Mixing 4 audio channels to 3?

On Fri, 11 Nov 2011 09:03:36 -0800, DaveC <invalid@invalid.net> wrote:

Here's the whole thing; DC coupled mixer with a charge pump for a
negative supply:
[J. Fields]

Thanks John. Nice design.

What part is used for SW1-3?
Click to expand...
---
A 4066 would work well.
---

Purpose?
Click to expand...
---
Anti thump
---

If these are "anti-thump" turn-on
suppression switches, I think the main amp is similarly turn-on delayed. Of
course it's important to know the timing of these to avoid any window through
which the sub's cone could launch (or lunch?)...
Click to expand...
---
The output delay, from turn-on, is determined by R11 and C5.

--
JF
 
WHAT COMPLICATED CRAP SHOTS......HAVE YOU TRIED APPLYING ONE SET OF
STEREO OUTPUT CABLES TO THE STEREO PORTION, THEN THE OTHER TWO TO THE
SUBWOOFER SECTION AS IS????
WHAT CAN POSSIBLY GO WRONG?
DON'T YOUR SPEAKERS HAVE A CROSSOVER NETWORK ALREADY ???
THEN ADDING ANY COMPONENTS TO THE MIX WILL ONLY DISTORT THE SIGNAL
COMING FROM THE SOUND CARDS AND YOU WILL HAVE TO RE-ATTENUATE IT ALL
OVER AGAIN BEFORE YOU ENERGIZE THE COMPLETED "REVAMPED" AUDIO SYSTEM
YOU PROPSE, NEEDLESSLY.

TGITM
 
On Fri, 11 Nov 2011 09:18:16 -0800, DaveC <invalid@invalid.net> wrote:

Here's the whole thing; DC coupled mixer with a charge pump for a
negative supply:

John, what's the reason behind choosing 6.2K for the feedback R on U3 (sub op
amp)?
Click to expand...
---
It makes the output of the sub channel approximately equal to the
outputs of the other channels with 1VPP inputs from the four sources.
---

And if that is to be a pot (the original idea), is 10K appropriate?
Click to expand...
---
Yes, if all the other resistors are 10k.
---
There should be some kind of sub volume control...
Click to expand...
---
That'll work.

--
JF
 
On Wed, 9 Nov 2011 12:31:43 -0800, DaveC <invalid@invalid.net> wrote:

with 1uF each. Replace C8 with 220uF, and
omit U4 & C9 entirely.

Leave C5-7 as is?

You dont want to use a 50k pot followed by a 10k load (R5-12).

Teach this man to fish: why don't I want to use 50K pot & 10K load
combination?
Click to expand...
---
OK, here's opamps 101:

You're using the inverting - or "summing" - amplifier configuration
because all of the sources are feeding loads connected to virtual
grounds and, consequently, can't interact and cause crosstalk between
channels.

Here's how it works: (view with a fixed-pitch font)

.. E2 E3
.. \ /
.. E1 +--[R2]--+
.. \ | |
.. +--[R1]--+--|-\ |
.. | | >--+
.. [GEN] +--|+/
.. | |
.. GND GND

Now, what the opamp's job is is to make the output voltage (E3)swing
to whatever it needs to be to make the voltage on the inverting (-)
input be the same as the voltage on the non-inverting (+) input.

In this case the + input is at ground, 0V, so if R1 is equal to R2,
and E1 is at 1V, then E3 has to go to -1V to make E2 = 0V.

What also happens is that since one end of R1 is sitting at 0V and the
other end is sitting at 1V, it's the same as if the end with 0V on it
was connected to ground and, indeed, the same current will flow
through the resistor in either situation.

But what does that have to do with a 50k pot feeding a 10k load?

Well...

Considering the 10k resistor to be grounded on one end and the other
end connected to a pot wired like a voltage divider, you'll have this:


..Vin>--+
.. |R1
.. [POT]<--+
.. | |R2
.. | [10K]
.. | |
..GND>--+-----+


Note that the portion of the pot's resistive element located between
the slider and ground is connected in _parallel_ with R2, so for a 50k
pot and a 10k load the total resistance will be:


.. R1 * R2 50k * 10k
.. Rt = --------- = ----------- ~ 8333 ohms
.. R1 + R2 50k + 10k


Now, with the opamp in there we'll have:


.. E2 E3
.. \ /
.. E1 +--[R3]--+
.. |R1 | |
.. [POT]<--[R2]--+--|-\ |
.. | | >--+
.. GND +--|+/
.. |
.. GND

If R1 is at 50k and R2 and R3 are 10k, then the circuit will look
like:


.. E2 E3
.. \ R3 /
.. E1 +-[10k]-+
.. \ R2 | |
.. +--[10k]--+--|-\ |
.. | | >-+
.. [50k] +--|+/
.. | |
.. GND GND


and the output voltage will be:

