for a single ended transformer, values are in mm, teslas
try them in your own designs and see, I have derived and simplified the more
complex but algebraically simple equations and boiled them down to VERY simple
perhaps TOO simple equations here, einstein said "make things as simple as possible, but no simpler "
based on that, these could be too simple, but they will work!!
1/ core area in mm (stack x tongue)= 450 x root of (power output )
eg 25 watt tx, 450 x 5 = 2250 mm squared for square core
area = 44.7 mm each side
2/ Primary turns = 44.7 x root of (primary impedance)
eg. for 2500 ohm primary ( hi power 300b)
primary turns = 44.7 x 50 =2235 turns
for 5000 ohm primary (lo power 300b)
turns =44.7 x 70.7 = 3161 turns
3/ primary turns is also
root of (primary inductance) / 1.99 x 10 ^ -3
4/ primary inductance Lp = 7.96 x 10 ^ -3 x primary impedance
eg. for 2500 ohm primary 7.96 x 10 ^ -3 x 2500 = 19.9 henries
note lp should be at 20 hz, 125.6 /20
this is nothing to do with Bac, the core saturation based upon Ac volts
the classic transformer equation
note also there are 3 lo frequency points based upon:
1/ the large signal -3db point this is the key one and highest value based upon inductance
2/ the small signal -3db point again based upon inductance, always less than 20 hz
3/ the core saturation, based upon ac flux,depends on ac volts, frequency, core area, no turns
also 4/ core saturation caused by dc flux
5/ Lp is also Np (squared ) x 3.98 x 10 ^-6
eg for 2235 turns as above, Lp = 2235 (squared ) x 3.98 x 10^ -6 = 18.38 henries
6/ permeability ( to calculate air gap)
Ue = 18972 / stack in mm
eg. 51mm stack, 18972 / 51mm = 372 no units
7/ Dc flux, Bdc =
2 x 10 ^ -5 x Ue x stack
in teslas
eg for 372 Ue and 51mm stack
2 x 10 ^ -5 x 372 x 51mm = .38 teslas, less than 0.8 teslas, approx half of silicon steel saturation
which is 1.6 T depending on who you take notice of
recall Bdc has unbalanced current in an S.E tx so Bdc + Bac = or less than 1.6T
8/ Bdc also 123.5 x root of (power) / Core Area
enjoy, this is for me as much as anyone seeking info, as I have it on a scrap of paper and its nice to have it here.
hope you can understand it all and find it useful.
try them in your own designs and see, I have derived and simplified the more
complex but algebraically simple equations and boiled them down to VERY simple
perhaps TOO simple equations here, einstein said "make things as simple as possible, but no simpler "
based on that, these could be too simple, but they will work!!
1/ core area in mm (stack x tongue)= 450 x root of (power output )
eg 25 watt tx, 450 x 5 = 2250 mm squared for square core
area = 44.7 mm each side
2/ Primary turns = 44.7 x root of (primary impedance)
eg. for 2500 ohm primary ( hi power 300b)
primary turns = 44.7 x 50 =2235 turns
for 5000 ohm primary (lo power 300b)
turns =44.7 x 70.7 = 3161 turns
3/ primary turns is also
root of (primary inductance) / 1.99 x 10 ^ -3
4/ primary inductance Lp = 7.96 x 10 ^ -3 x primary impedance
eg. for 2500 ohm primary 7.96 x 10 ^ -3 x 2500 = 19.9 henries
note lp should be at 20 hz, 125.6 /20
this is nothing to do with Bac, the core saturation based upon Ac volts
the classic transformer equation
note also there are 3 lo frequency points based upon:
1/ the large signal -3db point this is the key one and highest value based upon inductance
2/ the small signal -3db point again based upon inductance, always less than 20 hz
3/ the core saturation, based upon ac flux,depends on ac volts, frequency, core area, no turns
also 4/ core saturation caused by dc flux
5/ Lp is also Np (squared ) x 3.98 x 10 ^-6
eg for 2235 turns as above, Lp = 2235 (squared ) x 3.98 x 10^ -6 = 18.38 henries
6/ permeability ( to calculate air gap)
Ue = 18972 / stack in mm
eg. 51mm stack, 18972 / 51mm = 372 no units
7/ Dc flux, Bdc =
2 x 10 ^ -5 x Ue x stack
in teslas
eg for 372 Ue and 51mm stack
2 x 10 ^ -5 x 372 x 51mm = .38 teslas, less than 0.8 teslas, approx half of silicon steel saturation
which is 1.6 T depending on who you take notice of
recall Bdc has unbalanced current in an S.E tx so Bdc + Bac = or less than 1.6T
8/ Bdc also 123.5 x root of (power) / Core Area
enjoy, this is for me as much as anyone seeking info, as I have it on a scrap of paper and its nice to have it here.
hope you can understand it all and find it useful.
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error in point 4, where it says note, this is correct now....
Lp should be
XL = 2 x pi x freq. x inductance
thus L = XL (or Zprimary, effectively ) / 2 x pi x F
say 2500 ohms / 125.6 = 19.9 henries, based upon F being 20 hz.
note 2, in point 1/ where it says core size is 44.7mm each way,
your turns have to fit on the bobbin, the core size is the metal, and the bobbin slides in and will be less, perhaps 2mm less than the metal as the end cheeks take up space.
Thus, our 2250 turns have to fit in this area, the bobbin area and winding window.
if our primary wire for eg. is 1mm not in reality, then there will be around 42 turns that fit in the length of the bobbin.
thus we build the primary up in layers, too, and the primary will have many layers, obviously, or not, the primary will have to be around 50% of the window area, ie you build your coil up in the window area, many layers of primary, and then insulation, and then a layer or two of secondary, so a transformer is magical in that somehow the turns fit in the window, and you have to use a certain guage wire
the thinner the wire, the more resistance or winding resistance it will have and consequently heat up more and produce more insertion loss.
we seek for the coil to be just less than the 100% of the window height margin for error, insulation, too remember!!
Lp should be
XL = 2 x pi x freq. x inductance
thus L = XL (or Zprimary, effectively ) / 2 x pi x F
say 2500 ohms / 125.6 = 19.9 henries, based upon F being 20 hz.
note 2, in point 1/ where it says core size is 44.7mm each way,
your turns have to fit on the bobbin, the core size is the metal, and the bobbin slides in and will be less, perhaps 2mm less than the metal as the end cheeks take up space.
Thus, our 2250 turns have to fit in this area, the bobbin area and winding window.
if our primary wire for eg. is 1mm not in reality, then there will be around 42 turns that fit in the length of the bobbin.
thus we build the primary up in layers, too, and the primary will have many layers, obviously, or not, the primary will have to be around 50% of the window area, ie you build your coil up in the window area, many layers of primary, and then insulation, and then a layer or two of secondary, so a transformer is magical in that somehow the turns fit in the window, and you have to use a certain guage wire
the thinner the wire, the more resistance or winding resistance it will have and consequently heat up more and produce more insertion loss.
we seek for the coil to be just less than the 100% of the window height margin for error, insulation, too remember!!
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