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Old 21st March 2013, 03:12 AM   #11
JMFahey is offline JMFahey  Argentina
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OK.
a) I find so called "simulated log" a cheesy solution; getting the proper Log pots is worth it.
Worst case, I prefer the straight Lin pots, go figure.
b) some parts are unnecessary.
Remove Rlog1 , R1a , R1b , R1c, U1 .
Connect Pot 1a wiper > C1a > R1c .
Do the same on Ch2 .
c) if possible get Potvol (25K) in Log taper.
In that case remove Rvol.
Good luck.
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Old 21st March 2013, 12:12 PM   #12
picbuck is offline picbuck  United States
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Thanks for your interest, but you're suggesting the same circuit described by DUG in post #4. A perfectly legitimate mixer circuit, but not the one presented here.

The circuit (not my own) in my own post #6 is a factory circuit guaranteed (I'm confident) to work.

I personally think the log-fake pot idea is kinda neat. But I wouldn't say somebody is wrong if they prefer store-bought to roll your own.
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Old 21st March 2013, 02:42 PM   #13
JMFahey is offline JMFahey  Argentina
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Quote:
Originally Posted by picbuck View Post
Thanks for your interest, but you're suggesting the same circuit described by DUG in post #4. A perfectly legitimate mixer circuit, but not the one presented here.
Of course it's not the one in post #1.
I think the point is to suggest corrections , am I missing something?
And if 2 answers match, maybe it means something

By the way, you complain of
Quote:
And thanks DUG for pointing out the cheap, simple, and equally effective way, although with no amplification.
Well, yours doesn't provide amplification either , that's why we both suggested you remove those unneeded unity gain blocks.

Quote:
The circuit (not my own) in my own post #6 is a factory circuit guaranteed (I'm confident) to work.
It *does* work, who said it doesn't? ... but still has useless unity gain blocks.
To be more precise, *they* provide the marginal advantage of presenting 10K input impedance while using 5K pots ...but yours does not.

Quote:
I personally think the log-fake pot idea is kinda neat. But I wouldn't say somebody is wrong if they prefer store-bought to roll your own.
Roll your own would be to open the pot and paint a lower resistance track on the lower half; the crude idea of putting a lower resistance value from cursos to ground is *not* the same.
It has the terrible disadvantage of making the pot value "seen" by the previous 1/10th of what is expected, throwing calculations, frequency cutoff, etc. out of whack.
That "idea" does not work on:
> "Ampeg" (James) type tone controls
> "Fender" type tone controls
> "Vox" type tone controls
> Gain/Distortion controls as used on 95% of Distortion pedals
> Gain controls in live/recording Mixers
> parametric equalizers
> Audio Oscillator Frequency setting
.
.
> 1000000 other circuits where a *real* Log control is needed.

Yes, same pot may cost 1$ if linear and 3$ if log. So what?

By the way, I buy in bulk, straight from the Importer, and pay 85 cents any type.
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Old 21st March 2013, 03:48 PM   #14
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Old 21st March 2013, 06:47 PM   #15
picbuck is offline picbuck  United States
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Google never forgets, so this thread is going to come up in searches for years.

That being so, I'm going to refer future surfers to post #6 for a correct (I'm sure) mixer circuit, and otherwise just thank everyone for their input.
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