Overdrive effect

https://github.com/user-attachments/assets/1be89b33-93ce-4a43-86b3-0c42de0525c9

https://github.com/user-attachments/assets/2ba2562c-e011-4fdd-8f06-044c0ea2faf1

https://github.com/user-attachments/assets/6c816d3a-e20a-423a-b53f-dbef71fd2bd0

https://github.com/user-attachments/assets/7a0f3d32-2f00-4b13-9497-afda9a19c826

https://github.com/user-attachments/assets/31108739-4f4a-404e-89c9-3ae1a5144b9c

3.1 Derivation of cutoff frequency of coupling/decoupling capacitor

In this derivation we consider loaded decoupling capacitor which makes lowpass filter (integrating circuit). The last derived equation in 3.1 is identical for coupling capacitor working with resistance (differentiating circuit), but it is different on laplace's s-domain, and in fourier analysis (phase and magnitude plots).

Voltage divider equation:

$$U_{out}=U_{in} \cdot \frac{Z_L}{Z_L+Z_R}$$

$$U_{out}=U_{in} \cdot \frac{\frac{Z_C Z_{Ro}}{Z_C+Z_{Ro}}}{\frac{Z_C Z_{Ro}}{Z_C+Z_{Ro}}+Z_R} = U_{in} \cdot \frac{Z_C Z_{Ro}}{Z_C Z_{Ro}+Z_C Z_{R}+Z_RZ_{Ro}}$$

Laplace's s-domain impedances derivation

Definition of Laplace's transform:

$$F(s) = L[f(t)] = \int_{0}^{\infty} f(t) e^{-st} , dt \Rightarrow L\left[\frac{d^nf(t)}{dt^n}\right] = s^nF(s)$$

considering no initial conditions.

Impedance of resistance $Z_R$:

$$U=I\cdot R$$

$$L[U]=R\cdot L[I]$$

$$Z_R(s)=\frac{U(s)}{I(s)}=R$$

Impedance of capacitance $Z_C$:

$$I=C\cdot \frac{dU}{dt}$$

$$L[I]=C\cdot L[U']$$

$$I(s)=C\cdot sU(s)$$

$$Z_C(s)=\frac{U(s)}{I(s)}=\frac{1}{sC}$$

Substituting s-domain impedances to voltage divider equation:

$$G(s)=\frac{U_{out}}{U_{in}}=\frac{\frac{R_o}{sC}} {\frac{R_o}{sC}+\frac{R}{sC}+RR_o}=\frac{R_o}{R_o+R+sCRR_o}$$

$$G(s)=\frac{(\frac{R_o}{R_o+R})}{1+s(\frac{CRR_o}{R_o+R})}$$

Then:

$$\boxed{G(s)=\frac{K}{1+sT}},\qquad \boxed{K=\frac{R_o}{R_o+R}},\qquad \boxed{T=\frac{CRR_o}{R_o+R}}$$

Amplitude characteristic - Fourier analysis

Complex s-plane argument definition: $s = \sigma+j\omega$. To get amplitude characteristic we need to analyse holomorphic Fourier transform function. To achieve that we need to get rid of real part of s-plane (constant component $e^\sigma$).

$$G(j\omega)=\frac{K}{1+j\omega T}\cdot\frac{1-j\omega T}{1-j\omega T}=\frac{K-j\omega KT}{1+\omega^2 T^2}$$

$$G(j\omega)=\left[\frac{K}{1+\omega^2 T^2}\right] + j\left[\frac{-\omega KT}{1+\omega^2 T^2}\right]$$

$P(\omega)\left[Q(\omega)\right]$ characteristics:

(https://www.desmos.com/calculator/vwyactze44)

$$|G(j\omega)|=\sqrt{\frac{K^2}{(1+\omega^2 T^2)^2}+\frac{\omega^2 K^2 T^2}{(1+\omega^2 T^2)^2}}=\frac{K}{\sqrt{1+\omega^2 T^2}}$$

Finding -3dB frequency will be done based on magnitude.

$$A=20\cdot \log\left(\frac{K}{\sqrt{1+\omega^2 T^2}}\right)$$

Cutoff condition:

$$A(\omega_{-3dB})=A(0)-3$$

$$20\cdot \log\left(\frac{K}{\sqrt{1+\omega^2 T^2}}\right)=20\cdot \log\left(\frac{K}{\sqrt{1+0^2 T^2}}\right)-3$$

$$10\left[\log\left(\frac{K^2}{(1+\omega^2 T^2)}\right)-\log\left(K^2\right)\right]=-3$$

$$\log\left(\frac{1}{(1+\omega^2 T^2)}\right)=-0,3$$

$$\frac{1}{(1+4\pi^2f^2 T^2)}=10^{-0,3}$$

$$10^{-0,3}+10^{-0,3}4\pi^2f^2 T^2=1$$

$$f=\sqrt{\frac{\frac{10^{-0,3}}{1-10^{-0,3}}}{4\pi^2T^2}} \approx \frac{1}{2\pi T}$$

