Introduction To Loudspeaker Modelling and Design

 

Introduction to loudspeaker modelling & design

By Niels Elkjær Iversen

Technical University of Denmark DTU Electronics Group Department of Electrical Engineering September 2014 Copyright   ©2014 - Niels Elkjær Iversen

 

Contents 1 In Intr trodu oduct ctio ion n

 

2 Lou Loudspe dspeak aker er mode modellin lling g 2.1 Bas Basic ic lou loudspea dspeake kerr opera operatio tion n . . . . . . . . . 2.2 Infin Infinite ite baffl bafflee lou loudspea dspeake kerr   . . . . . . . . . . 2.3 Clo Closed sed box loud loudspeak speaker er   . . . . . . . . . . . . 2.4 Vented ented box lou loudspea dspeake kerr   . . . . . . . . . . . 2.4.1 2.4 .1 Gen General eral ste steps ps for alignme alignments nts   . . . . . 2.4.2 2.4 .2 4th orde orderr Butt Butterw erworth orth (B4 (B4))   . . . . . 2.4.3 2.4 .3 3rd orde orderr qu quasi asi But Butterw terworth orth (QB3 (QB3))   . 2.4.4 2.4 .4 Cheb Chebysh yshew ew equ equalal-ripp ripple le (C4) (C4)   . . . . . 2.4.5 2.4 .5 Des Design igning ing wit with h a giv given en d driv river er   . . . .

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1 1 3 4 5 6 7 7 8 9

3 Des Design ign ex examp ample le and si simu mulat lation ionss   12 3.1 Des Design ign and ssim imulat ulation ion of cl closed osed bo box x loud loudspeak speaker er   . . . . . . . . . . . . . . . 12 3.2 Des Design ign and si simu mulat lation ion of ve vent nted ed box loud loudspeak speaker er   . . . . . . . . . . . . . . 14 4 Co Conl nlus usio ion n

 

ii

17

 

1

In Intr trod odu uct ctiion

This article serves as an introduction to the fundamental operation of loudspeakers, how the can be modelled, how they can be designed using LTSPICE simulation in the design sig n proces process. s. Bas Basic ic lou loudspea dspeake kerr para paramet meters ers kno known wn as Thiele Thiele-Sm -Small all parame parameters ters will b bee described and loudspeaker models for closed box- and vent box-loudspeaker systems introduced. trodu ced. More Moreov over er vente vented d box align alignmen mentt theo theory ry will be pres presen ented ted and a desi design gn tool provid pro vided ed for vente vented d box desi designs. gns. Fin Finall ally y tw twoo desi design gn exam examples ples includ including ing sim simula ulatio tions ns of  the frequency response in LTSPICE will be presented.

2

Lo Loud udspea speak ker mod model elli ling ng

2.1

Basic Basic loudspe loudspeak aker er operatio operation n

Loudspeakers and loudspeaker models are well described in literature  literature   [2] [2] but the basic concepts are presented in this sections. A loudspeaker works by converting an electrical signal into motion of its’ diaphragm, creating pressure differences in air which we perceive as audio. Fig.   1 shows the conceptual elements of a loudspeaker unit.

Figure 1: Conceptual loudspeaker driver The signal enters the voice coil, this generates a magnetic field causing a displacement of  the voice coil due to the static magnetic field of the permanen p ermanentt magnet. This displ displacemen acementt is transferred to the diaphragm and emitted as sound. We can split the loudspeaker into three different domains: •

  Electrical domain

•

  Mechanical domain

•

  Acoustical domain

The electrical domain is characterised by the voice coil with a given DC resistance and self-inductance, R self-inductance, and L  R e   and  L e . As mentioned the electrical signal is converted to a mechanical motion. The strength of this coupling from electrical to mechanical domain is related to the force factor   Bl Bl,, which is the product of the magnetic field strength of the static 1

