So my first design, a ported 15 cubic foot model, was quickly and decidedly vetoed by the significant other.
I discovered the Linkwitz Transform the other day and proceeded to model sealed boxes with LT circuits. I was able to design a 2 cu. ft. model with a 15" Titanic driver in WinISD that looked pretty darn good.
In fact, it looks too good to be true. The model claims I can get 101 dB to 20Hz on 25 watts.
That can't be right. Can it?
I've posted my model at http://www.nogginboink.com/subwoofer_design3.htm. I'd appreciate any sanity checking that y'all might be able to provide.
Thanks!
-NogginBoink
I discovered the Linkwitz Transform the other day and proceeded to model sealed boxes with LT circuits. I was able to design a 2 cu. ft. model with a 15" Titanic driver in WinISD that looked pretty darn good.
In fact, it looks too good to be true. The model claims I can get 101 dB to 20Hz on 25 watts.
That can't be right. Can it?
I've posted my model at http://www.nogginboink.com/subwoofer_design3.htm. I'd appreciate any sanity checking that y'all might be able to provide.
Thanks!
-NogginBoinkBTW, put back the Le in your simulations hehe!
I checked, you would need a 800W amplifier to go down to 13 Hz.
Since you can afford an expensive driver like the Brahma, look at Ascendant Audio, they have nice high excursion 4 ohms drivers like the Avalanche 18. One Avalanche 18 in a 2 cu.ft box with a Linkwitz Transform would be a killer hehe!
I checked, you would need a 800W amplifier to go down to 13 Hz.
Since you can afford an expensive driver like the Brahma, look at Ascendant Audio, they have nice high excursion 4 ohms drivers like the Avalanche 18. One Avalanche 18 in a 2 cu.ft box with a Linkwitz Transform would be a killer hehe!
Okay, I added the "apparent load power" graph to my page and the graph is a lot more like what I would expect for this design.
I don't understand the difference between the "signal" tab of the model and the "Apparent load power" graph. I told WinISD to model a 25 watt input, but it's now saying the apparent load power at 17Hz is ~560VA. Since a watt is one volt-amp, WinISD seems to be contradicting itself.
The "apparent load power" graph is not explained in WinISD's help file.
Can anyone explain to me the difference between the Signal tab and this graph?
(Simon, thanks for reminding me about Le. I built two models of the Dayton driver and used the one without the Le parameter. Rather than changing all the graphs on my web page I stayed with the simpler model. I'll be sure to use the more accurate model before making any buying/design decisions.)
I don't understand the difference between the "signal" tab of the model and the "Apparent load power" graph. I told WinISD to model a 25 watt input, but it's now saying the apparent load power at 17Hz is ~560VA. Since a watt is one volt-amp, WinISD seems to be contradicting itself.
The "apparent load power" graph is not explained in WinISD's help file.
Can anyone explain to me the difference between the Signal tab and this graph?
(Simon, thanks for reminding me about Le. I built two models of the Dayton driver and used the one without the Le parameter. Rather than changing all the graphs on my web page I stayed with the simpler model. I'll be sure to use the more accurate model before making any buying/design decisions.)
Hi again, long time no see!
You asked for 25W imput, if you look at higher frequencies, you're really using 25W.
The Linkwitz Transform circuit is working like a very advanced equalizer. It's boosting the low end. When you boost the low end, you need more power.
This is a really rough illustration of the phenomenon. This circuit is playing with the impedance curve/phase, so the amplifier believe to see something else than the speaker, something that need more voltage and more current. Because of that, the amplifier send more power to compensate.
Just check for example your first design, the vented subwoofer. Compare the "apparent power load" graph with the "impedance" graph. The port is resonating, so the impedance is rising near the tuning point, because you need less power to achieve the same loudness. Also, the driver is resonating near Fs, so you need less power to achieve the same loudness again, the impedance is rising. You see two peaks in the "impedance" graph. You'll see two dips in the "apparent power load" graph. Higher impedance so less power is actually drawn from the amplifier at that frequency.
The imput power is based on power if the load was resistive. A speaker is not a resistive device. In the end, you ask a voltage to your amplifier. Let say you use a 4 ohms speaker, you ask 25W, so you ask 10V to your amplifier. When impedance is high, the power draw will be low. If you get a 50 ohms peak, the current will be 0.2 A so power drawn from the amplifier is 0.2 A x 10 V = 2W. At higher frequencies, the power drawn will be around 25W because it's a 4 ohms speaker over most of the range.
So, we could say roughly that a Linkwitz Transform circuit is working the other way around.
I'm not experienced much with the Linkwitz Transform circuit. Maybe someone with more knowledge than me will step in and explain the details hehe!
BTW, yes you'll definitely need a huge amplifier if you go with the Linkwitz Transform circuit. Huge amplifiers are quite cheap, and your girl prefer smaller boxes so you get the best of both worlds.
You asked for 25W imput, if you look at higher frequencies, you're really using 25W.
The Linkwitz Transform circuit is working like a very advanced equalizer. It's boosting the low end. When you boost the low end, you need more power.
This is a really rough illustration of the phenomenon. This circuit is playing with the impedance curve/phase, so the amplifier believe to see something else than the speaker, something that need more voltage and more current. Because of that, the amplifier send more power to compensate.
Just check for example your first design, the vented subwoofer. Compare the "apparent power load" graph with the "impedance" graph. The port is resonating, so the impedance is rising near the tuning point, because you need less power to achieve the same loudness. Also, the driver is resonating near Fs, so you need less power to achieve the same loudness again, the impedance is rising. You see two peaks in the "impedance" graph. You'll see two dips in the "apparent power load" graph. Higher impedance so less power is actually drawn from the amplifier at that frequency.
The imput power is based on power if the load was resistive. A speaker is not a resistive device. In the end, you ask a voltage to your amplifier. Let say you use a 4 ohms speaker, you ask 25W, so you ask 10V to your amplifier. When impedance is high, the power draw will be low. If you get a 50 ohms peak, the current will be 0.2 A so power drawn from the amplifier is 0.2 A x 10 V = 2W. At higher frequencies, the power drawn will be around 25W because it's a 4 ohms speaker over most of the range.
So, we could say roughly that a Linkwitz Transform circuit is working the other way around.
I'm not experienced much with the Linkwitz Transform circuit. Maybe someone with more knowledge than me will step in and explain the details hehe!
BTW, yes you'll definitely need a huge amplifier if you go with the Linkwitz Transform circuit. Huge amplifiers are quite cheap, and your girl prefer smaller boxes so you get the best of both worlds.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.