Calculating frequency rolloff in RC networks?

[PenPen]

I'm wondering if there's a quick and simple way to calculate the frequency rolloff for a given RC network. For example, the Rat has a dual RC network to ground on the inverting input. One pair is 47 ohms with 2.2uF and the other is 560 ohms with 4.7uF. What frequencies are these rolling off? Similarly, with a guitar tone control, I would like to be able to calculate the rolloff frequencies with a given cap. Anyone have a handy formula or chart for computing the rolloffs?

[idiot savant]

here's a few quick & simple java applets

low-pass
http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/lowpass/lpf.html

high-pass
http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/experiment/highpass/hpf.html

[twabelljr]

http://www.muzique.com/schem/filter.htm

[gez]

Incidentally, a really good intro to the world of AC can be found in John Clayton Rawlins 'Basic AC Circuits' (Newnes publishing).  Ideal for those who've covered the basics of biasing transistors and wish to take things a little further.

[PenPen]

Thank you all for the replies. These are exactly what I was after and will help massively.

[R.G.]

I have a profound disagreement with using special built calculators for figuring out RC rolloffs. I believe it interferes with one's understanding of the underlying electronics.

The rolloff frequency of *any* single RC network is F = 1/(2*pi*R *C). It does not matter whether it is highpass or lowpass, serial or parallel.

This comes out of Ohm's law and the definition of "rolloff". The rolloff frequency of a network is taken as the half-power point, or for voltages, the half voltage point.

If you have a two-resistor series network, then the voltage at the junction of the two is half the total voltage only when the two resistors are equal - right?

So we want to know in a resistor/capacitor network when the capacitor impedance equals the resistance. That happens when Xc = 1/(2*pi*F*C) = R.

It's simple to manipulate that to get the frequency where they are equal by multiplying both sides by F and dividing by R.

We then get F = 1/(2*pi*R*C).

Note that the same thing happens in parallel networks, but in current mode.

You would NEVER come to any understanding of the actions of the R's and C's if all you did was plug numbers into a magic calculator.

[gez]

Well said!  And for those of you who'd like to understand how that formula came about (and I swear I don't work for the publisher), that book I mentioned goes into the nitty-gritty...in an easy to understand way too!

[Lurco]

The rolloff frequency of a network is taken as the half-power point, or for voltages, the half voltage point.

the frequency at which the response of an equalizer or other audio device is reduced by 3dB,. This is also sometimes called the half-power point and can refer to both lowpass and highpass response curves. The rolloff frequencies of an amplifier are the frequencies where the output voltage drops to 0.707

from: http://www.dilettantesdictionary.com/index.php?search=1&searchtxt=rolloff%20frequency

don`t these 2 definitions bite each other?

[R.G.]

Not really. It's one of the oddities of half power versus half voltage (or half current), and similar to the way that dbs in power ratio is 10 times the log of the power ratio, but dbs in voltage ratios are 20 times the log of the voltage ratio.

This makes sense when you realize that power is proportional to voltage squared. Half power is -3db power and -6db voltage. The factor of 0.7071... is equal to the square root of two (1.414...) divided by two (0.7071).

[PenPen]

Thank you for explaining the formula and how it works, R.G. Very informative as usual.

Now I need to reconcile this with the gain formula for an opamp. My understanding is that the response curve is an inversion of the rolloff, since in this case the rolloff frequencies will have less resistance to ground, thus giving it a HIGHER gain in those frequencies. So there will be a boost of higher frequencies, the higher the cap value in the RC network the lower the cutoff frequency of the boost will be. What I am looking to do is put in some lower frequecies than the stock values provide. Therefore I should be increasing the cap value somewhat. I'm probably going to start with 6.8uF or 10uF and go from there.

Thanks again to all that responded.

[Guitar Toad]

Very good thread here, PenPen.
R.G. thanks for the formula and the explanation...great help.

If I may piggy back on this thread...I wanted to ask...
"What are typical values for about High and Low Pass Filter frequecy roll-offs?" My understanding is that the many stompboxes are simply a high pass and a low pass filters. I was trying use the Web-bench tool on National Semiconductor to use it for designing these filters and I wasn't sure what roll-off values I should plug in there. The default value on the high filter was 40kHZ, which I think is way too high for guitar usage, and and the LPF default is 50kHz, which again I thought was much too high. For Stompbox usage I'm thinking that 30kHz is reasonable for HPF and 20kHz for LPF. Thanks for allowing me to insert my question at this point.

Thank you.
Merci Beaucoup.

