Distortions with OpAmps

[Karkas]

Hello, I've been checkin' some distortions for electric guitars and have seen that some of them use common collectors in the first and last stages because of their input and output impedances. But there's something I don't understand, in this site I saw that they add a low pass filter and a high pass filter, but they put it in the feedback path, and is a way that I don't understand, as I know, a passive low pass filter is a resistance in series with a capacitor and you take the output from the capacitor, and for vice versa for the high pass filter. I don't know if that's some kind of band pass filter, but in the explanation they say that the low pass filter is made by C1 and R1, and the high pass filter by R2 and C2. But why are they in the feedback path? why are R2 and C2 in the high pass filter in parallel? from where do you take the output? and from where do you take the output in the low pass filter?

Thanks in advance.

 

[KeepItSimpleStupid]

R2C1 is usually a RF filter. It prevents things like radio stations from causing havoc.

R1C2 is rolls off the response of the amplifier which increases stability. a 40 kHz rolloff is common for power amplifiers and 100 kHz is common for very good preamplifiers. at f = 1/2*PI*fC is the -3 db point. The capacitor starts to short, reducing the gain of the amplifier. Pretend the cap is a wire at high frequencies.

 

[Karkas]

Hi KeepItSimpleStupid, thanks for your reply, this is the explanation in the site

"C1 = 1/(2 * PI * 0.001M * 40Hz) = 0,039µF = 39nF
39nF is not a standard value, so we'll have to use 50nF or 1µF. 22nF will have limit frequency of 72Hz, which is too high for downtuned guitar (low E is 82Hz). If I used bass guitar, I'd go for the 1µF with 15Hz roll-off frequency, otherwise I'd take the 50nF with 31Hz roll-off frequency.

C2 = 1/(2 * PI * 0.01M * 30,000Hz) = 0,00053µF = 530pF
Again, 530pF is not a standard value, but 470pF will do. It gives a roll-off frequency of 33863Hz = 33kHz. "

I inverted C1 and C2. What I don't understand is why do they work like filters? I mean are they really common low and high pass filter? because IU don't see them like that

 

[KeepItSimpleStupid]

You may have shown the circuit a little wrong. Usually there is a cap that sets the low frequency response, but it's not directly at the input. The ones at the input are usually for RF. If you take a look https://www.electro-tech-online.com/custompdfs/2011/07/LF351.pdf at the figure for a Hi-Z amp, you will notice a compensation capacitor and why it's used.

I need the full pic to verify.

There is another filter (high pass) that sets the low frequency response. e.g. makes the map start at 0.5 Hz and so it won't amplify DC.

Then there is the (low pass) in the feedback loop, that reduces the gain at high frequencies.

And finally with an amp there is a Zobel network or RC combination at the speaker terminals.

Better solid state amps will have an inductor at the ouput. Mine has one made from a resistor and a coil of wire wrapped around it.

This is the amp I have: https://www.electro-tech-online.com/custompdfs/2011/07/feb76feb77articles.pdf

On the schematic R3/C1 is the RF filter. R35/C11 is the high pass filter (0.5 hz) and R33/C10 is the low pass filter. Then there is C9/R32 which helps with driving the speaker.

 

[audioguru]

R2C1 is usually a RF filter. It prevents things like radio stations from causing havoc.

No.
An RF filter is at the input of an opamp. R2C1 are part of the negative feedback and cut low frequencies, not high frequencies.

R1C2 rolls off the response of the amplifier which increases stability.

No.
Opamps have a built-in compensation capacitor for stability. R1C2 cuts off high audio frequencies so that the output sounds muffled like a telephone.

 

[KeepItSimpleStupid]

No.
An RF filter is at the input of an opamp. R2C1 are part of the negative feedback and cut low frequencies, not high frequencies.


No.

AG: Suggest you put your glasses on. The input signal is applied DIRECTLY to C2 and C2 is connected to ground. I can't get any closer to the input than that. Can You?

