Help me understand Bridged-T RC Filters

[edvard]

OK, I studied around the Mr. EQ circuit trying to get ideas for tone controls to use with an Inverter-based circuit, but I don't quite understand how a notch filter like the Bridged-T layout can be used for Low-Pass and High-Pass as well.
However, the concepts are interesting and I'd like to know more about this kind of filter.

I read through John Hearfield's page, but the math got over my head pretty quickly.
Are there any easy ways to calculate component values?
I'm not a genius so please be gentle...

[R.G.]

The R-C notch filters like the Bridged-T and its more complex brother, the Twin T, are both a combination of a highpass and a low pass network. The same signal is applied to both the highpass and low pass at the same time, and then the results are added.

There can be no frequency selectivity without phase shift of some kind that I know of. So the high pass and low pass networks both shift phase, and in different directions. When you add the two together, there is a place where they tend to cancel. That's where the notch happens. Below the notch, the lowpass tends to dominate the response, above it the highpass does.

By adjusting the ratios of the highpass and lowpass signals, you get varying degrees of both notch and what happens above and below the notch.

However, I don't know any way to make the math a lot simpler than that page shows. That's a particularly clear and lucid explanation of how these work. Much better than my professor led me through. Maybe someone else has a better explanation of the math.

One simple way to do this is to go case up the math in a spreadsheet, and then you can easter-egg part values to see what happens to the responses. A circuit simulator lets you do much the same thing.

[edvard]

I was afraid you'd say that. 
Actually, I've learned quite a lot doing different simulations and have discovered that various loads at the head or tail can magically transform it into a Hi or Lo pass.
The trick is to learn what values alter the Q and what values move the band. 

So much experimentation, so little time...

[R.G.]

I was afraid you'd say that. 
Actually, I've learned quite a lot doing different simulations and have discovered that various loads at the head or tail can magically transform it into a Hi or Lo pass.
The trick is to learn what values alter the Q and what values move the band. 

I think we (the EEs of the world) have approached things a little wrong by not insisting that the driving impedance and the load impedance are integral parts of every circuit. They are; but we tell people to ignore them when they're starting out. It simplifies things. But the Real World waits at the end of the class. Source impedances and load impedances MATTER.

I'm afraid you'll find out that the SAME component values affect both F0 and Q. This is one reason the state variable filter is popular. You can actually isolate the F0 components from the Q components and not have them interact. Non interacting F and Q controls in a filter are sometimes hard to get.

[edvard]

OK, I'm a little closer...

Through much trial and error, I came up with this little equation and did some experiments.  
Assuming that:
1 - The resistors and capacitors (R and C) in the Bridged-T configuration are equal values;
2 - R is in Megohms and C in microfarads;

We can do:
f_c=.159/√(R²C²)
That is, notch frequency is equal to 159 one-thousandths divided by the square root of the product of R squared times C squared.

I checked this with the simulator and it holds true...
For example, (using component values that keep the math easy) with resistor values of 159k (.159 Megohms) and 1nf (.001 microfarads) capacitors, we get a notch of approximately -3.6db nicely centered at 1kHz... approximately.
From there, we can tweak the circuit to our liking...

Changing Capacitors:
Making C2 larger than C1* by proportionate factors (multiply C2 and divide C1 by the same number makes it easier - use a number between 2 and 5; up to 8 or 10 if you're adventurous) makes the notch narrower and deeper.
For example, changing them by a factor of 5, making C2 200pf and C1 5nf, we get a notch of approximately -22.5db.
The opposite happens when C1 is larger than C2, the notch gets shallower (and less useful IMHO).
Shifting values in the same direction moves the notch; larger values move the notch lower in frequency and smaller values move it up.
Changing C1 and C2 to 5nf brings the notch over to 200Hz, and changing both to 200pf puts the notch at 5kHz.