[MetalGuy]

Does all this mean that the 10k/10k-47nF combo after the first opamp stage in MT2 is a 338.8Hz low pass filter?

[R.G.]

Now I need to reconcile this with the gain formula for an opamp. My understanding is that the response curve is an inversion of the rolloff, since in this case the rolloff frequencies will have less resistance to ground, thus giving it a HIGHER gain in those frequencies.

While ideas like "the response is the inverse of the feedback" may be useful simplifications sometimes, you need to go back to the actual expressions for opamp gain.

The gain of an opamp from an inverting input is G = (-1)*Zfeedback/Zinput ; the -1 is an indication that the output will be inverted in sense from the input, and that the gain is the ratio of the impedances, whatever those are. If the impedances are RC networks, then the R will dominate the expression at low frequencies, and the cap at high frequencies, and the frequency where they have equal effect is F = 1/2*pi*R*C.

Some common examples:
If you have an R paralleled with a C in the feedback network, you may ignore that there is a cap even there until you reach frequencies near 1/2*pi*R*C. The C's impedance is so large that it has no practical effect. As frequencies increase above Frc (that being the magic term I just coined for the turnover frequency), then the capacitor's impedance decreases at a rate of 6db per octave, 20db per decade, and gain decreases monotonically. The R no longer matters much at such frequencies.

If you have an R in series with a C at the input, then the capacitor keeps gain at DC to zero. As frequency increases, gain increases at a rate of 6db/octave = 20db/decade until Frc, at which point the R increasingly makes the lower-impedance cap not matter, and gain becomes Zfeedback/Rin.

If you have BOTH of those (which is very common), then the opamp has a gain of 0 at DC to the inverting input, and the output DC is set entirely by the + input. The gain rises at a single RC slope of 6db/octave until it gets to the midband range of Rfeedback/Rin, which will happen at Frc of the input cap and resistor. It stays there until frequency gets near Frc for the feedback RC, and after that point, gain again decreases at -6db per octave.

The rate of -6db/octave or -20db per decade for voltages is know as a single time constant slope, for reasons that should now be obvious. Frequency rolloffs are quantized in unit multiples of single time constant slopes. A single time constant RC is a first order filter. Filters which make -12db/octave are second order filters, etc.

If the opamp has its input fed to the noninverting input, then you can have a series or parallel RC network there to. The opamp input is presumed to be so high impedance that it does not load an RC network so the input to the opamp will just be the voltage on the input RC network. As we've seen, that can vary with frequency as either a single RC slope up or down, at high or low freuqencies depending on which of the four possible arrangements you use.

From that + input, the gain of the opamp proper is then imposed on the input voltage. The gain is G = 1 + Zfeedback/Zinput where Zinput is the Z on the inverting input.  all of these are computed as per the examples.

What are typical values for about High and Low Pass Filter frequecy roll-offs?

Human hearing is generally taken as 20Hz to 20kHz, so filters outside that range are not useful in tone shaping.
Guitar's lowest note is 82Hz, bass is 41 Hz, so filters lower than those don't have any practical effect.
The fundamental of the highest note on most electrics is 1312Hz, which is the high e string at the 24th fret. Filters in that range affect the fundamentals of notes.
Electric guitar pickups stop producing useful outputs somewhere near 7kHz, so the range of 1312 to 7000Hz is where high frequency stuff on undistorted guitar has most effect.
Distortion processes make large amounts of high harmonics, so they can readily get up into the 7kHz-20kHz range. Generally these are harsh sounding harmonics, so many pedals have high frequency cuts for this area, and some of them even start cutting in the 1312 to 7kHz region.
Distortion in the bass region can cause ugly intermodulation in the bass region, so many distortion pedals have a low frequency cut.
For a sendup of how one distortion pedal does it, see "The Technology of the Tube Screamer" at GEO.
Frequencies around 2kHz affect "presence", around 3-4 "brightness" and 8k-20k "tinkle" or "sparkle". Or they would if those words had any real definable meaning.

Does all this mean that the 10k/10k-47nF combo after the first opamp stage in MT2 is a 338.8Hz low pass filter?

How are the resistors arranged, what drives them and what reads the output voltage? Those matter.

[Satch12879]

I programmed an Excel spreadsheet that calculates the voltage reduction as a percentage for both low pass and high pass filters and then plots the response on a log scale.  Try making one yourself; although the exercise was trivial compared to some of the seismic response analysis stuff I did in graduate school, it will really make the numbers stick in your mind.

