Schematic for Stereo to Mono Box
pe...@3-cities.com
I need a schematic for a simple stereo to mono box. Unbalanced in and
out. Portable consumer CD player will be the source and -10 mixer input
will be the destination (in mono). Thanx!
Perry
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2DJVengac
>
> Hi,
>
> I need a schematic for a simple stereo to mono box. Unbalanced in and
> out. Portable consumer CD player will be the source and -10 mixer input
> will be the destination (in mono). Thanx!
ummmmmm... a RadioShack Y-cable??
in a box?
John Noble
Tsk, tsk. The proper answer includes Jensen transformers and valves...
Seriously, though -- just plug L & R into (2) mono mixer channels and
pan them both center... or what he said.
-j
tm...@pe.net
Wye?"
Good advice from Rane on this topic, plus a schematic. And nearly no
math . . . .
TM
On Mon, 07 Jun 1999 19:03:43 GMT, pe...@3-cities.com wrote:
>Hi,
>
>I need a schematic for a simple stereo to mono box. Unbalanced in and
>out. Portable consumer CD player will be the source and -10 mixer input
>will be the destination (in mono). Thanx!
>
Bill
>
> In article <375c4b05...@news.pe.net>,
> tm...@pe.net wrote:
> > Good advice from Rane on this topic, plus a schematic. And nearly no
> > math . . . .
> >
>
> I wasn't sure if there would be a different set of resistor values for
> consumer stuff like cheesy cd players. I am actually just planning on
> building the circuit into a cable and keep it unbalanced. Just need to
> know the resistor values?
The circuit diagrammed in the Rane document claims an input impedance in
the neighborhood of 1000 ohms, which probably will not bother your cd
player in the least. And, as they point out, the values are fairly
non-critical, so you could proportionally increase the values if it made
you more comfortable.
Gordon Waters
2DJVengac <j.veng...@genie.com> wrote:
>pe...@3-cities.com wrote:
>> I need a schematic for a simple stereo to mono box. Unbalanced in and
>> out. Portable consumer CD player will be the source and -10 mixer input
>> will be the destination (in mono). Thanx!
>ummmmmm... a RadioShack Y-cable??
> in a box?
No... don't do this!!!
You may get lucky and have this work, but I've seen this actually blow
stuff up, without using a proper summing node device!
IOW, you need something to isolate one channel's output from the other (a
"summing node", as it's called), so that one output can't attempt to
"drive" the other! This needs to be at least an order of magnitude
higher impedence than the output itself, to avoid either possibly frying
or otherwise distorting the outputs!
IN this case, most unbalanced RCA outputs are in the 100-ohm ballpark...
so, the outputs should be buffered with at about 1K-ohm resistance, to
insure stability, and still be low enough in impedence to drive the input
impedence of the downstream stuff (which should be 20K-ohm or higher, in
most cases).
A simple box, with 3 RCA connectors wired with the grounds all in parallel,
and the input left and right jack hot terminals each wired through a 1k-ohm
resistor per channel in series, then combined to the single output RCA hot
terminal, would be what you need:
input L hot *----------/\/\/\/\---------*-------*
1K-ohm | |
input L ground *----* | *---* output mono hot
| |
*-------------------------------* output mono ground
| |
input R ground *----* |
|
input R hot *----------/\/\/\/\---------*
1K-ohm
That'll do the trick!
Regards,
Gordon.
--
GALAXY convention --------- Anime Weekend Atlanta 5- October 8-10,1999
/| || //| // /| ,, //~// //~// //~// ----- Marriott Gwinnett Hotel
//|| ||//||// //|| ./ //_// //_// //_// --- http://www.anime.net/~awa
//~~|| |/ |/ //~~|| / ,,_// ,,_// ,,_// Gordon Waters-...@crl.com
pe...@3-cities.com
> math . . . .
>
I read it, but its impedence is designed more for balanced 600ohm stuff.
