voltages, other pieces of audio gear haves specified ranges of input voltages(what they expect to see coming in). All have different impedances(something I have no hope of understanding!)


My question: Is there a way to measure the RMS or the peak voltage represented by the data on CDs and the waveforms of output of vinyl records, cassettes, etc?

[PStamler]

On Thursday, February 20, 2014 5:41:27 PM UTC-6, wrote:
> Some pieces of audio gear have specified ranges of output voltages, other pieces of audio gear haves specified ranges of input voltages(what they expect to see coming in). All have different impedances(something I have no hope of understanding!)
>
> My question: Is there a way to measure the RMS or the peak voltage represented by the data on CDs and the waveforms of output of vinyl records, cassettes, etc?

No, because CDs don'e put out any voltage. LPs don't put out any voltage (except static discharge), and cassettes don't put out any voltage. CD *players* put out voltage which used to be pretty standardized (0dBFS = 2V peak) but is now all over the map. LP *players* (turntables+arms+cartridges) used to put out about 5mV for a 5cm/sec velocity in the groove, but now they're all over the map too; that output will go through a preamplifier which boossts the signal and (usually) applies RIAA compensation; the compensation is specified but the degree of boost isn't, so even with a specified cartridge your preamp's output is not predictable. Cassette players have never had anything remotely resembling an output level standard.

[[email protected]]

I understand about the CD player output itself - 2V. Is that 2V per channel, or 1V(left) plus 1V(right)?


My question is - if CD #1 consists of material mastered at a low RMS with peaks every few seconds, VS CD #2, mastered very high RMS with constant full-scale peaks - how does that translate to the voltage the player sends to my receiver/mixer/other equipt?


Ditto for cassettes, Vinyl.

[John Williamson]

On 21/02/2014 11:40, wrote:
> I understand about the CD player output itself - 2V. Is that 2V per channel, or 1V(left) plus 1V(right)?
>
The standard, insofar as there is one is that the signal from the line
output of each output is 1 Volt RMS, aka +4dB, considering the reference
output to be 1 milliwatt into a 600 ohm impedance. Domestic equipment
tends to have an output set to -10dB of the reference power.
>
> My question is - if CD #1 consists of material mastered at a low RMS with peaks every few seconds, VS CD #2, mastered very high RMS with constant full-scale peaks - how does that translate to the voltage the player sends to my receiver/mixer/other equipt?
>
Any reference refers to output at full scale on the recorded material,
so that would be when the peak value of the waveform on the CD is at
full scale.

> Ditto for cassettes, Vinyl.
>
The reference for tape is normally given as a known magnetic field
intensity at the head surface, and for vinyl as a given stylus lateral
velocity, which are given in the standards, which this bloomin'
connection won't let me look up.

--
Tciao for Now!

John.

[[email protected]]

John Williamson wrote: "The standard, insofar as there is one is that the signal from the line output of each output is 1 Volt RMS, aka +4dB, considering the reference output to be 1 milliwatt into a 600 ohm impedance. Domestic equipment "


So by "each output" do you mean 1volt RMS per channel of a stereo pair(of RCAs)? Or 1 volt RMS combined(500mV rms per ch)?

[Luxey]

On Friday, 21 February 2014 13:08:15 UTC+1, wrote:
> John Williamson wrote: "The standard, insofar as there is one is that the signal from the line output of each output is 1 Volt RMS, aka +4dB, considering the reference output to be 1 milliwatt into a 600 ohm impedance. Domestic equipment "
>
>
>
>
>
> So by "each output" do you mean 1volt RMS per channel of a stereo pair(of RCAs)? Or 1 volt RMS combined(500mV rms per ch)?

per channel. Also difference btw +4 and -10 is not 14, because one is about Granny Smith, while the other is about Golden Delicious.

[[email protected]]

Luxey wrote: "per channel."


Thank youuuuuuu! :)


Just confirming what I did several months ago. I hooked my multimeter up to the opposite end of the left-channel RCA out cable from my Sony CD changer. This model has two sets of outs: fixed and variable(controlled by headphone vol. knob on the front).


I lowered the headphone knob fully clockwise, inserted

"10045-2-T SHEFFIELD / AUTOSOUND 2000 TEST CD" into the changer, track #8: "A 1 kHz sine wave at 0 dBFS", from page 4 of the booklet.


With that track on repeat, I slowly turned the headphone knob clockwise until the multi- read exactly "1.0" Vac. On this particular Sony changer, this reading resulted in the headphone vol. knob pointed approx. at 1 o'clock.

I penciled a mark onto the case to refer to if the knob was ever bumped.

I took a 500mV reading prior to all of this, the knob lamding at 11 o'clock.. I never did read the knob fully clockwise but would guess that position would equal full-out(approx level of the fixed outputs). I'm guessing full out using the same test tone is somewhere between 2-3V/ch.


Many of JVC's receivers from the time mine was built specify input sensitivity as "200mv @ 47k-ohms" so I'm assuming that's RMS?


I must say I do like the sound from this changer with the var. output at .5V better than I do at 1V or full/fixed out. Sure, I have to turn the receiver volume up higher, but the sound is more open, balanced, and alive.

[Scott Dorsey]

In article >,
> wrote:
>I understand about the CD player output itself - 2V. Is that 2V per channel, or 1V(left) plus 1V(right)?
>
>
>My question is - if CD #1 consists of material mastered at a low RMS with peaks every few seconds, VS CD #2, mastered very high RMS with constant full-scale peaks - how does that translate to the voltage the player sends to my receiver/mixer/other equipt?