Changing Resistors:
Changing resistors gets weird.
The more of a difference there is between the two resistors, the wider the notch.
In fact, it gets downright bowl-shaped.
However, making R1 larger than R2* by proportionate factors does make the notch deeper... a factor of 5 shift (R1=795k, R2=31.8k) brought the notch down to about -6db and making R2 larger than R1 proportionately pulled the notch up to -.9db
Making both resistors larger or smaller by the same factor does the same as raising and lowering both capacitor values.
Using 1.6 Megs for both resistors with 1nf capacitors pulls the notch over to about 100Hz, while using 31.8k shifts it up to 4kHz.

Well, that's all I can do for now... Mix and match value alterations as you deem useful.
Next time, I'll throw some simple RC Hi and Low pass filters in the mix at the head and tail and we'll see what happens, because as RG said, putting loads on either end throws in a monkey wrench, so be careful when designing with this simple (and probably more or less flawed to some degree) exposition...
Have Fun!

*for those who are not familiar with the Bridged-T Configuration, here's an ascii approximation:

   +-----C1-----+
   |                   |
in----R1--+--R2----out
             |
            C2
             |
-------ground------


P.S. I'm no good at algebraic substitution anymore, especially behind square roots and all.
Can anybody come up with equivalent formulas to plug in any two values, say put in R and f_c to get C and vice-versa?

[PRR]

> doing different simulations and have discovered that various loads at the head or tail can magically transform it into a Hi or Lo pass.

It's not magic.

Simulators give 7-digit answers but are not good tools for "understanding".

You can learn a lot by looking (and knowing how resistances/impedances interact).

Look at the extremes.

At zero Hz, all caps are open.

At infinite Hz, all caps are shorts.

At zero Hz, the caps drop out. You have a resistor feeding a load. Pretend it is infinite (a tube/FET grid comes close). The transfer is unity voltage.

Because of the caps from In to Out, at infinite frequency it is zero impedance. If source is zero impedance, then transfer function is unity voltage.

You can't get a boost. Only a loss.

So for "more highs", you really want "less lows" (plus maybe some broadband make-up gain). For "less lows", you hang a heavy resistance on the output. If resistors R and S are 10K each you have 20K. What is a "heavy load" for 20K? 200K gives 90% voltage transfer, that ain't heavy. 2K will give 10% voltage transfer, that's heavy loss.

So if it is 10% at zero and low frequency, unity for high and infinite frequencies, it is a "bass loss" or (with broadband gain) "more treble".

When you mock this up, you will find that the notch frequency has changed with the heavy load. The Classic Bridge Tee is really neat and "simple" when you want a simple notch between a zero source and an infinite load. You don't like that math, you sure don't want to hand-solve the un-simplified equations including source and load and non-elegant value ratios.

IMHO: if you want a high/low-pass, build a hi/lo-pass. If you want a dip, do a dip. Trying to get both in the same network is High-Pay work. If you gonna build a million, you do $400 of thinking to save 10 cents per copy; if you just build a few or one, it is usually best to assemble the most-obvious or most-plagiarizable plans even at the cost of a few "unnecessary" resistors and caps.

Don't forget that a dozen sharp guys have been down this path before you. James, Fender, Gibson, Sunn etc explored about every R-C and L-C network by 1970. For reasonable complexity, and not assuming any "perfect amplifiers" (which are now cheap), about anything you could want has been done before, if you can recognize it when you find it.

BTW: the Fender Tone Stack (the 5F6-A plan, where Fender mostly stopped and which EVERYbody copies) has a sort-of Twin-Tee hidden in it. And if you use Duncan's Tone Stack Calc, and diddles some values to 1 ohm, you can compute the General Twin Tee without a full-blow SPICE.

[B Tremblay]

This thread was very enlightening for me: http://www.diystompboxes.com/smfforum/index.php?topic=25788.0

[PRR]

> Through much trial and error.... 
> 1 - The resistors and capacitors (R and C) in the Bridged-T configuration are equal values;
> 2 - R is in Megohms and C in microfarads;
> f_c=.159/v(R²C²)

0.159 -- type the zero, the eye does not see leading dots.

