Mixing 4 audio channels to 3?

[DaveC[_2_]]

I want to connect the audio out (headphone jack) of 2 computer sound cards to
a desktop & woofer speaker-amp combination. (The original input was via USB
only and I'm modifying this for analog audio input.)

This is my guess at the necessary resistors to mix down these 4 outputs to
the 3 inputs in the amplifier (L, R, sub):

<http://i41.tinypic.com/97mpud.jpg>

The data sheet for the Philips TDA8510J amp IC shows an application for 2
channel input:

<http://i40.tinypic.com/5ets9w.jpg>

Suggestions welcome.

Thanks.

[DaveC[_2_]]

If I'm on the right track, what are the suggested values for resistors?

Thahks.

[NT]

On Oct 26, 7:58*am, DaveC > wrote:
> I want to connect the audio out (headphone jack) of 2 computer sound cards to
> a desktop & woofer speaker-amp combination. (The original input was via USB
> only and I'm modifying this for analog audio input.)
>
> This is my guess at the necessary resistors to mix down these 4 outputs to
> the 3 inputs in the amplifier (L, R, sub):
>
> <http://i41.tinypic.com/97mpud.jpg>

Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.


NT

[Don Pearce[_3_]]

On Wed, 26 Oct 2011 08:50:55 -0700 (PDT), NT >
wrote:

>On Oct 26, 7:58*am, DaveC > wrote:
>> I want to connect the audio out (headphone jack) of 2 computer sound cards to
>> a desktop & woofer speaker-amp combination. (The original input was via USB
>> only and I'm modifying this for analog audio input.)
>>
>> This is my guess at the necessary resistors to mix down these 4 outputs to
>> the 3 inputs in the amplifier (L, R, sub):
>>
>> <http://i41.tinypic.com/97mpud.jpg>
>
>Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.
>
>
>NT

Is there something wrong with my browser? I can't see any of these
component references. Actually I see what you mean. One problem here
is that R9 and R10 are going to cause crosstalk. Without some active
electronics there is no way around that apart from taking the sub from
one channel only. This is actually quite common.

d

[DaveC[_2_]]

> Is there something wrong with my browser? I can't see any of these
> component references.

Sorry. They didn't make onto the jpg file. Added here:

<http://i40.tinypic.com/5ets9w.jpg>

> Actually I see what you mean. One problem here
> is that R9 and R10 are going to cause crosstalk. Without some active
> electronics there is no way around that apart from taking the sub from
> one channel only. This is actually quite common.
> d

OK, can do that.

Suggested values for resistors?

Thanks.

[DaveC[_2_]]

> Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.
> NT

Callouts added:

<http://i40.tinypic.com/5ets9w.jpg>

R11 provides some adjustment for sub. Otherwise no way to have any difference
between L & R channel volume and sub volume.

The caps are suggested on the data sheet example:

<http://i40.tinypic.com/5ets9w.jpg>

Why would you do away with them?

What values for the resistors do you suggest?

Thanks.

[Don Pearce[_3_]]

On Wed, 26 Oct 2011 10:10:12 -0700, DaveC > wrote:

>> Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.
>> NT
>
>Callouts added:
>
><http://i40.tinypic.com/5ets9w.jpg>
>
>R11 provides some adjustment for sub. Otherwise no way to have any difference
>between L & R channel volume and sub volume.
>
>The caps are suggested on the data sheet example:
>
><http://i40.tinypic.com/5ets9w.jpg>
>
>Why would you do away with them?
>
>What values for the resistors do you suggest?
>
>Thanks.

For that particular circuit with that chip, the caps are necessary
because those inputs are not at ground potential. If you are feeding
normal hi fi unit inputs the caps aren't needed. Resistor values
around 5 to 10 k would be what you need.

d

[DaveC[_2_]]

> For that particular circuit with that chip, the caps are necessary
> because those inputs are not at ground potential. If you are feeding
> normal hi fi unit inputs the caps aren't needed. Resistor values
> around 5 to 10 k would be what you need.

And the value for the potentiometer?

Thanks.

[Don Pearce[_3_]]

On Wed, 26 Oct 2011 11:18:39 -0700, DaveC > wrote:

>> For that particular circuit with that chip, the caps are necessary
>> because those inputs are not at ground potential. If you are feeding
>> normal hi fi unit inputs the caps aren't needed. Resistor values
>> around 5 to 10 k would be what you need.
>
>And the value for the potentiometer?
>
That depends entirely on the input impedance of the subwoofer. You
could try a 100k pot and connect it as a normal volume control - one
end to the incoming signal, the other end to ground and the slider to
the subwoofer.

d

[DaveC[_2_]]

> That depends entirely on the input impedance of the subwoofer. You
> could try a 100k pot and connect it as a normal volume control - one
> end to the incoming signal, the other end to ground and the slider to
> the subwoofer.
>
> d

The data sheet says that the 2 single-ended inputs (R & L) are 50K impedance.
The bridge-tied load ("BTL") input -- used for the sub -- is 25K impedance.

Does this suggest any change to your recommendation of a 100K pot tied to
ground?

Thanks.

[Don Pearce[_3_]]

On Wed, 26 Oct 2011 11:52:49 -0700, DaveC > wrote:

>> That depends entirely on the input impedance of the subwoofer. You
>> could try a 100k pot and connect it as a normal volume control - one
>> end to the incoming signal, the other end to ground and the slider to
>> the subwoofer.
>>
>> d
>
>The data sheet says that the 2 single-ended inputs (R & L) are 50K impedance.
>The bridge-tied load ("BTL") input -- used for the sub -- is 25K impedance.
>
>Does this suggest any change to your recommendation of a 100K pot tied to
>ground?
>
>Thanks.

No, that would be about right.

d

[Michael Moroney]

(Don Pearce) writes:

>On Wed, 26 Oct 2011 08:50:55 -0700 (PDT), NT >
>wrote:

>>On Oct 26, 7:58*am, DaveC > wrote:
>>> I want to connect the audio out (headphone jack) of 2 computer sound cards to
>>> a desktop & woofer speaker-amp combination. (The original input was via USB
>>> only and I'm modifying this for analog audio input.)
>>>
>>> This is my guess at the necessary resistors to mix down these 4 outputs to
>>> the 3 inputs in the amplifier (L, R, sub):
>>>
>>> <http://i41.tinypic.com/97mpud.jpg>
>>
>>Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.
>>
>>
>>NT

>Is there something wrong with my browser? I can't see any of these
>component references. Actually I see what you mean. One problem here
>is that R9 and R10 are going to cause crosstalk. Without some active
>electronics there is no way around that apart from taking the sub from
>one channel only. This is actually quite common.

