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High Frequency Transformer Parameters Calculation Example
Apr 06, 2018


            High Frequency Transformer Parameters Calculation Example

In the traditional high frequency transformer design, due to the limitation of the core material, its operating frequency is relatively low, generally around 20 kHz. With the continuous development of power technology, the miniaturization, high frequency and high power of power systems have become an eternal research direction and development trend. Therefore, the study of more frequently used power transformers is a key factor in reducing the size of the power system and increasing the output power ratio of the power supply.

The converter in this example uses a single-ended flyback operation. The single-ended flyback converter is widely used in the design of low-power switching power supplies, and the multiple-output is more convenient. Single-ended flyback power supply has two operating modes: continuous current mode and current discontinuous mode. The former is suitable for lower power, and the secondary diode has no problem of reverse recovery, but the peak current of the MOS transistor is relatively large; the peak current of the latter MOS transistor is relatively small, but there is a problem of reverse recovery of the secondary diode. Diode absorption circuit needs to be added. These two modes of work can be selected according to actual needs. This paper adopts the latter

Most of the transformer design needs to consider the following issues: inverter frequency f (H2); primary voltage U1 (V), secondary voltage U2 (V); secondary current i2 (A); winding line parameters n1, n2; temperature rise τ(°C); winding relative voltage drop u; ambient temperature τHJ(°C); insulation material density γz(g/cm3)

1) Select the iron core according to the output power of the transformer. The value of the iron core selected should be equal to or greater than the given value.

2) Number of windings per volt

High-frequency transformer parameter calculation

one. Electromagnetic formula derivation:

1. Magnetic flux and flux density related formula:

Ф = B * S

B = H * μ

H = I*N / l


2. Inductive Inverse EMF and Current and Flux Correlation:


EL = ⊿Ф / ⊿t * N


 (4) EL = ⊿i / ⊿t * L

The following formula can be derived from the above two formulas:

⊿Ф / ⊿t * N = ⊿i / ⊿t * L deformation available:

N = ⊿i * L/⊿Ф Then by Ф = B * S, you can get the following formula:

N = ⊿i * L / (B * S) and can be obtained by direct deformation of (5):

⊿i = EL * ⊿t / L

Joint (1)(2)(3)(4) can also be derived as follows:

L = (μ* S )/ l * N2

This shows that in the case of a certain core, the inductance is proportional to the square of the number of turns (factor that affects the inductance)


3. The relationship between energy and current in the inductor:

QL = 1/2 * I2 * L

4. According to the law of conservation of energy and the factors that affect the inductance and the combination of (7)(8)(9), we can get the relationship between the ratio of the primary-secondary turns and the duty ratio:

N1/N2 = (E1*D)/(E2*(1-D))

two. 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-secondary turns ratio:

The secondary rectifier adopts two Schottky diodes with VRRM=100V forward current (10A). If the primary-secondary turns ratio is greater, the power is subjected to high backpressure and the turns ratio is lower than that of the power transistor. The following formula:

N1/N2 = VIN(max) / (VRRM * k / 2)

Here the safety factor takes 0.9

The resulting turns ratio N1/N2 = 340/(100*0.9/2) ≌ 7.6

3. Calculate the highest peak voltage of the power MOSFET:

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

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:

By (10) deformation available:

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


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

From these can be calculated the duty cycle D≌ 0.481

5. Calculate the primary inductance of the transformer:

For the sake of calculation, it is assumed that the primary current of the transformer is a sawtooth wave, that is, the amount of current change is equal to the peak value of the current. That is, ideally, the energy stored in the output tube during the conduction period is completely consumed during the cut-off period. The calculation of the primary inductance can then be analyzed with only one cycle of the PWM. In this case, the following inference process can be derived from (9):

(P/η)/ f = 1/2 * I2 * L

Substituting this formula into the (15) variant yields:

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 primary peak current:

Available from (7):

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

The peak value of the primary current is calculated as: Ipp ≌ 2.87(A)

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

6. Calculate the number of turns for the primary and secondary coils:

The magnetic core is selected as EE-42 (cross-sectional area of 1.76mm2). The magnetic flux density is 2500 gauss, or 0.25 Tesla, for preventing saturation. Thus, the number of turns of the primary inductor from (6) is:

N1 = ⊿i * L / (B * S ) = 2.87 * (0.558*10-3)/0.25* (1.76*10-4)

Calculate primary inductor turns: N1 ≌ 36 (匝)

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

7. Calculate the secondary current peak:

According to the law of conservation of energy, when the energy stored in the primary inductor when the power transistor is turned on is fully discharged on the secondary coil at the off time, it may have the following formula:

From (8)(9), we get: Ipp2=N1/N2* Ipp

         (17) Ipp2 = 7.6*2.87

The secondary peak current can be calculated as: Ipp2 = 21.812 (A)

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

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

Because the secondary output voltage is 23.5V, the excitation winding voltage takes 12V, so it is half of the secondary voltage.

The number of turns of the excited winding can be calculated as: N3 ≌ N2 / 2 ≌ 3 (匝)

Excitation winding current is taken: I3 = 0.1(A)

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