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Back to the Breadboard

VenomBallistics

Aug 30, 2018
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[mod edit: remove double quote]
I can't explain it technically, but I can from a. guitarists' perspective.

Anytime you use effects pedals (especially O/D or distortion), you get that noise. Most guitarists, including me, handle this as follows:

In a live performance, I use a Boss Noise gate pedal. It goes in line with your pedal chain and strips away the noise. You have parameters you can tweak to get just the right amount of elimination without eroding the effect too much.

When I record, I use noise gate software to do the same thing.

The thing I like about this builds I can make the circuit sound a lot different simply by playing with the Gain, Then and Volume pots, and also changing choice of pickups on the guitar. The guitar also has its tone and volume controls. So, with all these variables, it's easy to get different colors to your sound.
it's just additive noise. the amp has some noise, the guitar is picking up as much noise as it can find, pedals have some too ... then there's environmental noise from broadcast TV and radio, EMI from industrial operations .... as you stack them, the noise adds till you notice it, and keeps right on adding until you cannot ignore it. It starts to give you an appreciation of the efforts invested in any LA studio. It's a wonder anything was ever produced with it's RF bands saturated
 
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Audioguru

Sep 24, 2016
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Here is a video of a guitar pedal with soft clipping diodes and hard clipping diodes and different opamps:
 

SparkyCal

Mar 11, 2020
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Thanks guys. AG. What does the 1M resistor specifically do in the circuit I built?
 

VenomBallistics

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Thanks guys. AG. What does the 1M resistor specifically do in the circuit I built?
the 1M typical input load resistor ties the op amp to Vref, while offering a path for your signal.
Without it, the opamp sees whatever static charge is present as its reference and the signal has nowhere to go
 

Audioguru

Sep 24, 2016
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The 1M resistor biases the + input of the opamp to +4.5VDC then the gain pot provides DC negative feedback with a gain of 1 so that the opamp output will also be at +4.5VDC and can swing equally up 3.5V to 8V and down 3.5V to 1V with the audio.
 

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SparkyCal

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Thanks guys. I am just trying g to look back and understand why the circuit works.

Pin 5 of the second op amp is fed by a junction point where the two voltage divider resistors meet.

Whereas, Pin 3 ( the + In of the first op amp) is fed by the 1M resistor.

I am trying to understand what the difference is between both feeds. Why does Pin 3 get the 1M feed, whereas Pin 5 of the second op amp, gets the feed just prior to it running through the 1M resistor. Why is that?

Pleas try to tell me in simple English. ;-)
 

Harald Kapp

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Show us the circuit diagram so we have an idea which pins you do refer to.

In the circuit @Audioguru showed in post #45 the 1M resistor is required to set the "+" input of the opamp to 4.5 V (that's what "biasing" means). At the same time this high resistance prevents the input signal coming via the 10 nF capacitor from being short circuited to ground (note the 4.5 V are buffered by a 10 µF capacitor which would be a short circuit to ground for the AC input signal if no 1 M resistor were present).
Here the "+" pin needs both DC biasing and the AC input signal.
upload_2021-6-18_7-9-1.png
If at another point in the circuit you need only biasing but no AC signal at that point, you don't need the 1 M decoupling resistor. As I said, show your circuit diagram so we have a common base for discussion.
 

Audioguru

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A guitar pickup works best when loaded with 1M or higher. The input of an old vacuum tubes guitar amplifier always had a 1M resistor.
A TL081 single or TL082 dual opamp have Jfet transistors which have the same extremely high resistance of a vacuum tube. The old 4558 opamp has ordinary transistors that have a minimum resistance of 300k which is parallel with the 1M resistor producing a total of 230k then two inputs would produce a total of only 130k which is much too low so only the used input is connected to the 1M resistor.
 

