There are lots of books that cover this subject but it is not easy to pull the information together into a friendly format. Below is an example calculation of the electrical characteristics, number of turns and air gap for an output transformer suitable for use with a 300B. I should point out that the choice of core type and material is up to you. I have very limited data on cores so please don't email asking for data. However I will be pleased to help with the math if you have the data, also to receive helpful comments or corrections if you find any error or know a better technique / formula.
After the example, the theoretical basis of the equations is given.
SUMMARY OF FORMULAE (using CGS units):
Reactance of an inductor (Ohms):
X = 2.p.f.L
or L =
Equivalent permeability of air gapped magnetic circuit me:
is the material permeability relative to m0
for a given magnetic path length.
DC Flux (Gauss):
B = 1.257 x me.
N.I/MPL or N = B.MPL / 1.257 x me
L= 1.257 x me.N2.A
Peak AC Flux (Gauss)
BAC = Vr.m.s x 108/4.44.N.f.A
Vp/Vs = Np/Ns = (Zp/Zs)1/2 =
f Hz A sq
cm MPL cm
APPLICATION OF FORMULAE TO AUDIO
Flux density is the key consideration for the design of an audio transformer. We do not want the core to saturate at-all. In fact, we want to stay within the sensibly linear region of the B-H curve. The transformer equation shows that the AC flux density increases with decreasing frequency so we need to consider the lowest frequency of interest. If the transformer is to handle DC current, we need to consider not only the sum of the DC flux and the peak AC flux but also, the effect of the DC current on the primary inductance of the transformer.
Let IDC = 0.08A
Now we have a dilemma, we have to chose core material and dimensions. This is where either experience or a lot of iteration comes in. For the purposes of this example let’s use a core having the following parameters:
MPL = 26cm
m = 4000 for silicon steel C core with MPL = 26
A = 11.6cm2 (Note, this figure takes into account the stacking efficiency of the core.)
Now we come to the next dilema, we know that a gap is likely to be required because the DC current is comparable to the peak AC current in this application. In practice, it is a good idea to use a gap large enough that variations in DC current will not affect the primary inductance greatly. Experience suggests that a total gap length of 22mils (0.056cm) might be suitable.
lg = 0.056cm
1/ Determine a target value for the primary inductance, Lp. We know that the plate resistance of a 300B is 700W. The primary inductance controls the LF performance so let us make 10Hz the lowest frequency of interest.
A good ‘rule of
thumb’ (for a SE triode) is to
make the reactance at the lowest frequency of interest equal to 5 times the
plate resistance. For an inductor,
L = X/2.p.f
700 x 5 = 3500W
10Hz is the lowest frequency of interest:
L = 3500/2. p.10 = 56H
2/ Calculate the effective permeability of the core with an air gap of 0.056cm:
Using me = m / (1+m( lg/MPL));
= 4000 / (1+4000 x (0.056/26)) = 416
3/ Calculate number of turns permissible with the given DC current:
Using N = B.MPL / 1.257 x me .I;
For the given type of core, a good limit for the total flux density is 16,000 Gauss. At this level, the transformer will remain linear. We know that Idc = 0.08A. For a 300B, we can also see that the peak AC current will approach but not exceed the DC current and thus we can set a limit for the DC flux density as one-half of the total flux density, 8000 Gauss.
N = 8000 x 26 / 1.257 x 416 x 0.08
N = 4972, say 5000 turns.
Now we have another dilemma, how many turns are practical? For this we need to evaluate the core window against the chosen wire size, insulation and secondary turns. In this case, I know (again from experience) that 5000 turns will fit comfortably.
4/ And so we can calculate the primary inductance:
Using; L= 1.257 x me.N2.A x 10-8/MPL;
L = 1.257 x 416 x 50002 x 11.6 x 10-8 / 26
L = 58H This is just above the desired target of 56H.
(Yes, I did ‘cheat’ by doing all the necessary iteration before setting out this example. This is where the work lies……)
5/ Calculate the AC flux at 10Hz:
BAC = Vr.m.s x 108/4.44.N.f.A;
Also; Vp/Vs = Np/Ns = (Zp/Zs)1/2 = (Lp/Ls)1/2
Another good ‘rule of thumb’ is to make the primary impedance somewhat greater than 5 times the plate resistance*, say 3800W. This will give efficient power transfer while allowing for speaker impedance dips.
We want a primary impedance of 3800W and let’s say that the secondary is to be 8W, the turns ratio will be:
Np /Np = (Zp/Zs)1/2 = (3800/8) 1/2 = 22
We know that the turns ratio is also the voltage transformation ratio.
For a power of 10W into 8W: Vrms = (10 x 8)1/2 = 8.9Vrms.
Thus for 10W, the voltage across the primary will be 22 x 8 = 196Vrms
BAC = 196 x 108/4.44 x 5000 x 10 x 11.6 = 7600Gauss
6/ Check combined DC and peak AC flux:
The AC and DC flux densities sum to 15600Gauss. This is the maximum value at 10Hz at 10W.
Silicon steel C cores saturate at more than 16000Gauss so this is satisfactory.
* You should always confirm this by plotting the load-line and then make an allowance for load reactance which will cause the load-line to become elliptical.
BASIC THEORY OF ELECTROMAGNETIC INDUCTANCE.
Analogy between the magnetic circuit and the electric
S.I. definition of electrical and magnetic units and CGS conversions:
CGS units are used.
MAGNETIC FIELD STRENGTH.
H is the magnetic field strength due to a current flowing in a coil.
H = magneto motive force (mmf) per per unit length of the magnetic circuit.
