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PLEASE: Help with my distortion guitar effect!!!???

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wouter127

New Member
Okay so i decided to created a distortion effect that limits the signal.
If the signal looks like a sine, the upper and lower peaks should be
cut off.
I thought i'd do it with this:

Diode limiter..JPG

The 10V AC signal represents the input of the guitar. The guitar signal varies from
0-4V according to how hard you strum the strings, but this is just a simulation.

This part still works: if i adjust the two DC sources, i can vary the cutoff value of
the guitar signal. In the picture the sources are 3V and -3V so the signal is cutoff
at 3,7V and -3,7V(including the diode voltage drop).

What i want is to adjust the voltages at the same time using just one potentiometer.
Thats why i builded this adjustable symmetric supply:

symmetric supply&#.JPG

The supply consists of a 9V battery and a voltage divider.
By varying the potentiometer i can adjust the V+ and V-(referenced to ground) to
0 to 4,5 and 0 to -4,5. Which is perfect for my guitar signal.

I thought i could now just hook it up to the other circuit(V+ below the first diode and V-
below the second). It doesn't work??!!!

Maybe it's really stupid what i'm trying right now, i am pretty new at circuit design
but i have studied a lot of theory.

Somebody, please help me!!!
 

audioguru

Well-Known Member
Most Helpful Member
Your simple circuit has a high output resistance. It needs a couple of opamps for both polarities to have low output resistances like two batteries.
 
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wouter127

New Member
Thanks for your fast response!:D

Dont you mean i need opamps for a high output resistance?
The output resistances are the resisantce between +Vout and GND
and between -Vout and GND right?

Or is it the resistance from the 9V input to the V+ and from 9V
to V-?

I'm confused:confused:

BTW any cool ideas to make the guitar effect sound better are appreciated:D
 

Hero999

Banned
That's not an adjustable symmetric supply.

Do you need to use two batteries?

It's possible to use a single supply for this circuit.

I have an idea involving a few op-amps, it might look complicated but works on the same principle as your circuit.

Two of the op-amps could be eliminated but only if the load has a really high resistance.
 

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Bob Scott

New Member
BTW any cool ideas to make the guitar effect sound better are appreciated:D
You might try just limiting one polarity more than the other. ie: Just limit the top half.

Clipping both top and bottom generates odd order harmonics.

Clipping the waveform asymmetrically (eg: clip top half only) is the only way to generate even order harmonics. Even order harmonics sound more musical, and are the kind loved by audiophools who use class A tube amps.
 

wouter127

New Member
Thans for your schedule! It looks good(lot more components then mine):).
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?
You are also using capacitors, are those to alter frequency or to block signals?

I am trying to figure out why my divider circuit isn't a supply.
Because the output voltages are of the values i want.

Sorry for my avalanche of questions:D But i really want to learn!!!

I have been studying electrotechnics for 3 years now and i still cant design
a simple circuit cause everyone steals them from internet anyway...


@bob scott
Thank you for your musical suggestion, i will definitilly make use of it.
Though i doubt my three dollar circuit will sound like those valves:p.
 
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Hero999

Banned
Thans for your schedule! It looks good(lot more components then mine):).
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?
.
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.
 

wouter127

New Member
Great explanation!
I think i'll try your schedule with some adjustments.
(so i actually do something myself)
As bob scott said the output signal wil create a lot of
high order harmonics. I'll try to put an adjustable filter that
cuts of the higher frequencies behind your circuit.

I'll post with results!

And thanks again Hero for your effort:)
 

Hero999

Banned
Great explanation!
I think i'll try your schedule with some adjustments.
You mean schematic.:)

If you make modifications and ask more questions, please post what modifications you've made and why.
 
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MrAl

Well-Known Member
Most Helpful Member
Thanks for your fast response!:D

Dont you mean i need opamps for a high output resistance?
The output resistances are the resisantce between +Vout and GND
and between -Vout and GND right?

Or is it the resistance from the 9V input to the V+ and from 9V
to V-?

I'm confused:confused:

BTW any cool ideas to make the guitar effect sound better are appreciated:D

Hi there,


One of the other things you might consider to make the distortion effect even 'cooler' is
to add some compression.