[PenPen]

As usual, RG, you leak out more knowledge that it will take me a week to fully grasp.

Yeah, I had forgotten the bypass cap across the first resistor, I had thought about it after posting. To my understanding, the Rat circuit creates a band-amplification, that in stock form favors the high end. The curve of response would look more like a hump than a straight slope. The bypass cap sets the upper boundary for the curve, after it reaches that point it slopes back down due to the decreasing impedance at those frequencies in the cap. I will probably keep the stock value for this cap, since the ceiling response is just fine and preferred. My biggest goal is to adjust the lower boundary so that more bass content comes through. I have planned a BMP tone control that is somewhat close to the response curve of a Vox amp as per the Duncan TSC. So I want there to be the bass content available for the tone stack to be able to actually do something to it. I wish I could figure out spice enough to model both the opamp response filtered into the tone stack so I have a better picture of how it is going to sound, but I am content with experimentation.

Again, my understanding of the circuit is that by increasing the value of the cap, I flatten the hump a bit. I probably won't do this, since that isn't what I'm after. I want to be able to turn the tone toward the bass side and keep some high frequencies clear, instead of the "blanket over speaker" result you get when you turn up the filter in the stock rat. I figure, while I'm at putting in a new tone stack, I may as well experment with the response from the opamp itself as well.

[davebungo]

I have a profound disagreement with using special built calculators for figuring out RC rolloffs. I believe it interferes with one's understanding of the underlying electronics.

The rolloff frequency of *any* single RC network is F = 1/(2*pi*R *C). It does not matter whether it is highpass or lowpass, serial or parallel.

This comes out of Ohm's law and the definition of "rolloff". The rolloff frequency of a network is taken as the half-power point, or for voltages, the half voltage point.

If you have a two-resistor series network, then the voltage at the junction of the two is half the total voltage only when the two resistors are equal - right?

So we want to know in a resistor/capacitor network when the capacitor impedance equals the resistance. That happens when Xc = 1/(2*pi*F*C) = R.

It's simple to manipulate that to get the frequency where they are equal by multiplying both sides by F and dividing by R.

We then get F = 1/(2*pi*R*C).

Note that the same thing happens in parallel networks, but in current mode.

You would NEVER come to any understanding of the actions of the R's and C's if all you did was plug numbers into a magic calculator.

It's the -3dB point for voltages i.e. when the voltage gain is 0.7071 (with 45 deg phase lead or lag)

[MetalGuy]

How are the resistors arranged, what drives them and what reads the output voltage? Those matter.

Below is a link to the schematic. Check out the first opamp stage.

http://www.godiksennet.com/images/sch/mt2-06.jpg

[mac]

And what about a RL network?

The impedance of an inductor is XL = 2.pi.f.L, and the rolloff freq is f = R/(2.pi.L).
Noting that inductors are transparent to low freqs and filter highs, all what RG illustrated above apply.

The general case is a RCL series network of  impedance Z = sqrt{ r^2 + [ 2.pi.f.L - 1/(2.pi.f.C) ]^2 }.
When f tends to infinite or go to zero, Z tends to a very high value blocking high and low freqs. At f = 1/[ 2.pi.sqrt( L.C ) ] Z is minimal. This is called the resonant freq.
I have not seen RCL series network in pedals, simply because inductors needed are generally big. The only useful application I can think of right now is to place a RCL in a negative feedback loop of a transistor or opamp, as it reminds me a ( Twin-T )^(-1).

Once you understand how capacitors and inductor react to different freqs, and keeping in mind that the signal will follow the path of minimun resistance to a given freq, you could visualize a priori the behaviour of a mixed RCL network without the use of calculators.

... and a RG network??? 

mac

[WGTP]

What I have done as a quick rule of thumb, is set up a common formula and then I modify it to suit the application.  It is:

1uF & 1K = 160Hz
10uF & 100R = 160Hz
.1uF & 10K = 160Hz
.1uF & 1K = 1600Hz
.01uF & 1K = 16,000Hz
etc.
Then you can half and double the values.   

[davebungo]

What I have done as a quick rule of thumb, is set up a common formula and then I modify it to suit the application.  It is:

1uF & 1K = 160Hz
10uF & 100R = 160Hz
.1uF & 10K = 160Hz
.1uF & 1K = 1600Hz
.01uF & 1K = 16,000Hz
etc.
Then you can half and double the values.   

It is a good idea to remember a few basic combinations like this.  If only we all worked in radians it would be even easier i.e. 160Hz = 1000 radians/sec.