Well... think about it. 0dBFS corresponds to a certain output voltage... so
if you increase the peak value on the disc, it increases the peak voltage
of the output. The voltage can never go over that of the 0dBFS point.
Compression and limiting will increase the average voltage but it will not
increase the peak voltage over that point.

You might be better off asking questions like this in rec.audio.tech.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[[email protected]]

Scott Dorsey wrote: "You might be better off asking questions like this in rec.audio.tech. "


There's a reason why I post these questions here instead of in R.A.T. I think you can figure that out(!)


And thanks to you and everyone else for their explanations.


So the signal recorded on a CD(or other media) can never exceed the maximum peak output voltage of the device it is played in. Got it!


I guess then the real issue is the ever-rising RMS levels of commercial albums over the past 15 or so years. That's what I should be measuring out of my player next.

[Luxey]

петак, 21. фебруар 2014. 15.53.57 UTC+1, је написао/ла:
> Scott Dorsey wrote: "You might be better off asking questions like this in rec.audio.tech. "
>
>
>
>
>
> There's a reason why I post these questions here instead of in R.A.T. I think you can figure that out(!)
>
>
>
>
>
> And thanks to you and everyone else for their explanations.
>
>
>
>
>
> So the signal recorded on a CD(or other media) can never exceed the maximum peak output voltage of the device it is played in. Got it!
>
>
>
>
>
> I guess then the real issue is the ever-rising RMS levels of commercial albums over the past 15 or so years. That's what I should be measuring out of my player next.

Should I post thi link to that Zappa clip for the 2nd time today?

[Tom McCreadie]

, :

>I guess then the real issue is the ever-rising RMS levels of commercial albums over
> the past 15 or so years. That's what I should be measuring out of my player next.

Why not just rip your commercial CD to a wav file on your PC, then analyse that
wav file with a DAW such as Audition. That gives you everything you need, viz:

For each L & R channel, the dB values, relative to odBFS, for:
- Peak Amplitude
- Minimum RMS Power; Maximum RMS Power
- Average RMS Power; Total RMS Power.

You could probably do the entire operation in 15 minutes.

[Norbert Hahn[_2_]]

(Scott Dorsey) wrote:

>In article >,
> > wrote:
>>I understand about the CD player output itself - 2V. Is that 2V per channel, or 1V(left) plus 1V(right)?
>>
>>
>>My question is - if CD #1 consists of material mastered at a low RMS with peaks every few seconds, VS CD #2, mastered very high RMS with constant full-scale peaks - how does that translate to the voltage the player sends to my receiver/mixer/other equipt?
>
>Well... think about it. 0dBFS corresponds to a certain output voltage... so
>if you increase the peak value on the disc, it increases the peak voltage
>of the output. The voltage can never go over that of the 0dBFS point.

In theory output voltage can go over 0dB:
http://www.indexcom.com/tech/0dBFS+/
shows an example. However, nobody will normalize a 11.025kHz signal.

Norbert

[John Williamson]

On 21/02/2014 13:57, wrote:
> Many of JVC's receivers from the time mine was built specify input sensitivity as "200mv @ 47k-ohms" so I'm assuming that's RMS?
>
>
> I must say I do like the sound from this changer with the var. output at .5V better than I do at 1V or full/fixed out. Sure, I have to turn the receiver volume up higher, but the sound is more open, balanced, and alive.
>
That'll be because you're not massively overloading the receiver input
stages. You may find it sounds even better with the signal from the CD
player set to 200mv RMS which is the specified input signal level for
your unit. This should give the specified output power from your
receiver (Assuming your speakers are of the correct impedance) with the
volume control set to maximum.

--
Tciao for Now!

John.

[[email protected]]

On Friday, February 21, 2014 9:53:57 AM UTC-5, wrote:
> Scott Dorsey wrote: "You might be better off asking questions like this in rec.audio.tech. " There's a reason why I post these questions here instead of in R.A.T. I think you can figure that out(!) And thanks to you and everyone else for their explanations. So the signal recorded on a CD(or other media) can never exceed the maximum peak output voltage of the device it is played in. Got it! I guess then the real issue is the ever-rising RMS levels of commercial albums over the past 15 or so years. That's what I should be measuring out of my player next.

right, the peaks are about all the same...
the difference is...CDs that sound loud are hitting those peaks maybe 100's of times per second, even when the singer takes a breath might be hitting a peak, CDs that sound softer may hit the peak once a second during a drum hit for example , and CDs that really capture the full range of sound might hit the peak once every few seconds. Its sometimes called "density". Search on "loudness war" and Durroughs loudness meter.

Mark

[[email protected]]

4:45 wrote: "
right, the peaks are about all the same...
the difference is...CDs that sound loud are hitting those peaks maybe 100's of times per second, even when the singer takes a breath might be hitting a peak, CDs that sound softer may hit the peak once a second during a drum hit for example , and CDs that really capture the full range of sound might hit the peak once every few seconds. Its sometimes called "density". Search on "loudness war" and Durroughs loudness meter.

Mark "


It is the former of the two types of CD releases(the 'hotter' ones you described) that concerns me when it comes to setting/attenuating the output of my changer.


Tracks from Black Eyed Peas' "Elephunk" register 0.3 - 0.5V on the multimeter, while tracks from a variety of 1980s era CDS return 0.1 - 0.4V.


