LF353 Three-Band Active Tone Control

Thread Starter

bernardomarques

Joined Nov 14, 2019
14
Three Band Active Tone Control.png

Good afternoon, I'm currently working on a project where I have to build a three-band tone control based on the circuit given in this OpAmp datasheet (picture of this thread).

My question is, when you analyse the three bands separately how does the 11k resistor fit in the circuit (of just one of the bands, for example the treble part of the circuit)?

Thank you!
 

ci139

Joined Jul 11, 2016
1,088
Very approximately - the LF353 is a j-Fet input Op-Amp -- meaning -- it's input currents are almost Zero . . .
/// LF155/LF156/LF256/LF257/LF355/LF356/LF357JFET Input Operational Amplifiers
/// LF347,LF351,LF353,LF355,LF356,LF357,LF411,LF412,LF441,LF442,LF444,TL081,TL082 -- Application Note 447 Protection Schemes for BI-FET Amplifiers and Switches

. . . so -- whatever goes through that 11k must go back through it

the BASS/CUT basically set's a "DC" gain or just the General gain

// from the other hand what is been overseen while using the j-Fet input Op-Amp-s is the very fact that the input currents are neglectable
// what the resistorless biasing results is the input being "over-driven" by rough transitions -- so -- to get a "smoother" responce there should be "attenuation" (of the noise and steps) apx. 200kΩ or less for faster responce . . .

from what we can conclude that the (faster changing) higher frequencies won't need "attenuation" that much

what else we can study from the diagram is that the lower adjustment pots are being shunted for higher frequencies
 
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Thread Starter

bernardomarques

Joined Nov 14, 2019
14
So how could I analyse the treble band separately for example? I mean where would I connect the "loose" terminal of that 11k resistor/pot.

See picture below.

Thank you!Captura de ecrã 2019-12-16, às 19.02.54.png
 

AnalogKid

Joined Aug 1, 2013
8,256
The 11 K resistor circled affects only the bass adjustment. The opamp inverting input (pin 6) is a "virtual ground". A consequence of negative feedback is that the opamp does whatever it takes at the output to cause the - input to be equal (in voltage) to the + input. Various currents can be going in and out of the 11 K, 0.022 uF, and 0.005 uF components, but those currents do not interact with each other because the voltage node where they come together always is 0 V. This makes it relatively easy to adjust the component values in each of the three networks to get the frequency response performance you want.

Note, a virtual ground is not always at zero volts. it always is at the potential of the non-inverting input, whatever that is, and in this case it is the circuit ground.

https://en.wikipedia.org/wiki/Virtual_ground

ak
 

Jony130

Joined Feb 17, 2009
5,127
To the signal input ( left side of a 1.8k resistor).

And your circuit should look like this:
EQ12.png

Where in your circuit R1 = 1.8kΩ, Pot1 = PR1 = 500kΩ; R2 = Req = 83kΩ
Also, you can compare your results with mine.
 

Thread Starter

bernardomarques

Joined Nov 14, 2019
14
Ok I get how you get to the 83kΩ, now my question is how you analyse the circuit has you have it on your latest answer.

Knowing that the gain of the inverting amplifier is -Rf/Ri, what would be my Rf and my Ri? I know I have to analyse the circuit when the C1 pot is all the way to right and all the way to the left.

For example when the C1 pot is all the way to the left is the Ri the parallel between R1 and R2? And Rf the series of all the components in that net? I'm a bit confused on that particular part.

Thank you for your help.
 

Thread Starter

bernardomarques

Joined Nov 14, 2019
14
By doing that you're admitting the capacitor's impedance is irrelevant? How is that?

If you calculate the capacitors impedance for f=20kHz you get Z_c = 1/sC <=> Z_c = 1/ (2*PI*20000*0,005*10^-9) <=> 1591549,431Ω so I don't get why you would replace the capacitor for a short circuit.
 

Jony130

Joined Feb 17, 2009
5,127
First 0.005 = 0.005μF = 5nF and Xc at 20kHz is Xc ≈ 0.16/(20kHz * 5nF) ≈ 0.16/(20000Hz * 5E-9F) ≈ 1.6kΩ

And the gain will be equal to about:

Av_max ≈((500kΩ+1.8kΩ)||83kΩ)/1.8kΩ≈71kΩ/1.8kΩ ≈ 40 V/V ---> 32dB (we ignore mid-range influance) for a signal frequancy larger than:
F > 0.16/(1.8k*5nF) ≈ 17.8kHz

And circuit simulation confirm this results:
 

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