0.159 is (you may know but did not say) one over two Pi. Two Pi because the math works in radius and we like to think in whole cycles/circles.

Meg and Micro are both a million but going different ways. Yes, in tube amps it is often convenient to push the dot over a million; this gets annoying in low-impedance circuits. K-Ohm and milli-Farad would work, but we don't use milli-F caps.

The root is kinda-sorta "averaging" the part values. 1uF and 2uF average to about 1.5uF. Really 1.414uF.

If both Rs are equal and both Cs are equal, then the root is not needed.

If they are small-ratio like 1uF and 2uF, you can blur your eyes and work with a split-the-difference value. Your EE prof will flunk you, but in audio we seldom need great precision. 700Hz, 650Hz.... if we are trying to kill an annouing 699Hz whine, it matters, if we are trying to dip the middle of guitar to bring-out the bass and treble, 600Hz or 800Hz will work, and may be better for some specific axe and speaker.

So the formula is approximately 1/2*Pi*R*C, an exact formula you SHOULD know by HEART. It covers all the simple R-C plans exactly (if you know what exactly is), and is usually pretty-close for stuff with multiple similar reactances.

You should be able to bang-out 2 * 3.14 * 10,000 * 0.1EE-6, =, 1/x in a few seconds. Heluvalot faster than setting-up and interpreting SPICE.

> The more of a difference there is between the two resistors, the wider the notch.

The math is over my head; but I would assume that in a perfect case (chalkboard or SPICE) there is an optimum ratio of resistors for infinite null, and "equal" may be that ratio.

However in audio, _except_ the case where the smoke-alarm added a 1,234Hz tone which you want to remove, we very-very ever want an infinite null. (You might think that 60Hz would be useful, but 60 never comes without her 120 180 240 360Hz children, who may be smaller but are more annoying...)

BTW: SPICE is dumb. If you DO want an infinite null, it may slip between the cracks in SPICE. Be sure your points per decade number is higher than the infinitely-narrow width of an infinite null.

[edvard]

Thanks for the feedback, PRR.
I think you cleared out some of the cobwebs left over from working this out.
I'll be printing your comments for further rumination and distillation.
I must admit the math is still over my head a bit but this much is what I worked out with a good calculator and a sheet of paper or two in my quest to understand in some small way what was going on.

In short, this is the "ballpark" I was looking for.

I wasn't asking for ideals or infinite null or anything, just some sort of baseline math equation that I can plug in the part values and say "Oh, I see; that makes a -10db dip at 220Hz" (give or take depending on external loads, I know. Well... at least NOW I know).
I can do F_c=1/2ΠRC all day with Hi and Lo pass circuits, I was just looking for a simple way to understand this particular breed of notch filter so I can start USING the thing instead of being perpetually mystified by it as if it were some Lovecraftian cipher.
Hell, with a few tweaks I might be able to use it for a higher-Q hi/lo pass when it's needed.

Now that I have this, I can start real-world (i.e. breadboard) sims to hear just what that wacky math works out to actually sound like.
But like I said, I need the ballpark before I can do some serious pitching.

[edvard]

This thread was very enlightening for me: http://www.diystompboxes.com/smfforum/index.php?topic=25788.0

Aw, crap.
Wish I had found that thread before this.
Maybe would have saved some time.

Or maybe not...

[PRR]

> I can do Fc=1/2 Pi RC all day with Hi and Lo pass circuits

OK. That's an uncommon skill.

Do you do Bode Plots? While the actual response is a curve, you know the limits of the curve by drawing two lines, one straight and one slanted. That gives the approximate Magnitude response; in simple cases you can assume reality is 3dB down at the kink (Pole) and 1dB below the curve an octave either way.

Break the Twin Tee into highpass and lowpass. Plode the Bode. Add them. The result is little or no dip at all.

