Introduction to the calculation steps of the number of turns and the diameter of the coil of the high-frequency transformer

2019-04-29 11:13:36 JUKE CHINA ODM OEM Transformer factory Read

Introduction to the calculation steps of the number of turns and the diameter of the coil of the high-frequency transformer

First, the introduction of High frequency transformer
The high-frequency transformer is a power transformer with a working frequency exceeding the intermediate frequency (10 kHz). It is mainly used for high-frequency switching power supply transformers in high-frequency switching power supplies, and also for high-frequency inverters in high-frequency inverter power supplies and high-frequency inverter welding machines. Variable power transformer. According to the working frequency, it can be divided into several grades: 10kHz- 50kHz, 50kHz-100kHz, 100kHz-500kHz, 500kHz~1MHz, 10MHz or more.

The high frequency transformer is the most important part of the switching power supply. There are many topologies in switching power supplies. For example, a half-bridge power conversion circuit, in which two switching transistors are turned on in turn to generate a high-frequency pulse wave of 100 kHz, and then transformed by a high-frequency transformer to output an alternating current, and the ratio of the turns of each winding of the high-frequency transformer is determined. What is the output voltage? The most prominent of the typical half-bridge transformer circuits are three high-frequency transformers: the main transformer, the drive transformer, and the auxiliary transformer (standby transformer). Each transformer has its own measurement standard in the national regulations, such as the main transformer. As long as it is more than 200W, its core diameter (height) must not be less than 35mm. The auxiliary transformer, when the power of the power does not exceed 300W, the core diameter of 16mm is enough.
Second, the working principle of high frequency transformer

The high frequency transformer is the most important part of the switching power supply. The switching power supply generally adopts a half-bridge power conversion circuit. When working, two switching transistors are turned on to generate a high-frequency pulse wave of 100 kHz, and then stepped down by a high-frequency transformer to output a low-voltage alternating current, and each winding coil of the high-frequency transformer. The ratio of the turns determines the amount of output voltage.

The main cause of electromagnetic interference from high-frequency power transformers is the suction between the cores and the repulsive force between the winding wires. The frequency of these forces varies with the operating frequency of the high frequency power transformer. Therefore, a high-frequency power transformer with an operating frequency of around 100 khz has no special reason for not producing audio noise below 20 kHz.
high frequency transformer.jpg

Third, High frequency transformer design principle

In the design of high-frequency transformers, the leakage inductance and distributed capacitance of the transformer must be minimized because the high-frequency transformer in the switching power supply transmits a high-frequency pulse square wave signal. During transmission transients, leakage inductance and distributed capacitance can cause inrush currents and spikes, as well as top oscillations, resulting in increased losses. Usually, the leakage inductance of the transformer is controlled to be 1% to 3% of the primary inductance.

The leakage inductance of the primary coil—the leakage inductance of the transformer is caused by the fact that the magnetic flux between the primary and secondary coils is not fully coupled between the layers and between the layers.

Distributed Capacitor----Between the windings of the transformer windings, between the upper and lower layers of the same winding, between the different windings, the capacitance formed between the winding and the shielding layer is called the distributed capacitance.

Primary winding----The primary winding should be placed in the innermost layer, so that the length of the primary winding of the transformer can be minimized, so that the entire winding is minimized, which effectively reduces the distribution of the primary winding itself. capacitance.

The secondary winding----the primary winding is wound, and the (3~5) layer of insulating pad is added to rewind the secondary winding. This reduces the capacitance of the distributed capacitance between the primary winding and the secondary winding, and also increases the dielectric strength between the primary and secondary, meeting the insulation withstand voltage requirements.

The bias winding—the bias winding is wound between the primary and secondary, or the outermost layer, and the switching power supply is adjusted based on the secondary voltage or the primary voltage.
Fourth, high frequency transformer wire diameter calculation

The determination of the wire diameter of the high-frequency transformer can be calculated according to the formula D=1.13(I/J)^1/2, and J is the current density, and the calculated diameters are different for different values. Since the high-frequency current has a skin effect in the conductor, the penetration depth of the conductor at different frequencies is also calculated when determining the line.
High frequency transformer.jpg

Penetration depth formula: d=66.1/(f)^1/2

If the calculated wire diameter D is greater than twice the penetration depth, multiple strands or litz wires are required.

For example: 1A current, frequency 100K. Assume current density is 4A/mm^2

D=1.13*(1/4)^1/2=0.565mm Sc=0.25mm^

d=66.1/(f)^1/2=66.1/(100000)^1/2=0.209mm

2d=0.418mm

Using a 0.4mm line, the cross-sectional area of a single 0.4 is Sc = 0.1256mm^2. The cross-sectional area of two 0.4 is Sc = 0.1256 * 2 = 0.2512mm ^ 2

It can be seen that the 2*0.4 scheme can meet the calculation requirements.

