Re: Flyback transformer design confusion!!

lemonjuice wrote:
On Fri, 10 Nov 2006 13:32:13 +1300, Terry Given <my_name@xxxxxxxx>
wrote:


fastshot wrote:

On 6 Nov 2006 22:49:20 -0800, seegoon99@xxxxxxxxx wrote:



Hi to all.
I am looking to design a flyback transformer , for a small supply. I
have a couple design examples from ON-Semi that I'm looking at , and am
a little confused. Depending on what app note I use ,
I get wildly different results for the same design inputs. For example:
12V out 7W.
One app note shows a primary inductance of 3.3mH and the other app note
works
out at 1.8mH. I seem to come across this problem a lot. All the
different design procedures
seem to come out with totally different values for the same problem.
How do I know
which is the correct solution?


Energy entering the primary winding during each cycle of the core is from the definition of inductance E= 1/2* L * i^2
or more like 1/2 *L* (i peak - i min)^2 ; i peak and i min are the
peak and minimum currents flowing through the primary windings. if you want your result in watts multiply the above by the frequency.

also i pri = Vin * Ton ( time semiconductor is on) / Lpri

Though the output inductor may seem to be a voltage source it is
actually a current source whose output voltage is clamped by the
capacitor placed in parallel to the secondary inductor. so from E=1/2
*L* (i peak - i min)^2 one can reduce the size of the inductor and
maintain the same power output by increasing the peak current ...
that has its setbacks but that wasn't your question. That is why
different inductance values will work.

lemonjuice

oops. this might just be two identical typos, but its wrong.


OK ... I see you're in the mood for playing ...so here i come

dE = 0.5*L*(Ipeak^2 - Imin^2) is the amount of energy added to the core when current rises from Imin to Ipeak = Imin + dIp



dE is the symbol for an exact differential and that doesn't apply to
the case above or in other words you are using the wrong Mathematical
language so check on that.

OK, use the equation: FISH = 0.5*L*(Ipeak^2 - Imin^2) where FISH = incremental change in energy over the on-time, Imin = inductor current at beginning of on time, and Ipeak = inductor current at end of on time. I'd have used the capital delta symbol, if I had one on my keyboard, but I dont.


Its pretty obvious that the energy in the core at the start of the on time MUST BE

Estart = 0.5*L*Imin^2

and likewise, at the end of the on time, the energy in the inductor MUST BE

Efinish = 0.5*L*Ipeak^2

ergo the change in energy, FISH = Efinish - Estart = 0.5*L*(Ipeak^2 - Imin^2)




The current through a cycle is (I peak - I min) and the square of that
is NOT what you've written.
Obviously you have to derive the formula for power of an inductor
and hopefully you will be able to see your error. You can however express the power,as I did earlier, using E= 1/2* L * i^2 where i is the mean current but do you know how to
get that?



so, given your "reasoning", If L = 2mH and dI = Ipeak - Imin = 1A, then:


let Imin = 0A, Ipeak = 1A, then according to you,

FISH = 0.5*2mH*(1A - 0A)^2 = 1mJ


now let Imin = 100A, Ipeak = 101A, then according to you,

FISH = 0.5*2mH*(101A - 100A)^2 = 1mJ

in other words, you think that:

FISH = 0.5*2mH*(1A)^2 = 1mJ, *regardless* of Ipeak and Imin.




whereas I maintain that dE depends on Ipeak and Imin, not just their difference. the same numerical example:

let Imin = 0A, Ipeak = 1A, then according to you,

FISH = 0.5*2mH*(1A^2 - 0A^2) = 1mJ

now let Imin = 100A, Ipeak = 101A, then according to you,

FISH = 0.5*2mH*(101A^2 - 100A^2) = 201mJ


ROTFLMAO!

next you will be asserting that the change in energy stored in a cap is 0.5*C*(Vfinish - Vstart)^2 :)

whereas I can confidently state this is not the case; try throwing constant quanta of energy into a cap, and watch the incremental voltage get smaller each time.....

for Discontinuous Conduction Mode (DCM) all energy stored in core is transferred to the secondary, so Imin = 0 and the primary current is a sawtooth.

in CCM, not all the energy stored in the core is transferred to the secondary, so Imin > 0 and the primary current is a trapezoid.

regardless of CCM or DCM, the output power is:


CCM or DCM ... is totally irrelevant to the question raised.

Pout = n*dE*Fsmps where n = efficiency.


again you use dE ... wrong Mathematical symbol.

again, you are being ridiculously pedantic. I can call it anything I like, such as Pout = n*FISH*Fsmps, without invalidating the equation at all.


and dIp = V*Ton/Lp = V*D/(Lp*Fsmps)


for DCM one ends up with:

Pin = Pout/n

dE*Fsmps = Pin


dE??? wrong Mathematical symbol though E*Fsmps = Pin is repeating me.

LOL.



which ends up as

Pin = (V*D)^2/(2*Lp*Fsmps)


so so?
So do those formulae answer the Op's query. How so? Any average
intelligence engineer could write an entire book on a flyback
transformers(without basic errors) but the current problem is
answering the man's question.


best not to review that sentence given your glaring fundamental error!


Cheers
Terry

enjoyable play, till later if i have time.


lemonjuice


Cheers
Terry
.

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