Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Flux swing and flux density confusion

Status
Not open for further replies.

bogdanserban

Newbie level 4
Joined
Aug 3, 2019
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
59
Greetings,
I am designing a forward converter transformer. I have two books that I use as a reference, one is "Practical Switching Power Supply Design" by Marty Brown and the other is "Switching Power Supply Design" by Pressman.
To calculate the number of turns, the first book gives the formula: Np = Vin * 10^8 / f * 4 * Bmax * Ac. I have chosen Bmax to be 1600G and for a given transformer that results in about 60 turns.
The second book gives the formula Np = Vin * Ton * 10^8 / dB * Ac. In this case the flux swing dB is 1600G since the converter operates in the second quadrant only. That results in about twice the number of turns. So which is correct?
I have used the first formula for half-bridge converters with no issues, but I am unsure about the forward converter.

Thank you.
 

The fundamental equation these are based off of is V*dt=dB*Ac*N (where Ac is in square meters and dB is in Tesla). I'm guessing your factor of 10^8 comes from using units of gauss and square cm (lol).

It makes sense that the first formula is meant for half bridge converters. Since the flux swings positive and negative, the peak B will be equal to dB/2. And in a half bridge converter, the V applied to the primary will be Vin/2. So those together account for the factor of 4 in the first equation.

The second equation is correct for forward converters. It uses Ton instead of f, probably because the max Ton is often set to something other than 1/(2f) (often lower, but sometimes higher).

So neither equation is wrong, but you have to use each one correctly.
 

Rather than use a formula, sub in a value for Npri, and see what inductance it gives you, see what magnetising current peak, and hence Bpk you get....then look again at your wire tables, and the bobbin width, put in a new value and let the excel calculate it out for you again..continue "molding" it till its done...its very quick like that....
remember you really need to be doing full layers (or nearly full) of turns, and no formula will know whether thats the case or not.
 

Rather than use a formula, sub in a value for Npri, and see what inductance it gives you, see what magnetising current peak, and hence Bpk you get....then look again at your wire tables, and the bobbin width, put in a new value and let the excel calculate it out for you again..continue "molding" it till its done...its very quick like that....
remember you really need to be doing full layers (or nearly full) of turns, and no formula will know whether thats the case or not.
Most of the time I work with salvaged transformers (I have a box full of them) and I don't know any parameters of the core. 1500-1600G is a value that works for most ferrites. I still adapt the number of turns when I wind the transformer, depending on how many actually fit, and re-calculate the flux.
How do you choose Npri to start with? Do you have an excel table to calculate all these?
--- Updated ---

The fundamental equation these are based off of is V*dt=dB*Ac*N (where Ac is in square meters and dB is in Tesla). I'm guessing your factor of 10^8 comes from using units of gauss and square cm (lol).

It makes sense that the first formula is meant for half bridge converters. Since the flux swings positive and negative, the peak B will be equal to dB/2. And in a half bridge converter, the V applied to the primary will be Vin/2. So those together account for the factor of 4 in the first equation.

The second equation is correct for forward converters. It uses Ton instead of f, probably because the max Ton is often set to something other than 1/(2f) (often lower, but sometimes higher).

So neither equation is wrong, but you have to use each one correctly.
Never went over 0.5 Ton. Actually, in the book they give it as 0.45 (45% duty cycle). And the second formula correlates with the first only if flux swing is 0.32T (3200G).
 

Attachments

  • werthwrth.JPG
    werthwrth.JPG
    67.3 KB · Views: 119
Last edited:

Never went over 0.5 Ton. Actually, in the book they give it as 0.45 (45% duty cycle).
That's typical. Active clamp forward converters can exceed 50% without saturating, but these are used somewhat rarely.
And the second formula correlates with the first only if flux swing is 0.32T (3200G).
Could you show your work?
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top