Thans for your schedule! It looks good(lot more components then mine)
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The thing is i kinda want to do it myself so i can learn!
When creating a circuit, like the one you did, what kind of stuff do you have to consider?
I know in and output resistance are really important but how can i alter them to
my wishes? What are my wishes in this guitar circuit?
I forgot to number the op-amps in my schematic, for the purposes of my explanation below, the op-amps are numbered U1 to U4 from left to right.
R1 and R2 divide the supply voltage by two, giving 6V in this case. U1 is a unity gain follower with a high input impedance and a low output impedance, which buffers R1 and R2 thereby reducing the output impedance from 50k to near zero.
R3 and R4 provide the variable voltage you require. The output voltage will depend on R3's setting and will always be below half the supply voltage.
U2 acts as a unity follower to reduce the impedance from R3.
U3 is an inverting amplifier with a gain of 1, it inverts the signal from U2 so it's added to half the supply voltage.
For example, if the output voltage from U3 is set to 4V, 6V-2V, the output from U3 will be 6V+2V.
Look at it another way, U1 forms a virtual earth circuit, its output can be considered 0V and the supply rails +/-6V. If the voltage on R3 is -2V, the output voltage from U3 will be +2V. R3 and R4 could just've easilly been connected from the positive rail to the output of U3 to give a positive voltage but the diode polarities will need to be reversed, you'll see this later.
R5 and R5 bias the input signal around half the supply voltage (or 0V if you choose to see it as a virtual earth).
D1 and D2 will clip the input signal to 6V +/-2V +0.6V (assuming the output voltage of U3 is 6-2V).
You are also using capacitors, are those to alter frequency or to block signals?
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The capacitors just block DC and allow AC to pass through.
The following formula can be used to calculate the minimum capacitor value required to pass the lowest frequency of interest.
[latex]C = \frac{1}{2 \pi RF}[/latex]
F is the frequency, R is the load resistance and C is the capacitance.
For example
[latex] C = \frac{1}{2 \pi 10 \times 10^3 \times 20 = 795.8 \times 10^{-9}} = 795.8nF[/latex]
The nearest value up is 1μF.