center tap push-pull smps dc-dc converter transformer design

Hi all,

I want to design a centered tap push-pull smps dc-dc converter that has 24VDC input and 2x8VDC output (i use 7805 to get 5 V). I design it with SG3525 with %50 duty cycle ratio.

My transformer ratio is 3:1. I have already choose core (minimum body that has at least 9 pin) and it is EE1313S (https://blog.dianyuan.com/blog/u/65/1211428621.pdf) . I want to choose Np that has maximum acceptable output current without heatsink. f=50kHz.

How can i choose B not to get saturated and to get maximum current? From where should i start?

Hello

B should be kept on the cvasi-linear slope of the material of choice, below the knee point. But, before of that, the core must be chosen as a product of winding area and core area, at the very least, which, in turn, is calculated based on the power requirements of the design -- you haven't specified them.

But, considering that you have calculated these and, as a result, you have your desired magnetic core, then for your switching frequency and PL-5,7 core (sufficient for this frequency), in particular, you may choose B=0.2~0.25T in order to avoid current inrush saturation.


Good luck,
Vlad

Thanks for your reply bubulescu. I am new at these calculations so i want to learn at least push-pull design well.

bubulescu said:

But, before of that, the core must be chosen as a product of winding area and core area, at the very least, which, in turn, is calculated based on the power requirements of the design -- you haven't specified them.

Click to expand...

I want to get the maximum current which the core let me take without heatsink.

bubulescu said:

B should be kept on the cvasi-linear slope of the material of choice, below the knee point.

Click to expand...

In all EE13 datasheet's, there is no chart about B, all they give me AL graph (i will choose L after find the Np) How can i choose correct B? I have only the datasheet

bubulescu said:

But, considering that you have calculated these and, as a result, you have your desired magnetic core, then for your switching frequency and PL-5,7 core (sufficient for this frequency), in particular, you may choose B=0.2~0.25T in order to avoid current inrush saturation.

Click to expand...

Ok, i accept that B is 0.2T or 2000Gauss. So i use the above equation,
Eave=4*ΔB*Np*f*Ac*10^-8
Eave=48V (24*2 in primer)
ΔB=2000*2=4000Gauss (difference between +B - (-B) )
f=50kHz
Ac=0.124cm^2

so Np=48.39=48.
so primer Np=24+24
so Ns=8+8

(https://blog.dianyuan.com/blog/u/65/1211428621.pdf) according to this datasheet EE1313S, i can get 5W from push-pull configuration. So Is=0.625A. I have 2 8V output so each of the outputs have 0.3A approximetly.

Ip= 5W/24V= 0.208A ≈ 0.2A ; Al=810nH/N^2 so Lp=1.866uH=1.8mH

These equations are true or do i miss something??

To me, what you are doing seems to be hammering some data onto one's needs. You don't specify a current, thus no power, instead you go after the saying "as much current as it can deliver"; in that case, what's stopping you from making a 1MW power supply? Just use a 1m^2 cross-area transformer and you'll scare all the fuses around. Furthermore, you calculate the number of turns based on -- what it seems to be -- a rule of thumb formula, yet you don't verify if your design meets the core's geometrical size, what wire, etc. You simply can't go on like that. You must specifiy the power -- thus, the (maximum) current. Go over to this site: https://ntrs.nasa.gov/ and do a search for transformer design, you will find some papers there that will help you. I am sorry, but your approach isn't the correct one.

Good luck,
Vlad

Ok, fair enough. Then I will change my question and ask again.

I want to design a center-tap push-pull transformer that has 24V in primary and two 8V in secondary. Output current of each windings are 0.3A. I will use EE1313S core. f=50kHZ. Is my approach is acceptable now?

i can use B=0.2T for 50kHZ, as you said. Then ΔB=4000Gauss
Eave=24*2=48V (center tap configuration)
Ac=0.124cm^2 (for EE1313S core) Then,

Eave=4*ΔB*Np*f*Ac*10^-8 => Np=48.39 ≈ 48 (24+24)
Then,
Ns= 8+8 (Diode losses or Rds_on losses are neglected because i will get 5V output in fact)

Ap=(Pt*10^4)/(4*Kj*Ku*Bm*f)=0.019cm^4
J=Ip/(Kj*Ap)=3.56*(10^-4) cm^2=0.035mm^2 => diameter of the wire is 0.212mm

Is=0.3A => diameter of the wire is 0.3mm

Hi
I've a better suggestion for you ! and of course easier !
Read this thread completely ! then you'll be amazed ! ( i'll bet )
https://www.edaboard.com/threads/229615/

Best wishes and Good luck
Goldsmith