 

magnet in the voice coil gap,   B , and the length length of the voi voice ce coil (eg (eg.. the wi wire) re) in the static magnetic field,  field,   l. The mechanical domain is characterised by the mass,   M M D , of the diaphragm, the compliance,   C M S , of the suspension and a mechanical damping,   RM S . Th Thee mass mass,, the com com-pliance and the damper will introduce a resonance frequency,   f S  S , with a given Quality factor,   QM S , the mecha mechanica nicall Q-f Q-facto actor. r. At the reso resonanc nancee freq frequenc uency y the drive driverr wil willl reac reach h its’ maximum impedance. The electrical domain is also characterised by a Q-factor, Q Q-factor,  Q ES  which is dependent on  on   Bl Bl,,   R ,   M    and and   C  . Combining the mechanical and electrical e MD Q-factors results in a total Q-factor known M asS   QT S . The mechanical motion is converted as  to acoustical sound through the diaphragm and the strength of this coupling is related to the area of the driver diaphragm,  diaphragm,   S D . The acoustical domain is characterised by the acoustical impedance in front,   Z AF  AF , and behind ,  ,   Z AB diaphragm ragm.. Nor Normal mally ly a lou loudspea dspeake kerr wil willl be moun mounted ted in som somee kin kind d of  AB , the diaph enclosure and therefore a parameter known as the volume compliance, V  compliance,  V AS  AS , is introduced. The volume compliance corresponds to the equivalent volume of air which, when compressed by a piston having the same area as the driver diaphragm, will have the same compliance (mechanical spring) as the mechanical compliance of the driver suspension C M S . All these[3 areare known as in Thiele-Small parameters and well described in erature, [3parameters ] and [[4] 4]   and listed table   1. Thes These e paramet parameters ers aare normal normally ly giv given enlitin datasheets for drivers and are used when designing enclosures for the the loudspeaker driver, eg. cabinets.

Symbol Symbol Re   Le     Bl M M D   C M S    RM S   

Pa Param ramete eterr DC resista resistance nce of voi voice ce coil Self inductan inductance ce of voi voice ce coil Force factor Mass of diaphra diaphragm gm Compliance Compliance of suspension Mechanical damping

f S     Q M S  QES    QT S    S D     V AS  AS 

Resonance frequenc frequency Mechanical quality y factor Electric Electrical al qualit quality y factor Total quality factor Area of diaphra diaphragm gm Equivalen Equivalentt volume

Table 1: Thiele-Small parameters

A typical measurement of the impedance of a loudspeaker woofer is shown in fig.   2.  From the measurement one can indentify the resonance frequency, the DC resistance and the self inductance which causes the impedance to increase at high frequencies.

2

 

Figure 2: Impedance measurement of typical loudspeaker woofer

2.2

Infinit Infinite e baffle baffle loudspe loudspeak aker er

The infinite baffle loudspeaker is a loudspeaker standard where the driver is mounted in an infinite large wall (infinite baffle). This corresponds to the loudspeaker being mounted in an infinite large box. An equivalent electrical circuit can be implemented as shown in fig.   3.

Figure 3: Equivalent circuit for driver mounted in infinite baffle

The inductance,  inductance,   Lces , corresponds to the compliance of the suspension, the capacitance, C mes correspon esponds ds mes   corresponds to the mass of the diaphragm and the resistance,   Res , corr to the mec mechani hanical cal damp damping ing.. Toget ogether her thes thesee thre threee compo componen nents ts em emula ulate te the mechani mechanical cal characteristi chara cteristics cs of the driv driver er while while L  L e   and and R  R e  are the electrical self inductance and the DC resistance of the voice coil. Since the driver is mounted in free air the acoustical load will be the same on eac each h side of the diaph diaphragm ragm and there therefore fore it can be negl neglecte ected. d. Kno Knowin wingg the Thiele-Small parameters values for the components can be found by solving for   Lces , C mes and   Res   in: mes   and 3

 

QES   = 2πf SS  C mes mes RE 

 

(1)

QM S   = 2πf SS  C mes mes Res

 

(2)

f SS    =

2.3 2. 3

  1 2π C mes mes Lces

√ 

(3)

Clos Closed ed bo box x loud loudspe speak aker er

A closed box loudspeaker is a very simple loudspeaker configuration where the driver is mounted mounted in a sea seale led d enc enclo losur sure. e. Thi Thiss type type of loud loudspe speak aker er is one the mos mostt com commo mon n loudspeakers in the industry. When the loudspeaker driver is mounted in an enclosure the acoustical load on each side of the diaphragm is unequal and therefore the acoustical part must be taken into account. The volume of the enclosure, V  enclosure,  V AB AB , works as an acoustical compliance (spring) and can be modelled with the inductor,  inductor,   Lceb , in the electrical equivalent circuit shown in fig.   4.

