6.17 The Lossy Voice-Coil Inductance

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1 6.7. THE LOSSY VOICE-COIL INDUCTANCE Solution. The impedance function is given by (/3.3) (s/38.2) Z VC (s) = s +70 (s/38.2) 2 +(/3.3) (s/38.2) + Example 0 Solve for the element values in the equivalent circuit of the voice coil for the driver of Example. Solution. R ES =(Q MS /Q ES ) R E =70Ω, L CES = B 2`2C MS =0.53 H, C MES = M MS /B 2`2 = 34 µf. 6.7 The Lossy Voice-Coil Inductance If L e is a lossless inductor, its impedance is given by Z e (ω) =jωl e. To a good approximation, the impedance of a lossy inductor can be written Z e (ω) =(jω) n ω n L e L e = (6.50) cos (nπ/2) j sin (nπ/2) where the units of L e are no longer henries and 0 <m<. If we use this model for the voice-coil inductance, it follows that cos (nπ/2) sin (nπ/2) = Z e (ω) ω n j L e ω n = L e RE 0 (ω) j ωl E (ω) Thus RE 0 (ω) and L E (ω) are given by R 0 L e E (ω) = cos (nπ/2) ω n L E (ω) = L e sin (nπ/2) (6.5) ω n (6.52) For a given driver, the constant n can be determined from a log-log plot of the magnitude of the high-frequency voice-coil impedance versus frequency. Well above the fundamental resonance frequency, such a plot should be a straight line with a slope n dec/dec. If accurate impedance data for Z VC are known, n should be determined from the high-frequency plot of Z VC R E rather than the plot of Z VC. This requires knowledge of both the real and imaginary parts of Z VC.Amethod for doing this is described in Section 2.7. For many loudspeaker drivers, a typical value for n is in the range of 0.6 to 0.7. For example, let Z e be the value of Z VC R E at f and Z e2 be the value at f 2. If the log-log plot of Z VC R E versus f exhibits a straight line between f and f 2, it follows that n and L e are given by n = log ( Z e2 / Z e ) log (f 2 /f ) L e = Z e (2πf ) n = Z e2 (2πf 2 ) n (6.53) If Z e is measured at more than one frequency, a linear regression analysis can be used to determine L e and n. A method for doing this is described in Section Example Impedance measurements on a particular loudspeaker driver yield Z e = j4.6 Ω at f = 2242 Hz and Z e2 =36.6+j66.6 Ω at f 2 =20kHz. Calculate n and L e in Eq. (6.50). Solution. Z e =6.4, Z e2 =76.0. Thus n =log(76.0/6.4) / log (20000/2242) = and L e =6.4/ (2π2242) 0.7 = Example 2 Calculate the phase of Z e and Z e2 for the data in Example and the error in the phase of the approximating function for Z e at f and f 2.

