Home Forums DIY Speakers and Subwoofers Anyone familiar with the C-note bookshelf speaker kit?

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    kanaaudio
    Member

    Hi!

    Would it be a problem to put the woofer and tweeter of the c-note kit into a enclosure with a volume of 11-12 liters?

    The c-note bookshelf speaker has a volume of approx. 9 liters and they sound great! (https://www.soundimports.eu/en/c-note.html). 

    But the specs of the woofer recommends a vented volume of 0.47 ft3 ~ 13,3 liters. https://www.soundimports.eu/en/dayton-audio-dsa135-8.html

    What would happen to the quality of the bass and sound in general if I were to put the tweeter and woofer into another enclosure, with a volume of about 11-12 liters? I would keep the depth the same as the c-note bookshelf speaker, and the length of the tube the same also? 

    Thanks!

    Anyone familiar with the C-note bookshelf speaker kit?

  • Anyone familiar with the C-note bookshelf speaker kit?

  • 123toid

    Administrator
    September 16, 2020 at 8:38 am

    You would want to re-tune it.  Redesign the port for the larger cabinet.  Keep in mind, when PE states a recommended box size, that really is just that…a recommendation.  Sometimes, people change the box size for better power handling, tuning preference, etc.  For example, there have been quite a few times, where WinISD will spout off a recommended size for the box that is enormous. Interestingly enough, it isn’t uncommon to be able to cut the box size in half with little loss in response. 


    Anyone familiar with the C-note bookshelf speaker kit?

  • kanaaudio

    Member
    September 16, 2020 at 3:21 pm

    @123toid thanks! How does one calculate and design the port in good relation to a chosen enclosure size? (You are talking to a newbie :-)) (strong builder, but not so strong audio knowledge)

  • tvor-ceasar

    Moderator
    September 17, 2020 at 2:12 am

    Are you ready for some math? This is from the QBASIC routines I wrote going on 30 years ago.

    Ported enclosure – enclosure size given (existing box)
    Minimum required info:
    Internal volume of box in Cu.Ft. (Box vol)
    Vas of Driver
    Qts of Driver
    Fs (or Fo) of Driver
    Diameter of port in inches in decimal form (PD)

    F3 = ((Vas/Box Vol)^0.5)*Fs
    Fb = ((Vas/Box Vol)^0.32)*Fs
    Port length = (1.463*(10^7)*((0.5*PD) ^ 2))/(Fb^2*Box vol*1728)-(1.463*0.5*PD)

    This next section has to do with the peak response due to non-optimal box size. The peak may be positive or negative, indicating the level of the peak around the Fb in dB. There are 2 steps: the first is to calculate a temporary value (V) that will then be used to find the dB level. I will give you 2 ways to do the second part since QBASIC uses a different LOG function than standard calculators such as the TI35 and similar. (one is LOG and the other is Natural LOG, I forget which is which)

    V = ((Vas/Box vol)^0.35)*Qts*2.6
    Peak Response in dB = (LOG(V))*0.43429445*20 [this is the QBASIC version]
    Peak Response in dB = (LOG(V))*20 [this is the Calculator version]

    A helpful hint: most software calculators have conversion sections built into them. You just have to expand them and look under their menus. Often times they also include a Scientific layout that gives you access to the extra functions in the formulae.
    Btw, in case you (the reader of this post) wonder what the caret symbol (^) means, it is to raise the preceding number to the power of the following number. So 2^3 is 2 to the 3rd power, or 2^3=8

    Now, using all this info we start with the required info:

    Box Vol 0.388 CuFt (11L)
    Vas 0.28 CuFt. (7.93L)
    Qts 0.38
    Fs 51.81 Hz
    PD 1.375″ (1-3/8″)

    Plugging all that info into the formula gives you:
    For a box size of 0.388 Cu.Ft.
    Your target driver will have a
    F3 of 44.01257
    Fb of 46.67432
    a 1.375″ diameter port at 3.728515″ long
    A Peak response of -1.096577 dB above baseline

    Obviously, since the Peak response is negative, it is below baseline.

