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Push pull Transformer Winding Calculation

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Thamilarasan

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I need to design a high voltage transformer, 12V to input (Push pull Topology), Center tapped transformer, 2KV output.

Im having ETD29 ferrite core with me with BMAX = 490mT, Ac = 0.76 square centimetre, Duty Cycle of Maximum 40% (Because of Push pull Topology)


I need to know the number of turns and wire size for the transformer in the primary and secondary side. Primary Wire Size and Secondary Wire Size, And the Inductance tooo

Primary Side - Center tapped winding
Secondary Side - Normal Winding with 5 layers


The power is going to be 20W Maximum.


Please help me solving this.
 

The Kg method is a good solution. It considers all that you mentioned. It is well documented on the internet.
 

Set Bmax to 150mT go for a bigger core, say ETD44 - this will allow proper insulation of the 2kV winding - split this into 4 x 500V windings, rectify each and put in series.

push pull benefits greatly from peak current mode control - unless you have a guaranteed dead time of 10% of the max fet ON time for Tx reset...

dB/dt = V/NAe Bpk = E / ( 4 F N Ae ) when you know the turns you can worl out the wire length, then knowing the current you can work out the wire size for a given loss ( watts ) in the wire. L = Ur Uo N^2 Ae / lmag. Ae/lmag = AL - given in core tables ... in nH/T^2

good luck
 
Please let me know how to calculate the primary turns,.

As your advise, Ill select ETD44, And ill set Bmax as 150mT, Frequency 20000Hz, Max Duty Cycle - 40 Percent

Npri = ?
 

N = E / Bpk. 4. F. Ae Ae = m^2, F = Hz, Bpk = 0.15 Tesla, E = height of square wave, N = number of turns ...
 

ETD44: Ae = 173 mm^2 = 0.000 173 m^2 Thus for 20kHz, 12V on the input ( push pull ) & Bpk = E / ( 4 F N Ae )

We get Npri = 5.78 Turns, so anything above 6 turns will do

Nsec for 500V = 500/12 * 6 = 250 T, so four of these sec windings for 2kV DC ....
 

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