April 27th 1997

# Driver Analysis

## Terms

 Term Mm Rm Rv Ra Km Vas Qm Qe Qt t 3 ft Bl A Fm Pa n x X Y Z Ze Re Le Pe V I t B c j Units kg kg/s kg/s kg/s N/m m^2 radians/s radians/s 1/s N/A m^2 N Pascal Watts m m m/s m/s^2 Ohms Ohms Henry Watts Volts A s N/m^2 m/s kg/m^3 Description Mass of the cone Mechanical resistance of suspension Virtually equivalent mechanical resistance Acoustic resistance if the driver Spring constant of driver suspension Equivalent volume of acoustic suspension Mechanical Q Electrical Q Total Q of driver Total resonant frequency of driver Frequency of 1/2 acoustic output (-3dB) Total resonant frequency of driver (2f = t) Force/Current ratio Surface area of cone Force applied to cone Acoustic pressure at surface of the cone Acoustic intensity output of driver (SPL) Reference efficiency Cone displacement (+x is outward) Maximum cone displacement (complex valued) Maximum cone velocity (complex valued) Maximum cone acceleration (complex valued) Driver electrical impedance Electrical resistance of voice coil Voice coil inductance Power flowing into the driver Voltage across the voice coil Current flowing in the coil time Bulk modulus of air (1.42 * 10^5) Speed of sound (343, @20C) Density of air (1.205, @20C) The imaginary constant square root of -1

## Mass of the Cone:

When the cone moves back and forth it also brings air with it. This extra mass of air is also included in the Mm value because it is effectively part of the cone. Unfortunately the amount of air that sticks with the cone varies with frequency, so the mass does not appear to be a constant. This variation has a very small effect, and I will assume that the mass is constant for all my analysis. The application of a force (Fm) a free mass (Mm) causes it to accelerate according to

2
d x
Fm = Mm ---                                            (I1)
2
dt

## Mechanical and Virtual Resistance:

When the cone moves at a velocity dx/dt it experiences a force in the opposite direction due to resistance. Even if the velocity is held constant, driver non-linearity's make this force vary for different positions of the driver. Even so, it is useful to assume that the mechanical resistance (Rm) is constant. If the coil is connected to a voltage source, another braking force is provided by the back EMF; it is called virtually equivalent mechanical resistance (Rv).

dx
F = - Rm --                                            (I2)
dt

## Spring Constant:

The suspension of the driver exerts a restoring force proportional to the driver position (Km). Non-linearity's, again, make small variations in this ideal situation, and again, I will ignore these small imperfections.

F = - Km x                                             (I3)

## Definitions of Quality Factors and Resonant Frequency:

Total Parameters:

Assuming the driver is in free air, we define total Q of the driver (Qt) and total resonant frequency (t):

sqrt(Mm Km)                     / Km \
Qt = -----------           t = sqrt | -- |            (Q1)
Rm + Rv                       \ Mm /

Mechanical Q:

Ignoring the damping caused by the back EMF (Rv=0), we define mechanical Q (Qm):

sqrt(Mm Km)
Qm = -----------                                       (Q2)
Rm

Electrical Q:

If we assume the mechanical resistance is null, (Rm=0), we define electrical Q (Qe):

sqrt(Mm Km)
Qe = -----------                                       (Q3)
Rv

## Relationship Between Quality Factors:

It is a common occurrence to not have all three Q values of a driver. I guess they obey the same law that looses socks in a dryer. We can solve this problem; we can show the relationship between Qt, Qm and Qe:

1     Rv + Rm         Rv            Rm
-- = ----------- = ----------- + -----------
Qt   sqrt(Mm Km)   sqrt(Mm Km)   sqrt(Mm Km)

1    1    1
-- = -- + --                                           (Q6)
Qt   Qe   Qm

In it's more familiar form we have:

/ Qe * Qm \
Qt = | ------- |                                       (Q7)
\ Qe + Qm /

This is an excellent formula to check the parameters of a driver; to make sure the Q values agree with each other. But we also gain another insight from this analysis. Notice that if Qe < 1 or Qm < 1, Qt will be less than both of them.

## Driver Differential Equation:

We assume that the driver is going to be connected to a pure voltage source. It would be beneficial to relate the driver displacement (x) to the input voltage signal (V). We can easily relate all the mechanical forces on the cone to get a second order differential equation. The force applied by the voice coil (Fm) is balanced by the forces needed to overcome resistance, restoring forces and acceleration. Therefore we get the following:

2
d x      dx
Fm = Mm --- + Rm -- + Km x                             (D1)
2      dt
dt

The difficult part is to find the relationship between force (Fm) applied to the cone, and the voltage (V) applied to the coil. Lucky for us that this is all solved in Voice Coil Calculations (last line of). Remember that (D2) is only good for low frequencies because it does not take into account the voice coil inductance.

Bl        dx
Fm = -- V - Rv --                                      (D2)
Re        dt

We can set these two equations equal to each other.