I don't see component references, either. However, assuming R1-R4 are
the resistors from the inputs to ground, they may be necessary to provide
a proper impedance to the source driving them.

To reduce crosstalk, I'd replace the two resistors from the 220nF
cap junctions to the pot with four resistors from each of the four inputs
to the pot. Presumably, the output impedance of the amps is low compared
to the resistors reducing crosstalk.

For mixing signals there is a method using amps and feedback where the
connection between multiple sources is at a virtual ground, eliminating
crosstalk.

[NT]

On Oct 26, 5:09*pm, (Don Pearce) wrote:
> On Wed, 26 Oct 2011 08:50:55 -0700 (PDT), NT >
> wrote:
>
> >On Oct 26, 7:58*am, DaveC > wrote:
> >> I want to connect the audio out (headphone jack) of 2 computer sound cards to
> >> a desktop & woofer speaker-amp combination. (The original input was via USB
> >> only and I'm modifying this for analog audio input.)
>
> >> This is my guess at the necessary resistors to mix down these 4 outputs to
> >> the 3 inputs in the amplifier (L, R, sub):
>
> >> <http://i41.tinypic.com/97mpud.jpg>
>
> >Lose Rs 1-4, keep 5-8, lose R11, and lose C1,2,3, then itll be good.
>
> >NT
>
> Is there something wrong with my browser? I can't see any of these
> component references. Actually I see what you mean. One problem here
> is that R9 and R10 are going to cause crosstalk. Without some active
> electronics there is no way around that apart from taking the sub from
> one channel only. This is actually quite common.
>
> d

I was hinting that the OP might decide to add numbers to the
components, otherwise its sufficiently tedious to talk about them that
people often just wont bother.

Yes the caps are necessary, but for the amp IC, not for the mixer.
House them together and its the same difference, house them apart and
the distiction matters.


NT

[DaveC[_2_]]

> Yes the caps are necessary, but for the amp IC, not for the mixer.
> House them together and its the same difference, house them apart and
> the distiction matters.
> NT

This is unclear to me.

Do you mean that if the mixer resistor matrix is housed close to the amp IC
that the capacitors won't be needed? And if they are separated and connected
by shielded audio cables that the caps will be needed?

Thanks.

[DaveC[_2_]]

Latest rev:

<http://i40.tinypic.com/300feaq.jpg>

All resistors 5K-10K. Pot 100K.

Comments welcome.

Thanks.

[NT]

On Oct 27, 5:17*am, DaveC > wrote:
> > Yes the caps are necessary, but for the amp IC, not for the mixer.
> > House them together and its the same difference, house them apart and
> > the distiction matters.
> > NT
>
> This is unclear to me.
>
> Do you mean that if the mixer resistor matrix is housed close to the amp IC
> that the capacitors won't be needed? And if they are separated and connected
> by shielded audio cables that the caps will be needed?
>
> Thanks.

Nothing like that, no. The amp IC requires the caps, so theyre needed,
no question about it. But theyre needed for the amp, not for mixing
purposes, so if the 2 circuits can be disconnected by the end user,
then the caps need to be with the IC amp part, not the mixer part.


NT

[Tomi Holger Engdahl[_4_]]

DaveC > writes:

> If I'm on the right track, what are the suggested values for resistors?

10 kohms for all fixed resistor should work...


--
Tomi Engdahl (http://www.iki.fi/then/)
Take a look at my electronics web links and documents at
http://www.epanorama.net/

[John Fields]

On Wed, 26 Oct 2011 21:17:16 -0700, DaveC > wrote:

>> Yes the caps are necessary, but for the amp IC, not for the mixer.
>> House them together and its the same difference, house them apart and
>> the distiction matters.
>> NT
>
>This is unclear to me.
>
>Do you mean that if the mixer resistor matrix is housed close to the amp IC
>that the capacitors won't be needed? And if they are separated and connected
>by shielded audio cables that the caps will be needed?
>
>Thanks.

---
Since the chip is being powered by single positive DC supply, The
only way to get signals which swing below ground into and out of the
chip is to use capacitive coupling.

--
JF

[Arny Krueger[_4_]]

"John Fields" > wrote in message
...
> On Wed, 26 Oct 2011 21:17:16 -0700, DaveC > wrote:
>
>>> Yes the caps are necessary, but for the amp IC, not for the mixer.
>>> House them together and its the same difference, house them apart and
>>> the distiction matters.
>>> NT
>>
>>This is unclear to me.
>>
>>Do you mean that if the mixer resistor matrix is housed close to the amp
>>IC
>>that the capacitors won't be needed? And if they are separated and
>>connected
>>by shielded audio cables that the caps will be needed?
>>
>>Thanks.
>
> ---
> Since the chip is being powered by single positive DC supply, The
> only way to get signals which swing below ground into and out of the
> chip is to use capacitive coupling.

Or, as is not uncommonly done these days in portable digital players,
establish a floating ground for the purpose of providing and signal for the
(floating) load, so that the difference between the output and the floating
ground can become negative.

[DaveC[_2_]]

> Or, as is not uncommonly done these days in portable digital players,
> establish a floating ground for the purpose of providing and signal for the
> (floating) load, so that the difference between the output and the floating
> ground can become negative.

Which raises the question I was pondering recently: is it OK to connect 2
computers' sound cards' grounds together?

The PS for the powered speakers is via "wall wart" (not grounded).

Thanks.

[Les Cargill[_4_]]

DaveC wrote:
>> Or, as is not uncommonly done these days in portable digital players,
>> establish a floating ground for the purpose of providing and signal for the
>> (floating) load, so that the difference between the output and the floating
>> ground can become negative.
>
> Which raises the question I was pondering recently: is it OK to connect 2
> computers' sound cards' grounds together?
>

It should be. If you are concerned, check with a meter first.

