Introduction

Designing a sealed box speaker system would appear to be a very simple task but other not so obvious factors complicate matters. Variations in the system Q caused by changes in the total amount of electrical resistance in series with the driver and/or the effects of box lining and filling can result in a speaker system that doesn't behave as it was intended.

The volume of a sealed box, Vb is calculated by using this simple formula...

Vb = Vas / (( Qtc / Qts ) ² - 1)

where...

Vas is the volume of air having the same compliance as the suspension of the driver in litres and Qts is the Q of the driver at resonance ( fs ) in free air considering both its electrical and mechanical resistance.

Both Vas and Qts are listed in the drivers specifications and Qtc which is the Q of the driver in the box and at the resonant frequency of the closed box ( fc ) considering both its electrical and mechanical resistances and determined by you based on the "type" of sound you want the system to produce. There are various values of Qtc possible ranging from 0.5 all the way up to 1.2 which is about as high as any sensible sealed box design should go and summarised in the following table...

Qtc	Alignment	Box Size	Transient	Phase	Amplitude	Power	Overall Sound
0.5	Critically damped	Biggest	Best	.	.	.	Tight and dry
0.577	Bessel	.	.	Best	.	.	.
0.707	Butterworth	.	.	.	Best	.	.
0.8 - 1.2	Chebychev	Smallest	.	.	.	Best	Warm and loose

Picking a Qtc of 0.707 which will keep most people happy and using the following driver...

Driver	Qts	Qes	Qms	Vas	Re	fs
Morel MW267	0.33	0.40	2.08	113	6.6	25

We get...

Vb = 113 / (( 0.707 / 0.33 ) ² - 1) = 31.47 litres

Adding series resistance

Whilst Vb = Vas / (( Qtc / Qts ) ² - 1) does work it assumes a perfect system with no series resistances and in order to determine an accurate box volume we need to take all these series resistances into account otherwise the finished Qtc of the system will not be as calculated.

The first step is to calculate a new Qes which we call Qes` and...

Qes` = Qes (( Rg + Rx + Re ) / Re )

where...

Qes is the Q of a driver at fs in free air considering only its electrical resistance.

Rg is the DC output resistance of amplifier, and Rx is the total DC resistance of everything between amplifier and driver and Re is the DC resistance of a drivers voice coil.

Both Qes and Re are listed in the drivers specifications, as is Rg listed in the amplifiers specifications (or derived from the damping factor quoted) and you can measure Rx.

In general for a solid state amplifier Rg will be less than 0.1 Ohm and perhaps as much as 1.0 Ohms for a valve amplifier. Rx should be less than 0.1 Ohms unless you are using a passive crossover in which case the total DCR ( direct current resistance) of all inductors in series with the driver must be included.

Next step is to calculate a new Qts which we call Qts` and...

Qts` = ( Qes` x Qms ) / ( Qes + Qms )

where...

Qms is the Q of a driver at fs in free air considering only its mechanical resistance and is again supplied in the drivers specifications.

So our new box volume formula is now...

Vb = Vas / (( Qtc / Qts` ) ² - 1)

So back to our driver...

Driver	Qts	Qes	Qms	Vas	Re	fs
Morel MW267	0.33	0.40	2.08	113	6.6	25

Except this time we'll calculate Vb for three setups...

1 Solid State amp connected direct to driver with short leads Rg + Rx = 0.1 Ohm

Qes` = 0.40 (( 0.1 + 0.0 + 6.6 ) / 6.6) = 0.406
Qts` = ( 0.406 x 2.08 ) / ( 0.40 + 2.08 ) = 0.341
Vb = 113 / (( 0.707 / 0.341 ) ² - 1) = 34.25 litres

2 Solid State amp, long leads and a passive crossover where Rg + Rx = 1.0 Ohm

Qes` = 0.40 (( 0.1 + 0.9 + 6.6 ) / 6.6) = 0.460
Qts` = ( 0.460 x 2.08 ) / ( 0.40 + 2.08 ) = 0.386
Vb = 113 / (( 0.707 / 0.386 ) ² - 1) = 47.98 litres

