Why is my EQ distorted?

[xorophone]

Hello! I wanted to build a very simple EQ as a small side project while I'm waiting for some parts. I only had LM358 op-amps available at the time so I decided to go with this one: https://courses.cit.cornell.edu/bionb440/FinalProjects/f2005/asa27/ASAfinaldemo_files/image010.jpg

It's done. The EQ part of it works, but for some reason it's heavily distorted. A pretty horrible distortion too. I thought the chip might be broken so I tried changing it, but it didn't work.

I have two other theories:
1. There's a solder bridge or something somewhere.
2. This EQ can't handle the signal from a guitar/bass.

What do you guys think? Which part of this circuit could make a distortion? Would it be able to handle an instrument signal? If not, is there any way I can make it work with my guitar and bass?

Thanks!

[FiveseveN]

That circuit is set up for a bipolar power supply (not shown in the schematic).
Unrelated to bias/distortion issues, it also has a lower input impedance than you'd want to plug an instrument in. You have a dual op amp, use the other half to buffer/boost the input.

[xorophone]

That circuit is set up for a bipolar power supply (not shown in the schematic).
Unrelated to bias/distortion issues, it also has a lower input impedance than you'd want to plug an instrument in. You have a dual op amp, use the other half to buffer/boost the input.

Thanks for the reply! What is a bipolar power supply? Can my type of supplied power somehow be converted to bipolar?

[mth5044]

That circuit is set up for a bipolar power supply (not shown in the schematic).
Unrelated to bias/distortion issues, it also has a lower input impedance than you'd want to plug an instrument in. You have a dual op amp, use the other half to buffer/boost the input.

Thanks for the reply! What is a bipolar power supply? Can my type of supplied power somehow be converted to bipolar?

Sure can.

Put two resistors across your supply rails. Their junction is what we'll call a bias voltage or reference voltage. If your using 9V, it should be 4.5V. Connect the ground connections in your schematic, the 270 ohm resistor and the + input of the opamp, to this new reference voltage. Keep the opamp power pins on your V+ and ground supply.

[Blitz Krieg]

Can my type of supplied power somehow be converted to bipolar?

[xorophone]

Can my type of supplied power somehow be converted to bipolar?

That circuit is set up for a bipolar power supply (not shown in the schematic).
Unrelated to bias/distortion issues, it also has a lower input impedance than you'd want to plug an instrument in. You have a dual op amp, use the other half to buffer/boost the input.

Thanks for the reply! What is a bipolar power supply? Can my type of supplied power somehow be converted to bipolar?

Sure can.

Put two resistors across your supply rails. Their junction is what we'll call a bias voltage or reference voltage. If your using 9V, it should be 4.5V. Connect the ground connections in your schematic, the 270 ohm resistor and the + input of the opamp, to this new reference voltage. Keep the opamp power pins on your V+ and ground supply.

I think I'm starting to understand it a bit now. I made this little schematic by combining the information from both your posts:



Is this right or have i totally misunderstood it? The only concern I can think of is that the 0v wouldn't be 0v anymore, but I don't know. I'm guessing the -9v should say -4.5v or something too. If this is right, which value resistors would I have to use? And would it work with 1/2w resistors? I only have 1/2w and 1/4w right now.

[nocentelli]

This image explains how to get convert a single sided +9v supply to give a 1/2 supply reference voltage so you don't need a bipolar supply or two batteries AND it has an opamp input buffer. I would replace the first 100k bias resistor (to the non-inverting input) with a 1M and drop the second bias resistor altogether so the non-inverting input is connected directly to the 10k||10k node.


https://www.google.co.uk/search?q=baxandall+three+band&client=ms-android-sonymobile&prmd=sivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj16rWGiJHOAhWMLsAKHUUFBZsQ_AUICCgC&biw=598&bih=279#imgrc=FOXVCkbekNUW5M%3A

[anotherjim]

See this basic single supply scheme...

The equal voltage divider formed by R3 & R4 (we call Vref) is all you need to change in your filter, replacing the direct connection of the + input ground. Fortunately the filter components on the - input are all isolated from ground by capacitors, so that 270R can remain connected to ground where shown. The resistor values are not critical, most would pick 10k to 100k values. Less than 10k starts to eat battery current for no good reason.