> The PS for the powered speakers is via "wall wart" (not grounded).
>
> Thanks.
>

--
Les Cargill

[Don Pearce[_3_]]

On Thu, 27 Oct 2011 09:11:17 -0700, DaveC > wrote:

>> Or, as is not uncommonly done these days in portable digital players,
>> establish a floating ground for the purpose of providing and signal for the
>> (floating) load, so that the difference between the output and the floating
>> ground can become negative.
>
>Which raises the question I was pondering recently: is it OK to connect 2
>computers' sound cards' grounds together?
>
No. You will certainly create a ground loop which will result in mains
hum.

d

[[email protected]]

On Thu, 27 Oct 2011 17:59:59 GMT, (Don Pearce) wrote:

>On Thu, 27 Oct 2011 09:11:17 -0700, DaveC > wrote:
>
>>> Or, as is not uncommonly done these days in portable digital players,
>>> establish a floating ground for the purpose of providing and signal for the
>>> (floating) load, so that the difference between the output and the floating
>>> ground can become negative.
>>
>>Which raises the question I was pondering recently: is it OK to connect 2
>>computers' sound cards' grounds together?
>>
>No. You will certainly create a ground loop which will result in mains
>hum.

This is certainly likely but not a foregone conclusion. In any case, it's
"OK" (i.e. not dangerous). As long as the computers are plugged into the same
circuit, it's quite possible that he can get away with it, assuming line
levels. If hum is a problem, coupling capacitors or cheap audio transformers
will usually solve it.

[DaveC[_2_]]

>> is it OK to connect 2 computers' sound cards' grounds together?

> No. You will certainly create a ground loop which will result in mains
> hum.
> d

This kinda kills the project, doesn't it? (At least, the straightforward
solution.)

The alternative is...?

Thanks.

[Don Pearce[_3_]]

On Thu, 27 Oct 2011 11:12:00 -0700, DaveC > wrote:

>>> is it OK to connect 2 computers' sound cards' grounds together?
>
>> No. You will certainly create a ground loop which will result in mains
>> hum.
>> d
>
>This kinda kills the project, doesn't it? (At least, the straightforward
>solution.)
>
>The alternative is...?
>

Transformers. You can buy decent ones in Maplin for use in car audio.
Capacitors have been suggested, but they don't help. If they are big
enough to pass bass, the ground loop is still there. It doesn't
require a DC connection to function.

d

[NT]

On Oct 27, 5:11*pm, DaveC > wrote:
> > Or, as is not uncommonly done these days in portable digital players,
> > establish a floating ground for the purpose of providing and signal for the
> > (floating) load, so that the difference between the output and the floating
> > ground can become negative.
>
> Which raises the question I was pondering recently: is it OK to connect 2
> computers' sound cards' grounds together?
>
> The PS for the powered speakers is via "wall wart" (not grounded).
>
> Thanks.

I've yet to have a problem doing it.


NT

[DaveC[_2_]]

> 10 kohms for all fixed resistor should work...

Hi Tomi. Thank you for your advice. I have enjoyed your creative designs for
many years.

Do you have any comments on my revisions of the circuit (ie, are caps
recommended? should I be concerned about cross talk if I take resistors from
all 4 channels to drive the sub woofer? etc...)

Cheers,
Dave

[John Fields]

On Sat, 29 Oct 2011 00:49:32 -0700, DaveC > wrote:

>> 10 kohms for all fixed resistor should work...
>
>Hi Tomi. Thank you for your advice. I have enjoyed your creative designs for
>many years.
>
>Do you have any comments on my revisions of the circuit (ie, are caps
>recommended? should I be concerned about cross talk if I take resistors from
>all 4 channels to drive the sub woofer? etc...)
>
>Cheers,
>Dave

---
Why not go active and get rid of the caps altogether?


..LA>-+---[10k]-+-[10k]-+-------------> -(LA+LB)
.. | | |
.. | | +V |
.. | | | |
..LB>-|-+-[10k]-+--|-\ |
.. | | | >-+ 10K
.. | | +--|+/ +-[POT]-+
.. | | | | | |
.. | | GND -V | +5 |
.. | | | | |
.. | +------[10K]---+---+-|-\ |
.. | | | >--+--> -(LA+LB+RA+RB)
.. +--------[10K]---+ +-|+/
.. | |
.. +--------[10K]---+ GND
.. | |
.. | +------[10K]---+
.. | |
.. | |
.. | |
..RA>-+-+-[10k]-+-[10k]-+
.. | | |
.. | | +V |
.. | | | |
..RB>-+---[10k]-+--|-\ |
.. | >-+-------------> -(RA+RB)
.. +--|+/
.. | |
.. GND -V

Seems to work pretty well...

