Hello forum, Hello Perry,
one user pm`ed me, asking me to solve his problem to solder some
AQXM2-modules (don`t ask me why...🙂 )
Anyway, I faced the same problem in the past (even here in germany
they are hard to get and each modules cost around 5 EURO!!!) and I
want to share my workaround so far, as i allready did in another
german forum...
---------------------------------------------------------------------
as you allready know, because STEG keeped telling us all the time... 😉
The cut-off frequencies of the electronic crossover are selectable with
maximum precision (1% tolerance compared to <20% of potis) through
insertion of the optional AQXM2-module available at dealers in the values
indicated.
The AQXM2-module is inserted in the small opening located on the lower
panel of the amplifier.
Available AQXM2-modules with following cut-off-frequencies in Hz:
25,30,33,35,38,40,45,50,55,60,65,70,75,80,85,90,95,100,110,120,130,
150,180,200,250,300,350,400,450,500,600,700,800,900,1000,1100,1200,
1300,1400,1500,1800,2000,2500,3000,3500,4000,4500,5000,5500,6000,
6500,7000,7500,8000,8500.
---------------------------------------------------------------------
So, when you look at the following picture in the link, you will see that
these modules are just 4 resistors in parallel (resistor array with 4 resistors
and 8 legs):
http://www.klangfuzzis.de/download/file.php?id=1218&sid=811202fb09d07000502e9647785f0dad&mode=view
I started to measure and collect all data on AQXM2-modules I could get. This is what I have so far:
AQXM2-modules with:
f[Hz] R[Ohm] of each resistor (all 4 the same)
30 920000
50 255000
75 137000
80 124000
85 115000
100 93000
120 73000
150 56000
200 40000
250 31400
400 19000
500 15000
600 12300
1500 4840
2000 3600
3000 2430
3500 2100
4500 1600
5000 1470
Then I`ve build the following equation in a "quick and dirty way" to
calculate the missing modules:
with f (frequency in Hz) and R (resistance in Ohm)
f(R) = -3,037196759877982e+38 * 1/R^10
+8,184265391205442e+35 * 1/R^9
-8,932673700429702e+32 * 1/R^8
+5,0467987697890055e+29 * 1/R^7
-1,5550278602968464e+26 * 1/R^6
+2,46610221256473e+22 * 1/R^5
-1461279027298205700 * 1/R^4
-53885414305615 * 1/R^3
+7474954885 * 1/R^2
+ 6988861 * 1/R
+ 23
...using the linear regression method (thanks by the way Mr. Gauß 🙂 )
If you need more information on that, please click the following link:
Linear regression - Wikipedia, the free encyclopedia
Attached you will find the pdf file with the results in excel/openoffice calc.
green = measured correct resistance of modules
yellow = calculated resistance of modules (please post if available)
red = missing data (please post if available)
If you post the resistances of the modules in the red and yellow fields, I will improove the equation
and will also post the excel file. This will help STEG-owner to build their individual modules...
Thanks for reading...
one user pm`ed me, asking me to solve his problem to solder some
AQXM2-modules (don`t ask me why...🙂 )
Anyway, I faced the same problem in the past (even here in germany
they are hard to get and each modules cost around 5 EURO!!!) and I
want to share my workaround so far, as i allready did in another
german forum...
---------------------------------------------------------------------
as you allready know, because STEG keeped telling us all the time... 😉
The cut-off frequencies of the electronic crossover are selectable with
maximum precision (1% tolerance compared to <20% of potis) through
insertion of the optional AQXM2-module available at dealers in the values
indicated.
The AQXM2-module is inserted in the small opening located on the lower
panel of the amplifier.
Available AQXM2-modules with following cut-off-frequencies in Hz:
25,30,33,35,38,40,45,50,55,60,65,70,75,80,85,90,95,100,110,120,130,
150,180,200,250,300,350,400,450,500,600,700,800,900,1000,1100,1200,
1300,1400,1500,1800,2000,2500,3000,3500,4000,4500,5000,5500,6000,
6500,7000,7500,8000,8500.