The length of the magnetic circuit
is denoted by MPL
The mmf is the force caused by a current I flowing through N turns. In a coil it is the total current linked with the magnetic circuit.
The unit for H is the Oerstead which is equal to 1.257 ampere-turns per cm.
Thus H = 1.257xN.I/MPL (Oersteads)
DC FLUX DENSITY.
Consider a point C on magnetic field of radius r about a conductor A situated in a vacuum:
B is the flux
density at point C
The flux density at a point C on radius r is equal to
the permeability of free space m0
multiplied by the field strength at point C.
Substituting for H from above we have:
B = m0
x 1.257xN.I/MPL (Gauss)
NOTE, In CGS units, m0
= 1 Gauss/Oerstead
is defined as the number of flux linkages which exist when a current is flowing.
A coil possesses an inductance of 1
Henry if a current of 1
ampere through the coil produces a flux-linkage of
1 Weber-turn ş
F is the total magnetic flux produced by a current flowing in a coil.
The unit for F is the Maxwell or line.
F = B.A where;
B is the flux density. (Gauss)
A is the core cross sectional area. (cm2 )
Flux linkage is the linkage between the number of lines of flux with a single loop of wire.
Thus the total flux-linkage = F x N
Since flux is proportional to current and total flux linkage is proportional to flux, then it follows that flux-linkage is proportional to current thus;
Flux-linkage µ I.
Introducing a constant, k we have;
Flux-linkage = k.I
This constant, k is the self inductance of the circuit (electrical and magnetic) and is given the symbol L. Note that by definition, 1H results when 1A produces 108 lines-turn thus we must divide the result by 108; Making inductance, k (L) the subject we have;
L=Flux-linkage/I x 108
From above, flux-linkage = F x N, more usually written N.F, we have the result;
L= N.F/ I x 108
From above, we have F = B.A, B=m0.H and H = 1.257 x N.I/MPL
F = 1.257 x m0.N.I.A/MPL, substituting into the expression for L we get;
L= 1.257 x m0.N2.A x 10-8/MPL (H)
Observe that L µ N2 and so we can further state:
Lp/Ls = (Np/Ns)2
A further essential result can be derived at this point, the relationship between turns ratio and primary to secondary impedance.
Neglecting losses, the primary power will be transferred to the secondary. Ohm’s law gives,
P = V2 / R . In this case, R will be impedance, thus so:
Pp = Vp2 / Zp = Ps = Vs2 / Zs
Now, the voltage in the primary and secondary windings is directly proportional to the number of turns and so we can replace V with N which yields the important result:
Np2 / Zp = Ns2 / Zs Cross multiplying we get:
(Np /Np)2 = Zp / Zs which, from above is also equal to Lp/Ls thus:
Vp/Vs = Np/Ns = (Zp/Zs)1/2 = (Lp/Ls)1/2
Since we are considering iron (or other magnetic material) cored transformers, we need to modify the permeability from that of free space to that of the core. The permeability of magnetic materials is usually specified as relative permeability to that of free space. The permeability is also a function of magnetic path length. Permeability data for core materials is sometimes presented as a graph of permeability vs MPL.
Thus m = m0.mr
Hereafter, m will represent the product, m0.mr unless otherwise stated.
INDUCTOR WITH AN AIR GAP:
Consider an inductor having an air gap where subscript g indicates the gap and subscript m indicates the core:
Total reluctance = lg/mg.ag + MPL/mm.am
Now we can introduce equivalent permeability for the complete circuit, me
The total reluctance will be equal to lt /me.a
We can now equate these two formulae for total reluctance. Noting that the area of the gap is equal to the area of the core and so multiplying through by a;
lt /me = lg/mg+ MPL/mm
Note, mm is the product, m0.mr which we will refer to as m
Cross multiply to make me the subject and we have;
me = lt /(lg/mg + MPL/m)
lt = lg + MPL Ţ
me = lg + MPL /(lg/mg + MPL/m)
For any likely design, lg <<< MPL we can multiply both sides by m/MPL/m/MPL;
This is the classic equation for equivalent permeability of a core – air gap magnetic circuit. It is used to modify the equations for DC flux density and thus also, inductance.
AC FLUX DENSITY;
To calculate the flux density due to AC current, we need to use Faraday’s law of electromagnetic induction:
DEFINITION: The instantaneous value of e.m.f., in volts, induced in a coil is equal to the negative rate of change of flux-linkages, in Webers per second.
Consider a single sinusoidal magnetic flux wave wave of peak amplitude F and frequency f. The flux will change from +F to -F in ˝.f seconds thus;
Average rate of change of flux = 2F ¸ 1/2f = 4fF Webers/second.
By definition, 1 weber/second = 1 volt and noting that 1Wb ş 108 lines, we have; Average e.m.f. induced per turn = 4fF volts and the r.m.s. value is 1.11 times the average value and thus;
Vr.m.s. / turn = 4.44fF/108 volts and for N turns;
Vr.m.s = 4.44NfF/108 volts. Or;
N = Vr.m.s x108 /4.44.f.F This is the classic transformer equation (which is used to calculate the number of turns required for the primary of a power transformer for a desired maximum flux).
Re-writing the transformer equation to make F the subject and knowing that F = flux density times area we have the peak AC flux density;
BAC = Vr.m.s x 108/4.44.N.f.A (Gauss)
Terman, F. E. Radio Engineer’s Handbook
Hughes, E. Electrical Technology
Langford-Smith Radiotron Designer’s Handbook
McLyman, T Transformer & Inductor Design Handbook