Yet a moment's experiment shows this is incomplete; there IS a null.

Filters have pahse shift. At the corner, the hipass is 90 deg advance, the lopass is 90 deg late. The difference is 180 deg. And we know that two equal signals 180 deg apart will cancel. Hence the dip.

How to compute it? Figure the magnitude AND phase for every frequency.

How can you do this without maintaining and manipulating two equations? This is not clear to you or me because we are "no good at algebraic substitution". However each type of number has rules for algebraic substitution. There is a number which we can insert into the equation which allows the Phase information to be handled along with the Magnitude. That is the "imaginary operator" j. The j may be a strange thing; the point is that by using it we could (if we were good at algebraic substitution) solve these so-called "complex" equations by simple algebraic substitution.

Plagiarize. Understand how to estimate the rough impedance and frequency of a stolen design. Understand the simple (non-"complex") manipulations to transpose a 220K circuit into a 10K circuit, how to shift a 700Hz null to a 400Hz null.

[edvard]

Yes, that's exactly what I did.
I was looking at the filter networks used by Runoffgroove's Mr. EQ, hoping to "plagiarize" and learn something in the process.
I'm not up on Bode plots, but LTSpice does well enough to help me visualize things.

0.159 is (you may know but did not say) one over two Pi.

If both Rs are equal and both Cs are equal, then the root is not needed.

I just realized the implications of these two statements on the bus about two hours ago, and I may require intelligence testing after this...
I understand now that the final equation is the exact same as for Hi and Lo pass filters; namely 1/2ΠRC if R1=R2 and C1=C2.
The answer to many of my questions was right there the whole time, the only NEW thing I needed to discover was how Q was affected by the re-factoring of capacitor values.

Thank you for your patience, I think I need some coffee now...

[stringsthings]

I must admit the math is still over my head ..... I was just looking for a simple way to understand this ... instead of being perpetually mystified by it as if it were some Lovecraftian cipher ....

magic and math are mutually exclusive .... the purpose of magic is slight-of-hand .... one purpose of math is to provide a model for understanding a complex system ... i doubt if Lovecraft was too concerned about algegra ...

IMHO, get some coffee and enjoy the DIY experience instead of making it hard work.

[WGTP]

I'm not as far along as you guys in understanding all of this, but I frequently use the modified James Tone Stack from the Duncan Tone Stack to simulate notch filters, hi-pass, low-pass, shelving, etc.  If you think of it in reverse, feedback loop filters can be simed.  Is a pretty dumbed down way to do it, but it is helpful designing filters for distortions where the main thing is how they sound to your ears.  I'm pretty visual and seeing what is happening more or less, helps me understand.   

[edvard]

...
magic and math are mutually exclusive .... the purpose of magic is slight-of-hand .... one purpose of math is to provide a model for understanding a complex system ... i doubt if Lovecraft was too concerned about algegra ...

IMHO, get some coffee and enjoy the DIY experience instead of making it hard work.

But, but... I wasn't trying to make it hard work.
Remember Clarke's Third Law:
"Any sufficiently advanced technology is indistinguishable from magic."
I was simply trying to understand something that was previously a mystery (read: technology sufficiently advanced over my head...), and things are more fun when I understand them.
When I don't understand how a circuit works, it might as well be a monstrous inscription described in Lovecraftian adjectives which inspire similar feelings of loathing.
Now that I understand much more how the Bridged-T notch filter works, with math I can easily digest, my DIY experience is that much more convivial.
I have passed from magic to science, so to speak, and have come out the better for it.
So, there is a purpose to my madness, after all

I'm not as far along as you guys in understanding all of this, but I frequently use the modified James Tone Stack from the Duncan Tone Stack to simulate notch filters, hi-pass, low-pass, shelving, etc.  If you think of it in reverse, feedback loop filters can be simed.  Is a pretty dumbed down way to do it, but it is helpful designing filters for distortions where the main thing is how they sound to your ears.  I'm pretty visual and seeing what is happening more or less, helps me understand.  

