# 4th order bandpass calculator – With box design example

## How to build a 4th order bandpass subwoofer box in 7 steps?

The 4th order bandpass box design is not recommended for the first time builder. You need some experience before you get into these complicated designs. Actually, if you are a good listener and/or are savvy with enclosure design software, you shouldn’t have much of a hassle when making this kind of project. The problem is, even if you conceive a good 4th order bandpass design, but have slight building errors, the result will be sub-par. Now that you are all encouraged, let’s learn more about these bandpass enclosures.

**What is a 4th order bandpass box ?**

The term bandpass is also used in passive crossovers. Usually used for midrange drivers, to filter low frequencies and high frequencies at the same time, only to let the frequencies between the crossover points pass.How is this passive filter example relevant to the bandpass speaker enclosure ? It is, because when you look at the frequency response of a bandpass subwoofer, it looks exactly like the frequency response of a speaker with a bandpass filter on it. Imagine the frequency response of sealed or bass reflex enclosure. For the very low frequencies, when the resonant frequency is reached, the response starts to roll-off. Now, for the higher frequencies, the response will naturally start to roll-off, at different points, depending on the speaker. Because of the size of the woofer, it will be impossible for it to play high frequencies at some point, so it naturally rolls-off.

However, the 4th order bandpass design acts differently. It is composed by two chambers : a rear sealed chamber and a front vented chamber. You can think of it like a normal closed box system, but the vented chamber in front of the speaker acts like an acoustic filter, a low pass filter to be more precise, so higher frequencies are filtered depending on how it is designed. Because of this filtering, the frequency response bandwidth is quite narrow. You can make it broader, but you will sacrifice efficiency. Or you can make it even more narrower and increase efficiency by a lot. This seems appealing for those who want to play a narrow frequency range at very high volumes (like car audio SPL competitions). Sealed and bass reflex can make this kind of trade-off also, but bandpass has a better “yield”.

**How to make a bandpass box ?**

Further on, I will explain how to design your box without any design software. Your pen and paper will be the actual 4th order bandpass calculator. This will get a bit technical, but I’m sure you will figure it out. Before you start off your project, you should first select a woofer. Anything that will work well in a sealed box, will work well in 4th order bandpass design as well. If you want to crunch numbers when choosing your speaker, you can divide f_{s} / Q_{es} , and the smaller the number, the better (lower than 60). This means a good bandpass speaker will have a low f_{s}, or high Q_{es}, or both.

First of all, let’s define some terms, so we can speak the same language :

- f
_{B}– Tuning frequency of the port in the vented chamber. - R – The radius of the port in centimeters.
- L
_{v}– Length of the port in centimeters. - f
_{L}– f_{3}of the low frequency roll-off. - f
_{H}– f_{3}of the high frequency roll-off. - Q
_{bp}– Q of the sealed rear chamber. - V
_{f}– Volume of the front chamber (the acoustic filter chamber) in liters. - V
_{r}– Volume of the rear chamber (the sealed chamber) in liters. - S – Passband ripple (How many ±db does the frequency response deviate from linear frequency response).
- V
_{t}– Total volume (V_{f}+V_{r}) in liters.

As you can see from the graph above, f_{L} and f_{H} are positioned -3 db after the response starts to roll-off (for low and high frequencies). The passband ripple measures the amount of variation from linear response. High amount of ripple will result in peaks / dips in the frequency response. Judging from our graph the ripple looks like around ±0.5 db.

### 4th order bandpass calculator in 7 steps

We are going to explain, step by step, what are you supposed to do in designing a great sounding box. To make things more clear, we are going to use a real life woofer, and design a bandpass subwoofer box for it. The woofer I chose is the **JL 10TW3-D4** . There is no particular reason for my pick, other than it has good specs for a sealed enclosure, which means it will be good for bandpass as well.

I’m going to list the Thiele / Small parameters of the JL woofer for convenience :

- F
_{s}= 32.3 Hz. - Q
_{es}= 0.656 . - Q
_{ms}= 11.35. - V
_{as}= 19.82 L. - Q
_{ts}= 0.62. - X
_{max}= 15.2 mm.