With single supply, Vref is not ground. The amp -Vs supply pin becomes the 0v/Ground. It is perfectly possible to connect Vref to ground instead and let the supply negative "float" as a negative supply, but we don't do it that way  because you can't wire different circuits to a common power supply if you do.

As the input impedance of the filter is low, you can use the other half of a dual amp chip to make a high impedance input buffer...

You already have  the 10uF capacitor and Vr is the same as Vref.

[xorophone]

See this basic single supply scheme...

The equal voltage divider formed by R3 & R4 (we call Vref) is all you need to change in your filter, replacing the direct connection of the + input ground. Fortunately the filter components on the - input are all isolated from ground by capacitors, so that 270R can remain connected to ground where shown. The resistor values are not critical, most would pick 10k to 100k values. Less than 10k starts to eat battery current for no good reason.

So what you're saying is that I can just ignore everything on your picture except the things connected to non-inverting input on the op amp (R3, R4), right? Sorry if I'm misunderstanding. I'm not very experienced so it's very hard to know exactly what you mean.

With single supply, Vref is not ground. The amp -Vs supply pin becomes the 0v/Ground. It is perfectly possible to connect Vref to ground instead and let the supply negative "float" as a negative supply, but we don't do it that way  because you can't wire different circuits to a common power supply if you do.

The -Vs pin you're referring to is the ground pin (pin 4 on LM358), right?

As the input impedance of the filter is low, you can use the other half of a dual amp chip to make a high impedance input buffer...

You already have  the 10uF capacitor and Vr is the same as Vref.

I understand how this part works, but where does the Vref from this go? On the other blue/red schematic at the top there's no Vref marked, but if i understood it correctly, R3 and R4 is somehow the Vref?

This image explains how to get convert a single sided +9v supply to give a 1/2 supply reference voltage so you don't need a bipolar supply or two batteries AND it has an opamp input buffer. I would replace the first 100k bias resistor (to the non-inverting input) with a 1M and drop the second bias resistor altogether so the non-inverting input is connected directly to the 10k||10k node.


https://www.google.co.uk/search?q=baxandall+three+band&client=ms-android-sonymobile&prmd=sivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj16rWGiJHOAhWMLsAKHUUFBZsQ_AUICCgC&biw=598&bih=279#imgrc=FOXVCkbekNUW5M%3A

I'm having an easier time understanding this one. Would you replace both of the 100k resistors with an 1M or just one? Which resistor would you get rid of? I'm guessing you might mean that you'd get rid of both 100k resistors and then just use one 1M that both of the non-inverting inputs share. Is that right?

Are there any benefits/downsides if I use this schematic compared to the ones anotherjim shared? The biasing part looks pretty different.

Oh and the "arrows" under the 100uF cap, 10k resistor and 47uF cap are just normal ground connections, right? They don't look the same as in the rest of the circuit so I just want to make sure, but I'm guessing that's because they're two different schematics combined.

[anotherjim]

You can share the same Vref amongst all the amplifiers. So that "Vr" connects to Vref (R3/R4).

When a signal path must be connected to Vref, it needs to be given a lower impedance that the resistor divider, so a capacitor is added between Vref and Ground to provide that path. The capacitor will not change the Vref voltage, but will let AC signal "think" Vref is ground...

There's R9/R10 & C7 providing Vref - see how many parts it's feeding?

Yes, -Vs (negative voltage supply) is how the pin4 supply is often referred. Pin8 can be +Vs. You can also have +Vcc and -Vee  and other names used, but the + & - symbols tell us what we need to know.

Your LM358 is special case in that it is intended to be a single supply amplifier, so on it's data sheet it's pin4 is called Ground.

[xorophone]

You can share the same Vref amongst all the amplifiers. So that "Vr" connects to Vref (R3/R4).

When a signal path must be connected to Vref, it needs to be given a lower impedance that the resistor divider, so a capacitor is added between Vref and Ground to provide that path. The capacitor will not change the Vref voltage, but will let AC signal "think" Vref is ground...

etc

Wow! Thank you so much for the great explanation, Jim! I'm really starting to understand how it all works now. I'm going to start experimenting with it soon.

If someone has the answers to the questions I asked nocentelli previously I'd be very happy to hear them. I guess I can build it using the resistor values on the schematic and be fine but of course I want to make it as good as possible.