2 2 CHAPTER 6. MOVING-COIL LOUDSPEAKERS Solution. arg Z e =tan (4.6/7.54) = 62.7 o, arg Z e2 =tan (66.6/36.6) = 6.2 o. The phase of the approximating function is arg Z e = o =63 o. Thus the phase error at f is 0.3 o. At f 2,itis.8 o. The phase of Z e (ω) given by Eq. (6.50) is nπ/2 rad, which is a constant. To account for deviations from this value, an alternate model for Z e (ω) can be used. Let Y e (ω) =G e (ω)+jb e (ω) = Z e (ω) = R 0 E (ω) j X E (ω) (6.54) where X E (ω) =ωl E (ω). In the alternate model, separate functions are used to represent R 0 E (ω) and X E (ω). The representations are as follows: R 0 E (ω) = G e (ω) = R eω nr X E (ω) = B e (ω) = X eω nx (6.55) Let the real and imaginary parts of Z VC be known at two frequencies f and f 2 that are sufficiently high so thaththe motional impedance component of Z VC is negligible. G e (ω) and B e (ω) are given by G e =Re (Z VC R E ) i h and B e =Im (Z VC R E ) i. The constants R e, n r, X e,andn x are calculated as follows: n r = log [G e (f ) /G e (f 2 )] log (f 2 /f ) n x = log [B e (f ) /B e (f 2 )] log (f b /f a ) R e = (2πf ) n r G e (f ) = (2πf 2 ) n r G e (f 2 ) X e = (2πf ) n x B e (f ) = (2πf 2 ) n x B e (f 2 ) Example 3 Calculate R e, n r, X e,andn x for the voice-coil data given in Example. Solution. Y e =/Z e = j0.054, Y e2 =/Z e2 = j0.05. n r = log(0.0279/ ) log(20000/2242) n x = log(0.054/0.05) log(20000/2242) =0.678 R e = =0.706 X e = (2π2242) = (2π2242) =0.028 (6.56) Fig. 6.5 shows the plots of R 0 E and X E = ωl E versus frequency for a 0-inch driver having the small-signal parameters R E =5.08 Ω, f S =35.2 Hz, Q ES =0.443, andq MS =2.80. The measured values are plotted as circles. Only those values above khz are plotted. The solid curves represent the approximations calculated with the values R e =0.0728, n r =0.653, X e =0.0243, and n x = These values were derived from experimental data on a driver using a linear regression technique described in Section On-Axis Pressure Sensitivity The reference on-axis pressure sensitivity of a driver is deþned as the magnitude of the midband on-axis pressure at r =m for a voice coil voltage e g =V rms. We denote the pressure sensitivity by p V sens. It is obtained from Eq. (6.29) by setting e g =and G (jω) T u (jω) =and is given by p V sens = ρ 0 2π B` S D R E M AS = 2πρ0 c /2 f 3/2 VAS S (6.57) R E Q ES

3 6.8. ON-AXIS PRESSURE SENSITIVITY 3 Figure 6.5: R 0 E and X E versus frequency for a typical driver. The on-axis sensitivity is often speciþed as the SPL at r =m either for a voice-coil voltage of e g =Vrmsore g = R E V rms. The latter is the rms voltage required for a power of Wintoa resistor of value R E, i.e. the voice-coil resistance. We denote the two SPL sensitivities by SPL V sens and SPL W sens, respectively. They are given by SPL V =20logpV sens sens db SPL W =20logpV sens RE sens db (6.58) where p ref =2 0 5 Pa. p ref Example 4 For the driver of Example, calculate the on-axis pressure sensitivity p V sens and the SPL V sens without and with the resistor of Example 7. Solution. Without the resistor, the pressure sensitivity is evaluated in Example 3. It is p V sens = 0.32 Pa. The corresponding SPL is SPL V sens =20log 0.32/2 0 5 =83.9 db. From Example 7, the added resistor causes the pressure sensitivity to drop by the factor 7/9. to the value p V sens = 0.4 Pa. The corresponding SPL is SPL V sens =20log 0.4/2 0 5 =75. db. Thus the resistor attenuates the output by 8.8 db. Example 5 If the power ampliþer puts out W of average sine-wave power (sometimes erroneously called W rms), calculate the -meter on-axis SPL W sens for the driver of Example with and without the resistor of Example 7. Solution. Without the resistor, SPL W sens = 20log /2 0 5 = 92.3 db. With the resistor, SPL W sens =20log /2 0 5 =87.9 db. Although the output is less by 4.4 db with the resistor, the generator voltage for the two cases is different. The voltage output of the power ampliþer must be increased to keep its output power at W. Therefore, the db change does not correspond to the attenuation introduced by the resistor. Example 6 Calculate the -meter on-axis pressure sensitivity p V sens and the SPL V sens for the driver of Example with the new value of B` obtained in Example 6. Solution. The original pressure sensitivity is evaluated in Example 3. It is p V sens =0.32 Pa. Eq. (6.57) shows that p V sens is directly proportional to B`. Thusp V sens = /.4 = 0.89 Pa and SPL V sens =20log 0.89/2 0 5 =79.5 db. Example 7 If the power ampliþer puts out W of average sine-wave power, calculate the -meter on-axis SPL W sens for the driver of Example with the new value of B` obtained in Example 6. Solution. SPL W sens =20log /2 0 5 =88dB. p ref