### Step 1 : Know what do you want

You have to understand that there is no perfect bandpass enclosure for each kind of woofer. You have to make compromises between linear frequency response, how wide the frequency response is, and efficiency. General rule of thumb is that the louder it is, the narrower the frequency response is. Depending on what you want to achieve, you will have to find a balance between these 3 aspects, so that you will end up with a 4th order bandpass design that suits your needs.

### Step 2 : Determine how linear the frequency response should be

Remember, at the beginning of the article, we talked about some parameters that are relevant to this project. One of them is S, which is the passband ripple. This ripple, describes how many ±db, the frequency response will digress from linear response. In a perfect world the ripple would be ±0 db, which is achievable. However, that is at the expense of the other 2 characteristics we need to take care of, in the later steps.

There are 3 values of S, which give an acceptable amount of ripple, and have the following characteristics :

- Best transients for S = 0.7, and 0 db ripple.
- S = 0.6 , somewhat degraded transients , ±0.35 db ripple.
- S = 0.5 , worse transients than S = 0.6 , ±

Picking a certain value for S, will narrow your possibilities for the other factors we need to figure out. Another guideline which is useful when choosing S, is that if you go for a bigger value (0.7), the frequency response will be narrower. On the other hand, a lower value S (0.5), would translate into a wider frequency response. So, choose the ripple you like, but keep this in mind when doing so.

Here is a table with the values of S, and the values of the frequency response and sensitivity, corresponding to that certain S value. Get familiar with this table, as we will need it for the next step as well.

**For our woofer we will choose S = 0.6.**

### Step 3 : Determine the desired frequency response bandwidth and sensitivity

Now you have to determine the values of f_{L} and f_{H}. This means you will determine the – 3 db points, when the response starts to roll-off, for both low frequency and high frequency roll-offs. This will effectively set your frequency response bandwidth between the two values. You figured out the value of S, at the 2nd step, now you will chose the values of f_{L} and f_{H}, corresponding to that S. After that, the value of the sensitivity is chosen for you. If you choose a certain value for the sensitivity, then the values of f_{L} and f_{H} are chosen for you. This is the balance I talked about, that you need to make, to find the sweet spot.

To find the values of f_{L} and f_{H} you have to do the following :

- f
_{L}= F_{s}/ Q_{ts}* (f_{L}factor) - f
_{H}= F_{s}/ Q_{ts}* (f_{H}factor)

**Now let’s do some number crunching for our woofer :**

**I’m choosing the sensitivity to +5 db because I want a loud bandpass enclosure.****This means that the start and end of the frequency response are chosen for me.****f**_{L}= 32.3 / 0.62 * 0.7317 = 38 Hz.**f**_{H}= 32.3 / 0.62 * 1.6877 = 88 Hz.

**So, we have figured out that the frequency response will be from 38 Hz to 88 Hz, with a +5 db boost and **±**0.35 db ripple.**

### Step 4 : Calculate the volume of the front enclosure

Calculate the front chamber volume using the following formula:

**V _{f} = (2S * Q_{ts})^{2} * V_{as}**

**V**_{f}= (2 * 0.6 * 0.62)^{2}* 19.82 = 11 L

### Step 5 : Calculate the volume of the rear enclosure

Calculate the rear chamber volume using the following formula:

**V _{r} = V_{as} / ((Q_{bp} / Q_{ts})^{2} – 1)**

**Vr = 19.82 / ((1.1113 / 0.62)**^{2}– 1) = 19.82 / (3.21 – 1) = 9 L

### Step 6 : Calculate the tuning frequency of the front chamber

Calculate the tuning frequency using the following formula:

**f _{b} = Q_{bp} * ( F_{s} / Q_{ts})**

**f**_{b}= 1.1113 * ( 32.3 / 0.62) = 1.1113 * 52.1 = 58 Hz

### Step 7 : Calculate the dimensions of the port

The radius of the **port** (R), should be as large as possible, to minimize port non-linearity. Understand that making the port bigger, will mean that the length of the port will be longer. This means there are certain limitations to how large you can go.

