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Because LED lighting power requirements: civil lighting PF value must be greater than 0.7 , commercial lighting must be greater than 0.9 . For 10 ~ 70W LED driver power supply , generally using a single-stage PFC design . That saves space and cost savings. Then explore the design of a single-stage PFC high-frequency transformer .

In the instance of a 60W 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 One: Choose ic and core :

Ic with Queensland 's SA7527, with quasi-resonant output , efficiency, 0.87 should be no problem.

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

Po = 100 * Fs * Ve

Polonium : Output power ; 100: constant ; Fs: switching frequency ; Ve: core volume .

Here , Po = Vo * Io = 48 * 1.28 = 61.44; Working Frequency Selection : 50000Hz; then :

Ve = Po / (100 * 50000)

= 61.4 / (100 * 50000) = 12280 mmm

PQ3230 the Ve value : 11970.00mmm, here because it is the FM mode. Fully meet the requirements. Can be substituted into the formula needs to look at the actual operating frequency : 51295Hz.

Step two: calculate the primary inductance.

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

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

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

The maximum duty cycle options: wide voltage generally choose less than 0.5 , the narrow voltage is usually selected at about 0.3 . Considering the MOS tube voltage, generally do not choose more than 0.5 , 0.3 more appropriate choice when 220V 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

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.

The third step : Calculate the primary turns NP:

Richard core data , PQ3230 aluminum value : 5140nH / N ² , in the design of flyback transformer , to stay a certain atmosphere. Select aluminum 0.6 -fold more appropriate value . Here we take aluminum :

AL = 2600nH / N²

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

Step Four: 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

Step Five : Calculate the auxiliary winding NA

See IC 's datasheet, know VCC is 11.5 ~ 30V. In this election 16V.

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

Summary

Through testing, the sample results are as follows: overall efficiency 0.88, PF value : 176V when 0.989; 220V when 0.984; 265V when 0.975 . Transformer temperature rise 25K. Throughout the process, the transformer design . Simplify some things. Such diode drop . Compare , somewhat consistent with the general flyback transformer. Just because there is no large-capacity electrolytic capacitor connected to the rectifier bridge. Actual DC voltage is not the lowest 1.414 times .


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