SparkyCal

Mar 11, 2020
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Thank-you guys. Harald, the schematic you used is the one I was referring to. Sorry that I had not identified it.
If I understand it correctly, the 1M, in a way, acts like a "bouncer", preventing the 10nF from escaping to ground.
Questions:

1. I don't quite under stand the AC thing. Is the signal converted to AC by virtue of it being passed through the 10nF capacitor?

2. I understand that Pin 3 needs both the DC Bias (which comes from the signal being filtered through the 1M resistor), as well as the AC signal which comes from the signal being passed through the 10nF capacitor- but is this done because the op amp inherently needs this at Pin 3? If so, where in its data sheet

https://components101.com/ics/lm4558-dual-op-amp-pinout-datasheet

would I find that? Bearing in mind that I am using a 4558D dual op amp.

AG...forgive me. I know you posted your explanation before, but sometimes I have to read things like that more than once, and at different junctures in the project, before I begin to understand them. Thanks for posting it again.

3. Unrelated- If I keep the 9 volt battery connected, but there are no guitar cables going in or out of the box, will it still be draining the battery? Or is the battery used only when I have the guitar and amp plugged in?
 

SparkyCal

Mar 11, 2020
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Hi Guys- yes, I know you can hardly wait, so I am unveiling the first prototyper of the pedal, which is called the "Beratused-AudioGurued-Harladed Overdrive, version -1.0 (yes, the version is a negative number, based on the many mistakes I made along the way.

The backdrop is a world map because obviously, there will be worldwide demand for this pedal. What should the price point t be? Maybe I pay them $50 to take one ...lol

P.S. Sorry Venom B there was not enough room for one more name, but the VenonB name sounds like a future pedal ;-)
 

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Audioguru

Sep 24, 2016
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Questions:

1. I don't quite under stand the AC thing. Is the signal converted to AC by virtue of it being passed through the 10nF capacitor?
No. The input signal is already AC probably with an average of 0V so it goes positive and negative. Your opamp circuit has a single positive supply, not a positive and negative supply so it cannot use the negative part of the input signal. The opamp must be biased at half the supply voltage so its output can swing symmetrically up and down. The 1M resistor passes the 4.5VDC bias voltage to the opamp input. The 10nF coupling capacitor blocks the 0VDC at the signal input but passes the AC audio to the opamp input. Then the + input of the opamp averages +4.5VDC but the AC audio swings it a little higher and a little lower.

2. I understand that Pin 3 needs both the DC Bias (which comes from the signal being filtered through the 1M resistor), as well as the AC signal which comes from the signal being passed through the 10nF capacitor- but is this done because the op amp inherently needs this at Pin 3? If so, where in its data sheet?
The signal does not come through the 1M resistor. A filtered half the supply voltage cones through the 1M resistor.
The datasheet does not tell you the basics of all opamps. The opamp output needs to swing symmetrically up and down. To do this its output must idle at half the supply voltage. Since the gain pot feeds the output DC voltage to the opamp inverting input and the 1k resistor has a capacitor preventing the 1k resistor from connecting its DC to ground, then the opamp has a DC gain of 1. For its output to be at half the supply voltage then its + input must be biased at half the supply voltage by the 1M resistor.

3. Unrelated- If I keep the 9 volt battery connected, but there are no guitar cables going in or out of the box, will it still be draining the battery? Or is the battery used only when I have the guitar and amp plugged in?
Pedals usually use input and output jacks that have a switch on them. The switch disconnects the battery when both jacks have no plug in them. I do not know if you have switched jacks.
 
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Audioguru

Sep 24, 2016
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I am unveiling the first prototype of the pedal, which is called the "Beratused-AudioGurued-Harladed Overdrive, version -1.0 (yes, the version is a negative number, based on the many mistakes I made along the way.

The backdrop is a world map because obviously, there will be worldwide demand for this pedal. What should the price point t be? Maybe I pay them $50 to take one ...lol
The circuit is a very common pedal circuit that you are copying. Will you pay me $50 if I take one?:)
 

SparkyCal

Mar 11, 2020
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Hi guys. I will re-read your replies when I have time to truly think about them.