---------------------------------------------------------------------
So, when you look at the following picture in the link, you will see that
these modules are just 4 resistors in parallel (resistor array with 4 resistors
and 8 legs):
http://www.klangfuzzis.de/download/file.php?id=1218&sid=811202fb09d07000502e9647785f0dad&mode=view
I started to measure and collect all data on AQXM2-modules I could get. This is what I have so far:
AQXM2-modules with:
f[Hz] R[Ohm] of each resistor (all 4 the same)
30 920000
50 255000
75 137000
80 124000
85 115000
100 93000
120 73000
150 56000
200 40000
250 31400
400 19000
500 15000
600 12300
1500 4840
2000 3600
3000 2430
3500 2100
4500 1600
5000 1470
Then I`ve build the following equation in a "quick and dirty way" to
calculate the missing modules:
with f (frequency in Hz) and R (resistance in Ohm)
f(R) = -3,037196759877982e+38 * 1/R^10
+8,184265391205442e+35 * 1/R^9
-8,932673700429702e+32 * 1/R^8
+5,0467987697890055e+29 * 1/R^7
-1,5550278602968464e+26 * 1/R^6
+2,46610221256473e+22 * 1/R^5
-1461279027298205700 * 1/R^4
-53885414305615 * 1/R^3
+7474954885 * 1/R^2
+ 6988861 * 1/R
+ 23
...using the linear regression method (thanks by the way Mr. Gauß 🙂 )
If you need more information on that, please click the following link:
Linear regression - Wikipedia, the free encyclopedia
Attached you will find the pdf file with the results in excel/openoffice calc.
green = measured correct resistance of modules
yellow = calculated resistance of modules (please post if available)
red = missing data (please post if available)
If you post the resistances of the modules in the red and yellow fields, I will improove the equation
and will also post the excel file. This will help STEG-owner to build their individual modules...
Thanks for reading...
It appears that they may be using 0.018uf capacitors in the crossover. If so, you can calculate the resistors with the formula 1/(2*Pi*F*C).
For 80Hz:
1/(2*3.14*80*0.018^-6) = 110524 ohms or nearest available value
For 1kHz:
1/(2*3.14*1000*0.018^-6) = 8842 ohms or nearest available value
For 80Hz:
1/(2*3.14*80*0.018^-6) = 110524 ohms or nearest available value
For 1kHz:
1/(2*3.14*1000*0.018^-6) = 8842 ohms or nearest available value
Hello Perry,
yes you are right, i also started with rc circuit in series...
f = 1 / (2 * pi * R * C) --> R = 1 / (2 * pi * f * C)
but the result is always way far from the real resistors...
do i miss something?
yes you are right, i also started with rc circuit in series...
f = 1 / (2 * pi * R * C) --> R = 1 / (2 * pi * f * C)
but the result is always way far from the real resistors...
do i miss something?
Hello Perry,
here is the deviation chart... 🙂
f[Hz] R[Ohm] R[Ohm]calc deviation in %
30 920000 294881 -211,99
50 255000 176929 -44,13
75 137000 117952 -16,15
80 124000 110580 -12,14
85 115000 104076 -10,50
100 93000 88464 -5,13
120 73000 73720 0,98
150 56000 58976 5,05
200 40000 44232 9,57
250 31400 35386 11,26
400 19000 22116 14,09
500 15000 17693 15,22
600 12300 14744 16,58
1500 4840 5898 17,93
2000 3600 4423 18,61
3000 2430 2949 17,59
3500 2100 2528 16,92
4500 1600 1966 18,61
5000 1470 1769 16,92
here is the deviation chart... 🙂
f[Hz] R[Ohm] R[Ohm]calc deviation in %
30 920000 294881 -211,99
50 255000 176929 -44,13
75 137000 117952 -16,15
80 124000 110580 -12,14
85 115000 104076 -10,50
100 93000 88464 -5,13
120 73000 73720 0,98
150 56000 58976 5,05
200 40000 44232 9,57
250 31400 35386 11,26
400 19000 22116 14,09
500 15000 17693 15,22
600 12300 14744 16,58
1500 4840 5898 17,93
2000 3600 4423 18,61
3000 2430 2949 17,59
3500 2100 2528 16,92
4500 1600 1966 18,61
5000 1470 1769 16,92
Have you done a frequency sweep to see if the original modules produce the stated crossover point?
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