Is this EQ practical for my head amp?

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
I have constructed this negative feedback eq: https://instruct1.cit.cornell.edu/courses/bionb440/FinalProjects/f2005/asa27/ASAfinaldemo.htm

I am using it for my head amp now. Is this circuit pratical? It works, but I think it's for larger power. Right now I have it as a single chanel and the output of the single op amp I used feeds to another op-amp. Should I change anything? what sort od slope would it have? I am going to be looking at different frequencies with my oscope to get an idea but i'm curious if it is practical for a small head amp.

Also, the pots I am getting in will be 50k because I cannot find the same ones I want in 100k, what effect will this have?
 

bertus

Joined Apr 5, 2008
20,057
Hello,

The used circuit makes use of the Baxendall tone control ciruit.

In the circuit there is a 1nF capacitor and a 270 Ohms resistor at the output of the opamp.
This will cut the high frequencies a bit.

The use of 50K pots in stead of 100 K pots will probably shift the frequencies and control range a bit.

Bertus
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
Hello,

The used circuit makes use of the Baxendall tone control ciruit.

In the circuit there is a 1nF capacitor and a 270 Ohms resistor at the output of the opamp.
This will cut the high frequencies a bit.

The use of 50K pots in stead of 100 K pots will probably shift the frequencies and control range a bit.

Bertus
I didn't think about the 1nF cap...
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
The 1nF/270Ω hanging on the output will not make an audible difference in the output. I can't imagine why they are there.
I don't believe they as far as high frequency response goes becuase I do not notice any difference between having the pot positions "flat" and connecting my headphones directly into the music source which goes to show that the circuit does not alter the input's response unless the dials are moved from "rest" positions. Which is proper as far as that goes.

To calculate db boost/attenuation do I measure the voltage, and count how many times it doubles and add 3db per doubling? Oh and do I look at db from voltage or calculated power?
 
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