Single stage pfc design high-frequency transformer.

Let's take a 60W instance as an example:

Input conditions:

Voltage range: 176~265Vac 50/60Hz

PF>0.95

THD<25%

Efficiency > 0.87

Output conditions:

Output voltage: 48V

Output current: 1.28A

Step 1: Choose IC and magnetic core:

Ic uses Shilan's SA7527, which has a quasi resonant output and an efficiency of 0.87, so there should be no problem.  

Select the magnetic core based on power, according to the following formula:

Po=100*Fs*Ve ； (Po: output power; 100: Constant; Fs: Switching frequency; Ve: Volume of magnetic core. ）

here Po=Vo*Io=48*1.28=61.44； Operating frequency selection: 50000Hz; then:

Ve=Po/（100*50000） =61.4/（100*50000）=12280 mmm

The Ve value of PQ3230 is 11970.00mmm, as it operates in frequency modulation mode. It can fully meet the needs. You can substitute it into the formula to see if the actual required operating frequency is 51295Hz.

Step 2: Calculate the primary inductance.  

Minimum DC input voltage: VDmin=176 * 1.414=249V.  

Maximum DC input voltage: VDmax=265 * 1.414=375V.  

Maximum input power: Pinmax=Po/ef=61.4/0.9=68.3W (slightly higher than the overall efficiency was achieved when designing the transformer).  

Selection of maximum duty cycle: Wide voltage is generally chosen to be less than 0.5, and narrow voltage is generally chosen to be around 0.3. Considering the withstand voltage of MOS transistors, it is generally not advisable to choose a voltage greater than 0.5220V. It is more appropriate to choose 0.3 for power supply. Select here: Dmax=0.327.  

Maximum input current: Iinmax=Pin/Vinmin=68.3/176=0.39 A

Maximum input peak current: Iinmaxp=Iin * 1.414=0.39 * 1.414=0.55A

Maximum peak current of MOS transistor: Imosmax=2 * Iinmaxp/Dmax=2 * 0.55/0.327=3.36A

Primary inductance: Lp= Dmax^2*Vin_min/(2*Iin_max*fs_min)*10^3 =0.327*0.327*176/（2*0.39*50000）*1000 =482.55 uH ， Take 500uH.

Step 3: Calculate the number of primary turns NP:

According to the magnetic core data, the AL value of PQ3230 is 5140nH/N ^ 2. When designing the flyback transformer, a certain amount of air should be left. Choosing an AL value of 0.6 times is more appropriate. Here we take AL:

AL=2600nH/N^2 ， Then: NP=(500/0.26) ^ 0.5=44

Step 4: Secondary turns NS:

VOR=VDmin*Dmax =249*0.327=81.4

Turn ratio n=VOR/Vo=81.4/48=1.696

NS=NP/n=44/1.686=26

Step 5: Calculate the NA of the auxiliary winding

Check the datasheet of the IC and find out that VCC is between 11.5~30V. Choose 16V here.

NA=NS/(Vo * VCC)=26/(48/16)=8.67, take 9.