    Now running the box dimensions in the C-Note kit on PE, you end up with an internal volume of ~0.32 Cu.Ft. This calculates out to:
    F3 of 48.46382
    Fb of 49.64279
    a 1.375″ diameter port at 4.068575″ long
    A Peak response of -0.5108053 dB above baseline

    In your case, a slightly larger box results in slightly extended bass response that is just a tad stronger than originally designed, even though it may seem opposite. Say the baseline is 87 dB. The C-Note would be 86.5 dB at 49-1/2 Hz and 83.5 dB at 48-1/2 Hz, while your box would be 86 dB at 46-1/2 Hz and 83 dB at 44 Hz. So very close, but at lower frequencies.

    I hope I haven’t confused you too much. It can be a daunting task in the beginning. Once you dive in, it can be addicting. Take care, and ask away. I’ll share what I know.


  • 123toid

    Administrator
    September 17, 2020 at 5:36 pm

    @kanaaudio

    If you are a visual learner like me these will help you out. 

    How to input a new driver

    Once you do that, the ported version will help you figure out port tuning. 


  • kanaaudio

    Member
    September 18, 2020 at 7:32 am

    @tvor-ceasar @123toid wow, thanks guys! Seems pretty straight forward, but I might run into some questions along the way anyway 🙂

    Does a tweeter need an enclosed volume as well? I mean, are the woofer and tweeter sharing the box volume, or am I calculating the final box volume of both drivers with the above formulas?

    Thanks again, really appreciate it!

  • 123toid

    Administrator
    September 18, 2020 at 8:06 am

    @kanaaudio

    That is a great question.  Almost all tweeters are sealed units. So they do not need an enclosure.  So if you were to open up a two way speaker, you will see the tweeter sharing the same space as the mid.  That it perfectly acceptable.  However, if you are building a three-way the mid and woofer would need their own enclosure.  Typically the mid would be sealed and the woofer could be either sealed or ported.


  • kanaaudio

    Member
    September 18, 2020 at 8:15 am

    @123toid

    That makes perfect sense, thanks for that!

  • kanaaudio

    Member
    September 18, 2020 at 10:02 am

    @tvor-ceasar

    Some follow-up questions, if you don’t mind 🙂

    1) In your very educational calculations (thanks for that!) you conclude a port length of 3.73 inches. Is there any calculation / requirement for the distance from the “internal end” of the port tube to the front panel? See sketch

    2) In the assembly manual for the C-note kit the recommended port length is 7″. I imagine that’s the reason the speakers have a relative long depth of 9.5″?  Is it so that the longer port length you have, the more bottom you will get? Your calculation states a port length of 4.06″, which would allow me to make a more streamlined box, but what are the trade-offs?

     

    Overall, I guess what I am trying to ask is, if the depth of the box has any importance (other than the required length of the port, of course) or if it is just the overall volume H x L x D that matters?

    Thanks again 🙂

  • tvor-ceasar

    Moderator
    September 18, 2020 at 9:09 pm

    @kanaaudio

    Re: tweeter – what Nick (123Toid) said. Can’t add anything more. 🙂

    Re: port – you usually want at least the same distance (or more) as the diameter of the port from the wall to the port opening. So your ??” would at least be the same or more as the port diameter.

    Not sure why they have a 7″ port other than tuning or phase inversion due to rear porting. Maybe Nick can chime in. (Also I haven’t had much chance to play with WinISD, so I don’t know how to mess with port lengths. I tend to go with the calculated “optimum” lengths rather than “forcing it”.)


  • kanaaudio

    Member
    September 18, 2020 at 9:49 pm

    @tvor-ceasar you say phase inversion due to rear porting – not sure exactly what that means? Is there a difference in having the port in the rear contra the front? 

