How to calculate high-frequency transformer parameters in system design

1. Derivation of electromagnetic calculation formula:

1. Formulas related to magnetic flux and magnetic flux density:

Ф = B * S                    ⑴

Ф ----- Magnetic flux (Weber)
B ----- Magnetic flux density (Weber per square meter or Gauss) 1 Weber per square meter = 104 Gauss
S ----- Cross-sectional area of ​​the magnetic circuit (square meters)

B = H * μ ⑵

μ ----- Magnetic permeability (unitless, also called dimensionless)
H ----- Magnetic field strength (volts per meter)

H = I*N / l                     ⑶

I ----- current intensity (ampere)
N ----- number of turns of the coil (turns T)
l ----- length of the magnetic circuit (meter)

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

EL = ⊿Ф / ⊿t * N ⑷

EL = ⊿i / ⊿t * L ⑸

⊿Ф ----- flux change (Weber)
⊿i ----- current change (ampere)
⊿t ----- time change (seconds)
N ----- number of turns of the coil (turns T)
L ------- inductance of the inductor (henry)

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

⊿Ф / ⊿t * N = ⊿i / ⊿t * L Transformation yields:

N = ⊿i * L/⊿Ф

Then from Ф = B * S, we can get the following formula:

N = ⊿i * L / ( B * S )      ⑹

And by directly transforming formula (5), we can get:

⊿i = EL * ⊿t / L ⑺

Combining ⑴⑵⑶⑷, we can derive the following formula:

L =(μ* S )/ l * N2 ⑻

This shows that when the core is constant, the inductance is proportional to the square of the number of turns (factors affecting inductance)

3. The relationship between energy and current in inductance:

QL = 1/2 * I2 * L             ⑼

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

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

L ------- Inductance of the inductor (henry)

4. According to the law of conservation of energy and the factors affecting inductance and the combination of equations ⑺⑻⑼, the relationship between the primary-to-secondary turns ratio and the duty cycle can be obtained:

N1/N2 = (E1*D)/(E2*(1-D)) ⑽

N1 -------- Primary coil turns (turns) E1 -------- Primary input voltage (volts)
N2 -------- Secondary inductor turns (turns) E2 -------- Secondary output voltage (volts)

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-secondary turns ratio is large, the reverse voltage on the power tube is high. If the turns ratio is small, the reverse voltage on the power tube is low. This gives the following formula:

N1/N2 = VIN(max) / (VRRM * k  / 2)         ⑾

N1 ----- Primary turns VIN(max) ------ Maximum input voltage k ----- Safety factor
N2 ----- Secondary turns Vrrm ------ Maximum reverse withstand voltage of the rectifier Here the safety factor is 0.9

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

3. Calculate the maximum reverse peak voltage of the power field effect tube:

Vmax = Vin(max) + (Vo+Vd)/ N2/ N1              ⑿

Vin(max) ----- Maximum input voltage Vo ----- Output voltage Vd ----- Rectifier forward voltage

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

From this, the maximum voltage that the power tube can withstand can be calculated: Vmax ≌ 525.36(V)

4. Calculate PWM duty cycle:

By transforming formula ⑽, we can get:

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

D=(N1/N2)*(Vo+Vd)/Vin(min)+N1/N2*(Vo+Vd)  ⒀

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

From this, we can calculate that the 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 change is equal to the peak value of the current, that is, it is ideal to assume that the energy stored in the output tube during the conduction period is completely consumed during the cut-off period. Then the calculation of the primary inductance can be analyzed based on only one cycle of PWM. At this time, the following derivation process can be obtained from formula (9):

（P/η）/ f = 1/2 * I2 * L ⒁

P ------- Power supply output power (watt) η ---- Energy conversion efficiency f ---- PWM switching frequency Substitute equation ⑺ into equation ⒁:

(P/η)/ f = 1/2 * (EL * ⊿t / L)2 * L ⒂

⊿t = D / f (D ----- PWM duty cycle)

Substituting this formula into formula (1), we can get:

L = E2 * D2 *η/ ( 2 * f * P )        ⒃

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

The inductance at minimum input voltage is:

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

Calculate the primary inductance: L1 ≌ 558(uH)

Calculate the primary peak current:

From formula (7), we can get:

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

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

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

6. Calculate the number of turns of the primary coil and the secondary coil:

The core is EE-42 (cross-sectional area 1.76mm2) and the flux density is 2500 Gauss or 0.25 Tesla to prevent saturation. From formula (6), the number of turns of the primary inductor is:

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

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

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

7. Calculate the peak current of the secondary coil:

According to the law of conservation of energy, when the primary inductor stores energy when the power tube is turned on, it is completely released on the secondary coil when it is turned off. The following formula can be used:

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

Ipp2=N1/N2* Ipp                           ⒄

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 auxiliary winding):

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

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

The current of the exciting winding is: I3 = 0.1(A)