[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

[Bob E.]

What does U4 do?

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

[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.

[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

[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

[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.

[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

[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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FLAG -176 560 0
SYMBOL Misc\\NE555 80 176 M0
SYMATTR InstName U1
SYMBOL voltage -176 368 M0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 16
SYMBOL res 896 272 M0
SYMATTR InstName R3
SYMATTR Value 1000
SYMBOL cap 336 400 M0
SYMATTR InstName C1
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 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
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FLAG 464 -416 LA+LB
FLAG 464 -80 LA+LB+RA+RB
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FLAG 128 -336 0
FLAG 128 0 0
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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
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SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL res -112 272 R90
WINDOW 0 0 56 VBottom 2
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SYMATTR Value 10k
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WINDOW 0 -35 58 VBottom 2
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SYMATTR Value 10k
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SYMATTR Value 10k
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SYMATTR Value 6.2k
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SYMBOL sw 96 -80 R270
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SYMBOL cap 208 400 R0
SYMATTR InstName C5
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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

[[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.

[[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.

[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.

[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!

[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.

[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?

[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, 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_]]

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.

[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, 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_]]

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 12:19:18 -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