Figure 4: Equivalent circuit for closed box loudspeaker This introduces a new parameters known as the compliance ratio,   α.   α   is simply the ratio between the equivalent volume and the volume of the enclosure which the driver is mounted in. This corresponds to the ratio between the inductances.

α  =

  V AS    Lces AS  = V AB Lceb AB

(4)

This additional volume compliance will add up together the equivalent volume compliance and shift the resonance frequency and quality factor of the system. If it is assumed that the enclosure is lossless the resonance frequency and the quality factor of the closed box system becomes:   (5)  ≈ √ 1 + αf  √ 1 + αQ Q   (6)  ≈ It is seen that the system cannot achieve a quality factor which is lower than the total f C C  

S S  

T C 

T S 

quality factor of the driver. The -3dB cut-off frequency,  frequency,   f l , eg. the half power frequency, can be calculated from: 4

 

f l   =  f C C  

 

  1 2Q2T C 

−1



+

  

  1 2QT2 C 

1/2

  −  2

1

+1

(7)

(8)

2.4

Vented ented box loudspe loudspeak aker er

The vented box loudspeaker is also a very common speaker which uses an additional resonancee frequen resonanc frequency cy,, generated by a ven ventt in the box, to increase the low frequency output. There a many different kinds of vented box optimal alignments and the some the most common are described in this section. The vent in the box introduces an additional resonance frequency,  f B . The mas masss of the air in the vent will resonate with the volume volume of the enclosure. This type of resonator is known as a Helmholtz resonator. The mass of the vent can be modelled as an capacitance,  C mep mep in an electrical equivalent circuit. The vented box enclosure quality factor, Q factor,  Q L , is related to the enclosure leakage resistance which can be modelled as a resistance,   Rel , in the electrical equivalent circuit.   QL  will typically be small for big enclosures (Q (QL   = 5 10 10), ),

−

meaning high enclosure leakage and large for smallequivalent enclosures circuit (QL   = for (Q 10 the20 20), ), meaning low lo w encl enclosu osure re leak leakage. age. Fig Fig. .   5  shows the electrical vented box loudspeaker.

−

Figure 5: Equivalent circuit for vented box loudspeaker Knowing the Helmholtz resonance frequency,   f B , and enclosure volume,   V AB AB , one can assume a given Q given QL  value and the find the needed component values for electrical equivalent circuit by solving for  for   Rel   and and   C mep mep   in:

f B   =

QL   =

 

1

 

2π C mep mep Lceb   1 2πf B C mep mep Rel

(9)

(10)

For a given driver there will be an optimal alignment. These a described in the following subsections. Note that the procedure of calculating an alignment will result in a Q a  QT S  that the driver must have to be suited and not the other way around. 5

 

2.4.1 2.4 .1

Gen Genera erall st steps eps for ali alignm gnmen ents ts

Alignments1 of vented box loudspeakers systems is based on the magnitude squared function GV  ( j2  j 2πf  πf )) 2 as explained by [2] [2].. It’s given by:

 |

|

  (f /f 0 )8 GV  ( j2  j 2πf  πf )) = (f /f 0 )8 + A3 (f /f 0 )6 + A2 (f /f 0 )4 + A1 (f /f 0 )2 + 1

|

2

|

 

(11)

where   A1 ,   A2   and where and   A3  are coefficients dependent on three other coefficients:

A1   =  a 21

− 2a

 

2

A2   = 2 + a22 A3   =  a 23

(12)

 

− 2a

(13)  

2

(14)

The formula for determine the   a  coefficients are dependent on the choice of alignment. Once the coefficients are calculated, the alignments parameters can be calculated using the guide from  from   [2 [2]] page 140. 1. Fin Find d positi positive ve real root rootss of   d   in: d4

3

2

−A d −A d −A d−1=0 3

2

1

 

(15)

 

(16)

2. Fin Find d positi positive ve real root rootss of   r   in: r4

− (a Q 3

L )r

3

− (a Q )r − 1 = 0 1

L

3. Use the v valu alues es for for   r   and and   d  to determined the alignment parameters:   f B 2 h  = f SS   =  r q  =  =