.. -E1 * R3 -1V * 10k
.. E3 = --------- = ---------- = -1V
.. R2 10k

As the pot is rotated, the part of the element between the input
voltage and the wiper will appear in series with the parallel
combination of R2 and the element between the wiper and ground, so the
circuit now looks like this:


.. E1 E3
.. | R3 /
.. [Ra] +-[10k]-+
.. | R2 | |
.. E2-+----[10k]--+--|-\ |
.. | | >-+
.. [Rb] +--|+/
.. | |
.. GND GND


and, since R2 is effectively in parallel with Rb, that'll look like
this:


.. E1
.. |
.. [Ra] E2
.. | /
.. +-----+
.. | |
.. [Rb] [R2]
.. | |
.. GND GND


Now, since Rb and R2 are in parallel, their total resistance will be:


.. Rb * R2
.. Rt = ---------
.. Rb + R2


and the voltage across them will be:


.. E1 * Rt
.. E2 = ---------
.. Ra + Rt

Just for grins let's say we crank the pot so that Ra is 5k.

Then we can solve for Rt:

.. 45kR * 10kR
.. Rt = ------------- ~ 8182 ohms,
.. 45kR + 10kR

And E2:


.. 1V * 8182R
.. E2 = --------------- ~0.621 volt
.. 5000R + 8182R


Now, since that voltage appears across R2, and R2 is connected to a
virtual ground, a potential difference exists across the resistor and
charge must flow through it.

That current is supplied by the output of the opamp and, since it must
drive the virtual ground to zero volts, the sign of its output voltage
must be opposite to the sign of E2 while, since R2 and R3 are the same
value, the opamp's output will be the same magnitude as the input
voltage.


.. E1 -0.621V
.. | 0.621V R3 /
.. [Ra] / +-[10k]-+
.. | / R2 | |
.. +-+--[10k]--+--|-\ |
.. | / | >-+
.. [Rb] / +--|+/
.. | 0V |
.. GND GND

"OK", you may say, "but what on Earth does that have to do with a 50k
pot feeding a 10k load?"


If we make a table of changes in output voltage as a function of
successive 5kohm changes in pot resistance, we'll have:

.. R V dV
..--------------------
.. 50k 1.000 0.379
.. 45k 0.621 0.177
.. 40k 0.444 0.103
.. 35k 0.341 0.069
.. 30k 0.272 0.050
.. 25k 0.222 0.040
.. 20k 0.182 0.036
.. 15k 0.146 0.035
.. 10k 0.111 0.042
.. 5k 0.069 0.069
.. 0k 0.000 -----

You can see from dV that the change in voltage isn't very linear.

---

I'd go
with 10k pots and 100k for R5-12, adjusting the nfb Rs accordingly.
NT

"adjusting" means replace those with 100K's also?
Click to expand...
---
I'd leave the 10k fixed resistors in there and make the right and left
channel pots dual 1k's.

BTW, I plotted the difference between 50k pots into 10k loads and 1k
pots into 10k loads (same as 10k pots into 100k loads) and you can
find the PDF at:

news/1kdtb71h1pv0lqu9aovhjhhi39goobf988@4ax.com

If that link doesn't work it's over on abse as:

"Pots, loads, and linearity."

--
JF
 
On 11/12/2011 7:21 PM, John Fields wrote:
On Wed, 9 Nov 2011 12:31:43 -0800, DaveC<invalid@invalid.net> wrote:



. E1 -0.621V
. | 0.621V R3 /
. [Ra] / +-[10k]-+
. | / R2 | |
. +-+--[10k]--+--|-\ |
. | / |>-+
. [Rb] / +--|+/
. | 0V |
. GND GND

"OK", you may say, "but what on Earth does that have to do with a 50k
pot feeding a 10k load?"


If we make a table of changes in output voltage as a function of
successive 5kohm changes in pot resistance, we'll have:

. R V dV
.--------------------
. 50k 1.000 0.379
. 45k 0.621 0.177
. 40k 0.444 0.103
. 35k 0.341 0.069
. 30k 0.272 0.050
. 25k 0.222 0.040
. 20k 0.182 0.036
. 15k 0.146 0.035
. 10k 0.111 0.042
. 5k 0.069 0.069
. 0k 0.000 -----

You can see from dV that the change in voltage isn't very linear.
Click to expand...
Our hearing is more or less logarithmic, so it might be about right.

Further more most volume pots in audio are logarithmic.
 
Thanks for the great explanation JF.

news/1kdtb71h1pv0lqu9aovhjhhi39goobf988@4ax.com

If that link doesn't work it's over on abse as:

"Pots, loads, and linearity."

JF
Click to expand...
The link doesn't work, and my service doesn't carry binaries.

How about putting it here? ::

<http://www.tinypic.com>

Thanks!
 
