N400
I am building new filters for the Hartsfield ( E 145/8, 2445/16)
There are some points that I don’t understand in the N 400 Filter scheme, especially on the HF section:
If we calculate for the LF driver ( 150-4C / 16 ) L for low pass filter, the theoretical value of L for 500 Hz/6 Db slope/ 16 Ohms driver is 5.10 mH, close to the 4.3 mH in the scheme. With an 8 ohms driver, L shall be 2.55 mH, as for an E 145/8. We just can notice that in the technical sheet, 150-4C /16 must be replaced with E 145-8, and not 16, without changing the value of L!
About the HF section, I am dubious:
The scheme indicates 21 uf for C and 5.4 Mh for L, without regard to the tap on the inductor.
If I compute 500 Hz /12 Db Slope for a 16 Ohms driver, L shall be 10 mH, and C 10 Uf.
21 and 5.4 are double, and half of the precedent values, close to 19.90 and 5.10, the theoretical value for an 8 Ohms loudspeaker…
So N400 was made for an 8 ohms HF driver ( 2440 )?
I have found 2 explanations for this divergence:
1 – JBL made their filter based on the resistance of the diaphragm D16R…. of the 2440 which was close to 8 ohms. But on all the JBL filters, I’ve never seen a 4 Ohms scheme for the D8R…. diaphragms.
2 – As we can see on much JBL filters, the scheme was made based on a 8 Ohm speaker, and a final 18 or 20 ohm resistor was put in parallel to the 16 driver to make a 8 Ohm system. Then, the taped choke shall have an internal resistance of 18 or 20 Ohms…I will not use a CT choke, but a L.PAD (16 Ohms)
Maybe I’ am not right, and there is a third explanation. ! I tried to test the choke in my old N400 (the choke is not directly attainable, flood in resin), and have found various values, but apparently there is a residual value between 4 and 17 Ohms.
Any advice ?
Regards
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