(17)

√ 

  f l =  r d f SS  

  r2 QL QT S   = a1 rQL q 

−



  V AS  AS  α  = =  r 2 a2 V AB AB

−

  1 QL QT S 

2

 

(18)

 

(19)



−r −1

 

(20)

Having a specific driver with resonance frequency   f S  S , the Helmholtz resonance   f B , the -3dB cut-off frequency  frequency   f l  and the cabinet volume  volume   V AB AB  can be calculated using eq.  17, 18 and 2 and  20. 0. 1

Note that section 2.4.1, section  2.4.1, 2.4.2,  2.4.2, 2.4.3 and 2.4.3  and  2.4.4 is  2.4.4  is borrowed from [5] from  [5]  by the same author

6

 

QL   5 QT S    0.414 1444

8 0.401 0199

11 0.396 9655

14 17 20 0.3934 0.3915 0.3901

Table 2: B4  B4   QT S  values for different values of   QL

2.4.2 2.4 .2

4th ord order er B Butt utterw erwort orth h ((B4) B4)

The B4 alignment is the 4th the special case where   A1   =  A 2   =  A 3   = 0. Applying this to eq.   11, 11,   15 15   and and   16 16 we  we obtain the magnitude squared function for the B4 alignment and the   d   and the and   r  values.

  (f /f 0 )8 GV  ( j  ( j22πf  πf )) = (f /f 0 )8 + 1 2

|

 

|

(21)

d  = 1

 

(22)

r  = 1

 

(23)

Knowing that all  all   A  values is zero we obtain the  the   a  values from eq.   14 14::

√ 2a √  a  = 2 + 2 a1   =

2

   

2

a3   =

√ 2a

2

(24) (25)

 

(26)

Note that there for a given   QL  value only can be calculated one   QT S   value for the B4 alignme alignment. nt. In practi practice ce thi thiss mean meanss that a pure B4 ali alignme gnment nt rarely is imp implem lemen ented ted for a given driver. The Q The  Q T S  values have been calculated for Q for  Q L  values ranging 5 ranging  5 20 20.. Some of 

−

them are listed in table 2 table  2.. A butterworth alignment will result in a -3dB cutoff frequency equal to the resonance frequency of the driver.

2.4.3 2.4 .3

3rd o orde rderr qua quasi si Bu Butte tterw rwort orth h (QB (QB3) 3)

The QB3 alignment is used for drivers having a lower lower   QT S  value than needed for a B4 alignment. The magnitude squared function is obtained using   [2] [2] and is given by:   (f /f 0 )8 GV  ( j  ( j22πf  πf )) = 1 + B 2 (f /f 0 )2 + (f (f /f 0)8

|

2

|

 

(27)

where   B   is an alignment where  alignment parameter. For or   B   = 0  the magnitude square function reduces to eq.   21 21 and  and the alignme alignment nt becomes at pure B4 alignm alignment ent.. More Moreov over er we can see that B 2 =  A 1  while  while A  A 2   =  A 3  = 0. Using eq.   14 14 one  one can derive the the   a  values: 7

 

B 2 =  a 21

− 2a

 

2

(28)

  a22 + 2 a1   = 2a3 a2   > 2 + a3   =

(29)

√ 2

 

√ 2a

(30)

 

2

(31)

To design a QB3 alignmen alignmentt one must determine determine B  B and  and used the equations for the a the  a values  values to solve for  for   a2 , then calculate  calculate   a1   and and   a3  and then finally use the procedure described in section  2.4.1  2.4.1.. All QB3 align alignmen ments ts will resul resultt in a -3d -3dB B cutcut-off off freque frequency ncy that is hig higher her than the resonance frequency of the driver and a cabinet volume that is smaller than the equivalent volume, of the driver.