4 2.7. DRIVER PARAMETER MEASUREMENTS 305 By taking the product of f C and Q ECT given by the equations above, we have ρ f C Q ECT = = 0 c 2 µ + V AS 2πR AE C AT 2πR AE V AS V T (2.34) Similarly, we have for the product of f S and Q ES ρ f S Q ES = = 0 c 2 (2.35) 2πR AE C AS 2πR AE V AS The equation for V AS is obtained by taking the ratio of these two equations. It is given by V AS = V T fc f S Q ECT Q ES Conversion to InÞnite-Baffle Parameters (2.36) The above procedures give the parameters of the driver in free air, i.e. not in an inþnite baffle. Because only the air-load mass is different for the two cases, the parameters can be converted to the inþnite baffle parameters by dividing f S by the mass correction factor k M = M AS(ib) /M AS(fa) /2 and by multiplying Q MS, Q ES,andQ TS bythesamefactor,wherem AS(fa) is the acoustic mass of the diaphragm plus air load in free air and M AS(ib) is the acoustic mass of the diaphragm plus air load in an inþnite baffle. These are related as follows: M AS(fa) ρ 0 a = M AS(ib) 2 8ρ 0 3π 2 a where a is the equivalent piston radius of the diaphragm. Because (2.37) M AS(fa) = ρ 0 c 2 2πfS(fa) 2 VAS (2.38) it follows that the correction factor is given by k M = /2 MAS(ib) = M AS(fa) /2 2 V AS 2πf S(fa) c 2 (2.39) a After k M is calculated, the low-frequency parameters measured in free air can be converted to the inþnite baffle parameters as described above. The correction has a minor effect on the parameters in most cases Measuring the Voice-Coil Inductance The impedance of a lossy inductor can be written Z e (ω) =(jω) n L e (2.40) where L e and n are constants. It follows that Z e (ω) = ω n L e and arg [Z e (ω)] = nπ/2. This simple model for the lossy inductance often gives excellent results when applied to loudspeaker voice coils. To experimentally determine the constants L e and n.for a particular driver, the measurement test set shown in Fig. 2.8 can be used with the exception that a capacitor is connected in series with one lead of the driver. Because this capacitor must not be too large or too small, its value may have to be determined experimentally. Before connecting the capacitor, the Þrst frequency above the resonance frequency f S at which the Lissajous Þgure collapses to a straight line should be determined. This frequency is typically 4 to 5 times f S. Above this frequency, the inductance causes