It is for reasons like that Elephunk that I am considering of lowering the changer output to .5V per channel.

[[email protected]]

Dayam, Mark: Those Durrough meters are EXPENSIIIIIVE!! lol

But well laid out, esp. the ones that show rms, peak, clip all in one display.

I might just have to settle for one of their plug-ins. :)

[Mike Rivers[_2_]]

On 2/20/2014 6:41 PM, wrote:

> Since I am lucky to have variable outputs on this changer, would you
> deem it necessary to purchase a pair of RCA pads, and just plug those
> into the receiver end of my CD connection instead - and just go fixed
out?

Either way, if you play more than one CD (unless they're all compressed
for maximum loudness) you're going to have to adjust the volume at some
place just for your listening comfort. I'd suggest just padding down the
fixed output of the CD player so that a CD at maximum level won't clip
the input of the receiver.

If you want to make some measurements, create a WAV file with 0 dBFS
tone (lots of programs for doing this, including many DAWs) and burn a
CD with it. Play the CD, measure the fixed output level with your
voltmeter and note it. Then connect the variable output to your receiver
input and turn up the level until you hear clipping. It'll be pretty
obvious. Back it off just a tad, and measure that output level. The
ratio of the fixed to the variable level (just before clipping) will
tell you how much loss the pad needs to have.

You can calculate the ratio in dB if you want to try to find an
off-the-shelf in-line pad that's close enough to use, or you can build
an attenuator from a pair of resistors, or you could just nail the
variable output control in place and use the variable output.



--
For a good time, visit http://mikeriversaudio.wordpress.com

[[email protected]]

On Saturday, February 22, 2014 9:24:35 AM UTC-5, Mike Rivers wrote:
> On 2/20/2014 6:41 PM, com wrote:
>
>
>
> > Since I am lucky to have variable outputs on this changer, would you
>
> > deem it necessary to purchase a pair of RCA pads, and just plug those
>
> > into the receiver end of my CD connection instead - and just go fixed
>
> out?
>
>
>
> Either way, if you play more than one CD (unless they're all compressed
>
> for maximum loudness) you're going to have to adjust the volume at some
>
> place just for your listening comfort. I'd suggest just padding down the
>
> fixed output of the CD player so that a CD at maximum level won't clip
>
> the input of the receiver.
>
>
>
> If you want to make some measurements, create a WAV file with 0 dBFS
>
> tone (lots of programs for doing this, including many DAWs) and burn a
>
> CD with it. Play the CD, measure the fixed output level with your
>
> voltmeter and note it. Then connect the variable output to your receiver
>
> input and turn up the level until you hear clipping. It'll be pretty
>
> obvious. Back it off just a tad, and measure that output level. The
>
> ratio of the fixed to the variable level (just before clipping) will
>
> tell you how much loss the pad needs to have.
>
>
>
> You can calculate the ratio in dB if you want to try to find an
>
> off-the-shelf in-line pad that's close enough to use, or you can build
>
> an attenuator from a pair of resistors, or you could just nail the
>
> variable output control in place and use the variable output.
>
>
>
>
>
>
>
> --
>
> For a good time, visit http://mikeriversaudio.wordpress.com

______

Such tones already exist:

From Feb. 21 I mentioned the tone I used:

"10045-2-T SHEFFIELD / AUTOSOUND 2000 TEST CD" <snip>, track #8: "A 1 kHz sine wave at 0 dBFS", from page 4 of the booklet."

Using the multimeter, I adjusted the variable out to read 1V.

I suppose a 12dB pad as you mentioned would do the trick easily, but then I'd have to move my whole stereo two inches out from where it presently sits, as those pads are LONG! - 1.5 - 2".


I do understand about having to lower the volume for hotter modern CDs if I load them in the changer alongside older ones, but at least I can ensure, as you've state, that I'm not clipping my amp inputs.

[geoff]

On 23/02/2014 3:35 a.m., wrote:

>
> Using the multimeter, I adjusted the variable out to read 1V.
>
> I suppose a 12dB pad as you mentioned would do the trick easily, but then I'd have to move my whole stereo two inches out from where it presently sits, as those pads are LONG! - 1.5 - 2".
>
>
> I do understand about having to lower the volume for hotter modern CDs if I load them in the changer alongside older ones, but at least I can ensure, as you've state, that I'm not clipping my amp inputs.
>

All unnecessary. Just put on a hot CD, slowly wind up the volume until
your tweeters burn out, then just back up a tad !

geoff

[Sean Conolly]

> wrote in message
...
> I do understand about having to lower the volume for hotter modern CDs if
> I load them in the changer alongside older ones, but at least I can
> ensure, as you've > state, that I'm not clipping my amp inputs.

Well, there's clipping and there's distortion. Depending on the design of
the amp inputs yes you can cause noticible harmonic distortion before you
get to noticable clipping. And you can definitely start clipping the peaks
before your ears can really pick it out - it starts to sound a little
compressed or 'less open'.