These are the questions I'm talking about:

This image explains how to get convert a single sided +9v supply to give a 1/2 supply reference voltage so you don't need a bipolar supply or two batteries AND it has an opamp input buffer. I would replace the first 100k bias resistor (to the non-inverting input) with a 1M and drop the second bias resistor altogether so the non-inverting input is connected directly to the 10k||10k node.


https://www.google.co.uk/search?q=baxandall+three+band&client=ms-android-sonymobile&prmd=sivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj16rWGiJHOAhWMLsAKHUUFBZsQ_AUICCgC&biw=598&bih=279#imgrc=FOXVCkbekNUW5M%3A

I'm having an easier time understanding this one. Would you replace both of the 100k resistors with an 1M or just one? Which resistor would you get rid of? I'm guessing you might mean that you'd get rid of both 100k resistors and then just use one 1M that both of the non-inverting inputs share. Is that right?

Are there any benefits/downsides if I use this schematic compared to the ones anotherjim shared? The biasing part looks pretty different.

Oh and the "arrows" under the 100uF cap, 10k resistor and 47uF cap are just normal ground connections, right? They don't look the same as in the rest of the circuit so I just want to make sure, but I'm guessing that's because they're two different schematics combined.

[Kipper4]

I do hope I don't complicate things here. If I do guys put me right.

Heres a similar circuit. In it bottom left is the circuits DC power supply.
we use DC to power the componants and provide referanced voltages
the V+ supply. your case +9v
VBias (aka Vr,, Vref, signal ground, V/2 some other names too)
and finally V- aka Ground

now the V+ and V- we all know from our houses electric supplies they provide power.
but Vbias what's that for? Vbias (Vb) is there as a signal ground.

The signal from the guitar is AC not DC and if you look at it on an oscilloscope it looks like a wave with rounded tops and bottoms right (google sine wave).
This AC signal goes positive and negative above and below signal ground.
It has to have something to go above and below.
if we put this signal at V- it has nowhere to go below.
If we put it at V+ same thing but cant go above.
So we create a bias that is half the voltage between the V+ and V-
we use eqaul resistors (see schematic power supply 2x 10k resistors) to split the voltage of the V+ and V-.
A little math tells us the half of 9v=4.5v.
The 2 power supply capacitors (100uf and 47uf) are to smooth out any ripple and noise from the DC supply.

Also in the schematic is a buffer that Cozy talks about (top right 1/2 TL082)


http://img.photobucket.com/albums/v165/rickydon/3band.gif

I know I missed loads of stuff.
ps I'm not suggesting you use this circuit i'm posting it so you can see what the guys are trying to say. Use parts of it and insert your tonestack.
I hope that helps.
Have fun.

Edit; shoot I just realised Leo posted the very same pic. Gulp

[nocentelli]

http://img.photobucket.com/albums/v165/rickydon/3band.gif

I'm having an easier time understanding this one. Would you replace both of the 100k resistors with an 1M or just one? Which resistor would you get rid of? I'm guessing you might mean that you'd get rid of both 100k resistors and then just use one 1M that both of the non-inverting inputs share. Is that right?

Are there any benefits/downsides if I use this schematic compared to the ones anotherjim shared? The biasing part looks pretty different.

Just replace the 100k resistor that is shown going to the non inverting input of the first op amp (left hand side of schematic) with a 1M resistor: This gives the circuit a higher input impedance. You can replace the 100k resistor to the second op amp non-inverting input with a just a wire, I don't think a resistor is needed here.

The 10k/10k pair makes a voltage divider that provides a 4.5v reference signal for the opamp to operate on (i.e. the AC/audio signal fluctuates above and below 4.5 volts). The 200k/200k pair in anotherjim's diagram does the same. In theory, higher resistor values in the divider give a higher input impedance (good), but also more noise (bad); The 1M resistor delivers the reference voltage to the opamp, but increases the impedance (since the 10k/10k pair has a pretty low impedance, but also lower noise). In reality, I doubt you would notice much/any difference.

Oh and the "arrows" under the 100uF cap, 10k resistor and 47uF cap are just normal ground connections, right? They don't look the same as in the rest of the circuit so I just want to make sure, but I'm guessing that's because they're two different schematics combined.

Yep, down arrows are ground (and in this arrangement this is also zero volts, aka battery or power supply negative).

[xorophone]

I do hope I don't complicate things here. If I do guys put me right.