Version 4
SHEET 1 880 968
WIRE 112 -544 48 -544
WIRE 224 -544 192 -544
WIRE -96 -448 -528 -448
WIRE 32 -448 -16 -448
WIRE 48 -448 48 -544
WIRE 48 -448 32 -448
WIRE 112 -448 48 -448
WIRE 224 -432 224 -544
WIRE 224 -432 176 -432
WIRE 256 -432 224 -432
WIRE 112 -416 80 -416
WIRE -96 -352 -432 -352
WIRE 32 -352 32 -448
WIRE 32 -352 -16 -352
WIRE 80 -352 80 -416
WIRE -528 -240 -528 -448
WIRE -416 -240 -528 -240
WIRE 64 -240 -336 -240
WIRE 112 -240 64 -240
WIRE 240 -240 192 -240
WIRE -432 -192 -432 -352
WIRE -352 -192 -432 -192
WIRE 64 -192 64 -240
WIRE 64 -192 -272 -192
WIRE -272 -144 -336 -144
WIRE 64 -144 64 -192
WIRE 64 -144 -192 -144
WIRE -176 -96 -240 -96
WIRE 64 -96 64 -144
WIRE 64 -96 -96 -96
WIRE 128 -96 64 -96
WIRE 240 -80 240 -240
WIRE 240 -80 192 -80
WIRE 272 -80 240 -80
WIRE 128 -64 96 -64
WIRE 96 0 96 -64
WIRE 112 112 48 112
WIRE 224 112 192 112
WIRE -336 208 -336 -144
WIRE -96 208 -336 208
WIRE 32 208 -16 208
WIRE 48 208 48 112
WIRE 48 208 32 208
WIRE 112 208 48 208
WIRE 224 224 224 112
WIRE 224 224 176 224
WIRE 256 224 224 224
WIRE 112 240 80 240
WIRE -240 304 -240 -96
WIRE -96 304 -240 304
WIRE 32 304 32 208
WIRE 32 304 -16 304
WIRE 80 304 80 240
WIRE -528 688 -528 -240
WIRE -432 688 -432 -192
WIRE -336 688 -336 208
WIRE -240 688 -240 304
WIRE -144 688 -144 656
WIRE -48 688 -48 656
WIRE -528 832 -528 768
WIRE -432 832 -432 768
WIRE -432 832 -528 832
WIRE -336 832 -336 768
WIRE -336 832 -432 832
WIRE -240 832 -240 768
WIRE -240 832 -336 832
WIRE -144 832 -144 768
WIRE -144 832 -240 832
WIRE -48 832 -48 768
WIRE -48 832 -144 832
WIRE -528 928 -528 832
FLAG 144 192 +5
FLAG 144 256 -5
FLAG 80 304 0
FLAG 144 -464 +5
FLAG 144 -400 -5
FLAG 80 -352 0
FLAG -528 928 0
FLAG -144 656 +5
FLAG -48 656 -5
FLAG 160 -112 +5
FLAG 160 -48 -5
FLAG 96 0 0
SYMBOL Opamps\\LT1007 144 160 R0
SYMATTR InstName U1
SYMBOL voltage -240 672 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V3
SYMATTR Value SINE(0 1 1000)
SYMBOL res 0 192 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 0 288 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL res 208 96 R90
WINDOW 0 -43 58 VBottom 2
WINDOW 3 -37 60 VTop 2
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL Opamps\\LT1007 144 -496 R0
SYMATTR InstName U2
SYMBOL res 0 -464 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R4
SYMATTR Value 10k
SYMBOL res 0 -368 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res 208 -560 R90
WINDOW 0 -43 58 VBottom 2
WINDOW 3 -37 60 VTop 2
SYMATTR InstName R6
SYMATTR Value 10k
SYMBOL voltage -48 784 R180
WINDOW 0 -50 102 Left 2
WINDOW 3 24 16 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL voltage -144 672 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 5
SYMBOL res -80 -112 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R7
SYMATTR Value 10k
SYMBOL res -176 -160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R8
SYMATTR Value 10k
SYMBOL res -256 -208 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R9
SYMATTR Value 10k
SYMBOL res -320 -256 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R10
SYMATTR Value 10k
SYMBOL Opamps\\LT1007 160 -144 R0
SYMATTR InstName U3
SYMBOL res 208 -256 R90
WINDOW 0 -43 58 VBottom 2
WINDOW 3 -37 60 VTop 2
SYMATTR InstName R12
SYMATTR Value 5k
SYMBOL voltage -336 672 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V4
SYMATTR Value SINE(0 1 1700)
SYMBOL voltage -432 672 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V5
SYMATTR Value SINE(0 1 2300)
SYMBOL voltage -528 672 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V6
SYMATTR Value SINE(0 1 4260)
TEXT -562 952 Left 2 !.tran .005


--
JF

[DaveC[_2_]]

> Why not go active and get rid of the caps altogether?

Thanks John. Good simple design.

Can you suggest a good op amp for such an audio application?

Is it possible to use single-supply amps throughout?

Thanks,
Dave

[John Fields]

On Sun, 30 Oct 2011 02:26:06 -0700, DaveC > wrote:

>> Why not go active and get rid of the caps altogether?
>
>Thanks John. Good simple design.
>
>Can you suggest a good op amp for such an audio application?

---
I used an LT1007 in the simulation.

Did you run it?
---

>Is it possible to use single-supply amps throughout?

---
Yes, but you'd have to use caps on the inputs and outputs and
reference the opamp + inputs to Vcc/2.

--
JF

[Don Pearce[_3_]]

On Sun, 30 Oct 2011 02:26:06 -0700, DaveC > wrote:

>> Why not go active and get rid of the caps altogether?
>
>Thanks John. Good simple design.
>
>Can you suggest a good op amp for such an audio application?
>
>Is it possible to use single-supply amps throughout?
>
>Thanks,
>Dave

The best amps are pretty much all dual supply. The big advantage is
simplicity of the circuit - and of course not having the coupling caps
(you need to put another set at the output, by the way).

Generating a negative rail is pretty simple even if you only have a
positive one available. An oscillating op amp, a diode and a capacitor
does the job nicely.

d

[DaveC[_2_]]

>> Can you suggest a good op amp for such an audio application?

> I used an LT1007 in the simulation.
>
> Did you run it?

No tools.

I made the L & R op amps fixed gain and added input level control pots:

<http://i43.tinypic.com/2hh3y12.jpg>

Comments?

Thanks.

[DaveC[_2_]]

> <http://i43.tinypic.com/2hh3y12.jpg>

Should all pots be log taper?

Thanks.

[DaveC[_2_]]

> I used an LT1007 in the simulation.

Can't find them on Mouser.

Suggest another one that's good for audio?

How about LM833? ::

<http://www.national.com/ds/LM/LM833.pdf>

The SR and GBW beat the 1007... ;-)

Thanks.

[DaveC[_2_]]

Is the type of resistor important (wire wound, etc.)?

Thanks.

[John Fields]

On Sun, 30 Oct 2011 21:44:33 -0700, DaveC > wrote:

>>> Can you suggest a good op amp for such an audio application?
>
>> I used an LT1007 in the simulation.
>>
>> Did you run it?
>
>No tools.
---
They're free:

http://www.linear.com/designtools/software/
---


>I made the L & R op amps fixed gain and added input level control pots:
>
><http://i43.tinypic.com/2hh3y12.jpg>
>
>Comments?

---
Looks OK.


--
JF

[John Fields]

On Sun, 30 Oct 2011 21:47:34 -0700, DaveC > wrote:

>> <http://i43.tinypic.com/2hh3y12.jpg>
>
>Should all pots be log taper?

---
The input pots, OK. The feedback pot, probably not.

Plot the change in output voltage as a function of wiper position to
get a better handle on it.

--
JF

[John Fields]

On Mon, 31 Oct 2011 00:16:15 -0700, DaveC > wrote:

>> I used an LT1007 in the simulation.
>
>Can't find them on Mouser.

