Detailed steps for designing a single-stage PFC high-frequency transformer based on LED lighting power supply

Due to LED lighting Power supply requirements: the PF value of civil lighting must be greater than 0.7, and that of commercial lighting must be greater than 0.9. For 10~70W LED driver power supplies , single-stage PFC is generally used for design. This saves space and costs. Next, let's discuss the design of a single-stage PFC high-frequency transformer.

Let's take a 60W example to explain:

Input conditions:

Voltage range: 176~265Vac 50/60Hz

PF>0.95

THD<25%

Efficiency ef〉0.87

Output conditions:

Output voltage: 48V

Output current: 1.28A

Step 1: Select IC and core:

The IC uses Silan's SA7527, the output is quasi-resonant, and there should be no problem in achieving an efficiency of 0.87.

Select the core according to the power, according to the following formula:

Po=100*Fs*Ve

Po: output power; 100: constant; Fs: switching frequency; Ve: core volume.

Here, Po=Vo*Io=48*1.28=61.44; the operating frequency is 50000Hz; then:

Ve=Po/(100*50000)

=61.4/(100*50000)=12280 mmm

The Ve value of PQ3230 is: 11970.00mmm. Since it works in frequency modulation mode, it can fully meet the demand. You can substitute it into the formula to see that the actual required working frequency is: 51295Hz.

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 total efficiency when designing the transformer).

Selection of maximum duty cycle: Wide voltage is generally selected to be less than 0.5, and narrow voltage is generally selected to be around 0.3. Considering the withstand voltage of MOS tube, it is generally not recommended to select a value greater than 0.5. When the power supply is 220V, it is more appropriate to select 0.3. Here we select: 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

MOS tube maximum peak current: 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 primary turns NP:

According to the core data, the AL value of PQ3230 is 5140nH/N^2. When designing a flyback transformer, a certain margin should be left. It is more appropriate to choose an AL value of 0.6. Here we take AL ​​as:

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

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

NS=NP/n=44/1.686=26

Check the IC datasheet and you will know that VCC is between 11.5 and 30V. Select 16V here.

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

Winding method:

Summarize

The test results of the samples are: overall efficiency 0.88, PF value: 0.989 at 176V; 0.984 at 220V; 0.975 at 265V. The temperature rise of the transformer is 25K. In the whole transformer design process, some things are simplified, such as the voltage drop of the diode. In comparison, it is somewhat consistent with the general flyback transformer. It is just that there is no large-capacity electrolytic capacitor after the rectifier bridge. The actual DC minimum voltage is not 1.414 times.