Heres a similar circuit. In it bottom left is the circuits DC power supply.
we use DC to power the componants and provide referanced voltages
the V+ supply. your case +9v
VBias (aka Vr,, Vref, signal ground, V/2 some other names too)
and finally V- aka Ground

now the V+ and V- we all know from our houses electric supplies they provide power.
but Vbias what's that for? Vbias (Vb) is there as a signal ground.

The signal from the guitar is AC not DC and if you look at it on an oscilloscope it looks like a wave with rounded tops and bottoms right (google sine wave).
This AC signal goes positive and negative above and below signal ground.
It has to have something to go above and below.
if we put this signal at V- it has nowhere to go below.
If we put it at V+ same thing but cant go above.
So we create a bias that is half the voltage between the V+ and V-
we use eqaul resistors (see schematic power supply 2x 10k resistors) to split the voltage of the V+ and V-.
A little math tells us the half of 9v=4.5v.
The 2 power supply capacitors (100uf and 47uf) are to smooth out any ripple and noise from the DC supply.

Also in the schematic is a buffer that Cozy talks about (top right 1/2 TL082)


http://img.photobucket.com/albums/v165/rickydon/3band.gif

I know I missed loads of stuff.
ps I'm not suggesting you use this circuit i'm posting it so you can see what the guys are trying to say. Use parts of it and insert your tonestack.
I hope that helps.
Have fun.

Edit; shoot I just realised Leo posted the very same pic. Gulp

That's a simple and great explanation! I actually understand a lot more now, so thank you!

http://img.photobucket.com/albums/v165/rickydon/3band.gif

I'm having an easier time understanding this one. Would you replace both of the 100k resistors with an 1M or just one? Which resistor would you get rid of? I'm guessing you might mean that you'd get rid of both 100k resistors and then just use one 1M that both of the non-inverting inputs share. Is that right?

Are there any benefits/downsides if I use this schematic compared to the ones anotherjim shared? The biasing part looks pretty different.

Just replace the 100k resistor that is shown going to the non inverting input of the first op amp (left hand side of schematic) with a 1M resistor: This gives the circuit a higher input impedance. You can replace the 100k resistor to the second op amp non-inverting input with a just a wire, I don't think a resistor is needed here.

The 10k/10k pair makes a voltage divider that provides a 4.5v reference signal for the opamp to operate on (i.e. the AC/audio signal fluctuates above and below 4.5 volts). The 200k/200k pair in anotherjim's diagram does the same. In theory, higher resistor values in the divider give a higher input impedance (good), but also more noise (bad); The 1M resistor delivers the reference voltage to the opamp, but increases the impedance (since the 10k/10k pair has a pretty low impedance, but also lower noise). In reality, I doubt you would notice much/any difference.

Oh and the "arrows" under the 100uF cap, 10k resistor and 47uF cap are just normal ground connections, right? They don't look the same as in the rest of the circuit so I just want to make sure, but I'm guessing that's because they're two different schematics combined.

Yep, down arrows are ground (and in this arrangement this is also zero volts, aka battery or power supply negative).

Thank you so much nocentelli! You've been a big help. I think I've got enough information now to start modifying my pedal!

[PRR]

> down arrows are ground

Stickler....

The "arrow" used to be lines within a triangle representing a wire to a "slice of dirt".

Early radios really needed to be tied to dirt, as a reference for signals caught on an overhead long-wire antenna. In concept the aerial signal "returned" to the transmitted via the earth.

Dirt-contact is not needed in most audio, but "returning" power line buzz to the power line "neutral" reduces audible buzz. (It happens home power is also dirt-referenced, but that is for lightning. Consider a car or an airplane- no dirt contact at all.)

Any how.... somehow the "triangular slice of dirt" symbol devolved to a triangle, which can be seen as an "arrow" if you wish.

[xorophone]

> down arrows are ground

Stickler....

The "arrow" used to be lines within a triangle representing a wire to a "slice of dirt".

Early radios really needed to be tied to dirt, as a reference for signals caught on an overhead long-wire antenna. In concept the aerial signal "returned" to the transmitted via the earth.

Dirt-contact is not needed in most audio, but "returning" power line buzz to the power line "neutral" reduces audible buzz. (It happens home power is also dirt-referenced, but that is for lightning. Consider a car or an airplane- no dirt contact at all.)

Any how.... somehow the "triangular slice of dirt" symbol devolved to a triangle, which can be seen as an "arrow" if you wish.

A nice bit of history! Thanks!