Substitute T:

$$f=\frac{R_o+R}{2\pi CRR_o}=\frac{R}{2\pi CRR_o}+\frac{R_o}{2\pi CRR_o}$$

$$\boxed{f=\frac{1}{2\pi RC}+\frac{1}{2\pi R_oC}}$$

3.2 Derivation of voltage divider resistance values

We consider loaded symetric voltage divider

Impedance

$$R_{eq1}=\frac{RR_o}{R+R_o}$$

$$\lim_{R_o \to \infty} R_{eq1}= \lim_{R_o \to \infty}\frac{R}{\frac{R}{R_o}+1}=R$$

$$R_{eq}=2\cdot R$$

Flowing current loss

$$I=\frac{V_{cc}}{R_{eq}}=\frac{V_{cc}}{2R}$$

Thevenin's resistance

$$R_{TH}=\frac{RR}{R+R}=\frac{R}{2} \Rightarrow R = 2\cdot R_{TH}$$

Current loss vs Thevenin's resistance

Substituting both equations:

$$\boxed {I(R_{TH})=\frac{V_{cc}}{4\cdot R_{TH}}}$$

Ideal voltage source should have $R_{TH} \to 0$, to be as stable as possible. Ideal current loss should be $I \to 0$. Graphing this function allow to pick optimal resistance values.

$I(R_{TH})$ with $V_{cc}=9V$ characteristics:

(https://www.desmos.com/calculator/zl9quwhx7r)

3.3 Derivation of pullup resistor value

Considering methodology of choosing pullup resistors, low potential on signal line should not draw much current. On the other hand, operational amplifier has its bias current which will be drawn no matter what. It limits upper value of resistance, voltage drop on this resistor should not affect effect.

In project LM358 was used, so its bias current is around $45 nA = 45\cdot 10^{-9} A$

$$U = I\cdot R$$

$$R = \frac{U_{loss}}{45\cdot 10^{-9}}$$

Characteristics:

(https://www.desmos.com/calculator/fuyu1ywsum)

3.4 Operational amplifiers

Op amps has very high input impedance ($Z_{in} \to \infty$) and very low output impedance ($Z_{out} \to 0$). They can be analysed using 2 rules:

1. Op amps try to sustain same potential on both of its inputs,
2. Op amps bias current is negligible small (often $I \to 0$).

Using 2 rules, non-inverting amp can be modeled with:

$$U_{in} = U_{+} = U_{-} = Z_1 \cdot I$$

$$U_{out} = U_{Z_1} + U_{Z_2} = I\cdot Z_1 + I\cdot Z_2 \Rightarrow I = \frac{U_{out}}{Z_1 + Z_2}$$

$$G(s) = \frac{Z_2 (s)}{Z_1(s)} + 1$$

In this effect we are enabling user to regulate $Z_2$ (diode type, potentiometer and capacitor). We use capacitor $C_4$ to limit gain for DC current. Considering only $1M\Omega$ potentiometer and using information derived in 3.1:

$$G(s) = \frac{R_{V1}}{R_5+ \frac{1}{sC_4}} + 1 = \frac{sR_{V1} C_4}{sR_5 C_4 + 1} + 1 = \frac{1 + sC_4(R_{V1} + R_5)}{1 + sR_5C_4}$$

To calculate gain in somewhat same way that we derived 3.1 $s = j\omega$ :

$$G(j\omega) = \frac{1 + j\omega C_4(R_{V1} + R_5)}{1 + j\omega R_5C_4} = [\frac{1+\omega^2 C_4^2 R_5 (R_{V1}+R_5)}{1+\omega^2 R_5^2 C_4^2}] + j [\frac{\omega C_4 R_{V1}}{1+\omega^2 R_5^2 C_4^2}]$$

$$|G(j\omega)| = \sqrt{Re[G(j\omega)]^2+Im[G(j\omega)]^2} = \sqrt{\frac{1 + \omega^2 C_4^2(R_{V1} + R_5)^2}{1 + \omega^2 R_5^2 C_4^2}}$$

So impedance for DC current is:

$$\lim_{\omega \to 0} |G(j\omega)| = 1$$

So impedance for high frequency AC current is:

$$\lim_{\omega \to \infty} |G(j\omega)| = \frac{R_{V1}}{R_5} + 1$$

4.1 Voltage divider

Using $I(R_{TH})=\frac{V_{cc}}{4\cdot R_{TH}}\qquad$ (3.2). Voltage divider should not drain more than $0.5 mA$, so:

$$0.5 \cdot 10^{-3} > \frac{9}{4\cdot R_{TH}}$$

$$2 \cdot 10^{-3} \cdot R_{TH} > 9$$

$$R_{TH} > 4,5 \cdot 10^3 \Rightarrow R_{1,2} > 9 k\Omega$$

$$\boxed{R_{1,2} = 10 k\Omega}$$

4.2 Pullup resistor

Using $R = \frac{U_{loss}}{45\cdot 10^{-9}}$ (3.4). Maximal acceptable voltage drop is around $0.05 V$, so:

$$R_3 = \frac{0,05}{45\cdot 10^{-9}} \approx 1,111 \cdot 10^6 \Omega \approx 1 M\Omega$$

$$\boxed{R_3 = 1M\Omega}$$

4.3 Input resistor

It is standard current and voltage limiting resistor.