**For our project I’m choosing a port radius of 5 cm (10 cm in diameter).****R = 5 cm.**

Calculate the port length using the following formula:

**L _{v} = ((94250 * R^{2}) / (f_{b}^{2} * V_{f})) – (1.595 * R)**

**L**_{v}= ((94250 * 5^{2}) / (58^{2 }* 11)) – (1.595 * 5) = (2356250 / 37004) – 7.98 = 63.68 – 7.98 = 55.7 cm

### Results for our 4th order bandpass calculator

Now we have finished making a 4th order bandpass design box for our 10″ JL woofer, and the dimensions are as follows :

**Front chamber = 11 L.****Rear chamber = 9 L.****Port diameter = 10 cm.****Port length = 55.7 cm.**

Please bear in mind, that these are the net volumes. This means that you will have to add the volume displaced by the speaker, to the volume of the rear chamber. Also the volume displaced by the port needs to be added to the volume of the front chamber. Add any other elements to the total volume (like bracing).

The frequency response will look like the following graph :

### Variations of 4th order bandpass enclosure

In the example above, we used just one speaker, and made a 4th order bandpass design. If we use 2 speakers, we can place them in different positions so that we end up with different variations of bandpass enclosures. Here are some variations of the 4th order bandpass enclosure:

**Single driver bandpass**subwoofer box.**Dual driver**push / pull bandpass enclosure.- Push / pull compound bandpass box design.
- Triple chamber bandpass subwoofer box.
- Push / pull triple chamber bandpass box design.

For the dual driver push / pull variation, calculate the volume of the front and rear chamber, for each individual driver, and then add them up. For the triple chamber bandpass subwoofer box, the rear chambers are separated, and therefore calculate them normally. The 2 drivers are sharing the center chamber. So calculate the volume of the front chamber for each driver individually, and add them up, to get the volume of the center chamber. When there is a push / pull configuration, remember to connect one of the drivers out of phase electrically (reverse polarity).

### Conclusion on bandpass box design

The 4th order bandpass box design is definitely an interesting solution. If you do not need a wide frequency response, and want a boost in output, you should seriously consider it. However, the enclosure can get quite big, and you don’t have direct access to the speaker. As a result, if you need to replace the speaker, you have to tore open the enclosure. The design and build difficulty can be a let down for the inexperienced builder, but if done properly, the 4th order bandpass design can be quite impressive.

*References*

- Loudspeaker Design Cookbook 7th Edition by Vance Dickason (Audio Amateur Pubns, 2005).
- Introduction to Loudspeaker Design: Second Edition by John L. Murphy (True Audio, 2014).
- Image source : link.

## 64 comments

Nice write up

Thank you for breaking this down in easy to follow directions. I have always wanted to know how they were figured. Thanks to you, I will be building my first one today to enclose a JL Audio 13.5 shallow mount ironically enough. Thank you again!

prejudice compliments for the explanation. very clear the only past that I do not understand is this f L = F s / Q ts * (f L Factor)

f H = F s / Q tof H factor fl and fh factor have fixed values or has taken them from the woofer parameters. I have an audio jl 12w7

The values of fL Factor and fH Factor can be found in the alignments table, which can be found at step 2. Qts and Fs are woofer parameters and can be found in your 12W7 parameter spread sheet.

Let’s say you choose the ripple S=6 and the sensitivity of +3 dB. Then, the fL=0.6217 and fH=1.5778. You have the liberty to choose other values for fL and fH (from the alignments table), but then the sensitivity will be different.

Grazie mille per la spiegazione

these are the characteristics of my driver. S factor which I could use to have a good system

FS 27,2

Qes 0,514

Sgq 7,807

Qts 0,482

Vas 66,0 lt

Xmax 29mm

No 0,25%

(1w/1m) 86,2 db spl

Sd 0,0542 mq

That is a choice that you have to make, there is no perfect choice, it depends on your application. I would go for a +3 dB sensitivity and when you choose S, consider the following :

lower value S means : higher ripple, wider frequency bandwidth, worse transients, smaller box

higher value S means : low or no ripple, narrow frequency bandwidth, better transients, bigger box.

If you have a hard time choosing you can go for S=0.6 which is the middle point.

Grazie ancora

from the calculations I got back chamber volume 20.483 liters 22.080 liters anterior chamber diameter pipe length 10cm led 1901.420 cm. Now as I understand it should add to the rear volume the volume occupied by the driver, and front volume that occupied by the conduit?

The volumes are correct. The pipe is 10 cm in diameter and 38.7 cm in length. And yes, those are net volumes. You have to add the volumes displaced by anything (driver, pipe, bracing etc)

The length I 19 not 38. But if I put a tube with blunt ends I have to calculate as if it all straight?

