Guitar Fuzz Box

[MauriceM]

hi , i am new to forum and is studying electrical engineering 2nd year i m currently working on my project, as the title implies, a guitar fuzz box. i have figure out that D1 , D2 and R4 are the clipper circuits, C2 and C3 stabilizing oscillations, R1 and R2 as a reference voltage via the voltage divider rule, my question is what is C1 for ? and how does R1 and R2 affect the op-amp? i simulate my circuit in circuit maker with a sound signal 0.4v peak to peak, if i m not mistaken, diodes has forward voltage range of 0.6 - 0.7, but the output still produces a Fuzz effect? i think i m wrong with this, i hope you guys can help me with this thanks in advance !

 

[Nigel Goodwin]

hi , i am new to forum and is studying electrical engineering 2nd year i m currently working on my project, as the title implies, a guitar fuzz box. i have figure out that D1 , D2 and R4 are the clipper circuits, C2 and C3 stabilizing oscillations, R1 and R2 as a reference voltage via the voltage divider rule, my question is what is C1 for ? and how does R1 and R2 affect the op-amp? i simulate my circuit in circuit maker with a sound signal 0.4v peak to peak, if i m not mistaken, diodes has forward voltage range of 0.6 - 0.7, but the output still produces a Fuzz effect? i think i m wrong with this, i hope you guys can help me with this thanks in advance !

Most of your thoughts are unfortunately incorrect.

R1, R2 and C1 form a split supply for the opamp, rather than having two batteries - it's a simple potential divider and a decoupling capacitor. You should also have a decoupling capacitor across the main supply.

C2 and C3 are coupling capacitors, used mainly because of the previously mentioned virtual split-supply. They allow the AC signal to pass, but block the DC bias voltage on the opamp.

R3 sets the input impedance, and R4 is the feedback resistor that sets the gain of the opamp (but ONLY at low signal levels). At high signal levels D1 and D2 wil conduct and greatly increase the feedback (and thus reduce the gain at high levels).

If you remove the two diodes you have a standard inverting amplifier circuit.

 

[MauriceM]

thanks a lot ! and i was totally wrong damn , initially i thought the diodes were for clipping . 1 more question, the additional decoupling capacitor is added across the main supply, i added a diagram for the particulat question ! THanks again

 

[Jack Luminous]

Maurice, you might be interested in this article about creating guitar distortion circuits: http://www.generalguitargadgets.com/richardo/distortion/
If you scroll all the way down to the bottom of that page, there is a distortion circuit that is analyzed and explained.
I believe those diodes in your diagram are there for clipping.

Looks like the extra capacitor you drew is backwards. And I'm wondering if it's not better to have it placed directly across the V+ and gnd rails.

 

[Nigel Goodwin]

thanks a lot ! and i was totally wrong damn , initially i thought the diodes were for clipping . 1 more question, the additional decoupling capacitor is added across the main supply, i added a diagram for the particulat question ! THanks again

The supply decoupling capacitor should go across pins 4 and 7 of the opamp, the supply pins for it - not in either of the places you drew it.

The diodes ARE for clipping, just that they do so in the feedback loop by altering the feedback at high signal levels - conventionally the diodes would directly clip the signal, but this is a 'better' way of doing it.

 

[MauriceM]

Clipping Diodes

do u might elaborating more on the clipping diodes part ?i have some difficulties explaining it , thanks a bunch

 

[Nigel Goodwin]

do u might elaborating more on the clipping diodes part ?i have some difficulties explaining it , thanks a bunch

The gain of an inverting opamp is set by the feedback resistor divided by the input resistor. In this case the feedback resistor has the two diodes across it, so once the output of the opamp goes high enough (or low enough) to make the diode conduct the feedback resistor is effectively lowered, reducing the gain. So starting with a very low signal the opamp gives gain, but as the input signal increases, eventually the diodes start to conduct - reducing the gain and preventing the output going above the level set by the diodes. So once the output reaches a certain level it won't go any louder, effectively clipping the signal.

 

[MauriceM]

hi again this is my final circuit, i would like to ask if C2 and R5 is a low pass filter or high pass filter, and also C3 and R4 ? the R4 is to vary the roll frequency. IS there a way to derive the Vout for the circuit as i need it to obtain the gain , i got my Vout from the graph, but i need a formula to calculate it theoretically , THANKS for your HELp !!

