Op-Amp Basic Filtering

Started by hoshinotetsu, September 08, 2016, 11:54:57 PM

Previous topic - Next topic

hoshinotetsu

Hi everyone!
I'm trying to build simple op-amp recovery stage for some classic dirt boxes which suffer from volume drop (particularly now I'm building Drivemaster). Til now I just copied existing circuits, but now I'm really into trying to understand at least some basic principles of how they work.
I've read this thread http://www.diystompboxes.com/smfforum/index.php?topic=101398.20 , as well as National Instruments lessons http://www.ni.com/tutorial/13569/en/ , and many other related stuff in Google but there are still some questions left.

So, I made a simple non-inverting op-amp stage with Gain=2 and want to filter it and then couple it with the existing circuit of Drivemaster. By now I did the next filtering using formula f=1/2piRC for frequency calculations:
(Click for full-size)

1. Coupling cap C1 between recovery stage input and tonestack of Drivemaster. Will it affect Drivemaster's tonestack?

2. Negative feedback cap C2 let pass inaudible high frequencies above 23405 Hz to the inverting input of op-amp (blue path), where they "annihilate" with non-inverted inaudible high frequencies and gain for them tends to "1". Am I understand this principle right, or I'm completely wrong at understanding how it works?

3. Cap C3 filters frequencies above 23 Hz to ground (brown path), so they don't stay in the feedback loop, don't reach the inverting input of op-amp and do not annihilate with non-inverted signal in it. Inaudible low frequencies below 23 Hz though stays in the feedback loop and annihilate with non-inverted and gain for them tends to "1" too. So we've got only 23-23405 Hz bandwidth with gain=2. Again, am I right?

And now the question for me is why those high frequencies above 23405 Hz do not go from C2 to C3 and to ground (blue dotted path) but can actually make it to the op-amp inverting input?

And one really stupid question: is there any practical or theoretical difference in which order C3 and R3 are placed: C3->R3->GND or R3->C3->GND? And why? Shame for me, but I can't make a physical explanation.

4. Am I right about Volume pot? Will it actually be a variable filter or a constant one (7 Hz roll off no matter how much Volume pot is turned)?

Thanks in advance. BTW my first posting here, hope I did it right.
Excuse me for my "Engrish", it's not my native.

PRR

Welcome.

> 23-23405 Hz bandwidth with gain=2

Yes. (The 23KHz pole may be higher, but who cares?)

> why those high frequencies above 23405 Hz do not go from C2 to C3 and to ground

R3 is in the way.

> difference in which order C3 and R3 are placed

Parts in series with nothing connected in the middle: order does NOT matter.

> Volume pot? Will it actually be a variable filter

This part is incomplete. What is connected to the Output??

If the load is infinite (>>100K), then it is always 0.22uFd against 100K. 7Hz.

If the load were zero, it would be 7Hz full-down and infinite Hz full up.

If the load were 100K (a very possible value in guitar-cord work), then 0.22u+100K to 0.22u+50K, or 7Hz to 14Hz.

If you assume a goal of "-3dB at 50Hz" (doesn't dent 82Hz guitar much), then 0.22uFd could drive 14.5K total, or 17K in addition to the 100K. Few full-range guitar circuits have input as low as 17K.

If you face 10K studio line inputs, get serious. The opamp can easily drive 2K load. 10K pot is readily available. Minimum load could be 5K. While this suggests a couple of uFd, at this point you are looking at Electrolytic, and a 10uFd is as cheap as any smaller size. Output bass-loss is a non-issue, 3Hz at worst.
  • SUPPORTER

antonis

Quote from: hoshinotetsu on September 08, 2016, 11:54:57 PM
why those high frequencies above 23405 Hz do not go from C2 to C3 and to ground (blue dotted path) but can actually make it to the op-amp inverting input?
Quote from: PRR on September 09, 2016, 01:27:57 AM
R3 is in the way.
Well said but - for the shake of discussion - R3 value is MUCH smaller than Op-Amp Input impedance..

So it should be "easier" for signals above the frequecy of interest to "leak" through R3 & C3 to GND (at least to an amount  conversely proportional to paths respective resistances..) than to be by-passed from the NFB Gain loop..!!

JUst a food for thought (and dispute, of course..) :icon_wink:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

hoshinotetsu

Thanks PRR!
QuoteR3 is in the way.
But isn't that true for another part of feedback signal (0-23405Hz) that comes from R2? Or is it exactly because of it comes from resistor? Ugh, I think I can't see something really simple and essential here. I know it works, I can see it works, but I don't understand why. High-frequency signal from C2 sees R3 and can't pass. Signal from R2 sees R3 and can pass?

QuoteWhat is connected to the Output??
Uhmm, Drivemaster has a 22k resistor that goes from Volume pot wiper to output jack and 470pF cap then to ground, forming another filter, I believe, and an 1M pulldown resistor on the output. The latter also should serve as load, so it always will be 0.22uF+100K?