I wasn't sure if there would be a different set of resistor values for
consumer stuff like cheesy cd players. I am actually just planning on
building the circuit into a cable and keep it unbalanced. Just need to
know the resistor values?
Thanx,
Mike Rivers
In article <7jieda$o...@crl.crl.com> gwa...@crl.crl.com writes:
> >ummmmmm... a RadioShack Y-cable??
>
> No... don't do this!!!
>
> You may get lucky and have this work, but I've seen this actually blow
> stuff up, without using a proper summing node device!
>
> IOW, you need something to isolate one channel's output from the other (a
> "summing node", as it's called), so that one output can't attempt to
> "drive" the other! This needs to be at least an order of magnitude
> higher impedence than the output itself, to avoid either possibly frying
> or otherwise distorting the outputs!
Geez, you guys are a bunch of wimps. You can only blow stuff up when
there's power behind it, and there's no significant power behind the
output of a consumer CD player. (and, no, I didn't ask which brand)
People have been connecting these things together with Y cables ever
since there's been stereo. True, you can get a bit of distortion,
maybe, but nothing blows up. For $1.29, it's certainly worth a try.
The problem with putting resistors in series with the outputs is that
if the value is high enough to provide good isolation, you're building a
voltage divider which comes close to dropping the voltage in half, which
means you'll need more gain to record the mono signal, which will
increase the noise. But you can experiment if you wish.
Just to illustrate, the following circuit was suggested:
> input L hot *----------/\/\/\/\---------*-------*
> 1K-ohm | |
> input L ground *----* | *---* output mono hot
> | |
> *-------------------------------* output mono ground
> | |
> input R ground *----* |
> |
> input R hot *----------/\/\/\/\---------*
> 1K-ohm
But there are a couple of significant resistors missing. Remember, the
output impedance is assumed to be 100 ohms, so, looking back into an
output, we see 100 ohms between the hot side and ground. (the dreaded
load that will 'blow something up'), so we really have:
R1
L hot *----*-----/\/\/\/\---------*-------*
> 1K-ohm | |
RL < 100 | |
> | |
L ground *-*--* | *---* output mono hot
| |
*-------------------------------* output mono ground
| |
R ground *-*--* |
> |
RR < 100 |
> |
R hot *----*-----/\/\/\/\---------*
1K-ohm
R2
Look at the path from the L hot. The voltage goes through R1, R2,
and RR in series on its way back to ground. Since RR is almost
negligable when compared to the sum or R1 and R2, ignoring it means that
the voltage at the junction of R1 and R2 (where the output is taken)
will be divided in half.
In the grand scheme of things, this is probably insignificant, but then
most likely so is tying the two outputs directly together.
--
Mike Rivers (I'm really mri...@d-and-d.com)
Gordon Waters
geeks"... ^_^;;;)
In article <znr928845466k@TRAD>, Mike Rivers <mri...@d-and-d.com> wrote:
>In article <7jieda$o...@crl.crl.com> gwa...@crl.crl.com writes:
>> IOW, you need something to isolate one channel's output from the other (a
>> "summing node", as it's called), so that one output can't attempt to
>> "drive" the other! This needs to be at least an order of magnitude
>> higher impedence than the output itself, to avoid either possibly frying
>> or otherwise distorting the outputs!
>Geez, you guys are a bunch of wimps. You can only blow stuff up when
>there's power behind it, and there's no significant power behind the
>output of a consumer CD player. (and, no, I didn't ask which brand)
>People have been connecting these things together with Y cables ever
>since there's been stereo. True, you can get a bit of distortion,
>maybe, but nothing blows up. For $1.29, it's certainly worth a try.
I haven't seen many damage problems with home gear, but I remember seeing a
number of car audio components (electronic crossovers, etc) get blown
output preamp circuits from just Y-connecting things together, so I'm a
little more cautious than most in this regard... call it hard-learned
lessons, if you will...
Also, I've seen a couple of home audio components (for example, my old
Sanyo D3 cassette deck, may it R.I.P. after 10 years of good service),
that simply would shut down with the outputs paralleled... now, THAT'S an
inconvenient thing to have happen, in the middle of a gig, no?