  • tvor-ceasar

    Moderator
    September 18, 2020 at 10:39 pm

    When you look at how a speaker works, pushing air in first one direction and then the other in order to make the sound wave, and then knowing that the way the port works is to reinforce the target frequency/frequencies by being in phase with the sound coming from the front face of the speaker, and also realizing that no matter where the port is placed (front, rear, sides, top, bottom) the speaker is always acting the same on the port. That is, when the cone moves outward, it pulls air into the port from outside and when moving inward it will force air back into the port from inside. This is a constant fact. But where you place the port on the enclosure determines the phase shift the port needs to accomplish in order to have the port energy match the front face energy to actually reinforce each other. So, if the port is on the front, you want it to be in phase at the resonant frequency so they both push and pull in sync (technically the port is 180° since the back of the cone is pulling while the port is pushing). When you rear port, you want the rear port to be 180° (and again, technically this is 0° since the port is tracking the back of the cone) so that by the time it wraps around the speaker it becomes in phase with the front. The tuning of the port / length determines the output phase of the tuned frequency.

    Short enough, yet long enough to give a basic idea, I hope. It’s a deep subject that actually can require quite a few chapters to really nail down.


  • kanaaudio

    Member
    September 19, 2020 at 8:13 pm

    @tvor-ceasar

    Thank you for that, once again. I understand the first part fine, that makes total sense to me.

    But you lost me on this second part:
    “But where you place the port on the enclosure determines the phase shift the port needs to accomplish in order to have the port energy match the front face energy to actually reinforce each other. So, if the port is on the front, you want it to be in phase at the resonant frequency so they both push and pull in sync (technically the port is 180° since the back of the cone is pulling while the port is pushing). When you rear port, you want the rear port to be 180° (and again, technically this is 0° since the port is tracking the back of the cone) so that by the time it wraps around the speaker it becomes in phase with the front. The tuning of the port / length determines the output phase of the tuned frequency.”

    I am sure you have actually answered my question, I just don’t understand the answer 🙂 For example, if I place a port on the rear, say 1,375″ diameter and 4″ in length as per your calculation, that would be optimal for that given setup. What would be different if I placed the port on the front in stead? A 180 degree bending tube, a different length? A different calculation?

    I will be trying to install a port on the back of the speaker box. Are the calculations you have given me suitable for that? Do I need to take something additional into account, since you wrote that the reason PE had chosen a 7″ length for the port, even though the calculations states 4″, might be because of phase inversion due to rear porting?

    Sorry I keep pushing on this, I just really want to understand. I might be talking about something totally different 🙂

    – Martin

  • 123toid

    Administrator
    September 19, 2020 at 9:45 pm

    @kanaaudio

    To help with the two questions you have above. 

    Like @tvor-ceasar said, you typically will want about the port diameter away from the port wall.  This simply means, if you are using a 1.5″ port you will want it to be at least 1.” away from the front baffle.

    Your second question was what happens if you lengthen the port.  This will give you a lower tuning frequency.  However, that does not mean you will get more low end.  You could be tuning the enclosure too low, which would actually lose bass performance.  It also will affect cone excursion and rear port air velocity.  That is one of the reasons using a program like WinISD is so important.  It will allow you to see what would happen if you lengthen or shorten the port.  


  • kanaaudio

    Member
    September 19, 2020 at 10:12 pm

    @123toid thanks! I think I will play around with the numbers a lot more before trying to understand exactly what is going on. Seems like the right approach in this case, but at least now I understand the addiction 😉

    The videos you’ve linked me are great and very easy to follow step by step, but I have a habit of always trying to understand what happens behind the curtain. Not being able to interpret and validate computer readings and blindly following data output seems like a fast track to making bad designs 🙂

    That’s one of the reasons this forum rocks!

  • kanaaudio

    Member
    October 1, 2020 at 9:07 am

    Hello again – Just wondering, can WinISD and the excel spreadsheet Unibox 4.08 (attached) do the same? I mean, do they work / calculate the same?

    I have run into a problem where I cannot install WinISD on my laptop, because it is a company PC loaded with restrictions.

    All I need it to figure out a good box size for the C-note kit drivers related to a good port length, I guess. The excel sheet seems to be able to do that with some help from @tvor-ceasar ‘s math above. But I have never tried WinISD so I don’t know what I am comparing to.

    Sorry if I am asking an obvious question …

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