Baxandall audio amplifier to headphones

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I'm making a tone control audio amplifier for a pair of headphones.
I'm having trouble understanding the baxandall circuit.

So I understand that the baxandall circuit is a tone control circuit, able to configure a bass and treble through RC filters.

The general understanding I have from it, is that the capacitance of the capacitor can vary resistance depending on the frequency from the equation f = 1/(2piXC). By using this cutoff frequency you can determine resistors to put in the circuit and direct signal elsewhere.

Note: The circuit in the picture is not a final design, they just have varying values to experiment.

The questions are:

1. Even though the equation listed above is for a capacitor, RC filters usually have a resistor and a capacitor. I don't understand where the RC combination for low and high on this could be.
2. Assuming that bass is 200Hz and below and 2.5kHz and above is for treble, how could I change the values for these figures?
3. I understand that the op-amp is for gain however, is the gain configured by R3 and R2?
4. Without the op-amp the signal input and output is unchanged, why is that?
5. The signal output is out of phase by pi, do I need an inverter to correct this phase?
Here is my analysis of the circuit. Suppose a frequency of 1000Hz, and assume that there is a bass cut i.e X1 is max so 100K. (Using all values in circuit except the pots being 100k)

There are two routes the signal goes through.

Through R1 and R4. From R1 goes to C2 as the reactance of the capacitor is 3386ohms (1/(2pi*47nF*1000Hz)) which is smaller than the 100k, this then goes to R3 to negative input of op amp.

This is the thought process I have with my analysis, which leads to the questions above. Obviously not all current goes through a single route, but because the resistances are smaller I'm saying that the signal passes through them instead.


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> Assuming that bass is 200Hz and below and 2.5kHz and above is for treble, how could I change the values for these figures?

*Any* R-C filter, to change frequency: leave the resistors alone, change the caps bigger for lower frequency (or vice versa).

However your 10k pots with 22k/30k end resistors promise very little boost/cut. Assume the caps go short/open at infinite high/low frequency and compute the gain. Only about +/-3dB, while we usually give much more.
I'd suggest first understanding the Bandaxall active volume circuit, the tone circuit is basically two active volume control circuits with complementary freqeuncy-dependences. It is definitely not made of first order filters as the response has both poles and zeroes.

The opamp keeps the wipers at/near virtual ground, forcing the ratio between input and output to depend on the pot ratio. In the tone circuit the two parts are smoothly crossed over in frequency due to the capacitors. Yes it inverts. The R in the RC time constant is due to a network, not just one resistor.
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I see the point. Turn knobs to Bass -5 Treb -5. This is a 2-pole bandpass. Bass +5 Treb +5 is a 2-pole band reject. If you try to plot the interactions when both knobs are off zero you have to deal with two poles.

But we are Audio Folk. We normally run the midband fairly flat, and judge the response separately in bass and treble. Then the second pole is "negligible" (for some defintions of negligible) and the problem can be worked first-order.

The full-blown Bandaxall active volume circuit is a rather different plan with a fixed-gain amp inside an inverting loop. Better to look at the simple inverter (Bax did cover this on the way to his Volume Control).

EDIT: why is this in "Power Supplies"?? I suggest "Analog Line Level". (Tone controls are generally line-level, even if for headphones.)
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The Baxandall control is made from connected first order filters. It is essentially an adjustable lead-lag. Poles and zeroes do not necessarily imply higher order, if some of them are at the origin.

Yes, I was a bit confused there, but active circuitry isn't limited to first order behaviours combined, it allows for resonant higher order behaviour, which is what the circuit gives. The active element produces something more powerful, as the equations are of a different form allowing complex poles and zeroes.

Just as you wouldn't call a Sallen+Key filter as connected first order filters (it can be 2nd, 3rd, or higher), you don't think of the Baxandall circuit this way either.
The Baxandall circuit is quite different from the Sallen-Key. The S-K gets second-order behaviour with variable Q by using feedback (a form of partial bootstrap) to what would otherwise be a pair of similar first-order filters. The Baxandall stays with first-order but allows you to adjust whether it is LP or HP and by how much.

The Baxandall has an input impedance of Z+Ri into a virtual ground, a feedback impedance of Z+Ro and the constraint that Ri+Ro=a constant (the pot resistance). Z is some frequency-dependent impedance, almost always first order. Hence the gain is just (Z+Ro)/(Z+Ri) - which is a first order lead-lag except when Ro=Ri (the 'flat' position).
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