Opamps have a built-in compensation capacitor for stability. R1C2 cuts off high audio frequencies so that the output sounds muffled like a telephone.

I suggest you read this article: https://www.google.com/url?sa=t&sou...sg=AFQjCNFiyFjR4_yO5d2PaSBA1buWJh9jCg&cad=rja

The pic below from the LF 351 datasheet posted earlier and caption totally refute what you said

 

[Karkas]

Hi audioguru, Could you explain a little more so I don't get more confused?
R2C1 form a high pass filter, right? but my question is, ahy are they in the negative feedback? why they don't have an input and output, do you have some refernce where I can study that? And Same question with R1C2, It's a low pass filter, does that make the whole arrangement a bandpass filter? why are R1C2 in parallel?

 

[MrAl]

Hello, I've been checkin' some distortions for electric guitars and have seen that some of them use common collectors in the first and last stages because of their input and output impedances. But there's something I don't understand, in this site I saw that they add a low pass filter and a high pass filter, but they put it in the feedback path, and is a way that I don't understand, as I know, a passive low pass filter is a resistance in series with a capacitor and you take the output from the capacitor, and for vice versa for the high pass filter. I don't know if that's some kind of band pass filter, but in the explanation they say that the low pass filter is made by C1 and R1, and the high pass filter by R2 and C2. But why are they in the feedback path? why are R2 and C2 in the high pass filter in parallel? from where do you take the output? and from where do you take the output in the low pass filter?

Thanks in advance.

Hello there,

For your original circuit there...
For low frequencies C1 and C2 are open, which means you have a gain of 1.
For high frequencies, C2 is a short which means the gain is again 1.
For mid frequencies, you get some gain G that is greater than 1.
Depending on the value of the two caps, you either get a very wide range of mid frequencies being amplified or a more narrow range being amplified.

This makes it a wide bandwidth band pass filter.

The center frequency is:
Wc=1/sqrt(C1*C2*R1*R2)

and so
Fc=Wc/(2*pi)

and the gain at that Fc frequency is:
G=(C1*R2+C2*R1+C1*R1)/(C1*R2+C2*R1)

With these simple equations you can experiment to see what you get with various combinations.

 

[Karkas]

OK I get that perfectly right, but, let me see if I explain myself well.

When I first saw a lowpass filter, the explanation was this: "For high frequencies the cap presents very low impedance, and for low frequencies it presents very high impedance while the resistor presents the same for every frquency and the came the formula
fc = 1/(2 * PI * R * C)" But it always had an input and an output. My confusion here is why are they in the feedback path? would it work if I just make a conventional bandpass filter with the highpass in the input of the opamp and the los pass at the output?

 

[MrAl]

Hi again,

Various topologies are often found, not always the same exact circuit even if it does nearly the same thing.
Im not sure what you mean by 'conventional' bandpass, because there are different topologies.

The circuit you posted does have a wide, rather flat bandwidth so you have to consider that if you want to design a different circuit.

What's wrong with having components in the feedback path if you get the response you want from the circuit?

What exactly are you trying to do here? Just a bandpass or something else in mind too?

 

[Karkas]

No, I'm just trying to understand its functioning. This is what I mean by conventional, for me at least.

 

[KeepItSimpleStupid]

They would work stand-alone at the input, but you would need to adjust the input and output Z for them to work properly. The RC filter (series combo) would have to be driven by a low Z source. The output Z would also have to not effect the filter.

Elements placed within a feedback loop are more ideal. For, instance you can make a precision rectifier with a diode in the feedback loop.

The input filter, which I say is RF in your drawing or "compensation" in the LF351 application note,basically starts to shunt the signal to ground. At Fc, it has shunted ~30% of the signal to ground.

 

[KeepItSimpleStupid]

The problem with your conventional model is that if the source Z of Vin changes fro 100 ohms to 100K ohms, you have a different filter. The same is true on the output side. If your driving 600 ohms or 10K, the filter response changes.