---
You can buy them directly from Linear:

http://www.linear.com/purchase/LT1007
---

>Suggest another one that's good for audio?
>
>How about LM833? ::
>
><http://www.national.com/ds/LM/LM833.pdf>
>
>The SR and GBW beat the 1007... ;-)

---
Go for it; they're a lot cheaper!

--
JF

[John Fields]

On Mon, 31 Oct 2011 01:11:57 -0700, DaveC > wrote:

>Is the type of resistor important (wire wound, etc.)?

---
Don't use wirewound.

5% 1/4 watt carbon film is fine.

--
JF

[DaveC[_2_]]

>> Is it possible to use single-supply amps throughout?

> ---
> Yes, but you'd have to use caps on the inputs and outputs and
> reference the opamp + inputs to Vcc/2.
> [J. Fields]

Only a single supply (in the amplified speakers) is available to power this
circuit. I can tap this supply for my circuit:

<http://i41.tinypic.com/2vlo2t2.jpg>

I've added input & output caps. Are these values sound? ;-)

How do I go about getting a 1/2 Vcc ground reference? (See my non-EE
attempt.) What values to use for the divider resistors?

What needs to be reference to the new "ground"? Everything between the input
caps and output caps?

Thanks,
Dave

[Dave Platt]

In article >,
DaveC <newsgroups> wrote:

>Only a single supply (in the amplified speakers) is available to power this
>circuit. I can tap this supply for my circuit:
>
><http://i41.tinypic.com/2vlo2t2.jpg>
>
>I've added input & output caps. Are these values sound? ;-)
>
>How do I go about getting a 1/2 Vcc ground reference? (See my non-EE
>attempt.) What values to use for the divider resistors?

If you want to do it with just resistors, you could make R16 and R17
somewhere around 4k7 each, and add a few uF of bypass capacitance
around one or both resistors. 1k would give a stiffer reference if
you don't mind the additional power consumption.

If you can spare an op-amp section, you can get a better (stiffer) ground
reference than you'll get with just resistors, with lower power
dissipation (I think). Use R16 and R17, and a small cap from the
junction point, to create a Vcc/2 reference, but do not "ground" this
directly to your internal reference point (the rectangular-looking
ground symbol). Instead, feed this to the noninverting input of an
op-amp section, feed the op-amp output back to the inverting input
(i.e. create a unity-gain follower), and use the op amp's output as
your ground reference. In this arrangement R16 and R17 can be
high-value (100k?) as their junction point will be looking into a
high-impedance op amp input.

>What needs to be reference to the new "ground"? Everything between the input
>caps and output caps?

Pretty much... each of the op amps' noninverting inputs, and the
"bottom ends" of the potentiometers, as you have drawn them. *NOT*
the V- input to the op amp(s), of course.

You might want to add "pop preventer" resistors at the inputs and
outputs... say, 100k to DC ground, from the "outside" end of each of
the DC-blocking capacitors.

--
Dave Platt > AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[Dave Platt]

In article >,
DaveC <newsgroups> wrote:

>Only a single supply (in the amplified speakers) is available to power this
>circuit. I can tap this supply for my circuit:

One cautionary note with regards to this circuit, as drawn: be careful
when you wire up R15. If you use a standard three-terminal
potentiometer, make sure that you wire both the wiper, and one of the
two ends to U3's inverting input. Don't just wire up the wiper!

The reason: pots occasionally go "open" due to dirt or wear. If you
have only the wiper connected, and it goes open, you'll have no
feedback path around U3, and it'll immediately and enthusiastically
slam its output against one of the rails (or both in rapid succession
if there's a signal present). This will let out a really unholy
THWOMP from your subwoofer, and may pop the cone out of the cabinet or
at last shove the voice coil out of the gap. Expensive damage.

With a three-terminal wire-up, the resistance in this part of the
feedback loop will never be more than the bulk value of the pot (i.e.
open wiper == wiper all the way at one end) and this will limit the
maximum subwoofer volume. You can choose the maximum loudness by
setting the value of the pot.

--
Dave Platt > AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[DaveC[_2_]]

Latest rev:

<http://i40.tinypic.com/35m026h.jpg>

>> What needs to be reference to the new "ground"? Everything between the
>> input
>> caps and output caps?
>
> Pretty much... each of the op amps' noninverting inputs, and the
> "bottom ends" of the potentiometers, as you have drawn them. *NOT*
> the V- input to the op amp(s), of course.

You do mean each of the op amps' *inverting* inputs, yes?

> You might want to add "pop preventer" resistors at the inputs and
> outputs... say, 100k to DC ground, from the "outside" end of each of
> the DC-blocking capacitors.

Is this what you mean (see link)?

Thanks.

[John Fields]

On Tue, 8 Nov 2011 21:07:44 -0800, DaveC > wrote:

>Latest rev:
>
><http://i40.tinypic.com/35m026h.jpg>
>
>>> What needs to be reference to the new "ground"? Everything between the
>>> input
>>> caps and output caps?
>>
>> Pretty much... each of the op amps' noninverting inputs, and the
>> "bottom ends" of the potentiometers, as you have drawn them. *NOT*
>> the V- input to the op amp(s), of course.
>
>You do mean each of the op amps' *inverting* inputs, yes?

---
No.

Your drawing is wrong; signal goes to the inverting (-) inputs and the
Vcc/2 reference goes to the non-inverting (+) inputs.

--
JF

[DaveC[_2_]]

> Your drawing is wrong; signal goes to the inverting (-) inputs and the
> Vcc/2 reference goes to the non-inverting (+) inputs.
> JF

Thanks guys. Fixed:

<http://i44.tinypic.com/r1k8qa.jpg>

All else looks good?

Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
designs. Values for these?

Thanks.

[Michael Moroney]

DaveC > writes:

>> Your drawing is wrong; signal goes to the inverting (-) inputs and the
>> Vcc/2 reference goes to the non-inverting (+) inputs.
>> JF

>Thanks guys. Fixed:

><http://i44.tinypic.com/r1k8qa.jpg>

>All else looks good?

You still have the signal go to the non-inverting input. The way the
schematic is, U1-U3 will throw their output hard to a rail or oscillate
with positive feedback from R13-R15.

>Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
>designs. Values for these?

I worry that the RC time constant would have the reference be not at the
1/2 way point while C8 charges on powerup. I don't see a C9. Might do
something not so good to the subwoofer.

[DaveC[_2_]]

> You still have the signal go to the non-inverting input.

The current version of the drawing has signal going to the inverting input

>> Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
>> designs. Values for these?

> I worry that the RC time constant would have the reference be not at the
> 1/2 way point while C8 charges on powerup.

Suggestions?

> I don't see a C9.

That means you're not looking at the right version of the drawing. Copy &
paste this into a browser:

<http://i44.tinypic.com/r1k8qa.jpg>

> Might do something not so good to the subwoofer.
> [M. Moroney]

Suggestions?

Thanks.

[Don Pearce[_3_]]

On Wed, 9 Nov 2011 09:07:15 -0800, DaveC > wrote:

>> Your drawing is wrong; signal goes to the inverting (-) inputs and the
>> Vcc/2 reference goes to the non-inverting (+) inputs.
>> JF
>
>Thanks guys. Fixed:
>
><http://i44.tinypic.com/r1k8qa.jpg>
>
>All else looks good?
>
>Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
>designs. Values for these?
>
>Thanks.

What does U4 do?

d

[Dave Platt]

In article >,
DaveC <newsgroups> wrote:

>> Your drawing is wrong; signal goes to the inverting (-) inputs and the
>> Vcc/2 reference goes to the non-inverting (+) inputs.

Agreed.

>Thanks guys. Fixed:
>
><http://i44.tinypic.com/r1k8qa.jpg>
>
>All else looks good?

>Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
>designs. Values for these?

I'd eliminate C9. Some op amps aren't able to drive capacitive loads
without exhibiting instability.

If you do want some noise reduction on your reference, I'd add a
small decoupling resistor (say, 47R) between U4 and C9, and perhaps
use another .1 uF for C9. If you're using a good low-noise op amp,
you can probably just omit the filtering here and feed U4's output
directly to your "common".

I'd also recommend decoupling your 16-volt power supply, with a .1 uF
located as close as practical to the V+/V- pins of each op amp.

Remember to get the polarities of C1-C7 correct when you install them
(+ to the op-amp side, - to the outside world).

--
Dave Platt > AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[DaveC[_2_]]

> I'd eliminate C9. Some op amps aren't able to drive capacitive loads
> without exhibiting instability.

OK, done.

> If you're using a good low-noise op amp,
> you can probably just omit the filtering here and feed U4's output
> directly to your "common".
>
> I'd also recommend decoupling your 16-volt power supply, with a .1 uF
> located as close as practical to the V+/V- pins of each op amp.

Sound like basic good advice. :-)

> Remember to get the polarities of C1-C7 correct when you install them
> (+ to the op-amp side, - to the outside world).

I presumed that such coupling caps should be non-polar. No?

Thanks.

[Bob E.]

> What does U4 do?

Provides a Vcc/2 local "ground" so I can use these op amps with a single
supply voltage.

[NT]

On Nov 9, 5:07*pm, DaveC > wrote:
> > Your drawing is wrong; signal goes to the inverting (-) inputs and the
> > Vcc/2 reference goes to the non-inverting (+) inputs.
> > JF
>
> Thanks guys. Fixed:
>
> <http://i44.tinypic.com/r1k8qa.jpg>
>
> All else looks good?
>
> Are cap values reasonable? I added C8 & C9 out of habit of seeing in other
> designs. Values for these?
>
> Thanks.

First you can replace C1-5 with 1uF each. Replace C8 with 220uF, and
omit U4 & C9 entirely.
You dont want to use a 50k pot followed by a 10k load (R5-12). I'd go
with 10k pots and 100k for R5-12, adjusting the nfb Rs accordingly.


NT

[DaveC[_2_]]

> I'd also recommend decoupling your 16-volt power supply, with a .1 uF
> located as close as practical to the V+/V- pins of each op amp.

Since the V- pin is already PS ground, I need decouple caps only on the V+
pins, yes?

[DaveC[_2_]]

> 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?

> 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?

Thanks.

[Don Pearce[_3_]]

On Wed, 9 Nov 2011 12:10:19 -0800, Bob E. > wrote:

>> What does U4 do?
>
>Provides a Vcc/2 local "ground" so I can use these op amps with a single
>supply voltage.

More useful to let it oscillate as a square wave generator at 100kHz
or so, and rectify the output into a negative 15V rail. That way you
can run the op amps the way they are meant to be run.

d

[DaveC[_2_]]

> More useful to let it oscillate as a square wave generator at 100kHz
> or so, and rectify the output into a negative 15V rail. That way you
> can run the op amps the way they are meant to be run.
>
> d

Suggest a circuit...?

Thanks.

[Dave Platt]

In article >,
DaveC <newsgroups> wrote:

>> Remember to get the polarities of C1-C7 correct when you install them
>> (+ to the op-amp side, - to the outside world).
>
>I presumed that such coupling caps should be non-polar. No?

No need for that. You're going to have an 8-volt bias sitting on each
cap (half of your supply voltage), and the audio signals that they see
will only be a volt or two, peak-to-peak, so the caps will always be
polarized in the direction I indicated.

It's entirely usual and standard practice to use polar electrolytics
in this sort of situation. If you want to get fancy I'm sure you
could find an exotic 'lytic (like one of the new solid-electrolyte
types), but I see no need for that in this application.

You *could* use nonpolar 'litics if you have them around, but as
they're usually more expensive I don't see the point.

--
Dave Platt > AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[Dave Platt]

>> I'd also recommend decoupling your 16-volt power supply, with a .1 uF
>> located as close as practical to the V+/V- pins of each op amp.
>
>Since the V- pin is already PS ground, I need decouple caps only on the V+
>pins, yes?

Good practics is to put the bypass caps as close to the IC leads as is
practical, and run short traces (or wires) to the IC pins.

I wasn't suggesting one bypass cap from V+ to ground and another from
V- to ground... since you're using a single-sided supply and V- is DC
ground, that would be redundant (as you have noted).

--
Dave Platt > AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

[NT]

On Nov 9, 8:31*pm, 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?