Sean

[[email protected]]

On Friday, February 21, 2014 1:55:52 PM UTC-5, John Williamson wrote:
> On 21/02/2014 13:57, .com wrote:
>
> > Many of JVC's receivers from the time mine was built specify input sensitivity as "200mv @ 47k-ohms" so I'm assuming that's RMS?
>
> >
>
> >
>
> > I must say I do like the sound from this changer with the var. output at .5V better than I do at 1V or full/fixed out. Sure, I have to turn the receiver volume up higher, but the sound is more open, balanced, and alive.
>
> >
>
> That'll be because you're not massively overloading the receiver input
>
> stages. You may find it sounds even better with the signal from the CD
>
> player set to 200mv RMS which is the specified input signal level for
>
> your unit. This should give the specified output power from your
>
> receiver (Assuming your speakers are of the correct impedance) with the
>
> volume control set to maximum.
>
>
>
> --
>
> Tciao for Now!
>
>
>
> John.
_______________


Speakers are not a problem. This receiver specifies 8-16ohm speakers. Mine are 8.



I'd still like to know if the 12dB model of these: http://www.parts-express.com/harrison-labs-12-db-rca-line-level-audio-attenuator-pair--266-244 at the end of RCA cables coming from the player's fixed outputs would be better than using the player's volume knob regulating it's variable outputs.

[PStamler]

> I'd still like to know if the 12dB model of these: http://www.parts-express.com/harrison-labs-12-db-rca-line-level-audio-attenuator-pair--266-244 at the end of RCA cables coming from the player's fixed outputs would be better than using the player's volume knob regulating it's variable outputs.

Probably, because it's likely that there's an additional amplifying stage following the volume control, which adds a certain amount of additional distortion to the signal -- perhaps a small amount, but something nonetheless. Also a fixed, plug-in attenuator will be more reliable in he long run than a volume control, which will eventually get noisy.

Peace,
Paul

[[email protected]]

PStamler:


Well in that case I'll have to move my stereo out 2inches from the wall - the pads in that link are nearly 2" long! lol

[Peter Larsen[_3_]]

wrote:

> PStamler:

> Well in that case I'll have to move my stereo out 2inches from the
> wall - the pads in that link are nearly 2" long! lol

Or get a (short) extension type cable as just one of the solutions.

Kind regards

Peter Larsen

[Mike Rivers[_2_]]

On 2/24/2014 5:49 PM, wrote:
> Well in that case I'll have to move my stereo out 2inches from the wall - the pads in that link are nearly 2" long! lol

You could just knock a hole in the wall. Chances are the player isn't as
deep as the receiver. You could put them at that end of the cable. Or,
depending on the construction, you might be able to install a couple of
resistors inside the player's case to provide the attenuation that you
need.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[[email protected]]

Mike Rivers:


It is a carousel changer, about 1/2" deeper than receiver.


Also, the makers of these pads recommend either in-line or opposite end installation from the device you desire to attenuate. Something about resistance or loss in the cables.

[Mike Rivers[_2_]]

On 2/25/2014 9:33 AM, wrote:

> It is a carousel changer, about 1/2" deeper than receiver.

Maybe you need a deeper shelf, then. Your burdens are greater than mine.

> Also, the makers of these pads recommend either in-line or opposite
> end installation from the device you desire to attenuate. Something
> about resistance or loss in the cables.

Bah! Horse manure!



--
For a good time, visit http://mikeriversaudio.wordpress.com

[[email protected]]

Mike Rivers wrote: "Horse manure!"


So are you suggesting it's OK to put the pads on the back of the CD changer? I tought they were "directional".

I really don't know that much about such things, and so far the var. outputs sound pretty good.

[William Sommerwerck]

"Mike Rivers" wrote in message ...

On 2/25/2014 9:33 AM, wrote:

>> Also, the makers of these pads recommend either in-line or opposite
>> end installation from the device you desire to attenuate. Something
>> about resistance or loss in the cables.

> Bah! Horse manure!

Not so. The makers are correct. The best place is opposite-end (ie, near the
amplifier).

Why? Because the attenuators have a finite output impedance (that is, greater
than zero), and this impedance is probably higher than the CD player's output
impedance. You don't want that impedance driving the cable capacitance.

[Peter Larsen[_3_]]

William Sommerwerck wrote:

> "Mike Rivers" wrote in message ...

> On 2/25/2014 9:33 AM, wrote:

>>> Also, the makers of these pads recommend either in-line or opposite
>>> end installation from the device you desire to attenuate. Something
>>> about resistance or loss in the cables.

>> Bah! Horse manure!

> Not so. The makers are correct. The best place is opposite-end (ie,
> near the amplifier).

> Why? Because the attenuators have a finite output impedance (that is,
> greater than zero), and this impedance is probably higher than the CD
> player's output impedance. You don't want that impedance driving the
> cable capacitance.

A probable order of magnitude is a couple of kiloohms, that as source
feeding a cable that is 3 to 6 feet long does not seem to constitute An
Issue.

Kind regards

Peter Larsen

[Trevor]

"Peter Larsen" > wrote in message
k...
> wrote:
>> Well in that case I'll have to move my stereo out 2inches from the
>> wall - the pads in that link are nearly 2" long! lol
>
> Or get a (short) extension type cable as just one of the solutions.

30 years ago I simply soldered two 1/8W resistors as a divider into a
standard RCA connector at the end of a cable to do that. Cheap and simple.
Or open up the reciever and solder the resistors at the input connector.
Another solution for those with the capabilties is to reduce the gain of the
reciever preamp, rather than simply padding the input.

Trevor.