5 306 CHAPTER 2. A LOUDSPEAKER POTPOURRI the voice-coil impedance to increase with frequency. A series capacitor should be experimentally chosen to resonate with the voice-coil inductance well above this frequency. For example, a driver with a resonance frequency of 40 Hz may show a Lissajous Þgure that collapses to a straight line at 200 Hz. The capacitor might be chosen to resonate with the voice-coil impedance at 2 khz. If we assume that the frequency is high enough so that the back emf due to the voice coil velocity is negligible, the input impedance to the capacitor in series with the voice coil can be written Z (ω) = jωc + R E + Z e (ω) = jωc + R E + R e (ω)+jx e (ω) (2.4) where Z e (ω) is the impedance of the lossy voice-coil inductance, R e (ω) =Re[Z e (ω)], andx e (ω) = Im [Z e (ω)]. Z (ω) is real when ωc =/X e (ω). At this frequency, the Lissajous Þgure collapses to a straight line and Z (ω) =R E + R e (ω). Let capacitor C, determined by the method described above, be connected in series with the voice coil. Let the frequency f = ω /2π be the frequency at which Z (ω) is real. It is straightforward to show that R e (f ) and X e (f ) are given by R e (f )=Z(f ) R E = R S V b R E X e (f )= (2.42) V a 2πf C Next, replace C with capacitor C 2 which is chosen so that the frequency f 2 = ω 2 /2π at which Z (ω) is real is in the upper audio frequency range. For example, the capacitor might be chosen to resonate with the voice-coil impedance at 20 khz. R e (f 2 ) and X e (f 2 ) are given by R e (f 2 )=Z(f 2 ) R E = R S V b R E X e (f 2 )= (2.43) V a 2πf 2 C 2 The measured values of R e and X e are related to the constants L e and n by the equations where (2πf ) n L e = Z e (f ) (2πf 2 ) n L e = Z e (f 2 ) (2.44) Z e (f ) = p R 2 e (f )+X 2 e (f ) Z e (f 2 ) = p R 2 e (f 2 )+X 2 e (f 2 ) (2.45) These equations can be solved for L e and n to obtain n = log [ Z e (f 2 ) / Z e (f ) ] log (f 2 /f ) L e = Z e (f ) (2πf ) n = Z e (f 2 ) (2πf 2 ) n (2.46) The method described above is based on measurements at only two frequencies. A more accurate procedure is to make measurements at a series of frequencies and then plot the values of Z e (f) as a function of frequency on log-log scales. The plot should be approximately a straight line. A linear regression analysis of the data can be used to obtain the constants L e and n from the data and to test the accuracy of the resulting approximations. For a linear regression analysis, let Z e (f) = Z E (ω) R E be measured at N frequencies well above the fundamental resonance frequency where the impedance is dominated by the voice-coil inductance. Let k through k 4 be deþned by k = X log f i k 2 = X (log f i ) 2 k 3 = X f i k 4 = X f 2 i (2.47) The linear regression solutions for constants L e and n are h n = Nk 2 k 2 N X X i log f i log Z e (f i ) k log Ze (f i ) (2.48) h X X i log L e = Nk 2 k 2 k 2 log Ze (f i ) k log fi log Z e (f i ) n log (2π) (2.49)

6 2.7. DRIVER PARAMETER MEASUREMENTS 307 where all summations have the limits i N. It is straightforward to model the lossy inductor in SPICE with the analog behavioral modeling of PSpice. For example, let the inductor connect between nodes 2 and 3 in the circuit. The SPICE deck line for Z e is as follows: GZE 2 3 LAPLACE {V(2,3)}={/(L e *PWR(S,n))} where numerical values for L e and n must be used. A more elaborate model for Z e (ω) is to model RE 0 (ω) and X E (ω) with separate functions. Let Y e (ω) =/Z e (ω) =G e (ω)+jb e (ω), where µ G e (ω) =Re R e (ω)+jx e (ω) The approximations are as follows: µ B e (ω) =Im R e (ω)+jx e (ω) (2.50) RE 0 (ω) = G e (ω) = R eω nr X E (ω) = B e (ω) = X eω nx (2.5) If Y e (ω) is determined at two frequencies f and f 2 as described above, the constants R e, n r, X e, and n x are given by n r = log [G e (f ) /G e (f 2 )] log (f 2 /f ) n x = log [B e (f ) /B e (f 2 )] log (f 2 /f ) R e = X e = (2πf ) n r G e (f ) = (2πf 2 ) n r G e (f 2 ) (2πf ) nx B e (f ) = (2πf 2 ) nx B e (f 2 ) (2.52) (2.53) If a series of N measurements are made, the linear regression solutions for the constants are n r = Nk 2 k 2 N X X log f i log k log (2.54) log R e = X X k Nk 2 k 2 2 log k log fi log n r log (2π) (2.55) n x = Nk 2 k 2 N X log f i log X k log (2.56) log X e = X X Nk 2 k 2 k 2 log k log fi log n x log (2π) (2.57) Both RE 0 and jx E can be modeled with the analog behavioral modeling of PSpice. Letthelossy inductor connect between nodes 2 and 3 in the circuit. The SPICE deck lines for R 0 E and jx E are as follows: GREP 2 3 LAPLACE {V(2,3)}={/(R e *PWR(ABS(S),n r ))} GXE 2 3 LAPLACE {V(2,3)}={/(S*X e *PWR(ABS(S),n x ))} where numerical values for R e, n r, X e,andn x must be used. Note that jx E is implemented as jωx e ω n x.

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