2
Bl        dx      d x      dx
-- V - Rv -- = Mm --- + Rm -- + Km x                   (D3)
Re        dt        2      dt
dt

2
Bl        d x             dx
=>    -- V = Mm --- + (Rm + Rv) -- + Km x                    (D4)
Re          2             dt
dt

We can get rid of the differentials by assuming voltage (V) and displacement (x) vary in a sinusoidal fashion. Let V=Vexp(jt) and x=X*exp(jt).

Bl    jt        2  jt                 jt       jt
=>    -- Ve   = - Mm  Xe   + (Rm + Rv) j Xe   + Km Xe     (D5)
Re

Cancel the "e" terms.

Bl        /      2                   \
=>    -- V = X |- Mm  + (Rm + Rv)j + Km |                 (D6)
Re        \                          /

Using the equations (Q1) we can use the more conventional terms of resonant frequency (t) and total quality factor (Qt).

Bl         /   2  (Rm + Rv)     Km \
=>    ----- V = X |-  + ---------j + -- |                 (D7)
Re Mm        \          Mm        Mm /

Bl         /   2  t         2 \
=>    ----- V = X |-  + -- j  + t  |                     (D9)
Re Mm        \      Qt           /

The relationship provided by (D9) shows us that the total quality factor (Qt) is responsible for the driver response near resonance. The back EMF of the voice coil, along with the mechanical resistance of the suspension, provides the damping that contributes to Qt. If the driver was to be driven by a current source, the back EMF provided by the voice coil would be compensated for. That result is given in (Z5), and stated in conventional Thiele-Small parameters as (D10).

Bl        /   2  t         2 \
-- I = X |-  + -- j  + t  |                        (D10)
Mm        \      Qm           /

## Driver Impedance:

Driver impedance is the primary method for extracting driver parameters. It is therefore useful to derive it with respect to frequency (). We start with two formulas. The first relates mechanical forces to the force (Fm) provided by the voice coil (D1). The second is the relation between force (Fm) with respect to current through voice coil (I).

2
d x      dx
Fm = Mm --- + Rm -- + Km x                             (Z1)
2      dt
dt

Fm = Bl I                                              (Z2)

Set (D1) equal to (D2).

2
d x      dx
Bl I = Mm --- + Rm -- + Km x                           (Z3)
2      dt
dt

Just like in the derivation of driver differential equation above, we can get rid of the differentials by assuming current (I) and displacement (x) vary in a sinusoildal fashion. Let I=Iexp(jt) and x=X*exp(jt).

jt        2  jt          jt       jt
=>    Bl Ie   = - Mm  Xe   + Rm j Xe   + Km Xe            (Z4)

Cancel the "e" terms.

/      2             \
=>    Bl I = X |- Mm  + Rm j + Km |                       (Z5)
\                    /

Impedance of the driver (Ze) is defined as the ratio Ze=V/I. Keeping this in mind we will divide (D6) by (Z5).

Bl             /      2                   \
-- V        X |- Mm  + (Rm + Rv)j + Km |
Re      Ze     \                          /
----- = -- = -------------------------------           (Z6)
Bl I   Re        /      2             \
X |- Mm  + Rm j + Km |
\                    /

2
Ze   - Mm  + Rm j + Km + Rv j
=>    -- = ----------------------------                      (Z7)
Re            2
- Mm  + Rm j + Km

Ze               Rv j
=>    -- = 1 + ---------------------                         (Z8)
Re              2
- Mm  + Rm j + Km

/        /                2\ -1\
Ze   |     Rv |       t     t |   |
=>    -- = | 1 + -- | j  + -- - j -- |   |                  (Z9)
Re   \     Mm \       Qt       /   /

2 /                   \ -1
Bl  |       Rm      Km  |
=>    Ze = Re + --- | j  + -- - j ---- |                    (Z10)
Mm  \       Mm     Mm  /

To make our analysis appropriate for all frequencies, tack on an additional impedance due to voice coil inductance. Convert to conventional parameters; equation (Z11).

2 /                2\ -1
Bl  |       t     t |
Ze = Re + --- | j  + -- - j -- | + j Le
Mm  \       Qm       /

This is a graph of |Ze|. Notice the increase at resonant frequency due to back EMF of the voice coil. The rise on the right is the exaggerated effects of voice coil inductance (Le).

Some things that we should notice:

Ze = Re                                   (as  -> 0) (Z12)

2
Bl  / Qm \
Ze = Re + --- | -- | + j Le t          (when  = t) (Z13)
Mm  \ t /

2
Bl  / Qm \
Ze = Re + --- | -- |     (approximately, when  = t) (Z14)
Mm  \ t /

Ze = Re + j Le          (approximately, when  is    (Z15)
above resonant effects)

Ze = Re                  (approximately, when  is    (Z16)
not too high)

(Z14) makes the assumption that noticeable effects of the coil inductance will not appear until well above the effects of resonance. Most (sub)woofers are like this. There is an intermediate region where both effects are negligible, and Ze is close to Re. This is essential to assume when finding reference efficiency.