The James tone stack, you say?
I'll take a look... Thanks!

[Earthscum]

...by not insisting that the driving impedance and the load impedance are integral parts of every circuit. They are; but we tell people to ignore them when they're starting out. It simplifies things. But the Real World waits at the end of the class. Source impedances and load impedances MATTER.
...

R.G.
I've been wondering if there's some kind of general ratio of impedances we could use to get started out, say if the input impedance is  (x) times smaller than the output, then your filter circuit should have (x) amount of impedance? I noticed, for instance, when I use a bootstrapped T filter with inverters, I can get the same response by adjusting values up a decade (resistors) and down for caps. When I figured up my impedance values between stages (at the time), I was figuring somewhere around 12k out from the driving circuit, and about 24k to the next. I was using 270k resistors and .0022u cap to ground. When I used the same network in a circuit that didn't use feedback on the following stage, it attenuated the signal a bunch, so I went back to the 27k/.022u filter I was using in transistor and opamp circuits, and it worked just like normal.

Is that maybe just something I miscalculated, or had boarded wrong, or is this the proper response? I've noticed that if your in/out impedances are 1:10, you can use low impedance networks, but if they get to around 1:4 to 1:2, you have to increase the impedance of the RC network? Or am I backwards in many ways?

[R.G.]

I've been wondering if there's some kind of general ratio of impedances we could use to get started out, say if the input impedance is  (x) times smaller than the output, then your filter circuit should have (x) amount of impedance? I noticed, for instance, when I use a bootstrapped T filter with inverters, I can get the same response by adjusting values up a decade (resistors) and down for caps. When I figured up my impedance values between stages (at the time), I was figuring somewhere around 12k out from the driving circuit, and about 24k to the next. I was using 270k resistors and .0022u cap to ground. When I used the same network in a circuit that didn't use feedback on the following stage, it attenuated the signal a bunch, so I went back to the 27k/.022u filter I was using in transistor and opamp circuits, and it worked just like normal.

Is that maybe just something I miscalculated, or had boarded wrong, or is this the proper response? I've noticed that if your in/out impedances are 1:10, you can use low impedance networks, but if they get to around 1:4 to 1:2, you have to increase the impedance of the RC network? Or am I backwards in many ways?

You're close to understanding. Here's a helpful way to think about source impedance and load impedance.

First, **all** signal sources can be modeled accurately as a voltage source which acts like the signal source when there is no loading, in series with an impedance. The impedance encapsulates all of the funniness that the signal source really does with frequency. This is called the Thevenin model for a signal source.

So if you drive a filter network with any real signal source, the filter is "corrupted" by the addition of the source impedance in front of it. So a filter which works fine from a 50 ohm source impedance signal generator will act really badly when you drive it with the output of the same signal generator through a 100K resistor, right? Adding a 100K resistor in this case makes the signal generator act like it has a source impedance of 100K resistive. That's the equivalent of driving the filter from the collector of a transistor with a 100K collector resistor.

So you can calculate filter parts till you're blue in the face but if you then glom on other unseen and unexpected impedances in front of it by using a high impedance signal source, it's going to not match your calculations. You can get good results two ways: (1) make the signal impedance so low that the filter sees it as negligibly different from a zero impedance voltage source, or (2) calculate the source impedance in as part of the filter design.

Number 2 is what most RF stuff does, because it's moderately impossible to get parts which don't have parasitic impedances which interact with filters at high frequencies. So you use the parasitics as part of the filter. Making the parasitics be accurate or not too bad for the filter is another issue, and a tough one. Whenever you see Smith charts and "S parameters", or admittance diagrams, run the other way.

Number 1 is what you do when you don't want to spend days doing calculations. This gets us back to the 10:1 rule. In general, if the source impedance is 1/10 of the input impedance of the filter, you can ignore it; at least you can for your first pass. For high precision, you may have to make a couple of passes, or do more accurate modelling and math up front, taking into account more effects. Be aware, 10:1 is just a rule of thumb. It gets you into the ball park and lets you do stuff faster by inspection.

