Here's an equivalent circuit for a raw dynamic driver. It doubles as one for a sealed

(or enclosed driver)where “Rg” is speaker cable impedance & cross-over resistance & amplifier's output impedance. For what it's worth, most people ignore “Rms”.The magnitude of the impedance for real world driver is plotted below. The low frequency spike is modeled by the “ Ces”; “Res”; “Les” section of the circuit while the high frequency slope is the responsibility of “Le”(voice coil inductance).

From this we can glean Thiele-Small parameters, if not supplied. We will need the maximum impedance( Zmax ≈ 63Ω )atop the resonance spike & the minimum( Zmin ≈ 7Ω )in the well following. Now, we calculate the “R1” value which is the log mean value( R1 = Sqrt[Zmax×Zmin] ≈ 21Ω ). Now we look for “f1” & “f2”, which are frequencies about the spike where the impedance is “R1”(IE: 48Hz & 105Hz respectively). In lieu of “Re”, we will ignore “Rms” & subsequently equate “Re” with “Zmin”, becauseZmin = Re + Rms.R Now that we modeled the difficult part, let's find “_{es}= Z_{max}- R_{e}

( 56Ω )C

_{es}= Sqrt[Z_{max}/R_{e}]/( 2×p×(f_{2}- f_{1})×R_{es})

( 149.6 µF )L

_{es}= (f_{2}- f_{1})×R_{es}×Sqrt[R_{e}/Z_{max}]/( 2×p×f_{1}×f_{2})

( 33.6 mH )Le”. Let's take a read of 17Ω @ 10kHz where the inductance is significant.L Below is a reconstruction of the impedance curve in yellow._{e}= Sqrt[ Z_{10kHz}² - Z_{min}² ] / ( 2 × p × 10000Hz )

( 0.247 mH )

Here's the actual published data for the same driver.

R _{e}6 Ω L _{e}0.24 mH Q _{es}0.34 Q _{ms}4.27 f _{s}70.7 Hz

So when given this data:

R Here's the actual published values:_{es}= R_{e}× Q_{ms}/ Q_{es}

( 75.35Ω )C

_{es}= Q_{es}/( 2 × p × f_{s}× R_{e})

( 127.56 µF )L

_{es}= R_{e}/( 2 × p × f_{s}× Q_{es})

( 39.73 mH )

C _{es}125.0 µF R _{es}77.0 Ω L _{es}40.6 mH

Conjugate FiltersNow you know how to derive these values & you wonder why. Well, we can obliterate the spike by using a simple L;R;C conjugate filter. This is a series configuration whose entirety is wired in parallel across a dynamic driver's terminals. Therefore, the low frequencies look like a resistor as depicted by the red line in the above impedance graph.

0—————[ Ln ]—————[ Rn ]—————[ Cn ]—————0

L _{n}= C_{es}× R_{e}²R

_{n}= R_{e}×( R_{e}+ R_{es})/ R_{es}C

_{n}= L_{es}/ R_{e}²

Zobel CircuitSimilarly, a Zobel circuit nullifies the voice coil inductance. This R;C circuit is also wired across the driver's terminals. A graphed blue line illustrates what a Zobel does.

0———————[ Rz ]————————[ Cz ]———————0

R So what was this about? Well, if you want a cross-over to work, it depends on your driver resembling a resistor._{z}= R_{e}+ R_{ms}C

_{z}= L_{e}/R_{z}²

Acoustic ResponseIs that the only reason? No. Acoustics response is proportional to the voltage drop across “

Rms”.In this case, we have the published measured data:

What happens if all you have the is the impedance graph instead? Well, a well known low frequency response curve is based on “Qh” & “fh” which is equivalent to “Qtc” & “fb”(the enclosed version of Qts & fs), respectively.f Here's the actual published values:_{s}= Sqrt[ f_{1}× f_{2}]

( 71 Hz )R

_{0}= Z_{max}/ Z_{min}

( 9 )Q

_{ms}= f_{s}× Sqrt[R_{0}] /( f_{2}- f_{1})

( 3.74 )Q

_{es}= Q_{ms}/( R_{0}- 1 )

( 0.467 )Q

_{ts}= Q_{ms}× Q_{es}/( Q_{es}+ Q_{ms})

( 0.415 )

f _{s}70.7 Hz Q _{ms}4.27 Q _{es}0.34 Q _{ts}0.32 SPL 90.9 dB/m/W For those that understand complex equations ... the transfer fuction is :

(f/f _{h})² / { (f/f_{h})² - Sqrt[-1]×f/(f_{h}×Q_{h}) - 1 }

Associated group delay plotted in equivalent phase angle:

The equivalent electric circuit for a ported enclosure schematic is:

further elaboration