If you used the output topology, your filter would have to eat power if it was delivering into 8 ohms.

 

[MrAl]

No, I'm just trying to understand its functioning. This is what I mean by conventional, for me at least.

Hello again,

Well that filter will have a different response from the original one you posted in your first post. One of the differences is that this new filter has a different out-of-band response, which goes to a very low level for the new filter but stays close to 1 for the old filter. Thus, the two filters are somewhat different in overall operation even though they both might fall under the general category of a bandpass.

 

[ericgibbs]

hi kiss
What 'agu' says in post #5 about the circuit the OP posted in post #1, is correct.

The circuit you posted in post #6, is not the same as the OP's original.?

 

[KeepItSimpleStupid]

I backed up my post with references or at least I thought I did.


Post 6. You are right, it's not the same as the original. I did overlook that. The original is an odd configuration and possibly not correct?

In that application note the capacitance is shown as a parasitic capacitance and to compensate for that, you place a capacitance in the feedback loop. Don't ever try to put caps in the FB loop when the OP amp isn't unity gain stable. Been there fixing someone elses goof.

Not all op amps have internal compensation caps. The reference backs that up.

 

[ericgibbs]

I backed up my post with references or at least I thought I did.
Post 6. You are right, it's not the same as the original. I did overlook that. The original is an odd configuration and possibly not correct?

In that application note the capacitance is shown as a parasitic capacitance and to compensate for that, you place a capacitance in the feedback loop. Don't ever try to put caps in the FB loop when the OP amp isn't unity gain stable. Been there fixing someone elses goof.

Not all op amps have internal compensation caps. The reference backs that up.

hi k,
I did read the ref document that you posted, most if not all the OPA's I have used, which is lots, have some form of internal frequency compensation.
The document also dosn't say that the OPA, the OP is using, is not compensated.?

Why do you think the OP's circuit is odd and possibly incorrect.??

E.

EDIT:
Added a Sim of the OP's original circuit, used best guess cap values.

 

[KeepItSimpleStupid]

My fault, I didn't look at the original post carefully. The first RC determines the low response and the one in the FB loop determines the high response. I did mistake the location of the Input. I did confuse it with an RF input filter.

Way back when I worked with some CA3130, CA3140, LF351 and finally an LF14741 in I-V converter designs in an attempt of improving it which I did.. The first 3 oscillated occaisionally.

 

[MrAl]

hi k,
I did read the ref document that you posted, most if not all the OPA's I have used, which is lots, have some form of internal frequency compensation.
The document also dosn't say that the OPA, the OP is using, is not compensated.?

Why do you think the OP's circuit is odd and possibly incorrect.??

E.

EDIT:
Added a Sim of the OP's original circuit, used best guess cap values.

Hello there Eric,

There's got to be something wrong with that plot because that's too far off from the theoretical response. Try doing it again with an AC input of 1v instead of 0.1v. I dont think LTS understands AC analysis correctly. The low and high freq response on your plot is very very low, while the theoretical is 1. That's what makes that filter so special.

 

[audioguru]

Hi audioguru, Could you explain a little more so I don't get more confused?
R2C1 form a high pass filter, right?

Yes. KISS is looking at a different schematic than the one you posted.

but my question is, ahy are they in the negative feedback?

A highpass filter can also be at the input of the opamp but it is in the negative feedback to prevent the opamp from amplifying its own DC offset voltage, and while it is there it is a highpass filter.

why they don't have an input and output

Your opamp circuit is missing a power supply and is missing basing at the input.

do you have some refernce where I can study that?

I am not a teacher. The Art Of Electronics and many other good books explains it.

And Same question with R1C2, It's a low pass filter, does that make the whole arrangement a bandpass filter? why are R1C2 in parallel?

The circuit has a highpass filter and a lowpass filter then the entire combination is a bandpass filter.
R1 and C2 are in parallel so that R1 sets the gain of medium frequencies in combination with R2, and C2 bypasses high frequencies which reduces high frequency gain.