QuoteParts in series with nothing connected in the middle: order does NOT matter.
I believe that result is the same in the end, but the actual processes that occur over time in the resistor and in the capacitor should be slightly different?
I've made a simple DC circuit today from 9V Battery, Capacitor, Resistor and LED.
- LED behaves the same way in both cases, C->R->LED and R->C->LED. It lightens up and fades away as it should while capacitor charges.
- Voltage measured on the LED is the same too: 1.6V dropping to 1.4V as capacitor charges and LED fades off.
- Voltage drop across the resistor again is the same too in both cases, firstly it 7.5V and then it decreases to 0V. And because of that (I think) voltage on capacitor and resistor measures differently!
- When you turn on the C->R->LED circuit, voltage at capacitor rises instantly to 9V and stays this. Voltage at resistor rises instantly to 9V too and then falls down to 1.4V as capacitor charges.
- When you turn on the R->C->LED circuit, voltage at resistor rises instantly to 9V and stays this, but voltage at capacitor slowly rises up from 1.4V to 9V.

Should it be matter if we dealing, for example, with high-current / high-voltage DC circuits?
Excuse me for my "Engrish", it's not my native.

PRR

> "easier" for signals above the frequecy of interest to "leak"...
> sees R3 and can't pass. ... sees R3 and can pass?


Get off the vague, and absolute, language.

It's all more/less, not "leak" or "can't".

The questions imply you have not MASTERED "voltage dividers".

I'd take those shiny opamps away from you. Go back to pure resistor networks. If I ran the Boot Camp you'd spend weeks on Voltage Dividers.

The op-amp only does the "operation" given by the external impedances.

Voltage op-amp theory assumes the op-amp inputs ONLY respond to their voltages, and don't "leak". This is not true in practice, but any reasonable audio circuit with any part-decent chip WILL obey the passive components more exactly than a bench DMM can read.

Extract the out-in voltage divider. Draw it as a voltage divider.

The caps are not constant resistance with frequency. They go from infinite impedance at zero frequency to zero impedance at infinite frequency. Therefore they will pass through ALL intermediate values as you sweep frequency from zero to infinity.

So as a rough step, replace the caps with resistors of different values for the different frequency ranges.

For low-low frequency, the caps will be huge "resistance", pencil 10Meg. 10Meg is so much larger than the 68K resistors that we can assume 10Meg+68K= 10Meg, and 10Meg||68K is 68K.

For high-high frequency, the caps will be tiny "resistance", pencil 1K. 1K is so much larger than the 68K resistors that we can assume 1K+68K= 68K, and 1K||68K is 1K.

So for very low frequency the gain is 68K/10Meg or essentially unity. Divider ratio approaches "1" as frequency approaches zero.

At some low bass frequency, 0.1uFd becomes similar to 68K. Now the divider ratio drops toward 2/3rd. (But not quite, because caps are not resistors.)

At some high frequency, 0.1u is much less than 68K. Now we are nearing a divider ratio of 0.5.

At some even higher frequency 100p approaches 68K, and divider ratio gets back up toward unity. 68K/(68K+1K) is 0.98.

This will give you the correct trends for the sub-audio, audio, and super-audio bands. This is usually all you need.

For a simpler circuit this is a rise from DC to bass, flat from bass to treble, and a fall above treble.

The "corners" happen when cap reactance is equal to resistance. But! Equal resistors gives 0.5 ratio. Cap-resistor corner gives 0.707 response at the corner.

The Bode Plot way is to plot the straight lines, then sketch the "corners" 3dB down.
  • SUPPORTER

PRR

> Battery, Capacitor, Resistor and LED.

I said "two parts". The 3 series part answer is the same; but your reach has exceeded your grasp. Even if you had a 3-part solution, you have not proven your question about 2 parts.

What does the LED add to the experiment? As you say, it holds a nearly constant voltage, and the change is "small" compared to 9V.

The simple series R-C string is a VERY important system and you really should have mastered it before you were issued op-amps.

> voltage on capacitor and resistor measures differently!

Well, I did say "nothing connected in the middle".

If you are going to poke in there, you better be clear "reference to what??" Reference to a "ground"? Do floating parts know where ground is? Voltage *across* a part is sometimes illuminating.

> voltage at capacitor rises instantly

Voltage *ACROSS* a cap can not change instantly.

I bet your cap rose to 9V on *both* ends, zero V across. The rest of the 9V appears in the R and LED. Since there is voltage *across* the R, there is current, and the cap voltage will "slowly" increase toward 9V (or less, since there's that LED down there).

And if you flip the R and C the voltage at the internal node *does* "go the other way". If you have "nothing connected in the middle", you can't know that, so we do not care. If you are going to tap that point (changing from a 2-lead RC network to a 3-lead RC network), then you better decide which way you want to go.
===================

> Drivemaster has ...22k ...470pF ...1M ... the output.

And? Or is this output not going anywhere?

You have described like the power wires from the dam generator to the pole outside my house, but ignoring the load inside my house. I might have one 3-Watt lamp, or be running 40,000 Watts of laundry appliances. Makes a real difference: 125V to 111V at my meter.

Though the 22K series resistor means "any" external load will never cause more than 22K of added loading on the 0.22uFd cap.
  • SUPPORTER