Re: SMPS question

Robert Latest wrote: 
On Wed, 21 Sep 2005 08:56:58 +1200,
  Terry Given <my_name@xxxxxxxx> wrote
  in Msg. <wI_Xe.13255$iM2.1117632@xxxxxxxxxxxxxxx>

at such low voltages I would use a diagonal half-bridge primary - a switch at either end of the winding, one to +30V, one to 0V. And a pair of (schottky) diodes, one to 0V (at the +30V end), the other to +30V.

drive both switches simultaneously. Both on = energy storage. both off = energy transfer. the primary winding voltage is now clamped to the +30V bus, by the two diodes. Voila, no spike and almost no losses. 


Sounds good; I'll consider it.


sandwich the smallest winding (fewest turns) between the two halves of the largest winding (most turns). This'll give you a significant reduction in leakage inductance, and allow you to use thicker copper (the number of effective layers halves)


I must confess that this doesn't make sense to me. The way I understand it is: Leakage inductance is created by flux that goes through one winding but not fully through the other (which makes it synonymous with poosr magnetic coupling). In other words, the portion of flux that the primary created but that doesn't go through the secondary will obviously not be reset by the secondary (and be put to good use on the load side) but will cause voltage spikes on the primary side that need to be dealt with. In the "diagonal" setup that you described, it will be fed back into the primary-side storage cap. 

all very true.

its easiest to think of this stuff in terms of H=NI/l (l = path length).

consider 2 simple windings, one layer each. "look" into the end of the transformer, down thru the center leg. You "see" the two winding layers, edge-on. One close to the core, the other far away (ish). H ramps from 0 to Np*Ip/l across the thickness of the primary winding (lets pretend H is linearly distributed). H is then constant across the inter-winding insulation, and (remember, Is flows the opposite direction, and Np*Ip = Ns*Is) ramps back down to zero thru the secondary winding. All well and good.

Hpeak = Np*Ip/l


If you split the primary winding, this happens:

H ramps up to 0.5*Np*Ip/l thru the first half primary

H ramps down Ns*Is/l = Np*Ip/l thru the secondary. So H ends up at -0.5*Np*Ip/l

then H ramps back up thru 0.5*Np*Ip/l thru the 2nd half-primary, returning to zero.

Hpeak is now 0.5*Np*Ip/l

H is also zero at the centre of the 2ndary winding.

I cant recall if leakage is proportional to peak MMF or peak MMF squared, its one of them.


But before you go into explaining what "effective layers" means I really should read a book.


the "effective layers" are the number of layers between a plane of zero MMF and a plane of peak MMF. for a non-interleaved winding, its the number of layers. for a 1/2P-S-1/2P, its half the number of layers.



all first-order switchers have an L and a C. in a flyback, L=Lmag.

otherwise, the primary switches see a capacitive load, with either dV/dt or R setting the peak current (ignoring leakage....) 


Ooops .. I should have thought of that myself.


Keith Billings Switch Mode Power Supply Handbook, followed by Abraham Pressmans Switching Power Supply Design. both McGraw-Hill


A lot of money for a lot of knowledge. 

try Soft Ferrites, E.C. Snelling, its a *lot* more expensive.


Thanks,
robert


Cheers
Terry
.

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