2.4.4 2.4 .4

Che Cheby byshe shew w eq equal ual-ri -rippl pple e (C (C4) 4)

As the name indicates the C4 alignment allows the frequency response to have a ripple. This alignment is used for  for   QT S  val  values ues abov abovee the ones requir required ed for a B4 alignm alignmen ent. t. The magnitude squared function is given by:   1 + 2 GV  ( j  ( j22πf  πf )) = 1 + 2 C 24 (f n/f )

|

2

 

|

(32)

  is an alignment parameter related to another parameter k parameter k used  used for the calculation of the alignment and C  and C 24 (f n /f ) is the fourth order chebyshew polynomial. Moreover  Moreover   determines  determines the ripple allowed in the alignment.

k  = tanh 1 sinh−1 1 4 



 

8(f n /f )4 C 24(fn/f ) = 8(f 

2

− 8( 8(f  f  /f ) n

  +1

(33)  

 

dBripple   = 10 10log log 1 + 2

 

(34) (35)

The frequency  frequency   f n   is a normalization normalization frequency used in the chebyshew chebyshew polynomial. This frequency is related to the -3dB cutoff frequency  f l  (see [2] [2] p.  p. 142).   k  is used to calculate another parameter  parameter   D:   k4 + 6k 6k2 + 1 D  =

8

 

(36)

Having   D   and and   k,   the the   a  values can be determined using the following equations from [2 [ 2] p. 143: 8

 

a1   =

a2   =

 k

√  4+2 2

 

 

D1/4

1+k

2



 √  1+ 2

(37)

 

D1/2

(38)

2

a3   =  Da11/2 1

−

 1 k 2 2

−√ 



 

(39)

To design a C4 alignment one should specify     then calculate  calculate   k,   D  and the  the   a  coefficients and then proceed with the general procedure described in section  2.4.1  2.4.1.. A C4 alignment will result in a -3dB cutoff frequency that is lower than the resonance frequency of the drive dri ver. r. For man many y C4 alignm alignment ent the cabi cabinet net volum volumee wil willl be bigger bigger than the equiv equivalen alentt volume of the driver.

2.4.5 2.4 .5

Des Design igning ing wit with hag giv iven en d driv river er

As mentioned earlier in this section the general alignment procedure will generated   QT S  valu aluee whi which ch the dri drive verr mu must st comply with if the desig design n sho should uld work work.. In order to mak makee a design with a given driver one should calculate all possible alignments and find the one best suited. suited. Thi Thiss have have been done for enclo enclosure sure qual qualit ity y fact factor or of  of    QL   = 5,   QL   = 10 10   and QL   = 15  15   and fig.   6, 7   and   8  shows  shows   QT S   and the alignment parameters,   h   and   q , as a function of  α.  α . The blue graph corresponds to Q to  Q T S , the green curve corresponds to h to  h  an  and d the red curve corresponds to  to   q . The figures can be used as a design tool when designing with a given driver. One should just use the following steps. 1. Find the Q the  Q T S  value of the driver on the left side of the chart and draw a horizontal line that intersects with the  the   QT S  curve. 2. From the intersecti intersection on draw a vertical line that intersects intersects with the the h  h and  and q   q  curve   curve and the x-axis where  where   α  can be evaluated. 3. Draw two horizontal lines from the intersections between the vertical line and the h   and and   q  curves.   curves. 4. Read the value of the alignments parameters  parameters   h   and and   q  from   from the right y-axis. 5. Now the cabinet volume,   V AB AB , the Helmholtz resonance,   f B , and the -3dB cut-off  frequency can be evaluated from eq.  17, 18 18   and and   20 20.. Once a suited alignment is found and the Helmholtz resonance, f  resonance,  f B , and box volume, V  volume,  V AB AB , has been calculated the vent itself can be designed by using:

L p  =

   c 2πf B

2

S P  P  V AB AB

− 1.463

 

S P  P  π

 

Where L Where  L p  is of thesound. length of the ven vent, t, S   S P   is the cross section area of the vent and and c  c is the speed

9

(40)

≈ 345 345m/s m/s

 

Figure 6: Design tool for vented box -   QL  = 5

Figure 7: Design tool for vented box -   QL   = 10

10

 

Figure 8: Design tool for vented box -   QL   = 15

11

 

3

De Desi sign gn exam exampl ple ea and nd si sim mulat ulatio ions ns

This section presents design examples of a closed box- and vented box-loudspeaker including simulations simulations using the LTSPICE softwa software. re. The relev relevant ant Thiele-Sma Thiele-Small ll parameter parameterss for the loudspeaker driver used in this section are listed in table  table   3

Symbol Sym Re   Le   f SS     QM S    QES    QT S    S D   V AS    AS 

Val alue ue 6.1 6.1   Ω 0.11 mH 105 Hz 1.8 1.23 0.72 36.3 cm2 1L

Table 3: Thiele-Small parameters

3.1

Design Design and and simula simulatio tion n of closed closed box box loudspea loudspeak ker

The enclosure volume is chosen to be   V AB  liters ers as an initi initial al starti starting ng point. The AB   = 0.5  lit compression ratio, the closed box resonance frequency, the closed box quality factor and the -3dB cut-off frequency is then calculated using eq.  4, 5, 66   and and   8.

