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.

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.