I have made the calculation again, and even used a modeling software and it’s 38 cm. Check your calculations. If by blunt you mean flared, then the total length (38 cm) is the length of the straight pipe + half of the length of the flare.

Ho eseguito la formula sopra riportata. Quindi 38 cm da metá svasatura interna a metá svasatura esterna? Scusa tutte queste domande ma non vorrei fare dei danni

Corretta 🙂

La formula che ho eseguito è questa L v = ((94250 * R 2 ) / (f b 2 * V f )) – (1.595 * R)

E quindi L v = ((94250 * 25) / (3123,580* 22,080)) – (1.595 * 5)=26,189

La Fb=55,889

Vf=22,080

Yes, you are right. I somehow used the Qbp from S=0.7 and got an Fb of 47.9. But, taking a closer look now, you are correct. The length is 26 cm

help please , I can’t count the chamber and port , I don’t understand the subwoofer machete M15 D2 here the data of Thiele small SPL – 88.28 dB. vas-75.36 L. FS-34.10 Hz . BL-16.70 qts-0.63. I love when low bass.

For S=0.6 and a sensitivity of +4 dB you get the following :

Rear sealed chamber : 42.5 L

Front vented chamber : 43 L

Port diameter : 150 mm

Port length : 300 mm

For S=0.7 and a sensitivity of +3 dB (this will go very low, but the enclosure is big) :

Rear sealed chamber : 92.4 L

Front vented chamber : 58.6 L

Port diameter : 150 mm

Port length : 350 mm

thank you very much for your help !

Un’ultima domanda per determinare la lunghezza del condotto devo tener conto del diametro interno effettivo o di quello esterno perché internamente misura 9,6 cm ed esternamente 10 cm. Mille grazie

the internal diameter

Ok grazie mille proverò a realizzare questo progetto e ti farò sapere grazie ancora

Buongiorno! Volevo Avere un consiglio per la Realizzazione del box mio passa banda avrei Pensato di Mettere Sulla parte anteriore per Poter VEDERE il conducente del plexiglass vorrei Sapere di che spessore?Le altre Pareti Sono in MDF da 30 mm

Please stop speaking in Italian. I never used plexiglass before, so I can’t tell you how thick it should be. However 30 mm MDF is really thick. Totally overkill in my opinion. Don’t get me wrong, the thicker the better. But the enclosure will be very heavy. Most definitely you won’t be able to lift it alone.

Buon giorno! Volevo avere un consiglio per la realizzazione del mio box passa banda avrei pensato di mettere sulla parte anteriore per poter vedere il driver del plexiglass vorrei sapere di che spessore le pareti sono in MDF da 30 mm

I try with plywood 28 mm.

I am building a fourth order bandpass enclosure for 412 inch subwoofers and I am wondering if I will be able to do this with a space that is 36 x 37 x 15 please get back to me with answer or if you may even have a Nother suggestion if I could do a six order Banpass enclosure with 3 12s instead in that space

That’s a really big box. Calculate the volume for 1 speaker and multiply that by how many speakers you have. That’s your overall volume. My guess, it will surely work for 3 speaker, but you have to check for 4.

I have these factors:

Fs 30

Qes 0.53

Qms 7.07

Vas 67.8

Qts 0.49

Xmax 10

And with S:0.6 and -2dB i got Lv < 0, could you help me with that or am i doing something wrong

fL = 30 / 0.49 * 0.4052 = 24.8 Hz

fH = 30 / 0.49 * 1.3612 = 83.3 Hz

Vf = (2 * 0.6 * 0.49)^2 * 67.8 = 0.34 * 67.8 = 23.05 L

Vr = 67.8 / ((0.7427 / 0.49)^2 – 1) = 67.8 / (2.29 – 1) = 67.8 / 1.29 = 52.55 L

fb = 0.7427 * (30 / 0.49) = 45.5 Hz

I suspect you have a 12″ woofer so a 15 cm diameter port would be nice. R= 15 / 2 = 7.5 cm

Lv = ((94250 * 56.25) / (2070.25 * 23.05)) – (1.595 * 7.5) = 5301562.5 / 47719.26 – 11.96 = 111.1 – 11.96 = 99.14 cm

That’s going to be pretty hard to implement. Let’s try with a 10 cm diameter port R = 5 cm

Lv = ((94250 * 25) / (2070.25 * 23.05)) – (1.595*5) = 2356250 / 47719.26 – 7.97 = 49.37 – 7.97 = 41.4 cm

Good luck!