α  =   1L   = 2 0.5L

√   ≈ 1 + 2

QT C 

 ≈ √ 1 + 2

QT C 

182Hz f l  = 182Hz

 

·

105Hz 105 Hz =  = 182H 182Hz

·

0.72 = 1. 1.24

·

 

  1 2 1.242 ·

(41) (42)  

    − 1 +

f l  = 133H 133Hz

  1 2 1.242 ·

(43) 1/2

    − 2

1

+1

↓

 

(44) (45)

It is seen that the -3Db cut-off frequency is rather high and not satisfactory. satisfactory. BHy increasing the volume volume the low frequ frequency ency respon response se can be inc increas rease. e. By choosi choosing ng the encl enclosu osure re volume to be  be   V AB AB   = 1.5  liters we get: 12

 

α  =

  1L   = 0.67 1.5L

 ≈ √ 1 + 0.0.67  ≈ √ 1 + 0.0.67

 

QT C 

·

105Hz 105 Hz =  = 136H 136Hz

QT C 

·

0.72 = 0. 0.93

f l  = 136Hz 136Hz

·

 

  1 2 0.932 ·

(46) (47)  

    − 1 +

  1 2 0.932 ·

(48) 1/2

    − 2

1

+1

↓

 

f l  = 111H 111Hz

(49) (50)

Evidently a better low frequency response is achieved and therefore this design is used to construct a simulation model LTSPICE. The loudspeaker model from fig.   4  is used to built the simulation model in LTSPICE. The values of  L  L e   and and R  R e  can be directly inserted in the simulation model. The values of   Res ,   C mes and   Lceb  can be determined using mes ,   Lces   and eq.   1,   2,   3   and and   4.   C mes   = Res   = Lces   =

Lceb   =

2π

·

  2π

·

1.24 105Hz 105 Hz

306µF 6.1Ω  = 306µ

1.8   = 8.9Ω 105Hz 105 Hz 306 306µ µF

(51)  

(52)

·

  (2 (2π π

·

·

1 105Hz 105 Hz))2

·

306µ 306 µF

= 7.5mH

  7.5mH   = 11 11..2mH 0.67

(53)

(54)

Now the simulation model can be constructed with proper component values in LTSPICE and fig.   9  shows it.

Figure 9: LTSPICE simulation model for designed closed box loudspeaker Setting up an AC analysis in LTSPICE and measuring the current in   C mes mes   gives the frequency response curve shown in fig.   10 10.. From the LT LTSPIC SPICE E sim simulat ulation ion we can see that the response curve looks fine and with -3dB cut-off frequency close to the predicted.

13

 

Figure 10: LTSPICE simulation frequency response for designed closed box loudspeaker

3.2

Design Design and and simula simulatio tion n of ven vented box box loudspea loudspeake kerr

It is desi desired red to mak makee an optima optimall vente vented d box align alignmen ment. t. Sin Since ce the enclosure enclosure vol volume ume is expected to be fairly small, therefore an enclosure quality factor of   QL   = 15 15 is  is chosen. We can use fig.   8  to find the optimum alignment parameters using the steps presented in section 2.4.5 section  2.4.5.. Fig.   11 11 shows  shows how the alignment parameters are obtained from the graph.