>The problem with putting resistors in series with the outputs is that
>if the value is high enough to provide good isolation, you're building a
>voltage divider which comes close to dropping the voltage in half, which
>means you'll need more gain to record the mono signal, which will
>increase the noise. But you can experiment if you wish.
Actually, you'll have just as much of a voltage divider, if you connect
the outputs in parallel!
You've STILL got an identical impedence per channel, just less of it
(100 ohms vs. 1100 ohms)... which just means the channels will have to
sink more current in an out-of-phase or uncorrelated signal condition,
than with the resistors I suggested...
>But there are a couple of significant resistors missing. Remember, the
>output impedance is assumed to be 100 ohms, so, looking back into an
>output, we see 100 ohms between the hot side and ground. (the dreaded
>load that will 'blow something up'), so we really have:
> R1
> L hot *----*-----/\/\/\/\---------*-------*
> > 1K-ohm | |
> RL < 100 | |
> > | |
> L ground *-*--* | *---* output mono hot
> | |
> *-------------------------------* output mono ground
> | |
> R ground *-*--* |
> > |
> RR < 100 |
> > |
> R hot *----*-----/\/\/\/\---------*
> 1K-ohm
> R2
Actually, to get nit-picky, that really should be:
INSIDE CD PLAYER | OUTSIDE OF CD PLAYER
|
Effectively |
Zero-impedence |
voltage sources |
(preamp out driver |
transistors) |
| CD |
| output |
|\ V impedence |
| \ Ro1 | R1
|L hot \-----/\/\/\-*|*-------/\/\/\/\-----*-----*
| \ 100 | 1K-ohm | |
| (Eo left) \ ohms |RCA plug | |
| / | | |
|L ground /---------*|*---* | *---* output mono hot
| / | | |
| / | *----------------------*---* output mono ground
|/ CD | | |
output | | |
|\ impedence | | |
| \ Ro2 | R2 | |
|R hot \ --/\/\/\-*|*-----/\/\/\/\-------* |
| \ 100 | 1K-ohm |
| (Eo right) \ ohms | |
| / |RCA plug |
|R ground /---------*|*--------------------------*
| / |
| / |
|/ |
Note that the Kirchoff impedence loop of the new circuit is the same as in
your circuit... but it's much more effective a learning tool to consider
the effect of the voltage source in the circuit, as well (as it's what
we're trying to protect!)
>Look at the path from the L hot. The voltage goes through R1, R2,
>and RR in series on its way back to ground. Since RR is almost
>negligable when compared to the sum or R1 and R2, ignoring it means that
>the voltage at the junction of R1 and R2 (where the output is taken)
>will be divided in half.
Only if the signal is ZERO on one channel!
Look at the "revised", more precise circuit representation I did: If the
same signal is present on both channels, then there will be NO
attenuation, since the voltage at 'L hot' and 'R hot' will be identical.
However, if there is NO signal on one channel, but something on the
other, then YES, it will be attenuated by 6 dB, no matter whether you use
the buffer resistors or not! And this is EXACTLY a proper mono sum, I
submit... anything that's JUST on the left or right channel SHOULD be 6
dB down from the "center" image stuff, when folded to mono... there's
HALF as much energy there in the original stereo case, because of ONLY ONE
CHANNEL providing the information!
In the extreme case, if you have two completely out-of-polarity signals
on the channels, you will have ZERO output, as one voltage will cancel
the other (the resistors will be sinking ALL the signal, as one voltage
will be the inverse of the other, with a virtual ground at the output).
However, with any recording where the engineer has "done his job" and
insured PROPER MONO COMPATABILITY to his mix on whatever album is in the
CD player, you won't ever have an outright cancellation situation like
this...
>In the grand scheme of things, this is probably insignificant, but then
>most likely so is tying the two outputs directly together.
Well, if you want to take the chance, so be it... I personally rather be
safe than sorry, especially when the gig is important...