Bună seara, și felicitări din nou pentru toate marile sfaturile pe care le dau noi, cititorilor. pentru că am de gând să realizeze a patra comanda mea și a vrut să știe cum să calculeze volumul tubului reflex pentru a putea adăuga la volumul caseta. și conducta de formă circulară

Thank you for this write up! You’ve made it very easy to understand banpass box design

hi.can i ask u a question?

i got a sub with these parameters

Fs 43

Qes 0.5

Qms 5.3

Vas 12.15

Qts 0.46

Xmax 22mm

i want to use 12th of this sub in a 6th order box if u can please help me.

thank you

There isn’t much documentation out there for 6th order bandpass. Bose has a patent on the design and unfortunately it’s not available. You can find software to model the 6th order bandpass response, and you could make one for personal use, without any legal issues. However, the design has a very thin tolerance for error, and since you are vaguely asking for help, I’m advising you to chose another enclosure design.

hello

what’s you suggestion about regarding ti subwoofer parameters?

what’s you idea about 4th order

please advise us one technically channel virtual network and website or Email or another connection way.

best regards

Hi!

I’m calculating a 4th Bandpass Triple Chamber for Dayton Audio RSS210HF. I’m calculating a single woofer first to add the Vf values later. But I have some questions regarding the ports…

I want the frequency range to be really low, as the mid range is taken care of by other speakers. The woofers are supposedly able to handle the low frequencies quite alright. Low sensitivity is not going to be an issue, as the amp powering them will provide more than enough, so I went for 0,35dB ripple and -3dB sensitivity to get the frequency range where I want it, Fl 19,2 Hz & Fh 69 Hz.

However, I’m confused by the port. Using a 6 inch port will give a length of 340 (!) centimetres, and a 4 inch port will give a length of 148 cm, which isn’t possible to fit.

Also, how do I calculate more ports since I’m building a triple chamber with twice the volume of the front chamber Vf? I assume that two equal 4″ or 6″ ports won’t result in the same Vf as a single one?

Vf 11 litres

Vr 42,3 litres

Fb 37Hz

Lv 340cm/148cm ????

Link to woofer… http://www.daytonaudio.com/index.php/rss210hf-4-8-reference-hf-subwoofer-4-ohm.html)

Specs

Fs 29,6

Qts 0,57

Qes 0,69

Qms 3,25

VAS 23,71

Thanks

Looks like you already calculated the front and rear chamber volumes. In a triple chamber, the rear volume is the same (each speaker with its own chamber). And the front chamber is shared by the 2 drivers, and it has to be double the volume of Vf.

Unfortunately you calculated the port correctly. You can’t do nothing about that. The problem is that the Vf is only 11 liters. As the volume of the ported chamber get’s larger, it will demand a shorter port. Only thing you can do is go for a smaller diameter port. Going for 2 ports is not a solution. For ex : If 1 port of 4″ diameter needs to be 148 cm long, 2 ports of 4″ need to 300 cm EACH. So you can understand that makes things even worse.

Try to choose another alignment, where the ported box is bigger, or go for a smaller port (but you risk unwanted wind noise).

Hello,

I want to build band-pass enclosure as specified above with RCF L15S800 and i calculated Vf= 16,6 L which is 0.016 m3 and Vp=10,2 L which is 0.01 m3. I wonder isnt this a bit too little volume for 15″ transducer ?

I wait for Your help…

Your calculations look correct. Your driver has a very high Fs for a 15″ woofer. Very high sensitivity also. It’s probably designed for outdoor use, to be very loud. Because of this, it doesn’t go very deep in bass. Since it’s not playing the very low notes, it doesn’t demand a big enclosure.

Thanks a lot Marius 🙂

I’m obviously really thick as I don’t get any of that!

Is there anyone here can help me design a 4th order box as I don’t have a scoobie what any of those calculations mean??

thanks

can someone help me what a three chamber for 2 8″ kicker C8 old school model.