[Trevor]

"Peter Larsen" > wrote in message
k...
> William Sommerwerck wrote:
>> "Mike Rivers" wrote in message ...
>> On 2/25/2014 9:33 AM, wrote:
>>>> Also, the makers of these pads recommend either in-line or opposite
>>>> end installation from the device you desire to attenuate. Something
>>>> about resistance or loss in the cables.
>
>>> Bah! Horse manure!
>
>> Not so. The makers are correct. The best place is opposite-end (ie,
>> near the amplifier).
>
>> Why? Because the attenuators have a finite output impedance (that is,
>> greater than zero), and this impedance is probably higher than the CD
>> player's output impedance. You don't want that impedance driving the
>> cable capacitance.
>
> A probable order of magnitude is a couple of kiloohms, that as source
> feeding a cable that is 3 to 6 feet long does not seem to constitute An
> Issue.

Perhaps, but as a general rule it is better to attenuate the input of a
device rather than increase the output impedance of the source device with a
pad. I'd need a good reason to do it the other way simply for peace of mind.

Trevor.

[John Williamson]

On 25/02/2014 23:28, wrote:
> Mike Rivers wrote: "Horse manure!"
>
>
> So are you suggesting it's OK to put the pads on the back of the CD changer? I tought they were "directional".
>
They are in that you need to connect the player to the input and the
receiver to the output, or strange things can happen.

Mr. Sommerwerk's point about impedances is valid in theory, but in
practice, at the level of quality we're talking about here the
difference will be inaudible, and possibly just barely detectable using
a short, reasonable quality lead. It is also possible that any minor HF
rolloff caused by an impedance mismatch may make the music sound
smoother and "more analogue".

--
Tciao for Now!

John.

[[email protected]]

Thanks. I think I'll leave well enough alone and stick to the variable output setup currently employed in my system.


Reducing what gets sent to the receiver before reaching the receiver makes worlds more sense to me than reducing the sensitivity of the receiver's inputs while still feeding it a too-hot signal.


I may, however, puchase a pair of those pads for the RCAs opposite my DVD player.

[William Sommerwerck]

"John Williamson" wrote in message ...

> Mr. Sommerwerck's point about impedances is valid in theory, but in
> practice, at the level of quality we're talking about here the
> difference will be inaudible, and possibly just barely detectable using
> a short, reasonable quality lead. It is also possible that any minor HF
> rolloff caused by an impedance mismatch may make the music sound
> smoother and "more analogue".

Many years ago, Mr Sommerwerck made up a set of long cables -- using Neumann
coax -- for a Shure surround decoder. I didn't realize the Shure had a
relatively high output impedance (5K), and the cables relatively high per-foot
capacitance. The resulting sound was... well, lovely.

[Mike Rivers[_2_]]

On 2/25/2014 6:28 PM, wrote:
> So are you suggesting it's OK to put the pads on the back of the CD changer? I tought they were "directional".

Well, in a sense they are, but if they are, it's dictated by the gender
of the connectors rather than the actual circuit. There are two
resistors inside the little tube, one in series with the signal lead and
the other between the "output" end of that resistor and ground. So if
there's a male RCA plug on one end and a female one on the other end,
chances are they're built so that the shunt resistor is between the male
end and ground, and it's meant to be plugged into the receiving end. If
it's really an "in-line" attenuator, it'll go between two cables so
it'll have female RCA jacks at both ends and will probably have an arrow
pointing in the direction of signal flow.

> I really don't know that much about such things, and so far the var. outputs sound pretty good.

Contrary to the purists and skeptics, I don't see any reason why you
shouldn't use the variable outputs, as long as you adjust them once
based on your 0 dBFS test CD, and make your volume adjustments with the
volume control on the receiver. Purists and practical engineers would
call this "calibration," and once you have calibrated something you
don't want it to change.


--
For a good time, visit http://mikeriversaudio.wordpress.com

[Mike Rivers[_2_]]

On 2/26/2014 7:12 AM, wrote:

> Reducing what gets sent to the receiver before reaching the receiver
> makes worlds more sense to me than reducing the sensitivity of the
> receiver's inputs while still feeding it a too-hot signal.

By the way, I carry a pair of cables into which I built 6 dB pads in the
case with my portable recorder which has a fixed gain line input stage
(the record level control comes after this stage) which clips at +18
dBu. Since some mixing consoles can put out +24 dBu before clipping,
when I don't know the console or someone else is running it, I'll use
those cables to connect the recorder to the console's output. These have
1/4" plugs on both ends so there was plenty of room to stuff two
resistors inside the shell. And, yes, I have them marked as to which end
goes to the recorder.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[Mike Rivers[_2_]]

On 2/25/2014 8:02 PM, William Sommerwerck wrote:

> The best place is opposite-end (ie, near
> the amplifier).
>
> Why? Because the attenuators have a finite output impedance (that is,
> greater than zero), and this impedance is probably higher than the CD
> player's output impedance. You don't want that impedance driving the
> cable capacitance.

OK, you're the purist here. The shunt resistor is probably somewhere
around 1 or 2 kOhms. I wouldn't run that through 50 feet of cable, but I
wouldn't worry about 6 feet. At a conservative 35 pF/foot, that's about
38 kOhms at 20 kHz, hardly enough to worry about.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[Mike Rivers[_2_]]

On 2/26/2014 3:44 AM, Trevor wrote:
> as a general rule it is better to attenuate the input of a
> device rather than increase the output impedance of the source device with a
> pad. I'd need a good reason to do it the other way simply for peace of mind.

How about "It fits that way."

