Flyback transformer design confusion!!

There are an infinite number of solutions, and the subset containing correct solutions is also infinite. Forget all the design examples. Start at the beginning and *define* what you need: what is the desired voltage and what is the nominal load current; is there a required precision of that voltage (10%, 1%, etc); how much is the load going to vayt *and* how much can the voltage chang with that load change; what is the input voltage; what is the input voltage range; how much can the output voltage change when the input chamges over the maximum range; short circuit proof; open circuit proof; etc.

The inductance that you need in the primary depends on the voltage that you are using to charge up the primary, and the speed at which you want the current through the primary to increase.

A fly-back transformer works by applying the primary voltage acros the primary inductor, waiting until the current through the primary inductor has gotten close to saturating the core, then turning off the primary drive. The voltage across the primary inductor then reverses as the current charges up the stray capacitance of the coil until it reaches a level that allows it to charge up the capacitor or whatever that you want charged, and stays reversed and high until the current through the coil has decayed back to zero.

In both cases the behaviour of the current is given (roughly) by

di/dt=V/L

The voltage that charges up the coil is positive, while the reversed voltage that discharges it is negative.

You lose some voltage drop in the resistance of the coil, and you often have to worry about the parallel capacitance of the coil (most easily discoved by measuring the self-resonant frequnecy of the coil).

The lower the inductance, the faster the transistor/MOSFET doing the switching has to turn on and off, and the higher the dissipation in that device.

Hope this helps.

Not enough information is provided. The inductance required will depend not only upon the supply voltage and current (at maximum load) but also the design operating frequency.

Those inductances seem very high. The 1.8mH would be about right at

The 3.2mH comes directly out of the datasheet design. It runs at 67KHz. Vmax is 350 Vdc and Vmin is 140Vdc. Vout 12v Iout 560mA. If I take these Values and apply them to another design note equasions I come out at 1.8mH. Same input parameters , different outputs , depending on design methodology!!

Have a closer look at the methods and results. There is a very good chance that the result has been used to prove the method......... which should give you enough reason to question things.

In this case, the difference does look a little bit big.

Have a closer look at other factors...... one of them might be what you think the real operating frequency is. It's not unusual to find that these 'fixed' frequency controllers come in two flavours.... one that goes at

That would explain the difference, assuming my fingers worked my calculator properly.

DNA

OK; what the (&*^#@$! is an "unun" as in thier spiel "Ferrite Toroids. Iron Powder Toroids. Ferrite Beads. Split Cores. Cable clamp-ons. Balun Cores. W2FMI Baluns and Ununs." ?

That's Sevick speak for an unbalanced-to-unbalanced transmission line transformer.

Jerry Sevick's book "Transmission line transformers" is worth reading if you're into this kind of thing.

Jeroen Belleman

Thanks! I have been pushing electrons for over 40 years and this was the first time i have heard that term.

So now you can pull them too...

Would that be class "A" push-pull?

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

lemonjuice

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

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

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:

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

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

for DCM one ends up with:

Pin = Pout/n

dE*Fsmps = Pin

which ends up as

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

Cheers Terry

Could also be the input range. Orgs like PowerInt try to maintain incomplete energy transfer over the entire range; I assume this is to keep peak voltage stress low.

Some day they may figure out how to set a voltage clamp so that it doesn't depend solely on energy transfer during the slow clamp diode's reverse recovery. At piddling power levels, it woiks just fine.

The OP should try out both options. Just a matter of adjusting the gap, which he'll end up doing anyways.

RL

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

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.

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?

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

again you use dE ... wrong Mathematical symbol.

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

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.

enjoyable play, till later if i have time.

lemonjuice

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)

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

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.....

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.

LOL.

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

Cheers Terry

Ha! that explains. Ask lemonjuice what he thinks about amplifiers noise :-)

see news: snipped-for-privacy@g14g2000cwa.googlegroups.com and following. Worth its weight in peanuts...