## Acoustic Output:

This topic is fairly complex, and derivations are delegated to another of my pages. But there are some simplifying assumptions that can be used for bass drivers. We assume that all wavelengths are small with respect to the driver diameter. From this we take, as fact, that the acoustic resistance (Ra) of the driver, in an infinite baffle, is approximated by,

2  2
Ra = --- A                                           (A1)
2c

We also know the intensity of the sound waves (), coming from the front plus the back of the driver, with respect to cone velocity (|Y|) is

2
= |Y|  Ra                                           (A2)

Inspecting the substitution of (A1) into (A2) we realize that we have a ^2 term. To remove it we use Z = jY, implied from (9) in mass-spring equations.

Z 2      1     2  2   2     2    2
= --- Ra = --- --- A   |Z| = --- A  |Z|          (A3)
j       ^2 2c            2c

Acoustic output is directly proportional to the square of cone acceleration.

## Reference Efficiency:

Defined as the ratio of acoustic intensity output () to the electrical power input (Pe). First we assume that the sound wavelengths are very large with respect to the diameter of the driver. Secondly we take the measurement at some frequency above, and away from, the effects of resonance. Thirdly the driver in is an infinite baffle. From these we can use (A3) as a starting point.

2    2
= --- A  |Z|                                        (N1)
2c

Above the resonant region we can obtain a simplified approximation for the cone acceleration (|Z|).

/                2 \ -1
Fm |        t    t  |
Z = --- | 1 - j ---- - --- |                          (N2)
Mm  |       Qt      2 |
\                 /

Fm
=>    |Z| = ---           (approximately, above resonance)  (N3)
Mm

The force (Fm) on the cone is supplied but the current flowing (I) through the voice coil. The radiation impedance is assumed to be a negligible force on the cone.

Fm = Bl I                                           (N4)

Substitute (N4) into (N3). Then (N3) into (N1).

2  2
2 Bl  I
= --- A  ------                                     (N5)
2c        2
Mm

Assuming that the frequency is high enough to be out of the resonant region, but low enough to make the voice coil inductance negligible, we can use (Z16). The total power then, going into the driver is given by Pe = Ze * I^2.

2  2
Bl  A
=>     = --- --- -- Pe                                     (N6)
2c   2 Ze
Mm

Here is a graph of acoustic output () with respect to frequency at constant power input (Pe). Three Qt values are shown: 1, .707, and .5. Notice that, in the resonant region, the Qt effects the acoustic output. Outside that range Qt makes little difference. If the point for measuring acoustic output was too high, then the voice coil inductance (Le) would result in a lower measurement. You can see the reduction of acoustic output due to Le on the leftmost part of the graph. Our point of reference is a point that is neither affected by Le or resonant effects. Luckily, (sub)woofers have such a point; Ze approximately equal to Re (see Z16). Substitute Ze=Re in to (N6).

As mentioned at the beginning, the ratio of the power in (Pe) to power out () is the efficiency (n).

2  2
Bl  A
=>    n = -- = --- --- --                                  (N7)
Pe   2c   2 Re
Mm

We can simplify this to Theile-Small parameters. Rv=(Bl^2)/Re, Vas=(B*A^2)/Km, Qes=sqrt(Mm*Km)/Rv, t=sqrt(Km/Mm).

3                  3
Km             1   t Rv Vas      1   t
=>    n = ---- --- Rv Vas = ----- ---------- = ----- --- Vas (N8)
2cB   2          2c^3 sqr(Mm Km)   2c^3 Qe
Mm

Now we can see how reference efficiency relates to Theile-Small parameters. How efficient the driver is will be limited by the relationship between parameters. An important thing to remember: this analysis is only an approximation when the resonant frequency is low. Don't fool yourself into thinking that doubling t will really give you 8x (+9dB), improvement in efficiency. On the other hand, a lower Qes will give you the improvement you want, but that has bad effects at other points in the driver response. One last thing, to check (N8) against current popular knowledge. Remember that 2**ft=t.

2   3                   3
4  ft               -7 ft
=>    n = --- --- Vas = 9.78*10   --- Vas                  (N9)
c^3 Qes                 Qe

The coefficient is known to be 9.64*10^-7, that is a difference of 1.5%; an excellent agreement. The figure of 9.64*10^-7 was calculated at a different temperature, namely 23C.

## SPL:

In order to state anything in decibels, a reference level must be determined. There are a plethora of these reference levels. The one that we use will be, 10^-11.2 equals zero dB. The acoustic output of the driver, given Pe Watts of input is n*Pe. SPL is defined as

SPL = 10 log(n*Pe/10^-11.2)                          (S1)

We can simplify this,

/          11.2 \
=>    SPL = 10 log\n*Pe * 10     /

=>    SPL = 10 log(n*Pe) + 10 log(10^11.2)

=>    SPL = 10 log(n*Pe) + 112

If we assume the power input (Pe) is one Watt,

SPL = 112 + 10 log(n)                                (S2)