 

[akg]

the impdence of C2R5 combo and C3R4 combo will go down as freq goes high.
yes you can derive the gain equation , but it will contain frequency component and exponential componets. you need to calculate the impedence of the feedback loop and , input side separately and gain = Zfb/Zin.

 

[MauriceM]

can u elaborate more on the equation

How do i obtain the Gain Formula ? is it A = (OutpuT impedance/input impedance) ? the input impedance is C2 and R5 combo but how do we derived the output impedance? C3 ANd R4 only ? neglecting the DIodes ? thanks

 

[Nigel Goodwin]

How do i obtain the Gain Formula ? is it A = (OutpuT impedance/input impedance) ? the input impedance is C2 and R5 combo but how do we derived the output impedance? C3 ANd R4 only ? neglecting the DIodes ? thanks

It's not the output impedance, it's the feedback impedance, and it's basically R4 - likewise the input is basically R5, the capacitors will only start to take any effect at frequency extremes (C2 will roll off at low frequencies, and C3 will roll off at high frequencies).

As I mentioned at the beginning of this thread, the two resistors will set the LOW LEVEL gain of the stage, at higher signal levels the gain will be changed (reduced) by the two diodes, and can't really be calculated, as it will depend on the signal itself.

 

[audioguru]

Your final circuit shows a 10uF capacitor feeding an 8 ohm speaker. The value of the capacitor is so low that frequencies below 2kHz will be reduced.

It has AN OPAMP instead of a power amp driving the 8 ohm speaker.
Since the opamp has a low max output current then the max voltage across the speaker will be only about 0.16V peak and the clipping diodes won't work. You will hardly hear the speaker because its volume will be extremely low.

What is the function of the pot R3?

 

[Dr.EM]

I think R3 is for tone control, but i'm not sure. I hoped the speaker symbol merely represented a guitar amplifier, but if not, then audioguru is right and some changes will certainly need to be made to get any functionality out of this.

 

[audioguru]

You don't get a tone control function by shorting the high frequencies output of an opamp to a 1uF capacitor to ground. Nothing would happen until the pot's resistance is about 2k ohms then the opamp would current-limit the higher frequencies if the speaker wasn't there.

 

[Nigel Goodwin]

Your final circuit shows a 10uF capacitor feeding an 8 ohm speaker. The value of the capacitor is so low that frequencies below 2kHz will be reduced.

Never noticed the speaker there!, it makes nonsense of the thread title, and as you say would prevent it working at all!.

 

[audioguru]

Never noticed the speaker there!, it makes nonsense of the thread title, and as you say would prevent it working at all!.

The speaker's sound would be low and would be VERY fuzzy sounding.

 

[Jack Luminous]

I put this circuit together on a breadboard tonight, and with a few minor alterations, got it to sound very nice as a guitar fuzz effect.

1) Changed the speaker to an output jack.
2) Ran R3 lug 1 to ground.
3) R3 lug 2 (wiper) to output jack.
4) R3 lug 3 to C5 cathode.
5) Eliminated C4.
6) Used a 1M linear pot for R4 gain control.

Now R3 is working as a volume control. I used a 100K log pot.

*Edit* ...also substituted a TL081 for the LM741.

 

[PluginBaby]

Hi, I'm a complete novice when it comes to electronics, but I'm VERY keen to get into it and start building guitar effects. I don't want to start off by asking a whole lot of stupid questions, so where should I go to start learning about the stuff I need to learn about?

Thanks,
Andy

 

[HiTech]

visit these sites:

http://www.musicianstechcentral.com/schmatic.html#misc

http://www.diystompboxes.com/pedals/schematics.html That one is a real doozy of a site!!

 

[moody07747]

Maurice, you might be interested in this article about creating guitar distortion circuits: http://www.generalguitargadgets.com/richardo/distortion/
If you scroll all the way down to the bottom of that page, there is a distortion circuit that is analyzed and explained.
I believe those diodes in your diagram are there for clipping.

Looks like the extra capacitor you drew is backwards. And I'm wondering if it's not better to have it placed directly across the V+ and gnd rails.

I am also into guitar FX peal building now..just started out

heres a few links i have saved

http://www.diystompboxes.com/smfforum/
http://www.harmony-central.com/
**broken link removed**
**broken link removed**
http://www.geofex.com/

and the best of them all IMO

http://www.generalguitargadgets.com/index.php

im working on the ultra clean power supply from that one now, its going slow but almost done.