1uF

> > 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?
>
> > 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?
>
> Thanks.

that would work


NT

[John Fields]

On Wed, 09 Nov 2011 20:32:36 GMT, (Don Pearce) wrote:

>On Wed, 9 Nov 2011 12:10:19 -0800, Bob E. > wrote:
>
>>> What does U4 do?
>>
>>Provides a Vcc/2 local "ground" so I can use these op amps with a single
>>supply voltage.
>
>More useful to let it oscillate as a square wave generator at 100kHz
>or so, and rectify the output into a negative 15V rail. That way you
>can run the op amps the way they are meant to be run.

---
Amen to that!

The beauty parts are that it doesn't need to be regulated, it only
needs to be smooth enough so that its ripples don't bump into the
output's low peaks, and it gets rid of all those damned coupling caps
and their frequency response killing reactances.

I was going to suggest that, since the mixer is going to be external
to the amp, he use a couple of wall-warts to get the dual supplies,
but I like your solution a lot better. :-)

--
JF

[DaveC[_2_]]

> I was going to suggest that, since the mixer is going to be external
> to the amp, he use a couple of wall-warts to get the dual supplies,
> but I like your solution a lot better. :-)

Enough to suggest a nice circuit? ;-)

Thanks,
Dave

[DaveC[_2_]]

> No need for that. You're going to have an 8-volt bias sitting on each
> cap (half of your supply voltage), and the audio signals that they see
> will only be a volt or two, peak-to-peak, so the caps will always be
> polarized in the direction I indicated.
>
> It's entirely usual and standard practice to use polar electrolytics
> in this sort of situation. If you want to get fancy I'm sure you
> could find an exotic 'lytic (like one of the new solid-electrolyte
> types), but I see no need for that in this application.
> ...
> Dave P.

I'm learnin'! Thanks for the explanation. I'll use standard aluminum 'lytics
here, connected as noted.

Dave C.

[[email protected]]

On Wed, 9 Nov 2011 12:10:19 -0800, Bob E. > wrote:

>> What does U4 do?
>
>Provides a Vcc/2 local "ground" so I can use these op amps with a single
>supply voltage.

Hint: Don't use the chassis ground symbol as a "Vref" symbol. It's confusing
and someone along the line might get hurt. Grounds should be.

[DaveC[_2_]]

> Hint: Don't use the chassis ground symbol as a "Vref" symbol. It's
confusing
> and someone along the line might get hurt. Grounds should be.

Yeah, I wasn't sure about that. How should I have indicated 2 separate
grounds?

Thanks.

[isw]

In article
>,
DaveC > wrote:

> <http://i40.tinypic.com/35m026h.jpg>

Distortion performance might be better if you run the amps in inverting
mode. If you run the audio into the non-inverting inputs, the inverting
input follows and the amp has to operate throughout it's common-mode
range. If you use the amps in inverting mode, both the + and - inputs
will stay very close to ground.

Isaac

[isw]

In article >,
(Don Pearce) wrote:

> On Wed, 9 Nov 2011 12:10:19 -0800, Bob E. > wrote:
>
> >> What does U4 do?
> >
> >Provides a Vcc/2 local "ground" so I can use these op amps with a single
> >supply voltage.
>
> More useful to let it oscillate as a square wave generator at 100kHz
> or so, and rectify the output into a negative 15V rail. That way you
> can run the op amps the way they are meant to be run.

??

What is wrong with using an op-amp to "amplify" DC? Surely they are
"meant" to do that.

Isaac

[DaveC[_2_]]

> What is wrong with using an op-amp to "amplify" DC? Surely they are
> "meant" to do that.
>
> Isaac

I think he means that better audio results can be achieved by providing true
dual-voltage supplies and eliminating all coupling caps.

The "DC amplifier" is a fine design, but will result in inferior audio
performance due to the caps.

Now, if someone would volunteer such a negative voltage generator circuit...
;-)

Thanks.

[[email protected]]

On Wed, 9 Nov 2011 19:10:58 -0800, DaveC > wrote:

>> Hint: Don't use the chassis ground symbol as a "Vref" symbol. It's
>confusing
>> and someone along the line might get hurt. Grounds should be.
>
>Yeah, I wasn't sure about that. How should I have indicated 2 separate
>grounds?

It's not ground. It's Vcc/2. I generally call it Vref, or some such thing.

[Don Pearce[_3_]]

On Wed, 09 Nov 2011 20:49:28 +0000 (GMT), Stuart
> wrote:

>In article >,
> Don Pearce > wrote:
>> >Provides a Vcc/2 local "ground" so I can use these op amps with a single
>> >supply voltage.
>
>> More useful to let it oscillate as a square wave generator at 100kHz
>> or so, and rectify the output into a negative 15V rail. That way you
>> can run the op amps the way they are meant to be run.
>
>KISS

Exactly. Much easier than all these halved supplies, coupling caps and
multiple grounds.

d

[DaveC[_2_]]

> Exactly. Much easier than all these halved supplies, coupling caps and
> multiple grounds.
>
> d

So, how -- exactly -- would you create that negative voltage?

Dave

[John Fields]

On Thu, 10 Nov 2011 00:51:03 -0800, DaveC > wrote:

>> Exactly. Much easier than all these halved supplies, coupling caps and
>> multiple grounds.
>>
>> d
>
>So, how -- exactly -- would you create that negative voltage?

---
Version 4
SHEET 1 964 748
WIRE -32 80 -176 80
WIRE 960 80 192 80
WIRE -32 144 -80 144
WIRE 256 144 192 144
WIRE -32 208 -80 208
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WIRE 624 208 576 208
WIRE 672 208 624 208
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SYMBOL voltage -176 368 M0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
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SYMATTR Value 16
SYMBOL res 896 272 M0
SYMATTR InstName R3
SYMATTR Value 1000
SYMBOL cap 336 400 M0
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SYMATTR Value 1n
SYMBOL diode 736 224 M270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D2
SYMATTR Value 1N4148
SYMBOL diode 640 352 M0
SYMATTR InstName D3
SYMATTR Value 1N4148
SYMBOL polcap 800 352 M0
SYMATTR InstName C4
SYMATTR Value 1µ
SYMBOL polcap 512 192 M90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C2
SYMATTR Value 1µ
SYMBOL res 304 224 R0
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res 480 192 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 100
TEXT -40 536 Right 2 !.tran .01 startup uic

--
JF

[John Fields]

On Wed, 9 Nov 2011 14:45:13 -0800, DaveC > wrote:

>> I was going to suggest that, since the mixer is going to be external
>> to the amp, he use a couple of wall-warts to get the dual supplies,
>> but I like your solution a lot better. :-)
>
>Enough to suggest a nice circuit? ;-)

---
Sure, a 555 charge pump.

