# Push-Pull SMPS With EI50 Ferrite Core

#### mcmsat13

##### Member level 4
I am building DC DC boost stage for 340VDC, I want to know If I am right about the winding of the EI50 Ferrite transformer because I am no expert to magnetic components.

* Topology = Push-Pull

* VIn(min) = 10.5V

* Vin(nom) = 12V

* Vin(max) = 15V

*Vout = 340VDC

* Vaux = 25VDC

* Output Diode Vf = 0.5V

* Core = EI50

* Core Ae = 2.3cm2

* SW Frequency = 50KHz

* Max Duty Cycle = 94%

Please I need to know the Primary turns
Secondary turns
Auxiliary turns

#### Easy peasy

Bpk = E / 4 F N Ae, ( square wave ) for 50kHz, a conservative Bpk = 150mT peak each way, the core area = 230 mm^2

Thus for 340V out Ns = 50 turns, except the duty cycle must be taken into account so divide by 0.94 = 54 turns

10.5V is Vin min - so to get 340V ave out @ 94% applied square wave, Np/Ns = 10.5/340, giving Np = 1.66 Turns

Obviously we round this up to 2.00 turns ( + 2.00 turns for the other ) for the pri, giving 69T for the sec

[ check 10.5V * 69/2 = 362, * 0.94 = 340.5, maybe add a wee bit for the diodes and the wiring loss in the Tx ]

The aux is therefore 25/10.5 * 2T / 0.94 = 5 Turns

Last edited:

### mcmsat13

Points: 2

#### dick_freebird

I would only note that push-pulls have the
potential to "flux walk" with any volt-second
imbalance between the two legs (or winding
asymmetry) and you might need either extra
conservatism or a scheme that ensures
flux balance can't get out of hand. Some
controllers take a stab at that, I gather. If
the imbalance is too much then even being
conservative about the core vs expected
flux won't stop you from eventually saturating
on one phase. And "eventually" may not be
that long.

### mcmsat13

Points: 2

#### Easy peasy

Just to add further, 94% on time, gives 6% dead time ( minimum ) this allows the core to "fly-back" and self-reset to an extent in the dead time if the applied volts seconds are < 6% different in the ON times. Current mode control of the primary takes care of this completely.

### mcmsat13

Points: 2

#### mcmsat13

##### Member level 4
Bpk = E / 4 F N Ae, ( square wave ) for 50kHz, a conservative Bpk = 150mT peak each way, the core area = 230 mm^2

Thus for 340V out Ns = 50 turns, except the duty cycle must be taken into account so divide by 0.94 = 54 turns

10.5V is Vin min - so to get 340V ave out @ 94% applied square wave, Np/Ns = 10.5/340, giving Np = 1.66 Turns

Obviously we round this up to 2.00 turns ( + 2.00 turns for the other ) for the pri, giving 69T for the sec

[ check 10.5V * 69/2 = 362, * 0.94 = 340.5, maybe add a wee bit for the diodes and the wiring loss in the Tx ]

The aux is therefore 25/10.5 * 2T / 0.94 = 5 Turns

I appreciate your great help. The results of your calculations are same with mine. Though may not ply the road but arrived at the same destination.
Below is my own calculations. I used Microsoft Excel for my calculations. As I am not near expert in magnetics that's why I wanted to get this solved so that I can trust my own Excel calculations.
Please check out the Excel below. Only the parameters in Reds are to be edited. The rounded answers are in Green. The sheet is protected so the formula will not be messed up, so it is better viewed in Excel.

#### Attachments

• Ferrite Turns Calculator_DC_DC_Boost.xlsx.zip
15.1 KB · Views: 1

#### Easy peasy

I don't open odd zip files - apologies ...

#### mcmsat13

##### Member level 4
I don't open odd zip files - apologies ...

Sorry! I forgot that this COVID-19 of a thing can be funny!

Unless you can view it from Google Drive

Below is the screenshot.

#### Attachments

• Screenshot_20210130-104643.png
140.3 KB · Views: 7

#### treez

Treat it as a Buck converter. (here i assume CCM at max load)

Pick a max duty cycle....so pick say 0.7.
This will be in use at your min vin of 10.5V.

If it were a Buck, then you would want an input voltage of 340/0.7 = 486V.

So add in a transformer which converts your 10.5V to 486V...so thats with an NP/NS of 10.5/486

#### Easy peasy

respectfully ( Treez) the max duty cycle was quoted at 0.94 in the original post - and the min turns have been computed above ...
--- Updated ---

Re; screen shot above - you have used less than 10.5V in some of the calcs - and secondary is spelled wrong in one place

without seeing the math behind the calcs - there is no way of discerning whether the program is sound or not ...
--- Updated ---

the program should also give the peak volts on the sec side - 518V in this case ( 15Vin ) thus the diodes will need to be at least 600V with good snubbers to limit peak overshoot ...

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