If you can't get to 10:1, you have to consider the source impedance as part of the filter. And although I've been talking about impedances as though they're resistors, when you get much above midrange audio, this begins to be not true and the impedance may have to be considered to be a complex impedance with resistive, inductive and/or capacitive components. F'rinstance - what happens to your filter response if you block DC from a low impedance driving opamp which would otherwise be fine, but you use a capacitor which has too small a value? Yep, the capacitor is now part of your filter.

The same is true of loading on the filter. The input after the filter has an input impedance which looks just like a shunt impedance to ground. Just like you would prefer the source impedance driving a filter network to be less than 1/10 of the impedance of the filter, you would prefer the input after the filter to be MORE than 10x the filter impedance so the loading does not change the filter response.

All this is for voltage-mode filters. You can also do filters (and most electronics) as current mode. But most people think voltage mode.

[Earthscum]

That's a lot of info that just simplified a lot of my designs. I've been working around using buffers to drive my RC filters. I think I have a couple gain stages that are getting modified.

So, probably along the same lines, why do you see different resistor values on either side of the C in lowpass, and different C's in high pass? Are these kept so that (LP) R | 2C | R, where (R1+R2)/2 = R? and how does the capacitor side work? (C1+C2)*2 = C, right? Are the ratios selected because of the external impedances? I've always wondered about that and just stuck to your guides on t-filters to keep it simple (mainly the section about bridged twin-t's).

[R.G.]

That's a lot of info that just simplified a lot of my designs. I've been working around using buffers to drive my RC filters. I think I have a couple gain stages that are getting modified.

So, probably along the same lines, why do you see different resistor values on either side of the C in lowpass, and different C's in high pass? Are these kept so that (LP) R | 2C | R, where (R1+R2)/2 = R? and how does the capacitor side work? (C1+C2)*2 = C, right? Are the ratios selected because of the external impedances? I've always wondered about that and just stuck to your guides on t-filters to keep it simple (mainly the section about bridged twin-t's).

There's probably a simple, intuitive explanation for that. But I don't know it. 

I just went off to find my copy of "The Active Filter Cookbook" but it's apparently stored in one of my boxes of books.  My best understanding of the twin T is that it's really a phase shift network that shifts phase forward on one side, backward on the other, then sums the two responses. When the amplitudes are equal at the output and the phases add to 180 different, it makes an infinitely deep notch by the cancellation.

If you throw off the equal-amplitude result by imbalancing the components from the nominally perfect values, it makes the notch less deep and either the highpass or lowpass side more prominent. This may be useful for special situations. One such situation is where you're going to tune the notch frequency by changing one resistor to ground between the two caps. It broadens the range over which you can get a notch without it disappearing, but also makes the actual notch less deep, and of varying amplitude. But I've never dug all the way to the bottom of the twin T or bridged T to be able to do more than that. People more steeped in theory than I am  would immediately start writing Laplace transform equations and solving them. Could well be that imbalancing the T could let you include source or load impedance as part of the T setup as well.  Buffers in the right places are so cheap and easy to use and make the designs so much more robust that I tend to just put in the buffers. It's not elegant, but it makes for cast-iron designs.

Find yourself a copy of "The Active Filter Cookbook" if you can. If you do this electronics stuff you'll use it. My copy is over 25 years old as I remember.

[garybdmd]

     1
f = -------------
   19.9 RC

where
   --------C------
       |                 |
in------R-----+----R----out
                   |
                10C
                   |
             \/

this assumes both R's equal and bottom capacitor is 10 x's the top capacitor.
It's the typical default I would say.  The number comes from
                  1
--------------------------
2 pi sqrt(R1xR2xC1XC2)    where R1=R2 and C2 is 10x's C1 value. (That is a typical use I think)