$$\boxed{R_4 = 100k\Omega}$$

4.4 Filtering, gain and volume potentiometers

Potentiometers were chosen based on availability and functionality. Picking $10k\Omega$ potentiometers grants optimal selection of coupling and decoupling capacitors to work with (look 4.5). Also in main feedback loop there is main bass level control, so not to do too many knobs we stuck to a static highpass filter. It also works with $10k\Omega$ potentiometer as volume control. it gives relatively low internal resistance of an effect.

$$\boxed{R_{V2,V3} = 10k\Omega}$$

Gain potentiometer was also chosen based on construction standards, but to make effect more customizable we chose $1M\Omega$ which grants more boost.

$$\boxed{R_{V1} = 1M\Omega}$$

4.5 Coupling/decoupling capacitors

Decoupling $9V$ source and $4.5V$($V_{ref}$) after voltage divider is done by $C_1$ and $C_2$. Values were chosen using Thevenin's method.

Capacitor $C_2$ works with $5 k\Omega$ resistance, we want to cutoff frequencies all above $0.01 Hz$:

$$R_{TH} = \frac{10k \cdot 10k}{10k + 10k} = 5k\Omega$$

$$C_2 > \frac{1}{2\pi R_{TH} f} = \frac{1}{2\pi 5000 \cdot 0,01} \Rightarrow C_2 > 3,18 \mu F$$

$$\boxed{C_2 = 47 \mu F}$$

Based on that selection to unify values and stick with industry standards $C_1$ is also $47 \mu F$. It will work effectively with power supply and battery to supress instant voltage spikes.

Capacitor $C_3$ works with $>1M\Omega$ of resistance as highpass filter to block any DC current into amplification circuit. We can pass everything above $0,3 Hz$:

$$C_3 = \frac{1}{2\pi Rf} = \frac{1}{2\pi 10^6 \cdot 0,3} = 531 \cdot 10^{-9}$$

$$\boxed{C_3 = 470 nF}$$

Post amplification passive filtering is done by two segments.

Using buffer allows to separate 2 filters. Load resistance in approximation infinitely small. Derived equation looks like this (3.1):

$$\lim_{R_o \to \infty} f= \lim_{R_o \to \infty} \frac{1}{2\pi RC}+\frac{1}{2\pi R_oC} = \frac{1}{2\pi RC}+\frac{1}{2\pi \infty C} = \frac{1}{2\pi RC}$$

We want low pass filte maximal cutoff point to be 5 kHz:

$$5000 = \frac{1}{2\pi \cdot 10000 \cdot C }\Rightarrow C = \frac{1}{2\pi \cdot 10000 \cdot 5000} \approx 3,18 \cdot 10^{-9} F$$

$$\boxed{C_8 = 3,3 nF}$$

We want high pass filter cutoff point to be 15.9Hz (it will be fixed, look 4.4):

$$60 = \frac{1}{2\pi \cdot 10000 \cdot C }\Rightarrow C = \frac{1}{2\pi \cdot 10000 \cdot 10} \approx 3,18 \cdot 10^{-6}$$

$$\boxed{C_9 = 1 \mu F}$$

4.6 Active buffer

Buffer takes advantage of impedance characteristics of op amp. Using rule 1 we can derive:

$$U_{-in} = U_{+in} = U_{out}$$

Voltage on the output is the same as the voltage on the input. Buffer separates audio pipeline.

4.7 Main op amp feedback loop

To give wide range of gain we picked $R_{V1} = 1 M\Omega$, for $1000x$ voltage boost we need to calculate $R_5$ value:

$$R = \frac{R_{V1}}{k+1} = \frac{1 M\Omega}{1000+1} \approx 999 \Omega$$

$$\boxed{R_5 = 1k\Omega}$$

Based on this selection we choose $C_4$ value (using derivation from 3.1). Value of this capacitor controls the filtering before amplification. On guitar it can be described as pre amp bass control. To further customize the effect aditional switch was added. It has 3 possible cutoff frequencies. We chose typical values used in overdrives:

$$\boxed{C_4 = 220n, C_5 = 1\mu F, C_6 = 2,2 \mu F}$$

So based on chosen capacitance:

$$f_4 = \frac{1}{2\pi R_5 C_4} \approx 723 Hz$$

$$f_5 = \frac{1}{2\pi R_5 C_5} \approx 159 Hz$$

$$f_6 = \frac{1}{2\pi R_5 C_6} \approx 72,3 Hz$$

Additionally we added switch (SW3) to allow user to join capacitor $C_7$ to soften diodes behaviour. It modifies $Z_2$ value:

$$Z_2 = \frac{R_{V1} \cdot \frac{1}{sC_7}}{R_{V1} + \frac{1}{sC_7}} = \frac{R_{V1}}{sR_{V1} C_7 + 1}$$

So again value was chosen based on standard.

$$\boxed{C_7 = 100 pF}$$

Effect has 2 pair of diodes to choose from:

• LEDs
• Fast switching diodes

They can be enabled in soft clipping mode (in feedback loop), and in hardclipping (disconnected from loop, shortened to ground).