Figure 11: Use of design tool - Q -  Q L  = 15 The obtained alignment parameters are listed in table  table   4. From this we can calculate the needed enclosure volume, the Helmholtz resonance and the -3dB cut-off frequency using eq.  17, 18 18   and and   20 14

 

Symbol Sym α   h   q   

Val alue ue 0.28 0.6 0.51

Table 4: Obtained alignment parameters

f B   = 0.6

·

f l   = 0.51

105Hz 105 Hz =  = 63 63H Hz

·

(55)

105Hz 105 Hz =  = 53 53H Hz

(56)

  1L   = 3.6L 0.28

V AB AB   =

(57)

Now we can start building the a simulation model in LTSPICE similar to the one shown in fig.   5. The values of   Le   and and   Re  can be directly inserted in the simulation model. The values of   Res ,   C mes and   Lceb  can be determined using eq.  1, 2, 33   and and   4. mes ,  L ces   and

C mes mes   = Res   = Lces   =

Lceb   =

2π

 

·

  2π

·

1.24 105Hz 105Hz

 = 306µ 306µF

1.8   = 8.9Ω 105Hz 105 Hz 306 306µ µF

(58)  

(59)

·

  (2 (2π π

6.1Ω

·

·

1 105Hz 105 Hz))2

·

306µ 306 µF

= 7.5mH

(60)

  7.5mH   = 26 26..8mH 0.28

(61)

Moreover the component values for  for   C mep and   Rel  can be determined using eq.   9   and and   10 10.. mep   and C mep mep  =

Rel   =

  (2 (2π π

2π

·

·

1 63 63Hz Hz))2

·

26 26..8mH

  1 63 63Hz Hz 238µ 238µF ·

·

15

= 238µ 238µF

  = 0.71Ω

(62)

 

(63)

Now the simulation model can be constructed with proper component values in LTSPICE and fig.   12 shows 12  shows it. In order to measure the frequency response a small capacitor of  2 of  2µ µF is placed in parallel with   Lceb  as shown in fig.   13. with When performing an AC analysis the current in this capacitor will be the frequency response. The response curve is shown in fig.   14 14.. From the frequency response we can see that a little ripple indicating that the alignment is a C4 equal ripple alignment. The -3dB cut-off frequency frequency from the frequency response correlates quite well with the one calculated earlier in this section. For the vent design a rectangular shape is chosen. The shape could also be circular. The vent dimensions is chosen to be  be   1.5X 1.5cm2 thus obtaining a port cross section area of: 15

 

Figure 12: LTSPICE simulation model for designed vented box loudspeaker

Figure 13: LTSPICE simulation model for designed vented box loudspeaker - measurement configuration

S PP    = 0.015 015m m

·

0.015 015m m  = 0.225

·

10−3 m2

(64)

The length of the port can then be determind using eq.   40 40..

L p  =



  345 345m/s m/s 2π 63 63Hz Hz ·



2

−3

0.225 10 m 3.6 10−3 m3 ·

·

2

  − 1.463

16

 

0.225

10−3 m2   = 3.6cm π

·

(65)

 

Figure 14: LTSPICE frequency response for designed vented box loudspeaker

4

Conlusion

Throughout this article various loudspeaker models and loudspeaker designs has been presented. Theory concerning vented box alignments has been described and from this a design des ign tool has been dev develo eloped ped and prov provided ided.. It can be conc conclud luded ed that elect electric rical al equivalent models can be used to emulate the electrical, mechanical and acoustical part of a loudspea lou dspeake ker. r. More Moreov over thes thesee model modelssdesign. can be used in SPIC SPICE E sim simula ulation tionss to predict predict the frequency response ofer a loudspeaker

References [1] Rich Richard ard H. Small, "V "Vented ented-Box -Box Loudspeak Loudspeaker er Systems Part I: SmallSmall-Signal Signal Analysi Analysis", s", School of electrical engineering, The University of Sydney, 2006. [2] W. Mars Marshal halll Leac Leach, h, Jr. "Introd "Introducti uction on to Elec Electroa troacous coustic ticss and Au Audio dio Amp Amplifi lifier er Design", Kendall/Hunt Publishing Company, 2003. [3] A. N. Thiele, "Loudspeakers in vented boxes, Parts I and II", J. Audio Eng. Soc., vol. 19 pp. 382-392 (May 1971); pp. 471-483 (June 1971), [4] R. H. Small, "Closed-box loudspeaker systems", J. Audio Audio Eng. Soc., vol 20 pp. 383-395 (June 1972). [5] N. E. Ive Iversen, rsen, "Tu "Tuning ning of ven vented ted box b ox loudspeake loudspeakerr systems systems", ", Tec Technical hnical Univ Universit ersity y of  Denmark (May 2013).

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