what is better for flyback converter AC input and same power

[yassin.kraouch]

Hi
What is better for flyback converter :
-to drop AC voltage from 230AC to 15V, 350 mA
-or to drop AC voltage to 30V, 175 mA

 

[yassin.kraouch]

Can anyone help me and give me answer ?

 

[Jibby Benjamen]

or to drop AC voltage to 30V, 175 mA= Better

 

[yassin.kraouch]

why it is better ? can you please explain to me ?

 

[BradtheRad]

Better usually means whichever suits your needs better.

If better means easier or cheaper, then the reduced current (175mA) allows use of components which are smaller in size and rating. The transformer ratio is less and the wiring can be smaller (although its watt rating is the same).

If better means which can supply greater current, then the 15V@ 350mA supply is better.

If better means which can supply greater voltage, then the 30V@ 175mA supply is better.

 

[yassin.kraouch]

in fact the problem is that i want to put LED in series or in parallel, but i am confused which output is better,
i will use SMPS, a flyback topology, do you think the price difference will be interesting ?

 

[BradtheRad]

If it is an ordinary LED, then it only draws a small current and low voltage. You can reduce mains AC using the capacitive drop method.

(Caution: Since the wiring is not isolated from high voltage, you must insulate the project from risk of anyone touching it.)

Or several LED's can be strung together. It will use no more current than one LED. A single capacitive drop method will work here. And by arranging two strings in opposite polarity, you can utilize both directions of the mains AC.

 

[yassin.kraouch]

If it is an ordinary LED, then it only draws a small current and low voltage. You can reduce mains AC using the capacitive drop method.

(Caution: Since the wiring is not isolated from high voltage, you must insulate the project from risk of anyone touching it.)

Or several LED's can be strung together. It will use no more current than one LED. A single capacitive drop method will work here. And by arranging two strings in opposite polarity, you can utilize both directions of the mains AC.

Yes but what is the maximum current that i can have if i use capacitive drop ? it is not limited on current ? and how i calculate thrated current in order to measure the current output ?
and if the current is hight, whay using higher voltage output is better than using less output ?
thanks

 

[BradtheRad]

Capacitive drop is easiest to work with if you need 100 mA or less.

As for the proper Farad value, it can be calculated using the formula for capacitive reactance (at 50 Hz). You want a certain amount of impedance, in order to allow safe current through the led's. You'll also want to insert a certain amount of resistance, to reduce current surge on power-up.

Your thread title suggests you were first interested in a flyback. Was it because you want the led's to be isolated from mains AC?

 

[yassin.kraouch]

yes i need to have LED separated from the main, and i want to know what is better for me is higher voltage lower current or lower voltage higher current,
in order to estimate whch one is cheaper,

 

[Jibby Benjamen]

The word drop refers to linear voltage regulator and not SMPS .so
230-15=215 drop
215X0.35=75.2 Watts loss
Where as
230-30=200 drop
200X0.175=35 watts loss
But if it is SMPS makes no difference.

 

[yassin.kraouch]

The word drop refers to linear voltage regulator and not SMPS .so
230-15=215 drop
215X0.35=75.2 Watts loss
Where as
230-30=200 drop
200X0.175=35 watts loss
But if it is SMPS makes no difference.

i did not understand what are you talking about,

 

[Jibby Benjamen]

i did not understand what are you talking about,

You have ask to explain in detail on 3-10-2013 : why it is better ? can you please explain to me ?

I have explained in detail. : The word drop refers to linear voltage regulator and not SMPS .so
230-15=215 drop
215X0.35=75.2 Watts loss
Where as
230-30=200 drop
200X0.175=35 watts loss
But if it is SMPS makes no difference.