However, as I pointed out in another post, if plugging the attenuator
into the source put the shunt resistor across the output and left the
series resistor hanging out in the breeze, that would be incorrect.

You know why they make these attenuators only one way? Because if there
was more than one configuration, the customers wouldn't know what to buy
and the salesmen wouldn't know what to sell them. So they come here and
get distracted with abstract theory when there's a practical answer.


--
For a good time, visit http://mikeriversaudio.wordpress.com

[William Sommerwerck]

"Mike Rivers" wrote in message ...
On 2/25/2014 8:02 PM, William Sommerwerck wrote:

>> The best place is opposite-end (ie, near the amplifier).
>> Why? Because the attenuators have a finite output impedance (that is,
>> greater than zero), and this impedance is probably higher than the CD
>> player's output impedance. You don't want that impedance driving the
>> cable capacitance.

> OK, you're the purist here. The shunt resistor is probably somewhere around
> 1 or 2k. I wouldn't run that through 50 feet of cable, but I wouldn't worry
> about 6 feet. At a conservative 35 pF/foot, that's about 38k at 20 kHz,
> hardly enough to worry about.

One thing that puzzles me is the output impedance of the source. If it's an
emitter follower, the load should be about 10x the source impedance. What does
it run in "modern" consumer products?

[Scott Dorsey]

William Sommerwerck > wrote:
>"Mike Rivers" wrote in message ...
>On 2/25/2014 8:02 PM, William Sommerwerck wrote:
>
>>> The best place is opposite-end (ie, near the amplifier).
>>> Why? Because the attenuators have a finite output impedance (that is,
>>> greater than zero), and this impedance is probably higher than the CD
>>> player's output impedance. You don't want that impedance driving the
>>> cable capacitance.
>
>> OK, you're the purist here. The shunt resistor is probably somewhere around
>> 1 or 2k. I wouldn't run that through 50 feet of cable, but I wouldn't worry
>> about 6 feet. At a conservative 35 pF/foot, that's about 38k at 20 kHz,
>> hardly enough to worry about.
>
>One thing that puzzles me is the output impedance of the source. If it's an
>emitter follower, the load should be about 10x the source impedance. What does
>it run in "modern" consumer products?

All over the damn place, from less than 50 ohms to more than 2000.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

[Mike Rivers[_2_]]

On 2/26/2014 12:48 PM, William Sommerwerck wrote:
> One thing that puzzles me is the output impedance of the source. If it's
> an emitter follower, the load should be about 10x the source impedance.
> What does it run in "modern" consumer products?

Typically there's an op-amp with a lot of feedback driving the output
connector so the source impedance is only a couple of ohms. Usually
there's a 50 to 100 ohm resistor between the op amp and the connector,
and that's effectively the source impedance to limit the maximum output
current and keep it stable with a capacitive load. But because of the
amount of current that the op-amp can produce, the general setup is to
load it with 5 kOhms or more. A 5534 with a proper power supply can
drive 600 ohms, but most of the ICs they use in consumer gear these days
would prefer not to have to do that.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[Trevor]

"Mike Rivers" > wrote in message
...
> On 2/26/2014 3:44 AM, Trevor wrote:
>> as a general rule it is better to attenuate the input of a
>> device rather than increase the output impedance of the source device
>> with a
>> pad. I'd need a good reason to do it the other way simply for peace of
>> mind.
>
> How about "It fits that way."

You don't just connect a pad any which way it happens to fit.
(Except perhaps for a 6dB T pad which could be symmetrical.)


> However, as I pointed out in another post, if plugging the attenuator into
> the source put the shunt resistor across the output and left the series
> resistor hanging out in the breeze, that would be incorrect.

If you could then yes of course it would be.


> You know why they make these attenuators only one way?

Because an L pad can only be used one way with the attenuation as designed,
regardless of which end you place it.

Trevor.

[Trevor]

"Mike Rivers" > wrote in message
...
> On 2/25/2014 6:28 PM, wrote:
>> So are you suggesting it's OK to put the pads on the back of the CD
>> changer? I tought they were "directional".
>
> Well, in a sense they are, but if they are, it's dictated by the gender of
> the connectors rather than the actual circuit.

Rubbish. The common L pad circuit is NOT symmetrical as you correctly point
out below. The connectors are dictated by the circuit when they are M-F, not
vice versa.

>There are two resistors inside the little tube, one in series with the
>signal lead and the other between the "output" end of that resistor and
>ground. So if there's a male RCA plug on one end and a female one on the
>other end, chances are they're built so that the shunt resistor is between
>the male end and ground, and it's meant to be plugged into the receiving
>end. If it's really an "in-line" attenuator, it'll go between two cables so
>it'll have female RCA jacks at both ends and will probably have an arrow
>pointing in the direction of signal flow.

Trevor.