Compression acts like an automatic gain control, where the input signal gets amplified
by a variable amount in order to keep the output signal relatively constant (not perfectly
constant however).
One of the problems with the simple clipping circuit (for adding distortion) is that when the
string is first plucked the distortion is quite pronounced, but as the string damps out naturally
the signal level falls fast and the clipping level stays the same so the distortion seems to
diminish as the string damps out. Compression helps this quite a bit.
Another idea instead of compression might be to make the clip level variable instead, so
that the clip level remains the same relative to the signal level. That simply requires a
circuit to rectify the signal and provide a dc output that the clip level be adjusted with as
the signal dies out. Shouldnt be too hard to do.

Compression also goes by the other name of 'sustain' and you can probably read about
it all over the web.

Other parts of the signal envelope you may want to read about are:
Attack
Sustain
Decay
Release

and also:
Phase Shifter
Reverb

Some effect simulators do all of these and more.
 
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Hero999

Banned
That's a good idea, maybe the first op-amp could be better used as a precision rectifier to automatically adjust the clipping limit?

I'll have a look if the OP is still interested?
 

Hero999

Banned
Here's an adjustable clipper with a peak detector.

This circuit is not buffered like the one I posted previously.

If the source is high impedance and the load is low impedance a couple of buffers will need to be added.

You'll need to use quite a fast op-amp because the op-amps go in and out of saturation at double the fundamental frequency of the signal.
 

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MrAl

Well-Known Member
Most Helpful Member
Hi there Hero,


Looks interesting...did you by any chance do a simulation to see how it responds?
 

Hero999

Banned
I did a quick simulation in an old version Crocodile Technology, a kid's simulator which I like because it's easy and is real-time. I haven't simulated it with LTSpice of any real software.

It work all right, there's a slight glitch at the top of the positive cycles which is because of the peak detector but I don't see how it's a problem because it's a distortion circuit.
 

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wouter127

New Member
First of all i wanna thank you for all you've done.

Though you're circuit looks great i was having my doubts
about what makes a signal sounds good.

Looking on the net i saw that its really important to have just the right
amount of sustain, the reamain of instrument timbre, have ass less intermodulation
(sounds bad when playing chords), etc.

I found it take all these factors in consideration. Thats why i wanted to
post some effects that are already confirmed sounding good, and let you guys
analyze them. (if you feel like it)
 

wouter127

New Member
Here are the schematics (thanks hero:p)

a_procorat..gif
a_ibanezts..gif
a_mxrdist..gif

I found these at:
GM Arts - Guitar Amplifiers

The most notable difference to me with hero's schematic, is the way the diodes
are connected. They're in the feedback loop and thats probably why they apply
to all signal levels. The limiting isn't adjustable tough is at is in hero's;).

I'm hoping someone can explain these a little to me(espacially how the op-amp
does its thang:))

Thanks again:D
 

Hero999

Banned
The first and last schematics you posted are fixed clippers, all signals under 0.6V peak at the output of the op-amp won't be distorted at all.

The one with the diodes in the feedback look won't distort at all at lower frequencies <100Hz or so and will be greater at higher frequencies.
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,


It's been quite a while since i have looked at these kinds of circuits in detail, but i notice that one of the requirements in all of them is the ability to set the 'drive' level, and that it would be nice to have the same level of distortion for any level signal until actually adjusted by the user dynamically for the chords.
Taking all of these into consideration, it seems what we want is a precision rectifier where we can get a guaranteed clipping level which is the same for any level, in parallel with a simple adjustable gain stage with a mixer to add both the clipped signal and the pure signal. That would mean we can obtain virtually any degree of clipping from 100 percent (total square wave) to zero percent (total sine wave). The circuit should still be simple: a clipper, a gain stage, a mixer to mix the two outputs. If you want to experiment with asymmetrical clipping then that could be built into the clipper, or a second clipper only with asymmetrical clipping can be constructed and that output could be also presented to the mixer, which would then mix all three signals and allow the user to adjust for 100 percent regular distortion, 100 percent asymmetrical, 100 percent pure tone sine, or anything in between.

Hero, maybe you want to give this a shot?
 
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