Mixing 4 audio channels to 3?

--- I put in polarized from habit, (and, BTW, C2 is in backwards, oops) but there's no real reason for them to be.

Ceramic or polyester would also work.

---

--- Polyester or ceramic.

-- JF

Reply to
John Fields
Loading thread data ...

--
A 4066 would work well.
Reply to
John Fields

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

Reply to
The Ghost In The Machine

--
It makes the output of the sub channel approximately equal to the
outputs of the other channels with 1VPP inputs from the four sources.
Reply to
John Fields

--- 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]--+-----+

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]--+ . 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 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/ snipped-for-privacy@4ax.com

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

"Pots, loads, and linearity."

-- JF

Reply to
John Fields

Our hearing is more or less logarithmic, so it might be about right.

Further more most volume pots in audio are logarithmic.

Reply to
tuinkabouter

"mF" means microfarad?

Thanks.

Reply to
DaveC

Thanks for the great explanation JF.

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

How about putting it here? ::

Thanks!

Reply to
DaveC

On Sat, 12 Nov 2011 12:03:41 -0800, DaveC wrote:

--- No, "mF" means millifarad, = 1000µF

Here's the latest:

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WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R8 SYMATTR Value 10k SYMBOL res -80 -208 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R9 SYMATTR Value 10k SYMBOL res -176 -256 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R10 SYMATTR Value 10k SYMBOL Opamps\\LT1007 0 -144 R0 SYMATTR InstName U3 SYMBOL res 64 -256 R90 WINDOW 0 -37 62 VBottom 2 WINDOW 3 -37 60 VTop 2 SYMATTR InstName R12 SYMATTR Value 3K SYMBOL voltage -480 848 R0 WINDOW 3 24 96 Invisible 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 0 15 112 Left 2 SYMATTR Value SINE(0 2 1700) SYMATTR InstName RA SYMBOL voltage -656 848 R0 WINDOW 3 24 96 Invisible 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 0 15 112 Left 2 SYMATTR Value SINE(0 2 2300) SYMATTR InstName LB SYMBOL voltage -832 848 R0 WINDOW 3 24 96 Invisible 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 0 15 110 Left 2 SYMATTR Value SINE(0 2 4260) SYMATTR InstName LA SYMBOL Misc\\NE555 128 672 M0 SYMATTR InstName U4 SYMBOL voltage -80 848 M0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value 16 SYMBOL cap 400 896 M0 WINDOW 0 -19 1 Left 2 WINDOW 3 -21 57 Left 2 SYMATTR InstName C1 SYMATTR Value 1n SYMBOL diode 736 720 M270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL diode 656 848 M0 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL polcap 736 912 M180 WINDOW 0 24 57 Left 2 WINDOW 3 24 8 Left 2 SYMATTR InstName C2 SYMATTR Value 1µ SYMBOL polcap 544 688 M90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C3 SYMATTR Value 1µ SYMBOL res 368 720 R0 SYMATTR InstName R13 SYMATTR Value 10k SYMBOL res 528 688 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R14 SYMATTR Value 100 SYMBOL sw 128 -416 R270 WINDOW 0 32 15 Left 2 WINDOW 3 32 44 Left 2 SYMATTR InstName S2 SYMBOL sw 128 -80 R270 WINDOW 0 32 15 Left 2 WINDOW 3 32 44 Left 2 SYMATTR InstName S1 SYMBOL sw 112 208 R270 WINDOW 0 32 15 Left 2 WINDOW 3 32 44 Left 2 SYMATTR InstName S3 SYMBOL res -48 512 R180 WINDOW 0 36 76 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName R11 SYMATTR Value 10k SYMBOL cap 240 400 R0 SYMATTR InstName C5 SYMATTR Value 510n SYMBOL res 304 -400 R0 SYMATTR InstName R15 SYMATTR Value 1000 SYMBOL res 304 -64 R0 SYMATTR InstName R16 SYMATTR Value 1000 SYMBOL res 304 224 R0 SYMATTR InstName R17 SYMATTR Value 1000 SYMBOL res -848 672 R0 SYMATTR InstName R20 SYMATTR Value 500 SYMBOL res -768 672 R0 SYMATTR InstName R18 SYMATTR Value 500 SYMBOL res -672 672 R0 SYMATTR InstName R19 SYMATTR Value 500 SYMBOL res -592 672 R0 SYMATTR InstName R21 SYMATTR Value 500 SYMBOL res -496 672 R0 SYMATTR InstName R22 SYMATTR Value 500 SYMBOL res -416 672 R0 SYMATTR InstName R23 SYMATTR Value 500 SYMBOL res -320 672 R0 SYMATTR InstName R24 SYMATTR Value 500 SYMBOL res -240 672 R0 SYMATTR InstName R25 SYMATTR Value 500 SYMBOL zener 208 464 R180 WINDOW 0 23 69 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D3 SYMATTR Value 1N750 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

Reply to
John Fields

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!

Reply to
DaveC

--
They don't accept .pdf's.

Send me your email addy and I'll email you a copy.
Reply to
John Fields

--
Not true.

With R12 cranked down to minimum R the output of U3 goes down to about
600µVPP 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.
Reply to
John Fields

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

Thanks!

Reply to
DaveC

OH BROTHER :/ LETS HOPE SO. BOOWAHAHAHAHAHA ! GOOD LUCK WIT DAT.

PATECUM TGITM

Reply to
The Ghost In The Machine

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