On Sat, 12 Nov 2011 12:03:41 -0800, DaveC <invalid@invalid.net> wrote:

Here's the whole thing; DC coupled mixer with a charge pump for a
negative supply:

"mF" means microfarad?
Click to expand...
---
No, "mF" means millifarad, = 1000ľF

Here's the latest:

Version 4
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SYMATTR InstName U1
SYMBOL voltage -304 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
WINDOW 0 15 107 Left 2
SYMATTR Value SINE(0 2 1000)
SYMATTR InstName RB
SYMBOL res -112 176 R90
WINDOW 0 0 56 VBottom 2
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SYMATTR InstName R1
SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR InstName U2
SYMBOL res -112 -448 R90
WINDOW 0 0 56 VBottom 2
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR InstName R10
SYMATTR Value 10k
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SYMATTR InstName U3
SYMBOL res 64 -256 R90
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TEXT -616 1040 Right 2 !.tran .01 startup uic
TEXT -808 1080 Left 2 !.model SW SW(Ron=50 Roff=10Meg Vt=2.5 Vh=0)
TEXT -840 632 Left 2 ;1K POT
TEXT -648 632 Left 2 ;1K POT
TEXT -488 632 Left 2 ;1K POT
TEXT -296 632 Left 2 ;1K POT
--
JF
 
Here's the latest:
...
JF
Click to expand...
Almost! :)

I need some absolute attenuation for the sub's output and some relative to
the input signal.

In other words: I need to be able to change the input level to the mixer
(each of the 2 sound card's level will remain most of the time at max) so I
can change volume for each stereo input (from each sound card) and for the
*relative* level of the sub. The way it is now, the sub is outputting full
level (no pots in that circuit) and the L and R channels are variable.

So, the input to the sub mixer circuit needs to come from the input pots'
wipers, not directly from the input. And U3 needs variable gain.

The power amp (that this mixer is supplying the audio to) has a mute pin that
has a slow R-C tc on it, so we can eliminate S1-S3 and assoc. components, I
think.

Thanks!
 
On Sat, 12 Nov 2011 12:19:18 -0800, DaveC <invalid@invalid.net> wrote:

Thanks for the great explanation JF.

news/1kdtb71h1pv0lqu9aovhjhhi39goobf988@4ax.com

If that link doesn't work it's over on abse as:

"Pots, loads, and linearity."

JF

The link doesn't work, and my service doesn't carry binaries.

How about putting it here? ::

http://www.tinypic.com
Click to expand...
---
They don't accept .pdf's.

Send me your email addy and I'll email you a copy.

--
JF
 
On Sat, 12 Nov 2011 14:05:28 -0800, DaveC <invalid@invalid.net> wrote:

Here's the latest:
...
JF

Almost! :)

I need some absolute attenuation for the sub's output and some relative to
the input signal.

In other words: I need to be able to change the input level to the mixer
(each of the 2 sound card's level will remain most of the time at max) so I
can change volume for each stereo input (from each sound card) and for the
*relative* level of the sub. The way it is now, the sub is outputting full
level (no pots in that circuit) and the L and R channels are variable.
Click to expand...
---
Not true.

With R12 cranked down to minimum R the output of U3 goes down to about
600ľVPP out with 2VPP out of the sound card(s), and up to about 6VPP
with R12 at 10k so, if you use a 10k pot for R12, there's your center
channel volume control.
---

So, the input to the sub mixer circuit needs to come from the input pots'
wipers, not directly from the input. And U3 needs variable gain.
Click to expand...
---
Easy enough; if that's what you want, then just disconnect R7, R8, R9,
and R10 from the sound card outputs and connect them to the pot
wipers.
---

The power amp (that this mixer is supplying the audio to) has a mute pin that
has a slow R-C tc on it, so we can eliminate S1-S3 and assoc. components, I
think.
Click to expand...
---
The proof is in the pudding.

--
JF
 
With R12 cranked down to minimum R the output of U3 goes down to about
600ľVPP out with 2VPP out of the sound card(s), and up to about 6VPP
with R12 at 10k so, if you use a 10k pot for R12, there's your center
channel volume control.
---
JF
Click to expand...
Yeah, R12 needs to be a pot. That does it.

Thanks!
 
On Nov 12, 11:09 pm, DaveC <inva...@invalid.net> wrote:
With R12 cranked down to minimum R the output of U3 goes down to about
600ľVPP out with 2VPP out of the sound card(s), and up to about 6VPP
with R12 at 10k so, if you use a 10k pot for R12, there's your center
channel volume control.
---
JF

Yeah, R12 needs to be a pot. That does it.

Thanks!
Click to expand...
OH BROTHER :/ LETS HOPE SO.
BOOWAHAHAHAHAHA !
GOOD LUCK WIT DAT.

PATECUM
TGITM
 
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