The delicate relationship magnetic materials related to the transformer
Magnetic materials and the design of transformer, the main three, one is the industrial frequency transformer silicon steel sheet form, an iron silicon aluminum iron powder core magnetic ring, and is a kind of magnetic ring composed of manganese zinc nickel and zinc materials. Three kinds of applied in different situations, including silicon steel sheet is mainly used in power frequency transformer, because U value near the 1.5 K, moderate, Bsat value is big, up to 1.5 T, so the magnetic saturation strength. Iron, silicon aluminum iron powder core material U value is low, generally near hundred, B value relatively small silicon steel sheet, but a lot larger than high conductivity materials, mainly used for large dc component. Such as for BUCK continuous current circuit. Nickel, zinc and manganese core permeability is high, the highest up to 10 k, so coupling is very good, it is mainly used for small signal coupling transmission. Such as drive current and voltage signal samples. This kind of material is mainly around several turns can volume of satisfaction as well as the appropriate excitation current.
Tell me about the design of the transformer
First of all, we know the transformer is an ideal transformer excitation inductance and, of course, the leakage inductance of the primary level. But we can assume that the leakage inductance of the ignore don't remember. The transformer main parameters is the exciting current and the number of turns, namely magnetomotive force. It has to do directly and B values. Under other conditions unchanged, the bigger the NI, B value is larger, the more easy to magnetic saturation. So good, now discuss the value of NI how to make B values in a safe range.
I believe you know B = UH, this is defined, U is the magnetic permeability, is the ratio of B and H, U are not constant, but in a linear relationship with small H B and H, H = KNI, K is a proportionality constant, N is the number of turns, I was exciting current. Well, if you want to reduce the product of B value will have to reduce the NI. The exciting current I is inversely proportional to inductance. If larger inductance incentives will flow down, but N have to be increased, otherwise the inductance to rise. We know that is directly proportional to the inductance and N ^ 2, L ∝ N * N? Mu. And U = LI/T, the L value generation go U ∝ N * N? Mu? I/T. so
B = mu mu NI = k/H = k = KUT/N
This type is B ∝ 1 / N. So can reduce B value increases N, so we'd better let the value of N infinite in theory, that B is not easy to saturation, but the reality always have a degree, the first is our power transformer.
Because we always want to output a certain power, otherwise the transformer is lost. Since want to output power so there must be a current of winding, if got very thin, the line pressure drop is very big, line loss is very large. If line obtain coarse magnetic ring size limit, can't around so many winding, so the number of turns is limited. Unless the circular. , of course, we don't have to take the B value is too small, otherwise the circular utilization rate is low, so we're going to take a equilibrium value.
So the transformer design first consider the power output, and then determine how much wires. Began to choose the circular size after the wire diameter is given. According to the circular size to get a proper L value, under the value calculated, and then multiplied by N, look to whether out of B value, if more than the increase of N, until the B value within a certain range, if the wire wound not in bigger magnetic ring. Why not increase the conductivity can reduce N below with small circular and get high power output.
The same trend of circular and B values under different N:
By the above know B1 / B2 = N2 / N1 is derived. So increase one times the number of turns, the B value reduced by half.
(1) under the same shape same Ur different L. First, L values, so I value is also the same. High mu value will have to reduce the value of N to keep constant, L so L ∝ N ^ 2 * u know,
N1 mu = N2 ^ 1 ^ 2 * 2 * 2, mu
So u 1 / u 2 = ^ 2.
So the N1 N2 = *.
By mu H, B = K K when shape as the same. B1 / B2 =). So mu value twice the B value increases, the square root of 2 times. So although increased mu value can decrease N, but it is the price of B value more tends to saturation.
To BUCK circuit, winding on the circular is purely ACTS as inductance, so through the many exciting current is. So the BUCK circuit usually with high permeability magnetic ring, and you'd better use the iron if have dc component silicon or aluminum powder core, first they permeability is low, the second B value is big, it is not easy to magnetic saturation. The third work frequency can be tens of KHZ. And because of the silicon steel sheet is not suitable for high frequency, so BUCK rarely used the circular silicon steel sheet.
N try to increase the N, can use low mu value with low mu as far as possible. The premise is to ensure that good coupling. The same circular and increasing N decreases B, different magnetic ring L, same mu value is big, the B values.
L calculation formula:
L = [4 n ^ 2? H * 10 ^ * mu] / = 4 n ^ 2 * h)? 10 ^? mu
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