Electronics Repair (sci.electronics.repair) Discussion of repairing electronic equipment. Topics include requests for assistance, where to obtain servicing information and parts, techniques for diagnosis and repair, and annecdotes about success, failures and problems.

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Default Mixing 4 audio channels to 3?

On Wed, 9 Nov 2011 12:31:43 -0800, DaveC wrote:

First you can replace C1-[4?] 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?


---
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?


---
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:



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

"Pots, loads, and linearity."

--
JF
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Default Mixing 4 audio channels to 3?

On 11/12/2011 7:21 PM, John Fields wrote:
On Wed, 9 Nov 2011 12:31:43 -0800, 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.


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

Further more most volume pots in audio are logarithmic.




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Default Mixing 4 audio channels to 3?

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


"mF" means microfarad?

Thanks.

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Default Mixing 4 audio channels to 3?

Thanks for the great explanation JF.



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

Thanks!

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Default Mixing 4 audio channels to 3?

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

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


"mF" means microfarad?


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

Here's the latest:

Version 4
SHEET 1 1676 1124
WIRE -208 -528 -752 -528
WIRE -64 -528 -128 -528
WIRE -32 -528 -64 -528
WIRE 80 -528 48 -528
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WIRE 80 -416 80 -528
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WIRE 144 -416 80 -416
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WIRE -32 -400 -64 -400
WIRE 320 -384 320 -416
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WIRE -64 -240 -192 -240
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WIRE -64 -144 -64 -192
WIRE -64 -144 -192 -144
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WIRE -64 -96 -64 -144
WIRE -64 -96 -96 -96
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WIRE 144 -80 80 -80
WIRE 320 -80 224 -80
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WIRE -64 192 -128 192
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SYMBOL Opamps\\LT1007 0 144 R0
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
WINDOW 3 32 56 VTop 2
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SYMATTR Value 10k
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SYMATTR Value 10k
SYMBOL res 64 80 R90
WINDOW 0 -35 58 VBottom 2
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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
WINDOW 3 32 56 VTop 2
SYMATTR InstName R4
SYMATTR Value 10k
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SYMATTR Value 10k
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WINDOW 0 -38 58 VBottom 2
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SYMATTR Value 10k
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SYMATTR Value 10k
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WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R8
SYMATTR Value 10k
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WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R9
SYMATTR Value 10k
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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
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WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
WINDOW 0 15 112 Left 2
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SYMBOL voltage -656 848 R0
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WINDOW 123 0 0 Left 2
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WINDOW 123 0 0 Left 2
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WINDOW 0 -19 1 Left 2
WINDOW 3 -21 57 Left 2
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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
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SYMATTR Value 10k
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WINDOW 0 0 56 VBottom 2
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SYMATTR InstName R17
SYMATTR Value 1000
SYMBOL res -848 672 R0
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SYMATTR Value 500
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SYMATTR Value 500
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SYMATTR Value 500
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SYMATTR Value 500
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SYMATTR InstName R22
SYMATTR Value 500
SYMBOL res -416 672 R0
SYMATTR InstName R23
SYMATTR Value 500
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SYMATTR InstName R24
SYMATTR Value 500
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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


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Default Mixing 4 audio channels to 3?

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.

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!

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Default Mixing 4 audio channels to 3?

On Sat, 12 Nov 2011 12:19:18 -0800, DaveC wrote:

Thanks for the great explanation JF.



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


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

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

--
JF
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Default Mixing 4 audio channels to 3?

On Sat, 12 Nov 2011 14:05:28 -0800, DaveC 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.


---
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.
---

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.


---
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.


---
The proof is in the pudding.

--
JF
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Default Mixing 4 audio channels to 3?

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.
---
JF


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

Thanks!

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Default Mixing 4 audio channels to 3?

On Nov 12, 11:09*pm, DaveC wrote:
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.
---
JF


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

Thanks!


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

PATECUM
TGITM
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