Due to the LED lighting power requirements: the PF value of the civil lighting must be greater than 0.7, and the commercial lighting must be greater than 0.9. For 10~70W LED driver power supply, it is generally designed with single-stage PFC. That is to save space and save costs. Next, let's explore the design of a single-stage PFC high-frequency transformer.
Explain with a 60W example:
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:
Ic uses Silan's SA7527, the output has quasi-resonance, and the efficiency should be 0.87. There should be no problem.
Select the core by 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; working frequency selection: 50000 Hz; then:
Ve=Po/(100*50000)
=61.4/(100*50000)=12280 mmm
The Ve value of the PQ3230 is 11970.00 mmmm, which is due to the frequency modulation method. Can fully meet the needs. You can substitute the formula to see the actual working frequency required: 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 total efficiency when designing the transformer).
The choice of the maximum duty cycle: the wide voltage is generally chosen to be less than 0.5, and the narrow voltage is generally chosen to be around 0.3. Considering the withstand voltage of the MOS tube, generally do not choose to be greater than 0.5, and it is appropriate to select 0.3 when supplying 220V power. Choose 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 tube: 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 number of turns NP:
Check the core data, the AL value of PQ3230 is: 5140nH/N^2. When designing the flyback transformer, keep a certain breath. It is appropriate to select 0.6 times the AL value. Here AL we take:
AL=2600nH/N^2
Then: NP = (500/0.26)^0.5=44
The fourth step: secondary turns NS:
VOR=VDmin*Dmax
=249*0.327=81.4
åŒ ratio n=VOR/Vo=81.4/48=1.696
NS=NP/n=44/1.686=26
Step 5: Calculate the auxiliary winding NA
Look at the IC's datasheet and know that VCC is 11.5~30V. Choose 16V here.
NA=NS/(Vo*VCC)=26/(48/16)=8.67 Take 9.
Winding method:
to sum up
Through the test of the sample, the experimental results are: the whole machine efficiency is 0.88, the PF value is 0.989 at 176V, 0.984 at 220V, and 0.975 at 265V. The temperature rise of the transformer is 25K. Throughout the transformer design process. Simplified something. Such as the voltage drop of the diode. In contrast, it is somewhat consistent with a general flyback transformer. It is only because there is no large-capacity electrolytic capacitor after the rectifier bridge. The actual DC minimum voltage is not 1.414 times.
Editor: Li Jie
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