[geoff]

On 26/02/2014 9:44 p.m., Trevor wrote:
>
> "Peter Larsen" > wrote in message
> k...
>> William Sommerwerck wrote:
>>> "Mike Rivers" wrote in message ...
>>> On 2/25/2014 9:33 AM, wrote:
>>>>> Also, the makers of these pads recommend either in-line or opposite
>>>>> end installation from the device you desire to attenuate. Something
>>>>> about resistance or loss in the cables.
>>
>>>> Bah! Horse manure!
>>
>>> Not so. The makers are correct. The best place is opposite-end (ie,
>>> near the amplifier).
>>
>>> Why? Because the attenuators have a finite output impedance (that is,
>>> greater than zero), and this impedance is probably higher than the CD
>>> player's output impedance. You don't want that impedance driving the
>>> cable capacitance.
>>
>> A probable order of magnitude is a couple of kiloohms, that as source
>> feeding a cable that is 3 to 6 feet long does not seem to constitute An
>> Issue.
>
> Perhaps, but as a general rule it is better to attenuate the input of a
> device rather than increase the output impedance of the source device with a
> pad. I'd need a good reason to do it the other way simply for peace of mind.
>
> Trevor.
>
>


Yeah - the 'stronger' signal and lower Z being applied the length of
cable. A bit like gain-staging.

geoff

[Mike Rivers[_2_]]

On 2/27/2014 12:02 AM, Trevor wrote:
> Rubbish. The common L pad circuit is NOT symmetrical as you correctly point
> out below. The connectors are dictated by the circuit when they are M-F, not
> vice versa.

You missed my point, but you understand the issue. And I don't
understand your shorthand here. My point was that off-the-rack RCA in
line attenuators are built with the shut resistor across the male plug.
That's why that end of the adapter needs to go into the (female) jack on
the destination unit.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[Mike Rivers[_2_]]

On 2/26/2014 11:57 PM, Trevor wrote:
>> >You know why they make these attenuators only one way?
> Because an L pad can only be used one way with the attenuation as designed,
> regardless of which end you place it.

You're trying to be too studious about this. The practical side is that
off-the-shelf in-line attenuators are nearly all L-pads and they're
built with a female connector on the input side and a male connector on
the output side. If you turn it around so that you can plug it directly
into the CD player, you'll be tying the CD player output to ground
through the shunt resistor which isn't how it's designed to work.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[John Williamson]

On 27/02/2014 17:30, Mike Rivers wrote:
> On 2/26/2014 11:57 PM, Trevor wrote:
>>> >You know why they make these attenuators only one way?
>> Because an L pad can only be used one way with the attenuation as
>> designed,
>> regardless of which end you place it.
>
> You're trying to be too studious about this. The practical side is that
> off-the-shelf in-line attenuators are nearly all L-pads and they're
> built with a female connector on the input side and a male connector on
> the output side. If you turn it around so that you can plug it directly
> into the CD player, you'll be tying the CD player output to ground
> through the shunt resistor which isn't how it's designed to work.
>
Would there be any benefit in using a symmetrical Pi configuration? 600
ohms at each end to match impedances, and vary the top resistor to get
the right attenuation. Just a thought, I know it's used in RF a fair bit
using inductors and capacitors.

--
Tciao for Now!

John.

[PStamler]

On Thursday, February 27, 2014 12:16:07 PM UTC-6, John Williamson wrote:
> On 27/02/2014 17:30, Mike Rivers wrote:
>
> > On 2/26/2014 11:57 PM, Trevor wrote:
>
> >>> >You know why they make these attenuators only one way?
>
> >> Because an L pad can only be used one way with the attenuation as
>
> >> designed,
>
> >> regardless of which end you place it.
>
> >
>
> > You're trying to be too studious about this. The practical side is that
>
> > off-the-shelf in-line attenuators are nearly all L-pads and they're
>
> > built with a female connector on the input side and a male connector on
>
> > the output side. If you turn it around so that you can plug it directly
>
> > into the CD player, you'll be tying the CD player output to ground
>
> > through the shunt resistor which isn't how it's designed to work.
>
> >
>
> Would there be any benefit in using a symmetrical Pi configuration? 600
>
> ohms at each end to match impedances, and vary the top resistor to get
>
> the right attenuation. Just a thought, I know it's used in RF a fair bit
>
> using inductors and capacitors.

No real benefit, and with consumer gear, some detriment -- gear in that class isn't happy driving a 600 ohm load.

Peace,
Paul

[Mike Rivers[_2_]]

On 2/27/2014 1:16 PM, John Williamson wrote:
> Would there be any benefit in using a symmetrical Pi configuration? 600
> ohms at each end to match impedances, and vary the top resistor to get
> the right attenuation. Just a thought, I know it's used in RF a fair bit
> using inductors and capacitors.

600 ohms might be a little stiff of a load for a consumer HiFi output,
and if you made it symmetrical with input and output of 5k Ohms to make
it somewhat idiot resistant (a real idiot can find a way to screw up
anything) the source (output) impedance might be a little high. You
can't build one that's, say, 5k Ohms in and 600 Ohms out if you only
want 12 dB of attenuation. A 20 dB pad with 10k Ohms in and 600 Ohms out
is practical with real parts, but then you'd lose the idiot-resistance
since it's not symmetrical.

A 2k Ohm in-and-out 12 dB pad is possible and you might be able to get
away with that if the cable on the output side was pretty short.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[John Williamson]

On 27/02/2014 22:59, Mike Rivers wrote:
> On 2/27/2014 1:16 PM, John Williamson wrote:
>> Would there be any benefit in using a symmetrical Pi configuration? 600
>> ohms at each end to match impedances, and vary the top resistor to get
>> the right attenuation. Just a thought, I know it's used in RF a fair bit
>> using inductors and capacitors.
>
> 600 ohms might be a little stiff of a load for a consumer HiFi output,
> and if you made it symmetrical with input and output of 5k Ohms to make
> it somewhat idiot resistant (a real idiot can find a way to screw up
> anything) the source (output) impedance might be a little high. You
> can't build one that's, say, 5k Ohms in and 600 Ohms out if you only
> want 12 dB of attenuation. A 20 dB pad with 10k Ohms in and 600 Ohms out
> is practical with real parts, but then you'd lose the idiot-resistance
> since it's not symmetrical.
>
> A 2k Ohm in-and-out 12 dB pad is possible and you might be able to get
> away with that if the cable on the output side was pretty short.
>
I'm glad all my gear has matching connections, in that case. Except that
the input on one system I use needs +4db, and the only output I've got
for it is -10, and there's not enough make up gain, so it needs a pre-amp.
--
Tciao for Now!

