Anyone familiar with the C-note bookshelf speaker kit?
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?
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.
@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)
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)
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.
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.
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!
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.
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 🙂
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".)
@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?
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.