# High frequency transformer design and application

High frequency transformer design and application

High frequency transformer application and design; high-frequency transformer is also called pulse transformer. It is a power transformer with working frequency exceeding intermediate frequency (10kHz). It mainly has high-frequency transformer of switching power supply and high-frequency transformer of inverter power supply. According to the working frequency, it can be divided into several grades: 10kHz- 50kHz, 50kHz-100kHz, 100kHz~500kHz, 500kHz~1MHz, 1MHz or above.

1. High frequency transformer application

The most conspicuous of the typical half-bridge transformer circuits are three high-frequency transformers: power transformers, drive transformers and auxiliary transformers. Each type of high-frequency transformer has its own measurement standards in national regulations, such as power transformers, as long as it is For power supplies above 200W, the core diameter 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. High-frequency transformers are non-standard devices, so they are generally not available and can only be designed.

For example: How to choose a high frequency transformer for a 12V2A power supply?

First, use the formula or experience based on the operating frequency, circuit topology, and output power to select the type of high-frequency transformer core, and the size of the core, and sometimes also the working environment (such as temperature, use, sealing, etc.) Generally, it is a flyback circuit, frequency of about 60KHZ, and it can be done with EE25 or EE28. Inverter frequency up to 120KHz, must use a better quality iron core, such as iron silicon aluminum core. The first thing to look at is the shape of the product to be made, such as a toroidal transformer or an EI transformer. After determining the type of transformer to be used, first calculate the total apparent power Pt of the transformer. According to the total load power, calculate the core area product AP, and then find the type of core used according to the AP value. According to the parameters of the magnetic core model, the number of primary winding turns and current can be calculated. The secondary and bias turns and wire diameter are then calculated based on the primary turns, input and output voltage, and duty cycle. The next step is debugging and tuning. There are many kinds of high-frequency transformer parameters that can work normally, but there is only one best state. The calculation is to let it work first, and the adjustment is to find the best state.

High-frequency transformers are widely used; for example, mobile phone chargers, electronic ballasts, switching power supplies, color TV power supplies, computer power supplies, liquid crystal drivers and power supplies all use high-frequency transformers. Compared with the low-frequency transformer, the high-frequency transformer can reduce the space (the transformer is small) and improve the working efficiency. The main function of the high-frequency transformer is the transfer of energy or the storage of energy, depending on the operating mode of the switching power supply. The forward type is energy transfer, and the flyback type is energy storage and transfer. Of course, there is also the role of isolation and ratio. Take the energy, let’s take a list, a mouse and an elephant to carry things. Although the elephant is much transported, it is slow, and the mouse is the least transported, but the speed is fast, so the same thing As long as the speed is fast enough, the small size can also be completed. This is why the high frequency of the transformer can reduce the volume of the transformer. All high-frequency transformer applications convert electrical energy into magnetic energy, which is then converted into electrical energy, while using the turns ratio to obtain the required voltage.

2. High-frequency transformer parameter design

First, the electromagnetic calculation formula is derived:

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

Ф= B * S(1)

Ф-----Magnetic (Weber)

B -----Magnetic density (Weber per square meter or Gauss) 1 Weber per square meter = 104 Gauss

S -----The cross-sectional area of the magnetic circuit (m2)

B = H *μ(2)

μ-----permeability (no unit is also called dimensionless)

H -----Magnetic field strength (volts per meter)

H = I*N / l(3)

I -----current intensity (amperes)

N -----coil turns (circle T)

l -----Magnetic long road (m)

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

EL =⊿Ф/⊿t * N(4)

EL =⊿i /⊿t * L(5)

⊿Ф-----Magnetic flux change (Weber)

⊿i -----current change (amperes)

⊿t -----time change (seconds)

N -----coil turns (circle T)

L ------- inductance inductance (hen)

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

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

N =⊿i * L/⊿Ф

Then Ф = B * S can get 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)

QL -------- Energy stored in the inductor (Joules)

I -------- Current in the inductor (ampere)

L ------- inductance inductance (hen)

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)

N1 -------- Number of turns of the primary coil (circle) E1 -------- Primary input voltage (volts)

N2 -------- Number of turns of secondary inductance (circle) E2 -------- Secondary output voltage (volts)

Second, 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 ratio of primary to secondary turns:

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)

N1 ----- primary turns VIN (max) ------ maximum input voltage k ----- safety factor

N2 ----- secondary turns Vrrm ------ rectifier maximum reverse voltage

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)

Vin(max) -----Input voltage maximum Vo -----output voltage

Vd ----- rectifier forward voltage

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 ------- power output power (watts) η----energy conversion efficiency f ---- PWM switching frequency (7) is substituted into (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:

Available 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 value, which 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):

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

From this, the number of excitation winding turns can be calculated as: N3≌N2 / 2≌3 (匝)

The current of the excitation winding is taken as: I3 = 0.1(A)