Nov
24
How To Make Audio Output Transformers
Filed Under Articles |
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. - CroatiaBetter 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 / 8 ) = 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 diameter
For Ia=60 mA (tube data) see TABLE (row 12)
dp = 0,18 mm - with isolation 0,19 - 0,2mm
5b.) Secondary wire diameter Is=sqtr(N / R) = sqtr ( 19,7 / 8 ) = 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
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