John.

[Peter Larsen[_3_]]

John Williamson wrote:

> I'm glad all my gear has matching connections, in that case. Except
> that the input on one system I use needs +4db, and the only output
> I've got for it is -10, and there's not enough make up gain, so it
> needs a pre-amp.

A headphone output fits nicely if available.

Kind regards

Peter Larsen

[John Williamson]

On 28/02/2014 10:22, Peter Larsen wrote:
> John Williamson wrote:
>
>> I'm glad all my gear has matching connections, in that case. Except
>> that the input on one system I use needs +4db, and the only output
>> I've got for it is -10, and there's not enough make up gain, so it
>> needs a pre-amp.
>
> A headphone output fits nicely if available.
>
I've tried a few. None of them have enough output voltage available due
to the EU rules limiting volume on mp3 players.

The system in question is a Blaupunkt coach stereo, I have to (mis)use
the CD changer input, as I want to do more than just burn mp3 files to a
briefcase full of CD-R's, and nobody I've spoken to at the Blaupunkt
agency is admitting to knowing how to turn the gain up using the built
in functionality that I've used on previous models, or even knowing that
the "more" entry on the control menu does anything when the right code
is entered.

It's only going to be a small box with fixed gain, but it's a pain
replacing the battery every day.
--
Tciao for Now!

John.

[Mike Rivers[_2_]]

On 2/28/2014 5:22 AM, Peter Larsen wrote:

> A headphone output fits nicely if available.

I've used real headphone outputs, like on a mixer, successfully, but
most low power portable devices don't have enough voltage in their power
supply to drive a +4 dBu input to its full level. They get away with
this because the headphones that are supplied with those devices are
usually in the range of 15 to 30 ohms. As long as there's enough
current, they can get loud enough at a pretty low voltage.

--
For a good time, visit http://mikeriversaudio.wordpress.com

[Trevor]

"John Williamson" > wrote in message
...
> The system in question is a Blaupunkt coach stereo, I have to (mis)use the
> CD changer input, as I want to do more than just burn mp3 files to a
> briefcase full of CD-R's, and nobody I've spoken to at the Blaupunkt
> agency is admitting to knowing how to turn the gain up using the built in
> functionality that I've used on previous models, or even knowing that the
> "more" entry on the control menu does anything when the right code is
> entered.
>
> It's only going to be a small box with fixed gain, but it's a pain
> replacing the battery every day.

Why wouldn't you just run it from the 12V (or 24V) coach battery?

Trevor.

[John Williamson]

On 02/03/2014 04:31, Trevor wrote:
> "John Williamson" > wrote in message
> ...
>> The system in question is a Blaupunkt coach stereo, I have to (mis)use the
>> CD changer input, as I want to do more than just burn mp3 files to a
>> briefcase full of CD-R's, and nobody I've spoken to at the Blaupunkt
>> agency is admitting to knowing how to turn the gain up using the built in
>> functionality that I've used on previous models, or even knowing that the
>> "more" entry on the control menu does anything when the right code is
>> entered.
>>
>> It's only going to be a small box with fixed gain, but it's a pain
>> replacing the battery every day.
>
> Why wouldn't you just run it from the 12V (or 24V) coach battery?
>
Company policy forbids unauthorised (i.e. non-branded) modifications,
unfortunately. Even adding the connection was a bit iffy, although I'm
using official Balupunkt parts.

If I connected anything that drew power from anywhere other than the
provided power socket, I'd be in trouble. The power socket is already in
use for the phone, the handsfree unit and the satnav.


--
Tciao for Now!

John.

[Trevor]

"John Williamson" > wrote in message
...
>>> The system in question is a Blaupunkt coach stereo, I have to (mis)use
>>> the
>>> CD changer input, as I want to do more than just burn mp3 files to a
>>> briefcase full of CD-R's, and nobody I've spoken to at the Blaupunkt
>>> agency is admitting to knowing how to turn the gain up using the built
>>> in
>>> functionality that I've used on previous models, or even knowing that
>>> the
>>> "more" entry on the control menu does anything when the right code is
>>> entered.
>>>
>>> It's only going to be a small box with fixed gain, but it's a pain
>>> replacing the battery every day.
>>
>> Why wouldn't you just run it from the 12V (or 24V) coach battery?
>>
> Company policy forbids unauthorised (i.e. non-branded) modifications,
> unfortunately. Even adding the connection was a bit iffy, although I'm
> using official Balupunkt parts.
>
> If I connected anything that drew power from anywhere other than the
> provided power socket, I'd be in trouble. The power socket is already in
> use for the phone, the handsfree unit and the satnav.


No problem, just get a 4 or 6 way socket adapter. None of those items draws
much current and certainly won't overload a 10A-15A "cigarette lighter"
circuit.

Trevor.