Got LTspice yet?

--
JF

[John Fields]

On Thu, 10 Nov 2011 00:51:03 -0800, DaveC > wrote:

>> Exactly. Much easier than all these halved supplies, coupling caps and
>> multiple grounds.
>>
>> d
>
>So, how -- exactly -- would you create that negative voltage?
>
>Dave

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


Version 4
SHEET 1 1676 1124
WIRE -32 -528 -64 -528
WIRE 80 -528 48 -528
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WIRE -32 -96 -64 -96
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WIRE 80 -80 32 -80
WIRE 112 -80 80 -80
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WIRE 224 -32 176 -32
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SYMBOL Opamps\\LT1007 0 144 R0
SYMATTR InstName U1
SYMBOL voltage -240 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 1 1000)
SYMATTR InstName RB
SYMBOL res -112 176 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res -112 272 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL res 64 80 R90
WINDOW 0 -35 58 VBottom 2
WINDOW 3 -35 60 VTop 2
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL Opamps\\LT1007 0 -480 R0
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
SYMBOL res -112 -352 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res 64 -544 R90
WINDOW 0 -38 58 VBottom 2
WINDOW 3 -33 59 VTop 2
SYMATTR InstName R6
SYMATTR Value 10k
SYMBOL res -80 -112 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R7
SYMATTR Value 10k
SYMBOL res -176 -160 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R8
SYMATTR Value 10k
SYMBOL res -256 -208 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R9
SYMATTR Value 10k
SYMBOL res -320 -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 6.2k
SYMBOL voltage -336 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 1 1700)
SYMATTR InstName RA
SYMBOL voltage -432 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 1 2300)
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SYMBOL voltage -528 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 1 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 672 848 M0
SYMATTR InstName D2
SYMATTR Value 1N4148
SYMBOL polcap 768 848 M0
SYMATTR InstName C2
SYMATTR Value 1µ
SYMBOL polcap 560 688 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 560 688 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R14
SYMATTR Value 100
SYMBOL sw 96 -416 R270
WINDOW 0 32 15 Left 2
WINDOW 3 32 44 Left 2
SYMATTR InstName S2
SYMBOL sw 96 -80 R270
WINDOW 0 32 15 Left 2
WINDOW 3 32 44 Left 2
SYMATTR InstName S1
SYMBOL sw 80 208 R270
WINDOW 0 32 15 Left 2
WINDOW 3 32 44 Left 2
SYMATTR InstName S3
SYMBOL res 144 352 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R11
SYMATTR Value 10k
SYMBOL cap 208 400 R0
SYMATTR InstName C5
SYMATTR Value 100n
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
TEXT -296 1040 Right 2 !.tran .01 startup uic
TEXT -488 1080 Left 2 !.model SW SW(Ron=1 Roff=10Meg Vt= 8 Vh=0)
--
JF

[John Fields]

On Wed, 09 Nov 2011 21:07:44 -0800, isw > wrote:

>In article >,
> (Don Pearce) wrote:
>
>> On Wed, 9 Nov 2011 12:10:19 -0800, Bob E. > wrote:
>>
>> >> What does U4 do?
>> >
>> >Provides a Vcc/2 local "ground" so I can use these op amps with a single
>> >supply voltage.
>>
>> More useful to let it oscillate as a square wave generator at 100kHz
>> or so, and rectify the output into a negative 15V rail. That way you
>> can run the op amps the way they are meant to be run.
>
>??
>
>What is wrong with using an op-amp to "amplify" DC? Surely they are
>"meant" to do that.
>
>Isaac

---
Of course, but if an AC input and output is desired and no negative
supply is available, the input and output must be capacitively
coupled, which is, at best, kind of nasty.

--
JF

[DaveC[_2_]]

> 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? Purpose? 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?)...

Thanks,
Dave

[DaveC[_2_]]

> 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)? And if that is to be a pot (the original idea), is 10K appropriate?

There should be some kind of sub volume control...

Thanks,
Dave

[DaveC[_2_]]

C2, C3 are polarized, so I presume electrolytic?

What type should C1 be?

Thanks.

[John Fields]

On Fri, 11 Nov 2011 09:20:22 -0800, DaveC > wrote:

>C2, C3 are polarized, so I presume electrolytic?

---
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.
---

>What type should C1 be?

---
Polyester or ceramic.

--
JF

[John Fields]

On Fri, 11 Nov 2011 09:03:36 -0800, DaveC > 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?

---
A 4066 would work well.
---

>Purpose?

---
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?)...

---
The output delay, from turn-on, is determined by R11 and C5.

--
JF

[The Ghost In The Machine[_2_]]

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

[John Fields]

On Fri, 11 Nov 2011 09:18:16 -0800, DaveC > 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)?

---
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?

---
Yes, if all the other resistors are 10k.
---
>
>There should be some kind of sub volume control...

---
That'll work.

--
JF

[John Fields]

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

[tuinkabouter]

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.

[DaveC[_2_]]

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

"mF" means microfarad?

Thanks.

[DaveC[_2_]]

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!

[John Fields]

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:

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SYMATTR InstName U1
SYMBOL voltage -304 848 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
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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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SYMBOL res -112 80 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 10k
SYMBOL res 64 80 R90
WINDOW 0 -35 58 VBottom 2
WINDOW 3 -35 60 VTop 2
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL Opamps\\LT1007 0 -480 R0
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
SYMBOL res -112 -544 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res 64 -544 R90
WINDOW 0 -38 58 VBottom 2
WINDOW 3 -33 59 VTop 2
SYMATTR InstName R6
SYMATTR Value 10k
SYMBOL res -80 -112 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R7
SYMATTR Value 10k
SYMBOL res -176 -160 R90
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

[DaveC[_2_]]

> 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!

[John Fields]

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

[John Fields]

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

[DaveC[_2_]]

> 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!

[The Ghost In The Machine[_2_]]

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