3 Valve amp, long leads and a passive crossover where Rg + Rx = 1.5 Ohm

Qes` = 0.40 (( 0.6 + 0.9 + 6.6 ) / 6.6) = 0.491
Qts` = ( 0.491 x 2.08 ) / ( 0.40 + 2.08 ) = 0.419
Vb = 113 / (( 0.707 / 0.419 ) ² - 1) = 61.18 litres

Rg + Rx	Qts`	Qes`	Vb
0.0	0.330	0.400	31.47
0.1	0.341	0.406	34.25
1.0	0.386	0.460	47.98
1.5	0.419	0.491	61.18

Or by keeping the box volume the same at 31.47 litres and calculating the new Qtc using the following formula...

Qtc = sqrt (( Vas / Vb ) + 1 ) x Qts

Rg + Rx	Qts`	Qes`	Vb	Qtc
0.0	0.330	0.400	31.47	0.707
0.1	0.341	0.406	31.47	0.731
1.0	0.386	0.460	31.47	0.827
1.5	0.419	0.491	31.47	0.898

The above table shows that changes to Rg + Rx can have a marked effect on either the required box volume or the final system Qtc.

The effect of adding series resistance is to always increase the actual Qtc of the system and changes at lower starting values of Qtc are perhaps even more obvious.

This also explains why value amplifiers tend to "warm" the bass up a bit when driving sealed boxes and especially low Q speakers such as dipoles or very large sealed boxes. And not only is the Q of the system increased but there is no "low output impedance amplifier damping" to damp the back emf from the driver. Which is fine if you like your bass warm and loose but I reckon its a good reason why a valve amplifier should never go anywhere near a bass driver.

Filling the box

While calculating box volumes and final Qtc's for various values of Rg + Rx is simple and accurate the same cannot be said for the other variable which is the panel lining and fill material in the box. In general, lining a sealed box and filling it with absorbent material as well will always reduce the volume of the box required to achive a desired Qtc even though the fill material is reducing the available volume in the box!

The effects on Qtc of box lining and filling will depend on the type of material used as well as the density of the material. The only accurate way to determine these effects is to build, line and fill the box, fit the driver and then measure its resonance peak and from that determine Res and then Qms, Qes and Qts and finally Qtc.

Material with high specific heat such as fibreglass and long fibre wool at packing densities of 16gm/litre ( 1lb/ft³ ) can achieve reductions in box volume of around 20%, increasing to 30% at 32gm/litre( 2lb/ft³ ) but not appearing to increase with higher packing densities.

But as 16-32gm/litre is sufficient to achieve the other objectives of box filling such as reducing standing waves, damping the impedance peak at resonance and increasing efficiency then there would appear to be no advantage in higher packing densities.

Conclusions

Going back to our original design for a sealed box system with a Qtc of 0.707 and comparing the extremes in the following table...

Rg + Rx	Vb No Fill	Vb 16gm/L	Vb 32gm/L
0.0	31.47	25.18	22.03
0.1	34.25	27.40	23.97
1.0	47.98	38.38	33.59
1.5	61.18	48.94	42.83

Box volumes vary between...

23.97 litres for Rg + Rx = 0.1 Ohm (0.0 Ohms is never achievable) and with the box filled at 32gm/litre,

all the way up to...

61.18 litres for Rg + Rx = 1.5 Ohm (valve amp + long leads + passive crossover) and no filling!

Which I think you'll have to agree is a fair spread considering all we're changing is how we drive the system and how we fill the box.

Moral of this story, apart from the obvious which is to take all these things into account when designing sealed boxes, is save wood even if its only MDF. When it comes to bass, active crossovers, solid state amplifiers, short leads and heavy fill means that small boxes can do the same job as much bigger ones:).