Because LED lighting power supply requirements: civilian lighting PF value must be greater than 0.7, commercial lighting must be greater than 0.9. For 10~70W LED drive power supply, the general use of single-stage PFC design. Save both space and cost. Next, we will discuss the design of single-stage PFC high-frequency transformer.

Take a 60W example to explain:

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 magnetic core:

Step 1: select IC and magnetic core:Ic with SA7527, output with quasi resonance, the efficiency of 0.87 should be no problem.

According to the power to choose the core, according to the following formula:

Po Fs = 100 * * Ve

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

Here, Po=Vo*Io=48*1.28=61.44; Working frequency: 50000Hz; Is:

Ve = Po/(100 * 50000)

= 61.4 / (100 * 50000) = 12280 MMM

The Ve value of PQ3230 is: 11970.00 MMM. It can completely meet the demand. You can plug in the formula to see the actual required working frequency: 51295Hz.

Step 2: calculate the primary inductance.

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 the transformer is designed).

Maximum duty cycle: wide voltage is generally less than 0.5, and narrow voltage is generally around 0.3. Considering the voltage resistance of MOS tube, it is generally not recommended to choose a power supply greater than 0.5, and 0.3 is more appropriate for 220V power supply. Select here: Dmax=0.327.

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

Maximum peak input 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

The primary inductance: Lp = ^ 2 * on3dmax Vin_min/(2 * Iin_max * fs_min) * 10 ^ 3

= 0.327 * 0.327 * 176 / (2 * 0.39 * 50000 * 1000

= 482.55 uH

Take 500 uh.

Step 3: calculate the number of primary turns NP:

Step 3: calculate the number of primary turns NP:The AL value of core data, PQ3230:5140 nh/N ^ 2, in the design of flyback transformer, to be kept a certain smell. It is appropriate to choose 0.6 times AL value. Here, AL is taken as:

AL = 2600 nh/N ^ 2

Is: NP = ^ = 0.5 (500/0.26) 44

Step 4: secondary turns NS:

Step 4: secondary turns NS:VOR is = VDmin * on3dmax

= 249 * 0.327 = 81.4

N = VOR is/Vo turns ratio = 81.4/48 = 1.696

NS = NP/n = 44/1.686 = 26

Step 5: calculate NA of auxiliary winding

Looking at the datasheet of IC, we know that the VCC is 11.5~30V. Choose 16 volts here.

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

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