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Distortion (pedal?) circuit

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Charlie J

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I have a project in which I have to design/build an analog distortion circuit.
It is my aim to produce (harsh) symmetrical and asymmetrical clipping, with enough gain to also produce near-square output waveforms. I'm trying to keep it as simple as I can. A picture of the circuit I came up with is added as an attachment.

What I'd like to know is exactly how this circuit works (I have an idea which I'll write below) but I'm not 100% so all corrections/explanations welcome!.... ALSO, could this circuit work in a guitar distortion pedal. If not, why not? and how could I adapt it to do so?

*The following is what i think the circuit is doing but correct me if I'm wrong*

The 741 opamp (non-inverting) is used to amplify the input (voltage) signal. There is a pot controlling the gain of the opamp. A 9V battery is being used to power the opamp and provide a voltage divider (ref voltage?) for opamp input... The max output of the opamp will be limited (in amplitude) , hence once the opamp is gained sufficiently, the output signal will be clipped (evenly) top and bottom. A second pot is at the output end of the circuit to allow volume control. There are capacitors at either end of the circuit (coupling), as is the other capacitor(?). And there is a switch that can incorperate a diode into the circuit, which would produce asymmetrical clipping.

I built the circuit on MULTISIM using a signal generator for input signal to the circuit and used an oscilloscope to veiw the output waveform. The appeared to operate as I had hoped (clipping at approx +/- 3.6V and at -0.6V with diode) and the volume pot seems to work.

*Are there any obvious faults with this design?...*

Another concern I have is impedance matching...I read somewhere that the input/output of a typical distortion pedal is high impedance (which the opamp can take, right?, but the output impedance is low?) I'm not sure if/how to match it at the output. What is the result of not or incorrectly matching the impedences? Also, not 100% sure if my capacitor (and resistor) values are viable for this circuit...
 

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A 741 opamp is 42 years old and performs poorly with audio because it has trouble above only 9kHz and it is noisy (hiss).
Most audio opamps like the TL071 single, TL072 dual and TL074 quad work perfectly up to 100kHz and are low noise.
The 741 opamp does not have a minimum supply voltage like most other opamps. Some work with a supply as low as 12V and others don't. it is spec'd only with a 30V supply.

Many experts say that a guitar pickup should be loaded with 1M ohms minimum. Your circuit is 25k ohms which is much too low.
The TL07x opamps have Jfet inputs so your 50k resistors can be 2.2M ohms each then the input resistance is 1.1M ohms. Then the input capacitor value can be reduced.

Your circuit is missing an important supply bypass capacitor. I use 100uF.
 
Another victim of Multisim. We seem to be getting a few each week lately.

When you switch the diode in, the capacitor between the diode and the volume control will gradually charge until it stops drawing current even on peaks. The output will go silent. Multisim probably didn't notice that, because the capacitor is 10 farads.

10 farads is also a million times too much for the capacitor in the feedback loop. I would suggest 10µF to replace both of them, which may simulate accurately and will be much cheaper when you build it. The 10µF on the input is good.
 
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A 741 opamp is 42 years old and performs poorly with audio because it has trouble above only 9kHz and it is noisy (hiss).
Most audio opamps like the TL071 single, TL072 dual and TL074 quad work perfectly up to 100kHz and are low noise.
The 741 opamp does not have a minimum supply voltage like most other opamps. Some work with a supply as low as 12V and others don't. it is spec'd only with a 30V supply.

Many experts say that a guitar pickup should be loaded with 1M ohms minimum. Your circuit is 25k ohms which is much too low.
The TL07x opamps have Jfet inputs so your 50k resistors can be 2.2M ohms each then the input resistance is 1.1M ohms. Then the input capacitor value can be reduced.

Your circuit is missing an important supply bypass capacitor. I use 100uF.

Thanks for replying!
So you would recommend replacing the 741 with a TL071, and the 50k resistors for 2.2M ohms. I'll give that a go!

What is the purpose of the "supply bypass capacitor", and where would it go in relation to my original circuit?
 
Another victim of Multisim. We seem to be getting a few each week lately.

When you switch the diode in, the capacitor between the diode and the volume control will gradually charge until it stops drawing current even on peaks. The output will go silent. Multisim probably didn't notice that, because the capacitor is 10 farads.

10 farads is also a million times too much for the capacitor in the feedback loop. I would suggest 10µF to replace both of them, which may simulate accurately and will be much cheaper when you build it. The 10µF on the input is good.

Thank you for the advice! I'm gonna try out these changes and simulate them again (when I get the chance), then if all seems to work I'll built/test the circuit.

As for my 'theory' of how the original circuit works, am I on the right tracks or am I mostly talking nonsense?
 
The supply bypass capacitor filters the voltage from the battery. Then the circuit will not oscillate. Without a supply bypass capacitor the voltage from the battery jumps up and down with the signal or with oscillation.
 
The supply bypass capacitor filters the voltage from the battery. Then the circuit will not oscillate. Without a supply bypass capacitor the voltage from the battery jumps up and down with the signal or with oscillation.

I'm still unsure as to where in the circuit this capacitor would go. Would it go between the ground and the lower of the two resistors in the voltage divider?
 
The supply bypass capacitor that filters the voltage from the battery goes parallel to the battery, on the circuit board not at the battery.
 
I've redrawn the circuit with the suggested corrections. Is the supply bypass capacitor connected correctly here?
 

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  • distortion_circuit2.JPG
    distortion_circuit2.JPG
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Is the supply bypass capacitor connected correctly here?
Correct.
But the 10uf value of the input capacitor will take 55 seconds to fully charge. Reduce its value so that it passes the lowest frequency you want but charges must faster.
 
Thanks! I've been looking for info on capacitors to help me work out an appropriate value, (don't suppose you know any good sites?)

I ended up looking at schematics for existing pedals and the values 2.2uF and 0.1uF come up a lot for input capacitors.
 
The formula for calculating a coupling capacitor is "1 over 2 pi Rf".
The R is the input resistance of your preamp which is 1.1M if you use a Jfet input opamp and the f is the lowest frequency you want which might be 30Hz.
Then the input capacitor value is 5nF. Use 10nF or 0.01uF.

Your 10uf input capacitor has a value that is 1000 times too high.
 
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