I am in the process of designing a loudspeaker and have a question regarding the number of turns in the multilayered solenoidal coil and the speaker impedance of $8\:\Omega$. I understand that the Impedance is equal to the real part $R$ (resistance) added to the inductance and that the inductance is partly a function of the number of turn of the copper wire. I am having difficulty with understanding which equation describes this relationship between inductance and number of turns. Thanks.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Keep in mind that the speaker's "8 ohm" impedance only applies at one particular frequency. At other frequencies it will be higher. And the impedance is not determined purely by the coil, but also by the mechanical design of the speaker. $\endgroup$– endolithCommented Nov 29, 2011 at 19:26
-
1$\begingroup$ Your first mistake is to state that impedance is equal to inductance plus resistance. This is not correct. $\endgroup$– MagpieCommented Dec 16, 2012 at 4:16
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
a fairly nice read....wiki
In short, impedance $X$ is expressed as $$X=X_R+X_I+X_C$$ where the resistive load, $$X_R = R$$, the inductive load, $$X_I=j\omega L$$ the capacitive load, $$X_C=\frac{1}{j \omega C}$$
where, $L,C,R$ are inductance, capacitance and resistance.
Now since you have the complex impedance, find out the $|X|$ to get net impedance in Ohms.
So, to get $L$, we have a formula which as you correctly suspect is dependent on $N$, number of turns.
$$L=\frac{\mu_r \mu_0 N^2A}{l}$$ where, $A$ is circular cross section of solenoid, $l$ is length of solenoid (not of wire), $\mu_0$ permeability of air (a constant) and $\mu_r$ relative permability of core (iron, in your case; probabably; you can find the value here).
you can calculate it here itself.
answered Mar 14, 2012 at 5:51
-
$\begingroup$ I think he knows this, he's confused about the relation between L and N $\endgroup$ Commented Mar 14, 2012 at 17:45
-
$\begingroup$ ohh..ya...i read it again...and yes..he knows that... $\endgroup$ Commented Mar 15, 2012 at 10:10
-
$\begingroup$ No he was also confused about that too. Check again. This is a nice answer. $\endgroup$– MagpieCommented Dec 16, 2012 at 4:17
$\begingroup$
$\endgroup$
The one that comes to mind:
$L= μN^2 A/s ∴N= \sqrt{\frac{Ls}{μA}}$
$\text {where A is the cross sectional area, s is the length, and μ is the permeability }(4\pi \times 10^{-7} \frac {H} {m} \text{of air}) $
