High magnetizing current in Push-pull topology

Hello everyone,

I am desiging with a Push pull topology a 500W DC-DC converter (Vin=24V +- 10%, Vout=200 V at 100 kHz).
The calculations gave me that I can use a relationship n=Np/Ns=1/12 with a EPCOS-TDK E core 55/28/21 (https://en.tdk.eu/inf/80/db/fer/e_55_28_21.pdf) that has an AL=6400nH, where:

Imag=Vin_min*D_max/(L__mag*f_sw)=26.04A,

and

Iin_max=P_o/Vin_min=23.15A.

Is it ok that Imag>Iin? What do you think? What can you guys suggest me?

I am willing to use these Mosfets (FDH038AN08A1, https://www.mouser.com/ds/2/149/FDH038AN08A1-103477.pdf), and of course, every switch will have its own snubber network.

Thank you for your help. Have a nice day.

David.

You'll never achieve this magnetizing current amount with a correctly dimensioned push-pull transformer.

Seems like you are ignoring the minimal number of winding turns commanded by core Bmax. How did you calculate Np and Ns?

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I neither get the calculated 26 A, even for single turn primary.

Lmag = Uo.Ur N^2 Ae / Le for a ferrite core with no gap ( SI units)

di / dt = V / L where V = 24V and dt = 5uS, Ipk = di/2 and L is derived from above, and dB/dt = V / N Ae

where Bpk = dB/2, and Bpk at 100kHz should be less than 80mT for N87 material for reasonable losses,

thus N min for 80mT and 100kHz (Ton = 5us) = 2.11 turns

good luck ...

Hello FvM,

Thank you for your answer.

the way I am getting those values is:
N_pri=(Vin(min) 10^4)/(Kf * Bac * f * A_c) (Faraday's Law)
N_pri=ceil( (21.6V 10^4)/(4 * (0.75*320 mT) * 100 kHz * 3.54 cm^2) ) = ceil(0.63) = 1 turn

Where Bac is the 75% of Bsat, and Bsat is 320 mT.

N_sec=ceil( (Npri*Vo)/( Vin(min)*D_max) * (1+eta) )= ceil( (1*200V)/(21.6V*0.8) * (1+0.95) )=ceil(11.68)= 12 Turns
Where eta is the efficiency, and D_max was set to 0.4

What do you think? do you see the mistake?

Have a nice day.

David.

I forgot to tell. It could be obvious for you, but still...
Kf is the winding utilization factor, and was set to 0.4 for simplicity (McLyman).

Hello Easy peasy,

Thank you for your help.
I have the following questions.

You said: Lmag = Uo.Ur N^2 Ae / Le for a ferrite core with no gap ( SI units)
Whay is Vr?

At the end I think we get the very same equation (Faraday's Law):
N_pri=ceil( (Vin_min * 10^4)/(Kf * Bac * Fsw * A_e))

Where:
-Bac was set to the 75% of Bmax: 0.75*320 mT=240 mT (for Bac value see curves in https://en.tdk.eu/download/528882/3226013b0ed82a6a2af3666f537cbf83/pdf-n87.pdf and data in https://www.mouser.com/datasheet/2/400/e_55_28_21-1220353.pdf).
-Ae = 3.54 cm^2 (from core's datasheet).
-Kf is the winding utilization factor and was set to 4 (McLyman).

Npri=ceil( 21.6V*10^4/(4* (0.75*0.32T) *100kHz*3.54cm^2 ) ) = ceil(0.636) = 1 Turn.

Do you think I am taking Bac too high?

Best regards.

David.

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david.cano90 said:

Hello FvM,

Thank you for your answer.

the way I am getting those values is:
N_pri=(Vin(min) 10^4)/(Kf * Bac * f * A_c) (Faraday's Law)
N_pri=ceil( (21.6V 10^4)/(4 * (0.75*320 mT) * 100 kHz * 3.54 cm^2) ) = ceil(0.63) = 1 turn

Where Bac is the 75% of Bsat, and Bsat is 320 mT.

N_sec=ceil( (Npri*Vo)/( Vin(min)*D_max) * (1+eta) )= ceil( (1*200V)/(21.6V*0.8) * (1+0.95) )=ceil(11.68)= 12 Turns
Where eta is the efficiency, and D_max was set to 0.4

What do you think? do you see the mistake?

Have a nice day.

David.

Click to expand...

Sorry...

I forgot to tell. It could be obvious for you, but still...
Kf is the form factor, and was set to 4 for square wave signal (McLyman).

david.cano90 said:

Hello FvM,

Thank you for your answer.

the way I am getting those values is:
N_pri=(Vin(min) 10^4)/(Kf * Bac * f * A_c) (Faraday's Law)
N_pri=ceil( (21.6V 10^4)/(4 * (0.75*320 mT) * 100 kHz * 3.54 cm^2) ) = ceil(0.63) = 1 turn

Where Bac is the 75% of Bsat, and Bsat is 320 mT.

N_sec=ceil( (Npri*Vo)/( Vin(min)*D_max) * (1+eta) )= ceil( (1*200V)/(21.6V*0.8) * (1+0.95) )=ceil(11.68)= 12 Turns
Where eta is the efficiency, and D_max was set to 0.4

What do you think? do you see the mistake?

Have a nice day.

David.

Click to expand...

Hello FvM,

Thank you for your answer.

the way I am getting those values is:
N_pri=(Vin(min) 10^4)/(Kf * Bac * f * A_c) (Faraday's Law)
N_pri=ceil( (21.6V 10^4)/(4 * (0.75*320 mT) * 100 kHz * 3.54 cm^2) ) = ceil(0.63) = 1 turn

Where :
- Bac is the 75% of Bsat, and Bsat is 320 mT.
- Kf is the form factor and it is 4 for square wave signal (McLyman).

N_sec=ceil( (Npri*Vo)/( Vin(min)*D_max) * (1+eta) )= ceil( (1*200V)/(21.6V*0.8) * (1+0.95) )=ceil(11.68)= 12 Turns
Where eta is the efficiency, and D_max was set to 0.4

What do you think? do you see the mistake?

Have a nice day.

David.

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david.cano90 said:

I forgot to tell. It could be obvious for you, but still...
Kf is the winding utilization factor, and was set to 0.4 for simplicity (McLyman).

Click to expand...

Sorry, not winding factor but form factor, and it is 4 for square wave signals (McLyman)

Bpk is set by the losses at frequency. It is clear that this is new to you else you would not consider taking Bpk above 100mT for 100kHz. 80mT peak will give you core losses you can live with.

Also, more turns makes it easier to interleave your pri & sec windings in a way that reduces leakage L - very important in a push pull - else your turn off volt spikes will be very large ...

..yes

Lmag = Uo.Ur N^2 Ae / Le

Click to expand...

is gotten from Amperes Law...N.I = H.(le).....

.....or the reluctance equation ...Reluctance = N^2/L

..also subbing in B=uo.ur.H

..And reluctance = le/(uo.ur.Ae)

The point to remember is that unfettered cores are actually gapped......due to the non-smooth ferrite surface.......so the equation gets you in the zone, but as we all now, not perfect.
(i dont mean "unfettered", but i cant remember the word...to do with smoothing off a ferrite surface)

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sorry if this wasnt too helpful i was just skimming through the posts to see one that i could shovel some knowledge into

un-polished cores have a residual gap - can often be ignored ...