อยู่ที่ Link นี้ครับ
http://www.diyaudio.8m.com/Ot/ot_en.html#example*******************************
อ่านเองพอจะติดตามได้ครับ แต่ก็ยังสงสัยในเรื่องของค่า F ตามสูตรนี้ ที่ออกมาเป็นตารางเซ็นติเมตร แล้วเราจะระบุขนาดของแกนและเหล็ก EI ได้อย่างไรครับ******************************
Do It Yourself Forever back
by Elvis Rakic
HOW TO MAKE AUDIO OUTPUT TRANSFORMERS
Basic calculations enough for building transformers with examples.
(single ended amplifiers)
--------------------------------------------------------------------------------
R loudspeakers impedance Ohms
N tube power in W W
Ra plate load impedance Ohms
Ia plate current A (mA)
F core area cm2 n1 winding per Volt winding
Up AC on primary V
np primary winding winding
p prim/sec turns ratio
ns secondary winding windind
I secondary current A
d1 wire diametar - primary mm
d2 wire diametar - secondary mm
WARNING: be careful of high voltage on primary!
--------------------------------------------------------------------------------
1. Calculations for necessary transformer dimension (core area)
(1)
17.april 2000. suggestion Damir S. - Croatia
Better calculations for transformer dimension:
(EI core area)
F=20*SQTR(N/fd)
F = core area (cm2)
N = tube power in W
fd = the lowest freqency that you wont to accomplish
--------------------------------------------------------------------------------
2. Winding per Volt
(2)
--------------------------------------------------------------------------------
3. Calculation for primary winding (np)
a.) Before calculation for np - calculate AC on primary (Up),
(3)
b.) and primary winding (np) is:
(4)
--------------------------------------------------------------------------------
4. Calculation for secondary winding (ns)
a.) Before calculation for ns - calculate primary/secondary turns ratio (p)
(5)
b.) and secondary winding (ns) is:
(6)
--------------------------------------------------------------------------------
5. Wire diameter: dp (primary) and ds (secondary)
a.) primary: For Ia (tube data) - see TABLE
example: if Ia=50 mA - wire diametar is 0,16mm or 0,18 mm with isolation (row 11)
b.) secondary: current of secondary (Is) =
(7)
and see TABLE
example: if Is=1,3 A - wire diametar is 0,8 mm or 0,85 mm with isolation (row 31)
--------------------------------------------------------------------------------
E & C core
Isolation (gap) width highly effects final results
For that reason, specify the width experimentally (for a start: 0,2 mm )
Perform the necessary measuring with instruments (frequency response) and by listening !
--------------------------------------------------------------------------------
Allways put a thin isolation material (resistant to high temperatures)
between primary rows.
Secondary is not that critical, but wouldn't hurt.
--------------------------------------------------------------------------------
EXAMPLE 1e
Output transformers for triode 300B (plate voltage 400 V, DC gird voltage -89V)
Class A (single ended)
R ( loudspeakers impedance) = 8 ohms
N (tube power in W)= 11,5 W
Ra (plate load impedance)= 3000 ohms
Ia (plate current) = 50 mA
1.) F=2sqtrN F=2 sqtr11,5 = 2 * 3,39 = 6,78 cm2
2.) n1=70/F n1= 70/6,78 = 10,32
3a.) Up= sqtr(N*Ra) Up = sqtr (11,5* 3000) = 186 V
3b.) np = Up * n1 np= 186 * 10,32 = 1919
4a.) p=sqtr(Ra/R) p=sqtr(3000/8) = sqtr375 = 19,36
4b.) ns=np/p ns=1919 / 19,39 = 99
5a.)Primary wire diametar
For Ia=36 mA (tube data) see TABLE ( row 11)
dp = 0,14 mm - with isolation 0,16mm
5b.) Secondary wire diametar
Is=sqtr(N / R) = sqtr (11,5 /

= sqtr 1,43 = 1,19 A
For Ia=1,19 A see TABLE (row 31)
d = 0,8 mm - with isolation 0,85mm
--------------------------------------------------------------------------------
EXAMPLE 2e
Output transformers for triode 211 (plate voltage 1250 V, DC gird voltage -80V)
Class A (single ended)
R ( loudspeakers impedance) = 8 ohms
N (tube power in W)= 19,7 W
Ra (plate load impedance)= 9200 ohms
Ia (plate current) = 50 mA
1.) F=2sqtrN F=2 sqtr19,7 = 2 * 4,43= 8,87 cm2
2.) n1=70/F n1= 70/8,87 = 7,9
3a.) Up= sqtr(N*Ra) Up = sqtr (19,7* 9200) = 425 V
3b.) np = Up * n1 np= 425 * 7,9= 3363
4a.) p=sqtr(Ra/R) p=sqtr(9200/8) = sqtr1150 = 33,9
4b.) ns=np/p ns=3363 / 33,9= 99
5a.)Primary wire diametar
For Ia=60 mA (tube data) see TABLE ( row 12)
dp = 0,18 mm - with isolation 0,19 - 0,2mm
5b.) Secondary wire diametar
Is=sqtr(N / R) = sqtr (19,7 /

= sqtr 2,46 = 1,569 A (1,6 A)
For Ia=1,6 A see TABLE (row 31)
d = 0,9 mm - with isolation 0,95mm
--------------------------------------------------------------------------------
Example for AL4, 807, EL34 on croatian version Link
--------------------------------------------------------------------------------
Back to top