In the interim, I just plugged in the pedal again and tried all these different variations of tone, gain and volume settings. I was really amazed on the range of sounds this little pedal can produce. It seems to be far more dynamic and wide ranging than most pedals I have tried. You can go from very dark lows to very bright, almost screechy (in a good way) distorted sounds. I wonder if this is due to using the 1nF capacitor?

BTW- I popped it out of that black plastic case that proved too small, and it's inside a wooden case that is a little larger and seems to be a better fit.
 

Audioguru

Sep 24, 2016
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I just plugged in the pedal again and tried all these different variations of tone, gain and volume settings. I was really amazed on the range of sounds this little pedal can produce. It seems to be far more dynamic and wide ranging than most pedals I have tried. You can go from very dark lows to very bright, almost screechy (in a good way) distorted sounds. I wonder if this is due to using the 1nF capacitor?
Where is a 1nF capacitor? The circuit does not have one.
 

SparkyCal

Mar 11, 2020
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I based it on this stamenet you had made on the previous thread.


I think I mentioned before that the value of C23 (the input capacitor) is way too high at 470nF then it passes earthquake frequencies down to 0.34Hz that might clobber the signal. 10nF would reduce 16Hz and lower, 4.7nF would reduce 34Hz and lower, 1nF would reduce 160hz and lower. Do you have a solder slurper? Try changing C23.

I don't know if I made the right change, but it works great.
 

Audioguru

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The 1nF was calculated to cut 160Hz and lower for the Jfet-inputs TL081 opamp. The 4558 opamps have ordinary input transistors with a minimum input resistance of 300k ohms which in parallel with the 1M biasing resistor produces a minimum of 231k and cuts 693Hz and lower. The input resistance of a "minimum" 4558 also reduces the output level and cuts guitar high frequencies.

4558 opamps have a range of input resistance then each one will sound different. TL081 opamps will all sound exactly the same.
 

SparkyCal

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I can't pretend that I understood that, but I can say that the unit sounds great. Has a lot of range, for some reason.
 

SparkyCal

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I really hate to bring this up, but.....

As you know, I built this circuit correctly (finally).
I built it two more times after that, and it worked.

Now, I tried building it with greater care, and with a view toward making it look tidy, and it does not work. The op amp is getting all the right readings, so I aim assuming that it isn;'t the issue. ideas?
 

Harald Kapp

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Check all your wiring. You seem to have a tendency to forget a connection or a component. My favorite method is to make a copy of the schematic, then trace every component and connection and mark it on the copy as done. Once you're finished there should be no more components or connections that are not marked. Likewise yu should not have found any on the real circuit that aren't part of the schematic.
It all oils down to some meticulous checking.
 

SparkyCal

Mar 11, 2020
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Thank-you Harald. I think I found the problem, and then discovered another problem

I recently bought new guitar jacks because I was out of them. These jacks with three prongs arrived. Turns out I wired the In jack incorrectly. When I tested it with a multi meter for continuity between the lug and the tip, it failed. So, I used the correct lug and viola...I got sound! That's the good part.

But, I'm not getting a lot of distortion. in fact, it sounds like a clean signal coming through.

As you may know, I breadboarded the circuit and got it working. That breadboard model is still intact. So, before I discovered the issue with the jack, I began looking on what was different between the breadboard model and the circuit I am working on. One difference was the capacitor C220, which had a value of 220nF. On my circuit, I used the 3.3uF capacitor that AG had recommended. Just to eliminate any discrepancies, I changed the 3.3uF and inserted the 220nF. I am wondering if this could have caused me to lose the distortion? Or maybe I wired the capacitor replacement wrong, It is late at night here and I;'m sleepy. Will take a close look tomorrow.

Thank-you for the advice, Easier to make mistakes when sleepy ;-(
 
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