Five, high frequency transformer turns calculation
1. Derivation of electromagnetic calculation formula:

1) Correlation formula between magnetic flux and magnetic flux density:

Ф = B * S (1)

B = H * μ (2)

H = I*N / l (3)

2) The relationship between the back-induced electromotive force and the current and the magnetic flux in the inductor:

EL =⊿Ф / ⊿t * N (4)

EL = ⊿i / ⊿t * L (5)

From the above two formulas, the following formula can be derived: ⊿Ф / ⊿t * N = ⊿i / ⊿t * L Deformation is available: N = ⊿i * L/⊿Ф

Then Ф = B * S gives the following formula: N = ⊿i * L / ( B * S ) (6)

And directly deformed by (5) can be obtained: ⊿i = EL * ⊿t / L (7)

Union (1)(2)(3)(4) can also introduce the following formula: L =(μ* S )/ l * N2 (8)

This shows that the inductance is proportional to the square of the number of turns when the core is fixed (the factor that affects the inductance)

3) The relationship between energy and current in the inductor: QL = 1/2 * I2 * L (9)

4) According to the law of conservation of energy and the factors affecting the inductance and the combination of (7)(8)(9), the relationship between the primary and secondary turns ratio and the duty cycle can be obtained:

N1/N2 = (E1*D)/(E2*(1-D)) (10)
2. Calculate the transformer parameters according to the above formula:

1) High frequency transformer input and output requirements:

Input DC voltage: 200--- 340 V

Output DC voltage: 23.5V

Output current: 2.5A * 2

Total output power: 117.5W

2) Determine the primary to secondary turns ratio

The secondary rectifier uses two Schottky diodes with VRRM = 100V forward current (10A). If the primary and secondary turns ratio is large, the power is subjected to a high back-turn ratio and the power transistor is low. The following formula:

N1/N2 = VIN(max) / (VRRM * k / 2) (11) Here the safety factor is 0.9

From this, the turns ratio N1/N2 = 340/(100*0.9/2) ≌ 7.6

3) Calculate the highest peak voltage of the power FET:

Vmax = Vin(max) + (Vo+Vd)/ N2/ N1 (12)

Vmax = 340+(23.5+0.89)/(1/7.6)

This calculates the maximum voltage that the power tube can withstand: Vmax ≌ 525.36(V)

4) Calculate the PWM duty cycle

Deformed by (10):

D = (N1/N2)*E2/(E1+(N1 /N2*E2)

D=(N1/N2)*(Vo+Vd)/Vin(min)+N1/N2*(Vo+Vd) (13)

D=7.6*(23.5+0.89)/200+7.6*(23.5+0.89)

The calculated duty cycle D≌ 0.481

5) Calculate the primary inductance of the transformer:

For the convenience of calculation, it is assumed that the primary current of the transformer is a sawtooth wave, that is, the current variation is equal to the peak value of the current, that is, it is ideally considered that the energy stored in the output tube during the conduction period is completely consumed during the off period. Then the calculation of the primary inductance can be analyzed only in one cycle of the PWM. At this time, the following derivation process can be performed by the formula (9):

(P/η)/ f = 1/2 * I2 * L (14)

(P/η)/ f = 1/2 * (EL * ⊿t / L)2 * L (15)

⊿t = D / f (D ----- PWM duty cycle)

Substituting this formula into the (15) variant can be obtained:

L = E2 * D2 * η / ( 2 * f * P ) (16)

Here the efficiency is 85%, and the PWM switching frequency is 60KHz.

The minimum inductance at the input voltage is:

L=2002* 0.4812 * 0.85 / 2 * 60000 * 117.5

Calculate the primary inductance as: L1 ≌ 558(uH)

Calculate the primary peak current: from (7):

⊿i = EL * ⊿t / L = 200 * (0.481/60000 ) / (558*10-6)

Calculate the peak value of the primary current as: Ipp ≌ 2.87(A)

The primary average current is: I1 = Ipp/2/(1/D) = 0.690235(A)

6) Calculate the number of turns of the primary and secondary coils:

The core is selected as EE-42 (cross-sectional area 1.76mm2). The magnetic flux density is 2500 Gauss or 0.25 Tesla for the saturation control. This is obtained by (6).

The number of turns of the primary inductance is: N1= ⊿i * L / ( B * S ) = 2.87 * (0.558*10-3)/0.25*(1.76*10-4)

Calculate the number of primary inductance turns: N1 ≌ 36 (匝)

At the same time, the number of secondary turns can be calculated: N2 ≌ 5 (匝)

7) Calculate the peak current of the secondary coil:

According to the law of conservation of energy, when the energy stored by the primary inductor when the power tube is turned on is completely released on the secondary coil at the cutoff, there is the following formula:

From (8) (9), we can get:

Ipp2=N1/N2* Ipp (17)

Ipp2 = 7.6*2.87

From this, the secondary peak current can be calculated as: Ipp2 = 21.812(A)

The secondary average current is I2=Ipp2/2/(1/(1-D))= 5.7(A)

8) Calculate the number of turns of the excitation winding (also called the auxiliary winding):

Since the secondary output voltage is 23.5V and the excitation winding voltage is 12V, the half of the secondary voltage can be calculated from the number of excitation winding turns: N3 ≌ N2 / 2 ≌ 3 (匝), the current of the excitation winding is taken as: I3 = 0.1(A)
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