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[SOLVED] Half bridge SMPS xformer turns calculation

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codemaster11

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dear i 'm not good in english.
i have question that i have calculate the half bridge xformer
primary turns ≈ 40, using formula N(prim) = (Vin(nomi) * 0.5 * 10^8)/(4*fs*Bmax*Ae).

where Vin(nomi) = 311vdc
fs = 50KHz
Bmax = 1500 mT
Ae = 1.3 cm-sq.
Duty cycle = 0.98

with N(sec) ≈ 14 turns for 22 volt output Vsec = 22volt
& N(auxiliary) = 11 turns for 18volt.

is this calculation correct ? can any one help me regarding this calculations?
 
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Hi,

For Bmax of 1500mT, what core material are you using?
For duty cycle of 0.98, how did you arrive at that figure?
Also why are you calculating with the nominal value?

- - - Updated - - -

Always bear in mind that parts are best selected at worst case conditions. For this topology, you would be using Vin_min.
 

How did you select the core size? You need to follow through a series of steps to help you do a good selection.

You would have a lot of hits from a single internet search.
 

Hi,

For Bmax of 1500mT, what core material are you using?
For duty cycle of 0.98, how did you arrive at that figure?
Also why are you calculating with the nominal value?

- - - Updated - - -

Always bear in mind that parts are best selected at worst case conditions. For this topology, you would be using Vin_min.

i 'm using core EI33 with Ae = 1.3cm-sq from a readymade half bridge smps circuit. but i want to

wind it myself with.

i saw an example with duty cycle = 0.98. if it is not good what possible duty cycle i should use?

i take vin(nomi) = 311vdc. where in formula we will take half of this value for half bridge topology.

vin(nomi) = 311/2 vdc.

if i take vin(min) = 100*√2 = 141vdc then vin(min) = 141/2 vdc. calculate N(prim) ≈ 18 turns.

can this might be the right numbers of N(prim)?

also for calculating i N(prim) i will take half of Vin(min)?
 


SMPS transformer for Half bridge topology

Capture1.JPG

i enter the operating conditions according to my specification

fs = 50000Hz

Bmax = 1786mT.

& powerEsim calculate N0 = 10T (AWG24 x 6)

Nsec(N4 + N5 ) = (5 + 5) turns (AWG20 t/w x 3)

could i use this transformer according to my specifications?
 

To design a half bridge transformer, first design a virtual buck converter with your vout. Decide what is your max duty cycle (say 0.7).
So you then design a buck with whatever vin gives your vout with your chosen D.
Then think about that “virtual” vin value.
That virtual vin is what you must get out of your transformer when it has your minimum actual vin at its primary side.
(but remember that with ½ bridge your actual vin is half of the actual vin, so to speak…..because of the divider caps)
So you then do the transformer equation vpri/vsec = Npri/Nsec
….thats pretty much it.
If you are wondering about how many turns exactly to use, and what exact core etc, and you want an exact formula which spits out everything for you in one go then this is the wrong field for you……you have to do it iteratively and you converge to the solution.
But this is good as it enables you to use your artistic side.
You can use any core or wire you like as long as………………..
Solution is not too big
Wire loss not too high
Core loss not too high
Coupling not too poor.
Skin effect is accounted for
You can physically terminate the wire to the former pins.
Etc etc etc….
Here is my smps maths course to give you some of the equations for magnetics
https://massey276.wixsite.com/maths

and here is my electronics degree advisor
https://massey276.wixsite.com/electronicsdegree
 

Your Bmax is about factor 10 too large. Please notice that quoted Tahmid's blog is calculating in antique unit Gauss rather than mT.
 

i wrote mistakenly 1500mT. it's 0.1500T or 1500 gauss.
 
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