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cumesoftware 14th July 2012 07:16 PM

Question about preamp input impedance
 
I'm planning to build a simple preamp. Actually, it is more like an active attenuator, a volume control for monoblock amplifiers. I'm planning to use the TL072 for the task. I'm planning to use two, one for each channel. Each channel will have an RC network, followed by a voltage follower, the volume pot and the final output stage is another voltage follower.

My questions:
Is 100K a good value to be used to set the input impedance (in the input RC network that defines the RC constant)?

Between the voltage follower, is it adequate to use a 10K pot for the volume control? Or I'm better off using a higher value, like 100K?

sofaspud 14th July 2012 09:22 PM

It sounds like you are using op amp buffers for a volume pot. That shouldn't really be necessary. But as a general purpose unit, I would think those values are fine.
Note that this is a different situation than the one in this post.

jaycee 14th July 2012 11:12 PM

The standard is 47K but you will not have a problem with most equipment if you use 100K.

If you use an opamp, then volume pot, you should be OK with a 10K pot. Better still would be a 100K *linear* pot, with a resistor from the wiper to ground to "fake" a logarithmic law. Rod Elliot has a great article on this, and it works really well.

cumesoftware 15th July 2012 10:29 AM

Thanks sofaspud and jaycee!

I'll use a 100K pot with a 47K resistor to fake the logaritmic action. This will be located between the buffers. Thus, any variation on the volume won't affect the input impedance.

"The standard is 47K but you will not have a problem with most equipment if you use 100K."
When you say "you will not have a problem with most equipment if you use 100K", what kind of problems might arise? I'm leaving the option of using this preamp connected to the output of more sensible equipment, like tube preamps. That's why I want the impedance to be that high. But maybe 47K might be enough. What do you think?


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