I have a Rockford fosgate 12″t2. With a fosgate 1200rd amp, rms 1347 watts @ 1ohm. Would a 4th order box rock my Toyota t100?

Yes it would :). Probably you should go for bass reflex though. You have enough power and it’s easier to make.

Any secrets of the bass reflex or a helpful site?

Check out these 2 articles :

http://audiojudgement.com/bass-reflex-speaker-design/

http://audiojudgement.com/bass-reflex-alignments-explained/

I will read them! Thank you ! God Bless you.

Maybe one of you could help me out:

I am trying to design a small(ish) sub to fit under the sofa. In order to see what woofer could fit the bill, I have been trying a couple different ones.

With a Monacor SP-202PA (Q_ts= 1.1; V_as= 12 L; Q_es = 2.25, f_s= 68) I am getting a usable design even with S=0.7.

The problem is that I am getting a negative rear volume.

V_r = V_as /( (Q_bp/Q_ts)^2 )-1

V_r = V_as/ ( (0.8489/1.1)^2 )-1

Q_bp/Q_as is smaller than 1, so its square must also be. Subtract 1 and you end up with a negative coefficent to scale V_as.

I’m missing something very obviuos, ain’t I?

Your driver has a Qts of 1.1 which is very high. It is suitable for infinite baffle, and not much else. If you want a small enclosure you should take a look at sealed enclosures. You should definitely forget band pass.

Yeah, you’re probably right.

The damping on that particular chassis is terrible. The other ones that gave negative volumes are also very non-compliant drivers, so the math is obviously teeling me to keep my hands off cheap rubbish like that. :o)

How do you choose the S value?

I dont quite understand that.

A higher S number will translate into better transient response and better linearity. However, the response bandwidth will be narrower. (and vice-versa)

The PS48 subwoofer of bose (bass module from a Lifestyle V35 for exemple) comprises 3 event on the from face. With type of isobaric construction is it exactely ?

Because there is a small cavity inside the subwoofer box where the 2 speaker are mounted, but the cavity is not completely closed.

The box also presents two other opening, one on the upper part of the box (seems to be attached to the upper speaker) and a second at the lower part of the box (seems to be attached to the second speaker). What is the utility and functionning of all theses 3 openings ?

Thanks for explanations.

Can’t really tell for sure from pictures. But my guess it’s a 6th order bandpass enclosure (because bose has the patent on the design) and the extra cavity on the inside is an isobaric setup to reduce the volume of the box.

Thanks a lot for your reply.

On an isobaric construction the internal cavity is generally totally closed. But this is not the case here with the ps48 sub.

Do you know the reason ?

Do you think is it possible to design a sub similar to a PS48 from BOSE perfectely adapted to Home Theater use ?

Note that the particularity of this “bass-medium” enclosure from bose, it to work upto 260 hz ! (On the upper part of bandpass).

If you say that the inside cavity is open it does rise a lot of questions for me too. You kinda expect this from bose. This is their philosophy : very cheap speakers, but over-engineered enclosures and a lot of digital processing. If you try to copy the design, not knowing how it works, you will most likely end up with something that you won’t be proud about. The fact that it goes so high in frequency response means one of 2 things : the lower part of bandpass is high as well (not hitting the very low notes) or it does go low in frequency but has poor efficiency rating (not impressive in volume department).

I need some help. I can’t figure out the Qbp of my woofer. It’s a DDX-15D2. I have my front enclosure volume as 13.308L

I cannot tell you that. Qbp doesn’t depend on the speaker. It depends on how linear you want the frequency response (S), frequency response bandwidth (fH and fL), and efficiency of the enclosure. These are parameters that you choose (more or less) however you want. Read the article again and pay more attention.

So when designing the triple chamber how does port size work? Do we just double it in size as we do the volume of the box?

No. You just double the volume of the chamber (so Vf is now double). Set a diameter for the port as you would normally (go for a larger diameter now so you don’t end up with chuffing noise). Then, calculate the length of the port using the Lv equation (difference is now that Vf is larger so the length will be different).

I want to do a double 18” fourth order bandpass. I want both drivers to share the rear chamber in parallel. What do I do for calculating Vf and Vr?

I would like both magnets housed in the sealed chamber

It’s just like having one speaker but with double the Vas value. The rest of the parameters (which are relevant to this enclosure) remain the same (Fs, Qts).